source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34797 | A: 9
B: 12
C: 16
D: 18
E: 24
This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 03:08
1
2
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours
Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 6w widgets? | [
"36",
"6",
"8",
"10"
] | A | Let Y produce w widgets in y Days
Hence, in 1 day Y will produce w/y widgets.
Also, X will produce w widgets in y+2 days (given, X takes two more days)
Hence, in 1 day X will produce w/y+2 widgets.
Hence together X and Y in 1 day will produce {w/y + w/y+2} widgets.
Together X and Y in 3 days will produce = 3 * [{w/y + w/y+2}] widgets.
It is given that in 3 days together they produce (5/4)w widgets.
Equating, 3 * [{w/y + w/y+2}] = (5/4)w
Take out w common and move 3 to denominator of RHS
w{1/y + 1/(y+2)} = (5/12)w
Canceling w from both sides
{1/y + 1/(y+2)} = 5/12
2y+2/y(y+2) = 5/12
24y+24=5y^2 + 10y
5y^2-14y-24=0
5y^2-20y+6y-24=0
5y(y-4)+6(y-4)=0
(5y+6)+(y-4)=0
y=-6/5 or y=4
Discarding y=-6/5 as no of days cannot be negative
y=4
hence it takes Y, 4 days to produce w widgets.
Therefore, it will take X (4+2)=6 days to produce w widgets.
Hence it will take X 6*6=36 days to produce 2w widgets.
Answer : A |
AQUA-RAT | AQUA-RAT-34798 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
In a classroom, the average age of boys (b) is g and the average age of girls (g) is b. The average age of all is b+g. At the time of calculation, the class teacher, who is 42 years old was also present in the class. Can you find the value of b+g? | [
"8",
"9",
"10",
"11"
] | A | Solution:
b+g = 8
Explanation:
for b in range(1,99):
for g in range(1,99):
sum1=2*b*g+42
sum2=(b+g)*(b+g+1)
if(sum1==sum2):
print(b,g,sum1)
(b,g)=(3,5) => b+g=8
Answer A |
AQUA-RAT | AQUA-RAT-34799 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A can do a piece of work in 10 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in? | [
"15 days",
"10 days",
"12 days",
"19 days"
] | B | B
(x – 5)/10 + x/20 = 1
x = 10 days |
AQUA-RAT | AQUA-RAT-34800 | algorithms, dynamic-programming, np-hard
if left_is_greater(greater, precincts, n):
greater.append(precincts.popleft())
continue
p = precincts.pop()
# if either of the districts is "full", we don't have a choice
# and must select the opposite district
if len(right) == n / 2:
left.append(p)
elif len(left) == n / 2:
right.append(p)
# we always prefer the lesser summed district. if gerrymandering
# is feasible, we won't want to put all of our party's votes
# into one district, but prefer to have each district barely win.
elif sum(left) < sum(right):
left.append(p)
elif sum(right) < sum(left):
right.append(p)
# if the districts sum to the same amount, we prefer the district
# with less precincts, as this maximizes future decision branches
elif len(right) < len(left):
right.append(p)
elif len(left) < len(right):
left.append(p)
# we choose left as the default seed case, all else equal
else:
left.append(p)
if sum(left) > 0 and sum(right) > 0:
return True, left, right
else:
return False, left, right
def solve(precincts):
maximized = deque(sorted(precincts))
minimized = deque(sorted(x * -1 for x in precincts))
for polarized_precincts in [maximized, minimized]:
is_gerrymanderable, left, right = maximize(polarized_precincts)
if is_gerrymanderable:
print("It's gerrymanderable! Take a look: ", left, right)
return
print("Sorry, doesn't seem to be gerrymanderable.")
The following is multiple choice question (with options) to answer.
A certain city with a population of 180,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district What is the minimum possible population that the least populated district could have ? | [
"a) 10,700",
"b) 10,800",
"c) 10,900",
"d) 15,000"
] | D | Let x = number of people in smallest district
x*1.1 = number of people in largest district
x will be minimised when the number of people in largest district is maximised
10*x*1.1 = 11x = total number of people in other districts
So we have 11x + x = 180k
x = 15,000
Answer : D |
AQUA-RAT | AQUA-RAT-34801 | # What is the next number in this sequence: $1, 2, 6, 24, 120$? [closed]
I was playing through No Man's Sky when I ran into a series of numbers and was asked what the next number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
• What was the purpose of the question? – haqnatural Aug 16 '16 at 17:42
• @Battani I was trying to figure out what the next number in the sequence was. – Atom Aug 16 '16 at 17:43
• @Watson I did when I posted this, I was going to ask this last night but decided to work through it first and ended up solving it. When I saw that neither the question nor answer were on here already I selected the "answer your own question" option when posting the question. That way the question would be available online and I would instead be contributing instead of asking for an answer and providing a hodgepodge of behind the scenes work I was doing. I can delete this if that's not the proper way of doing it! – Atom Aug 16 '16 at 17:58
• oeis.org is a good resource. A search gives several hundred possibilities, but you'd want to go with the most comprehensible. – Teepeemm Aug 16 '16 at 20:30
The next number is $840$. The $n$th term in the sequence is the smallest number with $2^n$ divisors.
Er ... the next number is $6$. The $n$th term is the least factorial multiple of $n$.
No ... wait ... it's $45$. The $n$th term is the greatest fourth-power-free divisor of $n!$.
Hold on ... :)
Probably the answer they're looking for, though, is $6! = 720$. But there are lots of other justifiable answers!
The following is multiple choice question (with options) to answer.
Look at this series: 66, 66, 60, 60, 54, 54, ... What number should come next? | [
"48",
"15",
"17",
"19"
] | A | In this series, each number is repeated, then 6 is subtracted to arrive at the next number.
The next number should be 48
Answer : A |
AQUA-RAT | AQUA-RAT-34802 | # Number of rectangles with odd area.
We have a $10\times 10$ square.
How many rectangles with odd area are on the picture?
I say lets choose a vertex first, there are $11\cdot11=121$ possibilities.
Now, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\dfrac{121\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.
The width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\cdot 30=900$ odd area rectangles.
What you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.
The following is multiple choice question (with options) to answer.
A rectangular tiled patio is composed of 160 square tiles. The rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles. After the change in layout, the patio will still have 160 tiles, and it will still be rectangular. How many rows are in the tile patio before the change in layout? | [
"5",
"6",
"10",
"13"
] | C | Suppose there are c columns and there are r rows
Original Situation
So, Number of tiles = c*r = 160
Also. Reach column has r tiles and each row has c tiles
New Situation
Number of tiles in each column is r-2 and number of tiles in each row is c+4
So, number of rows = r-2 and number of columns is c+4
So, Number of tiles = (r-2)*(c+4) = 160
Comparing both of them we get
c*r = (r-2)*(c+4)
=> 4r -2c = 8
c = 2r - 4
Putting it in c*r=160
(2r-4)*r = 160
2r^2 - 4r - 160=0
r cannot be negative so r = 10
and c = 16
So, Answer will be C |
AQUA-RAT | AQUA-RAT-34803 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
In the first 20 overs of a cricket game, the run rate was only 4.8. What should be the run rate in the remaining 30 overs to reach the target of 302 runs? | [
"5",
"6.25",
"6.75",
"6.87"
] | D | Required run rate = 302 - (4.8 x 20) /30 = 206/30 = 6.87 Option D |
AQUA-RAT | AQUA-RAT-34804 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
Two successive price increase of 10% and 10% of an article are equivalent to a single price increase of: | [
"26(2/3)",
"25%",
"21%",
"33(1/3)"
] | C | Solution: 100----10%↑--→110---10%↑--→121.
Equivalent price increase = 21%.
Answer: Option C |
AQUA-RAT | AQUA-RAT-34805 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55 and 60, then the average marks of all the students is | [
"53.33",
"54.68",
"55",
"57"
] | B | Sol.
Required average = [55x50 + 60x55 + 45x60 / 55 + 60 + 45]
= [2750 + 3300 + 2700 / 160]
= 8750 / 160 = 54.68.
Answer B |
AQUA-RAT | AQUA-RAT-34806 | concentration
Title: Concentration of solutions I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%? The answer is that for a simple dilution the following formula applies:
$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$
Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).
The following is multiple choice question (with options) to answer.
A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture? | [
"16 liters",
"18 liters",
"10 liters",
"17 liters"
] | C | Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters.
Answer: C |
AQUA-RAT | AQUA-RAT-34807 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs.
3200. With the help of C, they completed the work in 3 days. How much is to be paid to C | [
"Rs. 300",
"Rs. 400",
"Rs. 500",
"Rs. 600"
] | B | Explanation:
C's 1 day's work =
13−(16+18)=(13−724)=124A:B:C=16:18:124=4:3:1CʹsShare=18
∗
3200=400
If you are confused how we multiplied 1/8, then please study ratio and proportion chapter, for small
information, it is the C raƟo divided by total ratio.
Answer: B |
AQUA-RAT | AQUA-RAT-34808 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
If Rs.7500 are borrowed at C.I at the rate of 4% per annum, then after 4 years the amount to be paid is? | [
"3377",
"2688",
"2688",
"8774"
] | D | A = 7500(26/25)^4 = 8774
Answer: D |
AQUA-RAT | AQUA-RAT-34809 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Albert borrowed a total of $ 5800 from Brian and Charlie. He paid simple interest at the rate of 7 % Per yr and 9 % per yr respectively.At the end of three years he paid $ 1400 as total interest. What is the interest paid to Brian at the end of three years? | [
"585",
"581",
"441",
"431"
] | B | Let x be the amount borrowed form Brian. So amount borrowed form Charlie = 5800-x.
1400 = 21/100x + 27/100(5800-x)
=>x = 16600/6.
Interest paid = 3*7/100*16600/6 = 581.
B |
AQUA-RAT | AQUA-RAT-34810 | Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.
Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is
$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$
So for $4$ and $7$ it would be one less than that which is $11$.
• so for 11 and 14 it would be 25? – simplton May 16 '13 at 19:56
The following is multiple choice question (with options) to answer.
Car X began traveling at an average speed of 35 miles per hour. After 84 minutes, car Y began traveling at an average speed of 42 miles per hour. When both cars had traveled the same distance, both cars stopped. How many miles did car X travel from the time car Y began traveling until both cars stopped? | [
"140",
"175",
"210",
"245"
] | D | In 84 minutes, car X travels 49 miles.
Car Y gains 7 miles each hour, so it takes 7 hours to catch car X.
In 7 hours, car X travels 245 miles.
The answer is D. |
AQUA-RAT | AQUA-RAT-34811 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a job in 8 days and B can do it in 16 days. A and B working together will finish twice the amount of work in ------- days? | [
"12 1/8 days",
"10 1/3 days",
"14 1/3days",
"2 1/2 days"
] | B | 1/8+ 1/16= 3/16
16/3 = 16/3 *2 = 10 2/3
Answer: B |
AQUA-RAT | AQUA-RAT-34812 | # Find the least next N-digit number with the same sum of digits.
Given a number of N-digits A, I want to find the next least N-digit number B having the same sum of digits as A, if such a number exists. The original number A can start with a 0. For ex: A-> 111 then B-> 120, A->09999 B-> 18999, A->999 then B-> doesn't exist.
You get the required number by adding $9$, $90$, $900$ etc to $A$, depending on the digits of $A$.
First Case If $A$ does not end in a row of $9$s find the first (starting at the units end) non-zero digit. Write a $9$ under that digit and $0$s under all digits to the right of it. Add the two and you get $B$.
Example: $A=3450$. The first non-zero digit is the 5 so we write a $9$ under that and a $0$ to its right and add:
\begin{align} 3450\\ 90\\ \hline 3540 \end{align}
There is a problem if the digit to the left of the chosen non-zero digit is a $9$. In this case we write a $9$ under that $9$ and $0$s to its right. And if there are several $9$ we put our $9$ under the highest one.
Example: $A=3950$. The first non-zero digit is the 5 but there is a $9$ to its left. We write a $9$ under that $9$ instead and $0$s to its right and add:
\begin{align} 3950\\ 900\\ \hline 4850 \end{align}
Second case If $A$ does end in a row of $9$s write a $9$ under the highest of the row of $9$s and $0$s under all digits to the right of it. Add the two and you get $B$. As you say, if $A$ is entirely $9$s there is no solution.
The following is multiple choice question (with options) to answer.
Take a number 'x' and follow these steps :
1. Find the sum of its digit
2. If the sum has only one digit , STOP
3. Else , go back to step 1 with the new number.
If x = 1684 ,what is the end result ? | [
"19",
"10",
"1",
"2"
] | C | 1+6+8+4=19
sum=19 two digits
so go to step 1
1+9=10 two digits
so go to step 1
1+0=1
ANSWER:C |
AQUA-RAT | AQUA-RAT-34813 | Quick way
Use Smart Numbers
Give 100 for the initial amount
Then you will have 50-0.25x = 30
x = 80
So % is 80/100 is 80%
Hope it helps
Cheers!
J
SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance
Re: If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
29 May 2014, 11:41
Or one can use differentials to slve
Initially 50% alcohol
Then 25% alcohol
Resulting mixture 30% alcohol
Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4
Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.
Hope this helps
Cheers
J
Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 463
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
02 Jul 2014, 17:12
Bunuel wrote:
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
The following is multiple choice question (with options) to answer.
A 6-liter solution is 30% alcohol. How many liters of pure alcohol must be added to produce a solution that is 50% alcohol? | [
"1.8",
"2.1",
"2.4",
"2.7"
] | C | Let x be the amount of pure alcohol required.
0.3(6) + x = 0.5(x+6)
0.5x = 3 - 1.8
x = 2.4 liters
The answer is C. |
AQUA-RAT | AQUA-RAT-34814 | Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13
The following is multiple choice question (with options) to answer.
What will come in place of question mark in the following equation ?
54.(?)3 + 543 + 5.43 = 603.26 | [
"6",
"1",
"9",
"8"
] | D | Explanation:
Let x + 543 + 5.43 = 603.26. Then, x = 603.26 - (543 + 5.43) = 603.26 - 548.43 =54.83
Missing digit = 8
ANSWER IS D |
AQUA-RAT | AQUA-RAT-34815 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
Mr. Bhaskar is on tour and he has Rs. 360 for his expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by Rs. 3. for how many days is Mr. Bhaskar on tour? | [
"10",
"20",
"30",
"40"
] | B | If his tour is for x days and he is spending Rs y per day, then
xy= 360 x= 360/y or y= 360/x
(x+4)*(y-3)=360
or
xy +4y -3x-12 =360
or
4y-3x-12=0
4*360/x - 3x -12=0
solving it, we get x=20
ANSWER:B |
AQUA-RAT | AQUA-RAT-34816 | # Math Help - help
1. ## help
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
name a fraction that is between 1/2 and 1/3....?
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
2. Originally Posted by BeBeMala
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
Originally Posted by BeBeMala
name a fraction that is between 1/2 and 1/3....?
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Originally Posted by BeBeMala
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
Originally Posted by BeBeMala
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
This one works just like the first one, above.
3. Originally Posted by stapel
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
This one works just like the first one, above.
sorry....
The following is multiple choice question (with options) to answer.
A bowl of fruit contains 14 apples and 20 oranges. How many oranges must be removed so that 70% of the pieces of fruit in the bowl will be apples? | [
" 3",
" 6",
" 14",
" 17"
] | C | Number of apples = 14
number of oranges = 20
let number of oranges that must be removed so that 70% of pieces of fruit in bowl will be apples = x
Total number of fruits after x oranges are removed = 14+(20-x) = 34-x
14/(34-x) = 7/10
=>20 = 34-x
=>x= 14
Answer C |
AQUA-RAT | AQUA-RAT-34817 | . . . . . . $\begin{array}{ccccccc}mn+1 &=& (2a+1)(2b+1)+1 &=& 4ab + 2a + 2b + 2 &=& 2(2ab + a + b + 1)\end{array}$
And we have: . $P \;=\;2(a + b + 1)\cdot2(2ab + a + b)\cdot2(2ab + a + b + 1)$
. . . . . . . . . . $P \;=\;8(a+b+1)(2ab + a + b)(2ab + a + b + 1)$
So we have a multiple of 8.
. . And here's where I hit the wall . . .
I tried various ways to find another factor of 2,
. . and settled on an exhaustive listing.
. . $\begin{array}{c|c|c|c}
(a,b) & (a+b+1) & (2ab + a + b) & (2ab + a + b + 1) \\ \hline
\text{even, even} & \text{odd} & \text{even} & \text{odd} \\
\text{even, odd} & \text{even} & \text{odd} & \text{even} \\
\text{odd, odd} & \text{odd} & \text{even} & \text{odd} \end{array}$
So for any combination of $a$ and $b$, at least one of the factors is even.
Therefore, $P$ is a multiple of 16.
Surely there must a more elegant method . . .
.
4. Soroban,
Whenever you said:
$
P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)$
The following is multiple choice question (with options) to answer.
The H.C.F of two numbers is 23 and the other two factors of their L.C.M are 15 and 16. The larger of the two numbers is: | [
"338",
"278",
"322",
"368"
] | D | Clearly, the numbers are (23 * 15) and (23 * 16). Larger number
= (23 * 16) = 368.
Answer: D |
AQUA-RAT | AQUA-RAT-34818 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
Suzie’s Discount Footwear sells all pairs of shoes for one price and all pairs of boots for another price. On Monday the store sold 22 pairs of shoes and 16 pairs of boots for $460. On Tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $560. How much more do pairs of boots cost than pairs of shoes at Suzie’s Discount Footwear? | [
"15",
"18",
"21",
"23"
] | A | Let X be pair of shoes and Y be pair of boots.
22x+16y = 460 ... eq1
8x +32y = 560....eq 2.
Now multiply eq 1 by 2 and sub eq 2.
44x =920
8x = 560.
36x = 360 => x = 10.
Sub x in eq 2.... we get 80 + 32y = 560... then we get 32y = 480 then y = 25
Differenece between X and Y is 15
Answer : A |
AQUA-RAT | AQUA-RAT-34819 | A short computer code shows there are: 24,000 5-digit integers that are NOT divisible by all the elements in the set {1, 2, 3, 4, 5, 6}. They begin like this:
(10001 , 10003 , 10007 , 10009 , 10013 , 10019 , 10021 , 10027 , 10031 , 10033 , 10037 , 10039.......and end like this:
99953 , 99959 , 99961 , 99967 , 99971 , 99973 , 99977 , 99979 , 99983 , 99989 , 99991 , 99997)
Therefore, the probability is: 24,000 / 90,000 ==4 / 15
Feb 4, 2023
The following is multiple choice question (with options) to answer.
How many integers are between 9 and 67/5, inclusive? | [
"5",
"7",
"9",
"10"
] | A | 67/5 = 13.xx
We are not concerned about the exact value of 67/5 as we just need the integers.
Since the values are small, we can write down the integers.
The different integers between 9 and 67/5 would be 9, 10, 11, 12,13
Total number of integers = 5
Option A |
AQUA-RAT | AQUA-RAT-34820 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
Speed of a boat in standing water is 14 kmph and the speed of the stream is 1.2 kmph. A man rows to a place at a distance of 4864 km and comes back to the starting point. The total time taken by him is: | [
"600",
"500",
"700",
"900"
] | C | Speed downstream = (14 + 1.2) = 15.2 kmph
Speed upstream = (14 - 1.2) = 12.8 kmph
Total time taken =4864/15.2+4864/12.8 = 320 + 380 = 700 hours
Answer is C. |
AQUA-RAT | AQUA-RAT-34821 | $⌊100/2⌋+⌊100/4⌋+⌊100/8⌋+⌊100/16⌋+⌊100/32⌋+⌊100/64⌋=50+25+12+6+3+1=97 .$
Hence, 2 divides $S$ to an odd power. So we need to divide $S$ by $k!$ which 2 divides to an odd power to get a perfect square quotient. This reduces the possibilities for $k$ to 50 or 51. Since
$S={2}^{99}·{3}^{98}·{4}^{97}\dots {99}^{2}·100=\left(2·4\dots 50\right)\left({2}^{49}·{3}^{49}·{4}^{48}\dots 99\right){}^{2}=50!·{2}^{50}\left(\dots \right){}^{2} ,$
The following is multiple choice question (with options) to answer.
The value of 99^(98/99) x 99 is: | [
"9989",
"9899",
"9890",
"9809"
] | B | (100 – 1/99) x 99 = 9900 – 1
= 9899.
ANSWER:B |
AQUA-RAT | AQUA-RAT-34822 | # Getting everyone to meet everyone else
There are 25 students in a class who sit in five rows of five. Each week they sit in a different order.
After a number of weeks every student has sat next to every other student, next meaning side by side, one behind the other, or sitting diagonally together. What is the fewest number of weeks in which this can happen?
The case for 16 students sitting in four rows of four has been dealt with at:
https://puzzling.stackexchange.com/questions/83720/my-sixteen-friendly-students?noredirect=1#comment243063_83720
• What are your thoughts on the problem? – Servaes May 8 '19 at 12:58
• @Servaes: Far more difficult than it seems! – Bernardo Recamán Santos May 8 '19 at 13:02
• Each student must sit together with $24$ other students, so $\tfrac{24\times25}{2}=300$ pairs must be made. Each day a total of $$\frac{8\times9+3\times4+5\times12}{2}=72,$$ pairs are made, which shows that we will need at least $5$ weeks for every student to have sat next to every other student. In $5$ weeks a total of $5\times72=360$ pairs are made, so this gives quite a bit of leeway to make the necessary $300$ pairs. A bit of trial and error would go a long way? Perhaps some affine transformations will do the trick? – Servaes May 8 '19 at 13:03
• @Servaes Why did you delete your answer? I checked your solution, it worked! – Mike Earnest May 9 '19 at 1:14
• @MikeEarnest Ha! It was late last night, I was tinkering with it a bit more in an attempt to get it down to six weeks (to no avail), and then reverted to the wrong configuration, apparently, so I believed I made a mistake. Thanks for checking :) I'll recheck it for myself as well later today. – Servaes May 9 '19 at 11:21
It can be done in 5 weeks:
The following is multiple choice question (with options) to answer.
The average attendance in a school for the first 4 days of the week is 30 and for the first 5 days of the week is 32. The attendance on the fifth day is | [
"32",
"40",
"38",
"36"
] | B | Attendance on the fifth day = 32 × 5 – 30 × 4
= 160 – 120 = 40
Answer B |
AQUA-RAT | AQUA-RAT-34823 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
A car is traveling 72 kilometers per hour. How many meters does the car travel in one minute? | [
"1,200 meters / minute",
"1,300 meters / minute",
"1,450 meters / minute",
"1,550 meters / minute"
] | A | Convert hour into minutes ( 1 hour = 60 minutes) and kilometers into meters (1 km = 1000 m) and simplify
72 kilometers per hour = 72 km/hr
= (72 * 1000 meters) / (60 minutes) = 1,200 meters / minute
correct answer A |
AQUA-RAT | AQUA-RAT-34824 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A is twice as good a workman as B and they took 6 days together to do the work B alone can do it in. | [
"18 days",
"12 days",
"29 days",
"25 days"
] | A | WC = 2:1
2x + x = 1/6
x = 1/18 => 18 days
ANSWER:A |
AQUA-RAT | AQUA-RAT-34825 | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
For an employee to qualify for early retirement at a certain company, the sum of the employee’s age and years of service must be at least 60, If Sue was K years old when she was hired by the company, what is the minimum age at which she could possibly qualify for early retirement. | [
"K+35",
"2K+35",
"(60+K)/2",
"(70-K)/2"
] | C | Say Sue was hired when she was already 60 years old (k=60), then she could retire right away, at the age of 60. Now, plug k=60 in the answer choices and see which yields 60. Only C fits.
Answer: C. |
AQUA-RAT | AQUA-RAT-34826 | Correct. This is a much easier problem, isn’t it?
In particular, I would explain what you’ve done in this way:
Since one particular team can be singled out as “the one with the four friends”, making the teams no longer indistinguishable, we need only to count ways to fill out the rest of this one team, and do not need to divide by 2 in the end. There are 8 other students, from which we choose 2 in addition to the four friends, so there are 8C2 = 28 ways.
We could instead choose the 6 for the other team, putting the remaining 2 on the team with the friends; since 8C6 = 8C2, this gives the same result.
We have either two slots to fill from the 8 remaining students, or six slots (on the other team):
## Problem 3: every school on each team
Here is his answer to the third part:
Meanwhile, for the third part, I am solving it this way:
Since we want at least one student from each school, Team 1 will be (3C1 × 4) × 8C2 and Team 2 will take the remaining students. Hence, we have 336 ways. But this includes the combinations with the second team that might not have all students from all 4 schools.
So we subtract by (3C1 × 4) × 4C1 = 48. Thus my final answer is 336 – 48 = 288 ways.
What do you think about my solutions and answers?
### A subtractive method
My response continued:
This problem is much more complicated, and this is a good attempt, though flawed. You’ve done it a different way than I did, and got a different answer. So let’s see who’s right! We won’t want to check by listing this time, so we’ll just compare our answers.
You’ve used a subtractive method, which can be good. (I used addition.) You focus on one team (which suggests we may have to divide by 2 at the end, since the teams are indistinguishable). But you haven’t explained the details of your (3C1 × 4) × 8C2. I think you are saying that for each of the 4 schools, you are picking 1 student, and then picking 2 more from the remaining 8. There are a couple problems here.
The following is multiple choice question (with options) to answer.
Jill is dividing her 14-person class into two teams of equal size for a basketball game. If no one will sit out, how many different match-ups between the two teams are possible? | [
"1110",
"1125",
"1716",
"2152"
] | C | With 14 players, the process of figuring out how many groups of 7 can be formed is pretty straight-forward....
14C7 = 14!/(7!7!) = 3432 possible groups of 7
Once forming that first group of 7, the remaining 7 players would all be placed on the second team by default.
The 'twist' is that the two teams of 7 canshow upin either order:
C |
AQUA-RAT | AQUA-RAT-34827 | 1. ## More Probability
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.
2. Originally Posted by hansel13
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
The following is multiple choice question (with options) to answer.
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good. | [
"125/128",
"125/125",
"125/126",
"125/121"
] | C | Required probability = 1 - 1/126
= 125/126
Answer: C |
AQUA-RAT | AQUA-RAT-34828 | python, combinatorics
Title: Print all "balanced" sequences of 'A' and 'B' the aim of the code below is to print all balanced sequences with equal number of 'A' and 'B'. A sequence is balanced if every prefix of length \$\geqslant 4\$ is balanced. A prefix is balanced if the ratio of 'A' in this prefix is between 0.4 and 0.6 inclusive.
For example, ['A', 'B', 'B', 'A', 'A', 'B', 'A', 'B'] is balanced because:
Prefix
Ratio of 'A'
['A', 'B', 'B', 'A']
2/4 = 0.5
['A', 'B', 'B', 'A', 'A']
3/5 = 0.6
['A', 'B', 'B', 'A', 'A', 'B']
3/6 = 0.5
['A', 'B', 'B', 'A', 'A', 'B', 'A']
4/7 = 0.57
['A', 'B', 'B', 'A', 'A', 'B', 'A', 'B']
4/8 = 0.5
On the other hand, ['A', 'B', 'B', 'A', 'A', 'A', 'B', 'B'] is not balanced, because the ratio of 'A' in the prefix ['A', 'B', 'B', 'A', 'A', 'A'] is 4/6 = 0.67, which is too high.
The code:
from itertools import combinations
n = 8
MIN_RATIO = 0.4
The following is multiple choice question (with options) to answer.
If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 6-letter code words to the number of 5-letter code words? | [
"5 to 4",
"3 to 2",
"2 to 1",
"5 to 1"
] | D | Number of ways to form 6-letter code: 10!/4! = 10*9*8*7*6*5
Number of ways to form 5-letter code: 10!/5! = 10*9*8*7*6
Ratio: 5 to 1
Answer : D |
AQUA-RAT | AQUA-RAT-34829 | $(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
The following is multiple choice question (with options) to answer.
How many terms are in the G.P. 3, 6, 12, 24, ......., 384,768 | [
"8",
"9",
"67",
"5"
] | B | Explanation:
Here a = 3 and r = 6/3 = 2. Let the number of terms be n.
Then, t = 768 => a * r^(n-1) = 768
=> 3 * 2^(n-1) = 768 => 2^(n-1) = 256 = 2^(8)
=> n-1 = 8 => n = 9.
Answer: B |
AQUA-RAT | AQUA-RAT-34830 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
If the selling price of 50 articles is equal to the cost price of 40 articles, then the loss or gain percent is? | [
"20",
"98",
"66",
"54"
] | A | Let C.P. of each article be Re. 1.
Then, C.P. of 50 articles = Rs. 50;
S.P. of 50 articles = Rs. 40.
Loss % = 10/50 * 100 = 20%
Answer: A |
AQUA-RAT | AQUA-RAT-34831 | Why are things "weird" here? Let's think of it like this: If I travelled at $$8 \frac{\text{mi}}{\text{hr}}$$ for an hour, then travelled at $$12 \frac{\text{mi}}{\text{hr}}$$, I'd definitely agree that the average speed is $$10 \frac{\text{mi}}{\text{hr}}.$$ Things are pretty normal here: If I want to calculate the total distance traveled, it's just $$8 + 12 = 20.$$ Then, if I want to calculate the total time traveled, it's just $$1 + 1 = 2.$$ So the average or overall speed is just
$$\frac{20 \text{mi}}{2 \text{hr}} = 10 \frac{\text{mi}}{\text{hr}}.$$ It works!
The following is multiple choice question (with options) to answer.
If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is: | [
"50 km",
"56 km",
"60 km",
"70 km"
] | A | Let the actual distance travelled be x km.
x/10 = (x+20)/14
14x = 10x + 200
4x = 200
x = 50 km.
answer :A |
AQUA-RAT | AQUA-RAT-34832 | Veritas Prep Reviews
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts [#permalink] 10 Apr 2012, 03:29
2
KUDOS
For anybody who is used to solving mixture probs using the cross method
Attachment:
IMG.JPG [ 11.34 KiB | Viewed 8177 times ]
Hence, the ratio of the two gives 1/7 i.e 1 part of 1st bar and 7 parts of second.
Please notice that the question can be easily twisted to ask for the weight of first bar for minimum possible integer weight of final bar (Figure 8kg not provided) in which case this method would certainly help.
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts [#permalink] 24 Jul 2012, 22:05
can some one solve it using cross method ?
A portion of the 85% solution of chemicals was replaced with an equal amount of 20% solution of chemicals. As a result, 40% solution of chemicals resulted. What part of the original solution was replaced?
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Re: PS: Melted Gold-Silver [#permalink] 24 Jul 2012, 22:20
VeritasPrepKarishma wrote:
You can easily use the scale method here. The scale method is explained here:
tough-ds-105651.html#p828579
Focus on any one of the two elements say Gold.
First bar has 2/5 gold. Second bar has 3/10 gold and mixture has 5/16 gold.
Make the fractions comparable for easy calculation i.e. give them the same denominator. LCM of 5, 10 and 16 is 80.
First bar has 32/80 gold. Second bar has 24/80 gold and mixture has 25/80 gold.
Attachment:
Ques3.jpg
The following is multiple choice question (with options) to answer.
Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these be mixed to get an alloy 15 times as heavy as water? | [
"1:2",
"3:2",
"4:1",
"5:2"
] | B | G = 19W
C = 9W
Let 1gm of gold mixed with x gm of copper to get 1+x gm of the alloy
1gm gold + x gm copper = x+1 gm of alloy
19W+9Wx = x+1 * 15W
19+9x = 15(x+1)
x = 2/3
Ratio of gold with copper = 1:2/3 = 3:2
Answer is B |
AQUA-RAT | AQUA-RAT-34833 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
_________________
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Posts: 1059
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GPA: 3.82
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
The average of four consecutive even numbers is one-fourth of the sum of these numbers. What is the difference between the first and the last number? | [
"4",
"6",
"2",
"Cannot be determined"
] | B | Let the four consecutive even nos. be 2x, 2x + 2, 2x + 4 and 2x + 6 respectively.
Reqd difference = 2x + 6 – 2x = 6
Answer B |
AQUA-RAT | AQUA-RAT-34834 | of a circle and therefore its area is a fraction of the area of the whole circle. com Area and Perimeter of Circle and Semi-Circle Perimeter. 1) A circle has a radius of 5 cm. Area of rectangle = 4 × 8 = 32 m² A circle of radius 8 cm is cut into 6 parts of equal size, as shown in the diagram. A button is made in the shape of a circle, with two congruent rectangles cut out, as shown in the diagram. The area of a partial circle is the area of the circular segment equal to the area of the circular sector minus the area of the triangular portion formed from the. So cropping is quick, highly secured and consumes less bandwidth. The dimensions are m = 80. To find the perimeter of a rectangle, add the lengths of the rectangle's four sides. Cut a rectangle (more or less about 8 X 16 inches) of your 'swimsuit' stretch fabric. Tie the ribbon to zipper pull. The perimeter of a plane 2-D figure is the length of the outer periphery of the figure. **Many of our plants are listed as Not a Good Fit for Kid Traffic/Dog Traffic/Around Pools. Geometry calculator solving for arc length given central angle and circle radius. – Project-Based Learning. But we could get a better result if we divided the circle into 25 sectors (23 with an angle of 15° and 2 with an angle of 7. Semi-circular arcs are drawn with the sides of the rectangle as diameters. 99) Find great deals on the latest styles of Half circle area rugs. To find the total area, find the area of each part and add them together. The figure can be any regular or irregular polygon; it does not have to be a regular geometric figure. Angles Games. 68 cm 2 Example #2: Find the surface area of a cylinder with a radius of 4 cm, and a height of 3 cm. A filled rounded rectangle with b=0 (or a=0) is called a stadium. 14 for pie , and do not round your answer. cropping is much Faster, since we are not uploading your images to our server. The rectangle has a dimensions 5 length and 4 width, the semicircle has a radius of 2 using 3. Hence the diameter of a circle with area equal to 40 square cm is found to be 7. The round cut-out
The following is multiple choice question (with options) to answer.
A 25 cm wide path is to be made around a circular garden having a diameter of 4 meters. Approximate area of the path is square meters is? | [
"3.34",
"3.36",
"3.39",
"3.35"
] | A | Area of the path = Area of the outer circle - Area of the inner circle = ∏{4/2 + 25/100}2 - ∏[4/2]2
= ∏[2.252 - 22] = ∏(0.25)(4.25) { (a2 - b2 = (a - b)(a + b) }
= (3.14)(1/4)(17/4) = 53.38/16
= 3.34 sq m
Answer: A |
AQUA-RAT | AQUA-RAT-34835 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A car covers a distance of 526 km in 9 hours. Find its speed? | [
"104 kmph",
"289 kmph",
"58 kmph",
"277 kmph"
] | C | 526/9
= 58 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-34836 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A group of men decided to do a work in 40 days, but 5 of them became absent. If the rest of the group did the work in 60 days, Find the original number of men? | [
"60",
"15",
"40",
"10"
] | B | Original number of men = 5*60 / (60-40) = 15
Answer is B |
AQUA-RAT | AQUA-RAT-34837 | $$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\
The following is multiple choice question (with options) to answer.
6 + 6+ 6 + 2 × 6^2 + 2 × 6^3 + 2 × 6^4 + 2 × 6^5 + 2 × 6^6 + 2 × 6^7 = | [
"6^3",
"6^8",
"6^2",
"6^1"
] | B | We have the sum of 9 terms. Now, if all terms were equal to the largest term 2*6^7 we would have: sum=9*(2*6^7)=18*6^7=~6^9, so the actual sum is less than 6^9 and more than 6^7 (option E) as the last term is already more than that. So the answer is clearly B.
Answer: B |
AQUA-RAT | AQUA-RAT-34838 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Hari purchased 20 dozen notebook at Rs. 48 per dozen. He sold 8 dozen at 10% profit and remaining 12 dozen with 20% profit. What is his profit percentage in the transaction? | [
"8%",
"15%",
"16%",
"none of the above"
] | C | Explanation :
C.P of 20 dozen = Rs.(48 x 20) = Rs.960
C.P of 8 dozen = Rs. (48 x8) = Rs. 384.
C.P of 12 dozen =Rs. (960 384) = Rs. 576
Total S.P = Rs. (110/100 x 384 +120/100 x 576) = Rs.1113.60
Profit % = (153.60/960 x 100)% =16%
Answer : C |
AQUA-RAT | AQUA-RAT-34839 | Similar Questions
1). How many different signals , can be made by 5 flags from 8 flags of different colors?
A). 6270 B). 1680 C). 20160 D). 6720
-- View Answer
2). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets?
A). 12 B). 64 C). 256 D). 60
-- View Answer
3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together?
A). 10! B). 14!/(4!) C). 151200 D). 3628800
-- View Answer
4). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible?
A). 120 B). 100000 C). 6720 D). 30240
-- View Answer
5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible?
A). 72 B). 144 C). 288 D). 234
-- View Answer
6). Find the number of ways, in which 12 different beads can be arranged to form a necklace.
A). 11! / 2 B). 10! / 2 C). 12! / 2 D). Couldn't be determined
-- View Answer
7). 20 persons were invited to a party. In how many ways, they and the host can be seated at a circular table?
A). 18! B). 19! C). 20! D). Couldn't be determined
-- View Answer
8). A committee of 5 members is going to be formed from 3 trainees, 4 professors and 6 research associates. How many ways can they be selected, if in a committee, there are 2 trainees and 3 research associates?
A). 15 B). 45 C). 60 D). 9
-- View Answer
9). In how many ways, a committee of 3 men and 2 women can be formed out of a total of 4 men and 4 women?
The following is multiple choice question (with options) to answer.
If a 3-member subcommittee is to be formed from a certain 4-member committee, how many different such subcommittee are possible? | [
"4",
"18",
"20",
"108"
] | A | Another way:
1st member can be selected in 4 ways
2nd can be selected in 3 ways
3rd can be selected in 2 ways
So total ways : 24
But to avoid the similar scenarios 24/3!=4
A |
AQUA-RAT | AQUA-RAT-34840 | => $$\frac{74}{100} \times x$$ = 1,11,000
=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000
$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000
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The following is multiple choice question (with options) to answer.
Anand and Deepak started a business investing Rs. 3500 and Rs. 7000 respectively. Out of a total profit of Rs. 2000, Deepak's share is: | [
"1333.3",
"2687.1",
"1200.6",
"1580.2"
] | A | Ratio of their shares = 3500 : 7000 = 1:2
Deepak's share = 2000* 2/3 = Rs. 1333.3.
Answer: A |
AQUA-RAT | AQUA-RAT-34841 | Say colour 1 is used twice.
There are (5×4) /2 ways of painting 2 out of the 5 buildings.
Now there are 4 colors, so the above is true for each of the 4 colors.
We have 4 × [(5×4)/2] ways of painting 2 out of the the 5 buildings with the same color.
3 remaining buildings still need to be painted with the remaining 3 different colors.
For each of the ways where 2 equal colors have been used on 2 out of 5 buildings we can paint the remaining 3 buildings in 3×2×1 ways
Altogether: 4 × [(5×4)/2] × (3×2×1) = 240.
The following is multiple choice question (with options) to answer.
Before lunch, a painter can paint a wall at a rate of r square feet per minute. The same painter, after lunch and a cup of coffee, can paint the same wall at a rate r + 1 square feet per minute, and the same wall would be finished in 30 seconds less time. What is the value of r? | [
"1/3",
"3",
"1/2",
"1"
] | D | Convert 30 seconds to 1/2 minute
Equation: 1/r = 1/(r + 1) + 1/2
Plug options into the equation:
D fits: 1/1 = 1/(1 + 1) + 1/2
Answer: D |
AQUA-RAT | AQUA-RAT-34842 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 10 metres apart, how many poles will be needed ? | [
"28",
"56",
"57",
"58"
] | A | Solution
Perimeter of the plot = 2(90 + 50) = 280 m.
∴ Number of poles = [280/10] =28m Answer A |
AQUA-RAT | AQUA-RAT-34843 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
The value of a machine depreciates at 20% per annum. If its present value is Rs. 1,50,000, at what price should it be sold after two years such that a profit of Rs. 20,000 is made? | [
"328897",
"120000",
"877888",
"116000"
] | D | The value of the machine after two years = 0.8 * 0.8 * 1,50,000 = Rs. 96,000
SP such that a profit of Rs. 20,000 is made = 96,000 + 20,000 = Rs. 1,16,000
Answer: D |
AQUA-RAT | AQUA-RAT-34844 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
A truck covers a distance of 240 km at a certain speed in 8 hours. How much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 0 km more than that travelled by the truck ? | [
"6 hours",
"5 hours",
"7 hours",
"8 hours"
] | B | Explanation :
Speed of the truck = Distance/time = 240/8 = 30 kmph
Now, speed of car = (speed of truck + 18) kmph = (30 + 18) = 48 kmph
Distance travelled by car = 240 + 0 = 240 km
Time taken by car = Distance/Speed = 240/48 = 5 hours.
Answer –B |
AQUA-RAT | AQUA-RAT-34845 | Quick way
Use Smart Numbers
Give 100 for the initial amount
Then you will have 50-0.25x = 30
x = 80
So % is 80/100 is 80%
Hope it helps
Cheers!
J
SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance
Re: If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
29 May 2014, 11:41
Or one can use differentials to slve
Initially 50% alcohol
Then 25% alcohol
Resulting mixture 30% alcohol
Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4
Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.
Hope this helps
Cheers
J
Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 463
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
02 Jul 2014, 17:12
Bunuel wrote:
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
The following is multiple choice question (with options) to answer.
Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 150 milliliters of solution X to create a solution that is 25 percent alcohol by volume? | [
"250/3",
"500/3",
"400",
"450"
] | D | we know that X is 10% , Y is 30% and W.Avg = 25% . what does this mean with respect to W.Avg technique ?
W.Avg is 1 portion away from Y and 3 portion away from X so for every 1 portion of X we will have to add 3 portions of Y.
If X = 150 then Y = 450
ANSWER:D |
AQUA-RAT | AQUA-RAT-34846 | Also say that L is the length of the square.
You can at least state these equations :
• S1+S2+S3+S4+S5+S6 = $$L^2$$
• S1+S6 = $$\frac{1}{2}\pi\left(\frac{L}{2}\right)^2$$
• S1+S2+S5+S6 = $$\frac{\pi L^2}{4}$$
• S1+S2+S3 = $$\frac{1}{2}\pi\left(\frac{L}{2}\right)^2$$
• S3+S4 = $$\frac{(2L)^2-\pi L^2}{4}$$
• S2+S5 = $$\frac{\pi L^2-\pi\left(\frac{L}{2}\right)^2}{4}$$
Alas these are not independent, but I'm pretty sure that you can find six independent ones like this.
• This approach cannot possibly work unless you find an equation involving $\tan^{-1}(1/2)$. – Rahul Dec 3 '18 at 4:17
Let's assume we have 4 semi circles inside ABCD. There will be 4 intersections and all will be equal to DP.So we can calculate DP by 4 times area of DP + the area of square = 2 times the area of circle (r=5) so DP= 50/4 (pi=3) İF we assume one more semi circle in ABCD (APB) and one more quarter circle in ABCD (CDHB) so area of 2 quarter circles - the area of the square will give us the area of intersection DB 10*10*3/2= 150 10*10=100 150-100= 50 In half the area of intersection there are 2 red areas and DP so 50/2= 25 and (25-(25/2))/2=7.5 will be our answer
## protected by Community♦Apr 21 at 1:17
The following is multiple choice question (with options) to answer.
If the side length of Square B is three times that of Square A, the area of Square B is how many times the area of Square A? | [
"9",
"8",
"6",
"3"
] | A | Let x be the side length of Square A.
Then the area of Square A is x^2.
The area of Square B is (3x)^2=9x^2.
The answer is A. |
AQUA-RAT | AQUA-RAT-34847 | ## 35.
The Correct Answer is (1/29) — The probability of choosing one student wanting to enter finance is 6/30. The probability of choosing another student wanting to enter finance is then 5/29. The probability of both of these events occurring is therefore (6/30)(5/29) = 30/870, which can be reduced to 1/29.
## 36.
The Correct Answer is (40) — Since the small triangle and the entire triangle share two angles (the one on the right, and the right angle), they must share all three angles, and therefore are similar triangles. You can use the Pythagorean Theorem on the bigger triangle to find that the length of the horizontal side is the square root of 1002 - 602, which is 80. (This is a special 3-4-5 triangle with the side lengths multiplied by 20). Since the ratio of the vertical side to the horizontal side of the big triangle is 60/80 = 3/4, this must be the same as the corresponding ratio in the small triangle. Since the vertical side has length 30, the length marked x must have length 40.
## 37.
The Correct Answer is (10) — There are 6 full cages of mice, which means there are 6 × 5 = 30 mice. If we let n be the number of male mice, then the number of female mice is twice that at 2n. There are 30 mice in total, so n + 2n = 30. Solving for n gives you n = 10. Since n is the number of male mice, there are 10 male mice.
## 38.
The Correct Answer is (17) — The reduced cost of maintaining each cage is 0.5($1.25) =$0.625 per cage per day. The student needs 21 cages to house 102 mice, so her daily cost is $0.625 × 21 =$13.125. She has a budget of $225, so she can afford to maintain the cages for$225/\$13.215 = 17.14 days. Rounding this to the nearest whole day gives you 17 days.
The following is multiple choice question (with options) to answer.
5 cats are about to jump on 5 different mice. What is the probability that exactly one cat will jump on a single mouse? | [
"5*5",
"1/5!",
"5!/5^5",
"5!/5^2"
] | C | Each cat out of 5 has 5 options, hence total # of outcomes is 5^5;
Favorable outcomes will be 5!, which is # of ways to assign 5 different mice to 5 cats:
1-2-3-4-5 (mice)
A-B-C-D-E (cats)
B-A-C-D-E (cats)
B-C-A-D-E (cats)
...
So basically # of arrangements of 5 distinct objects: 5!.
P=favorable/total=5!/5^5
Answer: C. |
AQUA-RAT | AQUA-RAT-34848 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A dishonest dealer professes to sell goods at the cost price but uses a weight of 800 grams per kg, what is his percent? | [
"29%",
"25%",
"95%",
"26%"
] | B | 800 --- 200
100 --- ? => 25%
Answer: B |
AQUA-RAT | AQUA-RAT-34849 | # How can I find the minimum number of tiles?
Consider an $$n\times n$$ chessboard whose top-left corner is colored white. But Alice likes darkness, so she wants you to cover those white cells for her. The only tool you have are black L-shaped tiles each of which covers $$3$$ unit cells.
Formally, each tile covers unit cells satisfying the following:
1. Two of the cells are adjacent to the third (shares a side).
2. All three of the cells do not lie on the same row or same column.
3. No two tiles should overlap (cover the same cell) or go outside the board.
Since these tiles cost a lot, you have to cover all the white cells using the minimum number tiles.
### Example: $$1\times 1$$
Answer: Impossible, there's a single cell which is white. Since one tile needs $$3$$ empty cells, there's no way to cover this cell.
### Example: $$4\times 4$$
Answer: $$4$$ ($$4$$ tiles can be placed as shown)
### Example: $$7 \times 7$$
If each tile can be represented by a number, and each uncovered piece of board can be represented by 'zero', then the answer for a $$7 \times 7$$ board is $$16$$:
$$\begin{bmatrix} 16& 16& 15& 15& 14& 14& 13 \\ 16& 12& 15& 11& 14& 13& 13 \\ 12& 12& 11& 11& 10& 10& 9 \\ 8& 8& 7& 6& 10& 9& 9 \\ 8& 7& 7& 6& 6& 2& 2 \\ 5& 5& 4& 3& 3& 1& 2 \\ 5& 0& 4& 4& 3& 1& 1\\ \end{bmatrix}$$
### Question
For any given $$n$$, what will be the minimum number of tiles?
(Note: Answer exists for odd value of $$n \geq 7$$)
An image of the 7×7 solution as mentioned by OP:
And one possible way to extend this to a 9×9 solution (and further):
The following is multiple choice question (with options) to answer.
What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 06 cm? | [
"153",
"187",
"540",
"748"
] | A | The marbles used to tile the floor are square marbles.
Therefore, the length of the marble=width of the marble.the length of the marble=width of the marble.
As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 06. And it should be the highest factor of 5 m 78 cm and 3m 06.
5 m 78 cm = 578 cm and 3 m 06 cm = 306 cm.
The HCF of 578 and 306 = 34.
Hence, the side of the square is 34.
The number of such square marbles required,
=578×306/34×34
=153 marbles
Option(A) |
AQUA-RAT | AQUA-RAT-34850 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A man rows downstream 30 km and upstream 18 km, taking 5 hours each time. What is the velocity of the stream (current)? | [
"1.2 km/hr",
"1.5 km/hr",
"2.5 km/hr",
"1.8 km/hr"
] | A | 30/(B+5) = 18/(B-5) = 5
? B+5 = 6 km/hr
B - 5 = 3.6 km/hr
? 5 = 1.2 km/hr
Answer: A. |
AQUA-RAT | AQUA-RAT-34851 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Boy purchased two flats for Rs.5,00,000. On one he gains 15 % while on the other he losses 15%. How much does he gain or lose? | [
"2%",
"2.12%",
"2.25%",
"2.30%"
] | C | Generally in such cases, there is always loss.
So always remember, when two materials are sold and if one material gets profit and the other gets a loss, then use the trick shown below to calculate the loss.
Loss% = Common loss and gain% 2 = x 2
10 10
Therefore, here common loss and gain % = 15%
Hence,
Loss% = 15 2 = 2.25%
C |
AQUA-RAT | AQUA-RAT-34852 | ### Show Tags
22 May 2018, 02:57
belagerfeld wrote:
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
This type of sums can also be solved using Fermet's Theorem approach.
Refer photo attached below:
Attachments
WhatsApp Image 2018-05-22 at 3.18.54 PM.jpeg [ 90.53 KiB | Viewed 2464 times ]
Re: What is the units digit of 2222^(333)*3333^(222) ? &nbs [#permalink] 22 May 2018, 02:57
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
If a is a positive integer, and if the units' digit of a^2 is 4 and the units' digit of (a + 1)^2 is 1, what is the units' digit of (a + 2)^2 ? | [
"\t0",
"\t2",
"\t4",
"\t6"
] | A | A for me.
_1² = _1
_2² = _4
_3² = _9
_4² = _6
_5² = _5
_6² = _6
_7² = _9
_8² = _4
_9² = _1
_0² = _0
On the list, the only number that squared has the units digit = 4 preceding a number that squared has the units digit = 1 is the _8.
So, a = _8, that has square 4. The next square is 1 andnext nextsquare is 0. |
AQUA-RAT | AQUA-RAT-34853 | human-biology, human-anatomy, human-genetics
Title: Are males taller than females in humans? Is there any scientific evidence that in humans males are taller than females? And if so, what is the reason that they are taller (please include genes or hormones that accounts for human growth and how they are affected in males and females)? Are males taller than females?
Best data I could find come from the Statistical Abstract of the United States (1999) > Section 3. Here is a table reporting the percentage of the male and female population which height is lower than a given threshold
Note that this data collection was done among students in US universities and is therefore not representative of the whole world or even the whole country.
Does height follow a bimodal distribution?
A difference in height between males and females is often used as a classical example in introductory statistic class to exemplify a bimodal distribution as seen in this picture
and on these (a priori fake) data
However, Schilling et al. 2002 argued that while the difference in mean height between the sexes is real, this difference is too small relative to the variance in height within each sex to be clearly depicted on a graph.
Note that I found this non-peer-reviewed paper which shows real data that display a truly bimodal distribution of height.
Genetics of height
The question why are they taller? is very broad. I will just focus to give you some hints about the genetics of height in humans. First, you want to make sure you understand the concept of heritability.
Evoy and Vissher 2009 report a heritability coefficient of 0.8. This estimate is impressively high - only a few phenotypic traits have such high heritability. They also review articles discussing that 50 loci are correlated with variation in height (actually, today, more than 500 loci are known to contribute to height, see the link in AlexDeLarge's comment to this answer). However, these loci are not sufficient to explain the whole heritability observed (common missing heritability issue). Yang et al. 2010 provide evidence that the remaining heritability is due to incomplete linkage
disequilibrium between causal variants and loci of weak effects. In short, height is a highly polygenic trait.
Related post
You should have a look at Is there a genetic reason explaining the difference of the height of male and female? for more information.
The following is multiple choice question (with options) to answer.
On my sister's 15th birthday, she was 159 cm in height, having grown 6% since the year before. How tall was she the previous year? | [
"150 cm",
"140 cm",
"142 cm",
"154 cm"
] | A | Explanation :
Given that height on 15th birthday = 159 cm and growth = 6%
Let the previous year height = x
Then height on 15th birthday = x ×100+6/100=x×106/100
⇒159 = x × (106/100)
⇒�x=159×100/106=1.5×100=150;cm
Answer : Option A |
AQUA-RAT | AQUA-RAT-34854 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg. | [
"1 : 3.",
"2 : 3.",
"2 : 4",
"1 : 4"
] | B | Cost of 1 kg of 1st kind=720 p
Cost of 1 kg of 2nd kind=570 p
Mean Price=630 p
Cost of 1 kg of 1st kind - Mean Price=720-630=90
Cost of 1 kg of 2nd kind- Mean Price=570-630=60
Required ratio = 60 : 90 = 2 : 3.
Answer:B |
AQUA-RAT | AQUA-RAT-34855 | Hey, thanks for your help guys. For a minute there, I thought that this theoretical person could not safely expect to live to be 82 years old.
9. Jun 16, 2012
### SW VandeCarr
In fact, on a purely probabilistic basis, for any finite time no matter how large, there is a non zero probability that a person would survive that long. So for a sufficiently large population, there would be a theoretic person that would live 100,000 years. This, of course, has no basis in biology.
In terms of the probability of being murdered, the model would not hold for the 100,000 year old person. In terms of the model, probably the best one can do is assume the proportion of causes of death would be constant. The calculation above needs to be corrected for overall survival in terms of death from any cause.
Last edited: Jun 16, 2012
10. Jun 16, 2012
### viraltux
Interesting... but 0.37% is not that small percentage, don't you think? That means, roughly speaking, that a community of around 300 persons can expect that one of them will be murdered.
If you consider that the number of people we know plus acquaintances can easily be around 300 persons that would mean that most 82 year old persons know of someone in their circles who has been murdered. Mmm... that might be an interesting survey.
11. Jun 16, 2012
### SW VandeCarr
As I said in my previous post, this is a misapplication of statistics. You have to consider survival in terms of all cause death. If you just consider the murder rate, then at some point nearly everyone gets murdered.
12. Jun 16, 2012
### moonman239
I know that.
This person will not die until he reaches age 82, if he is not murdered. As mentioned before, this person has a 68% chance of living to be 82.
13. Jun 16, 2012
### D H
Staff Emeritus
The probability of living to 82 per this problem is 99.63%, not 68%. You missed the decimal point on the 0.37%.
14. Jun 16, 2012
### SW VandeCarr
The following is multiple choice question (with options) to answer.
The last day of a century cannot be | [
"Monday",
"Wednesday",
"Tuesday",
"sunday"
] | C | 100 years contain 5 odd days.
Last day of 1st century is Friday.
200 years contain (5 x 2) 3 odd days.
Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 1 odd day.
Last day of 3rd century is Monday.
400 years contain 0 odd day.
Last day of 4th century is Sunday.
This cycle is repeated.
Last day of a century cannot be Tuesday or Thursday or Saturday.
Answer: Option C |
AQUA-RAT | AQUA-RAT-34856 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A train starts full of passengers at the first station it drops 1/3 of the passengers and takes 280 more at the second station it drops one half the new total and takes twelve more. On arriving at the third station it is found to have 246 passengers. Find the no of passengers in the beginning? | [
"292",
"180",
"192",
"282"
] | D | Let no of passengers in the beginning be X
After first station no passengers=(X-X/3)+280=2X/3 +280
After second station no passengers =1/2(2X/3+280)+12
1/2(2X/3+280)+12=246
2X/3+280=2*234 = 468 - 280 = 188
2X/3=188
X=282. So the answer is option D) 282. |
AQUA-RAT | AQUA-RAT-34857 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man goes downstream at 12 kmph, and upstream8 kmph. The speed of the stream is | [
"2kmph",
"4kmph",
"16kmph",
"2.5kmph"
] | A | Speed of the stream = 1/2(12-8)kmph
=2kmph.
correct option A |
AQUA-RAT | AQUA-RAT-34858 | • Wooow, nice answear man, I really appreciate that and it actually opened my eyes to this economics field. But, your billion dollars example just take into account 1 iteration, I too would choose $10^9$, because is guaranteed, and because, as you said, $10^9$ to $10^{10}$ is nothing in this order of magnitude. In my example, I want to take into account multiple operations, not just one; and favorable odds, not a coin flip. Nov 27, 2021 at 7:03
• As I said, I made a Python simulation, and using x = 100 (100 USD), in 1 iteration, 10 interations, 1,000,000 iterations (that took a while to compute, almost burned my computer hahaha), the income for the first option was getting closer to 148.02 USD and the second option to 115.19 USD. Nov 27, 2021 at 7:03
• @rookie Well, I believe your mathematics is correct, so I'm guessing your code is the one that is wrong. Could you provide the code? Nov 27, 2021 at 7:29
• Okay, I see the problem. I thought that when you "fail", you'll lose all the interest not just for that month, but all the interest for all the months. This is what your mathematics do. Nov 27, 2021 at 7:52
The following is multiple choice question (with options) to answer.
Siddharth wants to borrow Rs.9000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction? | [
"s.21.60",
"s.32.40",
"s.21.18",
"s.21.98"
] | B | Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.9,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth's income = C.I - S.I
= p[1 + r/ 100]n - p - pnr/100
= p{ [1 + r/ 100]2 - 1 - nr/100
= 9,000{ [1 + 6/100]2 - 1 - 12/100}
= 9,000 {(1.06)2- 1 - 0.12} = 9,000(1.1236 - 1 - 0.12)
= 9,000 (0.0036) = 9 * 3.6
= Rs.32.40
Answer: B |
AQUA-RAT | AQUA-RAT-34859 | finite-automata, regular-expressions
Title: Regular Expression representing the following language I am having trouble understanding how to write a regular expression for the set of words that contain at least two b's and at least two a's, where the alphabet is {a,b}.
I understand that set of words that contain at least two b's is: [1] (a+b)*b(a+b)*b(a+b)* and the set of words that contain at least two a's is: [2] (a+b)*a(a+b)*a(a+b)*. In addition, I understand that the set of words that contain at least one a and at least one b is: [3] (a+b)*(ab+ba)(a+b)* (or at least I think that is correct).
I know the easy solution would be to use an intersection between [1] and [2]; however, I would like to understand how to accomplish this without using an intersection.
I know I would need to have (a+b)* at the beginning and at the end to say that any string made up of a's and b's can be at the beginning and at the end. But I am having trouble understanding the logic between those two points.
Just looking for some direction as I do not know where to go from there. You have written (a+b)(ab+ba)(a+b)
This contains atleast 1 a & at least 1 b.
So what are ab & ba ? ab & ba are permutations of string you can form with at least one a & at least one b !
As you are asking for hint,
You can write something like
(a+b)* (All Permutations of aabb ) (a+b)*
Hope this helps.
There are just 6 permutations (4!/2!2!) of aabb here, so it wont be hard to calcualate them ! .
The following is multiple choice question (with options) to answer.
How many such pairs of letters are there in the word EXPRESSION which have as many letters between them in the word as in the alphabet ? | [
"4",
"5",
"3",
"2"
] | A | (E,N)
(P,S)
(O.N)
(X,S)
ANSWER:A |
AQUA-RAT | AQUA-RAT-34860 | Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
To put it in words, think of it this way:
Is x^2 + y^2 > 100?
(2) (x + y)^2 > 200
which means: x^2 + y^2 + 2xy > 200
The following is multiple choice question (with options) to answer.
What is the value of 100p2 ? | [
"9801",
"12000",
"5600",
"9900"
] | D | Explanation :
100p2 = 100 x 99 = 9900. Answer : Option D |
AQUA-RAT | AQUA-RAT-34861 | ## Solution 2
With investing just a little bit of time, we can manually calculate 19!. If we prime factorize 19!, it becomes $2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19$. This looks complicated, but we can use elimination methods to make it simpler. $2^3 \cdot 5^3 = 1000$, and $7 \cdot 11 \cdot 13 \cdot = 1001$. If we put these aside for a moment, we have $2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19$ left from the original 19!. $2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192$, and $3^8 = (3^4)^2 = 81^2 = 6561$. We have the 2's and 3's out of the way, and then we have $7 \cdot 17 \cdot 19 = 2261$. Now if we multiply all the values calculated, we get $1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000$. Thus $T = 4, M = 8, H = 0$, and the answer $T + M + H = 12$, thus (C).
## Solution 3
The following is multiple choice question (with options) to answer.
What is the greatest value of n such that 18^n is a factor of 17! ? | [
"1",
"2",
"3",
"4"
] | C | Another way to solve this question :
We know that 18 = 2 *3^2.
Hence, we need to find total number of 3s in 18! ( As out of 3 and 2 , we will have count of 3 least)
No. of 3s = 18/3 =6/3 =2.
Add the numbers in Bold, you will get total number of 3s = 8.
Hence, total number of 3^2 = 8/2 = 4.
Hence, answer is 3. C |
AQUA-RAT | AQUA-RAT-34862 | your kids. The volume formula for a rectangular box is height x width x length, as seen in the figure below: To calculate the volume of a box or rectangular tank you need three dimensions: width, length, and height. Find the dimensions of the box that minimize the amount of material used. The calculated volume for the measurement is a minimum value. Rectangular Box Calculate the length, width, height, or volume of a rectangular shaped object such as a box or board. This is the main file. You must have a three-dimensional object in order to find volume. Volume is the amount of space enclosed by an object. Our numerical solutions utilize a cubic solver. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides. Volume of a Cuboid. The largest possible volume for a box with a square bottom and no top that is constructed out of 1200$\mathrm{cm}^2$of material is 4000$\mathrm{cm}^3$. For example, enter the side length and the volume will be calculated. 314666572222 cubic feet, or 28. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. CALCULATE VOLUME OF BOX. So: Answer. A container with square base, vertical sides, and open top is to be made from 1000ft^2 of material. A cuboid is a box-shaped object. 1) Enter a valid Start value into text box below, default is "1", 2) Select an increment value from the list below, default is 1, 3) Select an accuracy (significant digits) value from the list below, default is 5, 4) Most cases the result will appear immediately, 5) Click on the "Create Table" button. Everyone has a personal profile and you can use yours to choose colours that really suit your face. Volume of a square pyramid given base side and height. Volume of a cube - cubes, what is volume, how to find the volume of a cube, how to solve word problems about cubes, nets of a cube, rectangular solids, prisms, cylinders, spheres, cones, pyramids, nets of solids, examples and step by step solutions, worksheets. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. For
The following is multiple choice question (with options) to answer.
A rectangular box of volume V has width, depth and height in the ratio of 2:2:1 (in that order). What is the width as a function of V? | [
"(3V/2)^(1/4)",
"(7V/6)^(1/4)",
"(9V/2)^(1/4)",
"(16V/4)^(1/4)"
] | D | We are given width = 2, depth = 2, height = 1.
Then Volume V = Width x Depth x Height = 2*2*1 = 4
i.e. V = 4
The correct option should result in 2 (the width) on substituting the value of V=4.
Checking options with V=4:
A) (3V/2)^(1/4) = (6/2)^(1/4) = 3^(1/4) INCORRECT ANSWER
B) (7V/6)^(1/4) = (28/6)^(1/4) INCORRECT ANSWER
C) (9V/2)^(1/4) = (36/2)^(1/4) = 18^(1/4) INCORRECT ANSWER
D) (16V/4)^(1/4) = (64/4)^(1/4) = 16^(1/4) = 2 CORRECT ANSWER
E) (4V/2)^(1/4) = (16/2)^(1/4) = 8^(1/4) INCORRECT ANSWER
Answer: Option D |
AQUA-RAT | AQUA-RAT-34863 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
A man covers a distance of 1200 km in 70 days resting 9 hours a day, if he rests 10 hours a day and walks with speed 1½ times of the previous in how many days will he cover 840 km ? | [
"25 days",
"75 days",
"35 days",
"15 days"
] | C | Explanation:
Distance d = 1200km
let S be the speed
he walks 15 hours a day(i.e 24 - 9)
so totally he walks for 70 x 15 = 1050hrs.
S = 1200/1050 => 120/105 = 24/21 => 8/7kmph
given 1 1/2 of previous speed
so 3/2 * 8/7= 24/14 = 12/7
New speed = 12/7kmph
Now he rests 10 hrs a day that means he walks 14 hrs a day.
time = 840 x 7 /12 => 490 hrs
=> 490/14 = 35 days
So he will take 35 days to cover 840 km.
Answer: C |
AQUA-RAT | AQUA-RAT-34864 | 3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100
The following is multiple choice question (with options) to answer.
A boy gets some rupees from his mother and spends them on 5 stores. He spends one rupee more than half of the money. How much money he had at the time of entering the shop. | [
"60",
"61",
"62",
"63"
] | C | (x+2)/(2^n)=2
where n=no. of shops
so we get ,(x+2)/(2^5)=2
x=62
ANSWER:C |
AQUA-RAT | AQUA-RAT-34865 | So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. _________________ Intern Joined: 26 May 2012 Posts: 21 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 15 Jul 2012, 02:22 Bunuel wrote: farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?
(A) 8
(B) 12
(C) 16
(D) 20
(E) 24
Say $$x$$ is the original price of an orange, then:
$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.
So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. are there any other ways? Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 377 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 22 Oct 2016, 11:32 Set up: 12/p = x 12/0.60 = x+4 Manipulate and plug the first equation into the second --> you'll find p =$2
Thus $24/$2 per orange = 12 oranges
VP
Joined: 07 Dec 2014
Posts: 1128
Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
If apples cost x dollars per m dozen, how many dollars will it cost to buy n apples? | [
"xn/(12m)",
"xn/m",
"m/(xn)",
"12m/(xn)"
] | A | let it 2 dollars for 2 dozens(24) and we are asked to find the value n apples...let n be 4...
In order to buy 24 apples it took 2 dollars...then for 4 apples it'll take 4*2/12*2 = 1/3.. i.e. n*x/12*m.
Answer: option A is correct answer... |
AQUA-RAT | AQUA-RAT-34866 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
Interest Rate: 4% monthly Starting Balance: 124 Time Passed: 4 months How much interest has accrued if calculated as compound interest? What is the new total balance? | [
"21 & 145",
"22 & 146",
"23 & 145",
"20 & 140"
] | A | Compound Interest: Total Balance = P(1 + R)T P = principle = starting balance = $124 R = interest rate = 4% T = time = 4 years Total balance = principle × (1 + interest rate)time = 124 × (1 + (4 / 100))4 = 145 Interest accrued = total balance - starting balance = 145 - 124 = 21
Answer A |
AQUA-RAT | AQUA-RAT-34867 | To solve more problems based on Heron’s … Question Bank Solutions 4046. Similarity to Heron's Formula. We can easily draw many more quadrilaterals and we can identify many around us. 2 + base. Important Solutions 2865. Chapters. where is the semiperimeter, or half of the triangle's perimeter. 4. Ex 12.2, 2 Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. Program to find the angles of a quadrilateral in C++; Maximum area of rectangle possible with given perimeter in C++; Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts in C++; Find maximum volume of a cuboid from the given perimeter and area in C++; Maximum area rectangle by picking four sides from array in C++ side a: side b: side c: side d: sum of angles θ 1 +θ 2: area S . % Progress . Calculates the area and perimeter of a quadrilateral given four sides and two opposite angles. Sides of triangle are 12 cm, 13cm and 15cm. … or own an. Find its area. Now what I need is to implement a method get_area() which calculates the area of my quadrilateral, but I have no idea how. Watch Area and Perimeter of Quadrilaterals Videos tutorials for CBSE Class 8 Mathematics. Progress % Practice Now. Another construction of the point T A is to start at A and trace around triangle ABC half its perimeter, and similarly for T B and T C. Syllabus. Franchisee/Partner Enquiry (North) … Questionnaire. Change Equation Select to solve for a different unknown Scalene Triangle: No sides have equal length No angles are equal. A splitter of a triangle is a cevian (a segment from a vertex to the opposite side) that divides the perimeter into two equal lengths, this common length being called the semiperimeter of the triangle. It is classif The sides a n of regular inscribed polygons, where R is the radius of the circumscribed circle = a n = 2R sin 180 o /n; Area of a polygon of perimeter P and radius of in-circle … Properties of a quadrilateral. Find the Length
The following is multiple choice question (with options) to answer.
Find the perimeter and area of a square of side 19 cm. | [
"324",
"361",
"400",
"441"
] | B | We know that the perimeter of square = 4 × side
Side= 19 cm
Therefore, perimeter = 4 × 19 cm = 76 cm
Now, area of the square = (side × side) sq. units
= 19 × 19 cm²
= 361 cm²
ANSWER : B |
AQUA-RAT | AQUA-RAT-34868 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
The value of a machine depreciates at 20% per annum. If its present value is Rs. 1,50,000, at what price should it be sold after two years such that a profit of Rs. 28,000 is made? | [
"Rs. 1,10,000",
"Rs. 1,20,000",
"Rs. 1,24,000",
"Rs. 1,21,000"
] | C | The value of the machine after two years = 0.8 * 0.8 * 1,50,000 = Rs. 96,000
SP such that a profit of Rs. 24,000 is made = 96,000 + 28,000 = Rs. 1,24,000
ANSWER:C |
AQUA-RAT | AQUA-RAT-34869 | gene, genetics
Title: How long does it take for a person to lose all offsprings due to inheritance? From this I know we will only inherit some genetic informations from parents, which is about 50 percent. But the problem is, gene has finite size, after some generations a person leaves only $0.5 \times 0.5 \times 0.5$ parts of genes to the offsprings, and it will be casted into zero. My question is, how long does it required for a person to lose all genetic information in the world? I think your question conveys some misunderstanding.
A child is related to each parent by a factor of ½. Humans have a diploid genome, meaning they have two copies of each chromosome (see: autosome). When two humans reproduce, they each contribute one copy of each chromosome to the offspring, in other words, they contribute a haploid genome to make a diploid child. Genetic information is not "lost" - the genome is not shrinking by a factor of ½ every generation.
However, relatedness does decrease from generation to generation. You are related to each of your parents by a factor of ½, each of your grandparents by a factor of ½ $\times$ ½, your great-grandparents by a factor of ½ $\times$ ½ $\times$ ½... You are also related to your children by a factor of ½, you are related to your grandchildren by a factor of ½ $\times$ ½... You get the picture, right?
For example, imagine the genome carries just one gene. Your father carries alleles $AA$ at that locus, and your mother $aa$. You would then be $Aa$ and, because half of your alleles came from your father and the other half from your mother, be related to each by a factor of ½, but all three of you have the same number of genes (1) and that gene is the same length (in nucleotides, barring mutations) in all three.
The following is multiple choice question (with options) to answer.
A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 65 years, that of the parents is 32 years and that of the grandchildren is 5 years. The average age of the family is | [
"29 1/7 years",
"29 6/7 years",
"39 6/7 years",
"19 6/7 years"
] | B | Total age of the grandparents = 65 × 2=130
Total age of the parents = 32 × 2=64
Total age of the grandchildren = 5 × 3=15
Average age of the family = (130+64+15)/7
=209/7
=29 6/7 years
ANSWER:B |
AQUA-RAT | AQUA-RAT-34870 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
A rectangular lawn of length 200m by 120m has two roads running along its center, one along the length and the other along the width. If the width of the roads is 5m what is the area W covered by the two roads? | [
"400",
"1550",
"1575",
"1600"
] | C | Area Covered by Road Along the Length = 5*200 = 1000 Square Meter
Area Covered by Road Along the Width = 5*120 = 600 Square Meter
Common Area in both Roads (where the roads intersect) = Square with Side 5 meter = 5*5 = 25
Total Area of the Roads W = 1000+600-25 = 1575
Answer: option C |
AQUA-RAT | AQUA-RAT-34871 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row his boat with the stream at 20 km/h and against the stream in 8 km/h. The man's rate is? | [
"1 kmph",
"7 kmph",
"6 kmph",
"5 kmph"
] | C | DS = 20
US = 8
S = ?
S = (20 - 8)/2 = 6 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-34872 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The owner of a furniture shop charges his customer 32% more than the cost price. If a customer paid Rs. 5400 for a computer table, then what was the cost price of the computer table? | [
"Rs.4090",
"Rs.4067",
"Rs.6290",
"Rs.6725"
] | A | CP = SP * (100/(100 + profit%))
= 5400(100/132) = Rs.4090.
Answer: A |
AQUA-RAT | AQUA-RAT-34873 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
The difference between compound interest and simple interest on a certain amount of money at 5% per annum for 2 years is 15. Find the sum : | [
"4500",
"7500",
"5000",
"6000"
] | D | Sol.(d) Let the sum be 100.
Therefore, SI = 100×5×2100=10100×5×2100=10
and CI = 100(1+5100)2−100100(1+5100)2−100
∴ =100×21×2120×20−100=414=100×21×2120×20−100=414
Difference of CI and SI = 41⁄4 - 10 = 1⁄4
If the difference is 1⁄4 , the sum = 100
=> If the difference is 15, the sum
= 400 × 15 = 6000 Answer D |
AQUA-RAT | AQUA-RAT-34874 | $$C^2_w$$ # of selections of 2 women out of $$w$$ employees;
$$C^2_{10}$$ total # of selections of 2 representatives out of 10 employees.
Q is $$\frac{C^2_w}{C^2_{10}}>\frac{1}{2}$$ --> $$\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}$$ --> --> $$w(w-1)>45$$ --> $$w>7$$?
(1) More than 1/2 of the 10 employees are women --> $$w>5$$, not sufficient.
(2) The probability that both representatives selected will be men is less than 1/10 --> $$C^2_{(10-w)}$$ # of selections of 2 men out of $$10-w=m$$ employees --> $$\frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10}$$ --> $$\frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$ --> $$w>6$$, not sufficient
(1)+(2) $$w>5$$ and $$w>6$$: $$w$$ can be 7, answer NO or more than 7, answer YES. Not sufficient.
Hope it's clear.
_________________
Intern
Joined: 20 Feb 2012
Posts: 40
Re: If 2 different representatives are to be selected at random [#permalink]
### Show Tags
27 Feb 2012, 09:11
10
37
If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ?
The following is multiple choice question (with options) to answer.
C and D work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that C and D will both be chosen? | [
"A)1/3",
"B)1/4",
"C)1/15",
"D)3/8"
] | C | Total number of people = 6
Probability of selecting C first and then D is 1/6∗1/5=1/30
Probability of selecting D first and then C is 1/6∗1/5=1/30
Therefore probability of selecting Cand D for the review is 1/30+1/30=1/15
Answer is C |
AQUA-RAT | AQUA-RAT-34875 | Kudos [?]: 5 [0], given: 54
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
16 May 2013, 01:13
Bunuel wrote:
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
Hi Bunuel,
I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?
Thanks a lot.
Manager
Joined: 22 Apr 2013
Posts: 88
Followers: 1
Kudos [?]: 26 [0], given: 95
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
17 May 2013, 21:18
pranav123 wrote:
These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
I tried uploading the diagram but unsuccessful.
That's one of those tips that can make my life easier. I'm book marking this page. Thanks.
_________________
I do not beg for kudos.
Manager
Joined: 13 Jul 2013
Posts: 75
GMAT 1: 570 Q46 V24
Followers: 0
Kudos [?]: 8 [0], given: 21
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
26 Dec 2013, 23:04
Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played?
Then is 28*2 the correct approach?
Math Expert
Joined: 02 Sep 2009
Posts: 34057
Followers: 6082
Kudos [?]: 76440 [0], given: 9975
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
How many ways are there to award a gold, silver and bronze medal to 15 contending teams? | [
"15 × 14 × 13",
"10!/(3!7!)",
"10!/3!",
"360"
] | A | we clearly know that there can be only one winning team which deserves the gold medal. we can do the selection in 15 ways
if gold medal is given to 1 of the teams then only possible 14 teams can be considered for silver medal. we can do selection in 14 ways
similarly if gold and silver medals are awarded then only remaining 13 teams can be considered for a bronze medal. we can do the selection in 13 ways
Total number of ways to select the 3 possible medal winners = 15 * 14 * 13
Correct answer - A |
AQUA-RAT | AQUA-RAT-34876 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The tax on a commodity is diminished by 16% and its consumption increased by 15%. The effect on revenue is? | [
"2%",
"3.8%",
"3.4%",
"3.6%"
] | C | 100 * 100 = 10000
84 * 115 = 9660
-----------
10000-----------340
100-----------? => 3.4% decrease
Answer: C |
AQUA-RAT | AQUA-RAT-34877 | # Thread: show for a if a square matrix A...it will hold for A^T
1. ## show for a if a square matrix A...it will hold for A^T
Show that if for a square matrix $A$ satisfies $A^3+ 4A^2 - 2A + 7I = 0$ then so does $A^T$
I suspect that this is not always true so I need to do an example not the general case ( I partially suspect this cause the next question asks me if this is always true or sometimes false so I need to find a counter-example as well)
not really sure where to start this problem.
2. ## Re: show for a if a square matrix A...it will hold for A^T
Originally Posted by Jonroberts74
Show that if for a square matrix $A$ satisfies $A^3+ 4A^2 - 2A + 7I = 0$ then so does $A^T$
I suspect that this is not always true so I need to do an example not the general case ( I partially suspect this cause the next question asks me if this is always true or sometimes false so I need to find a counter-example as well)
not really sure where to start this problem.
$\left(A^T\right)^2=\left(A^2\right)^T$
$\left(A^T\right)^3=\left(A^3\right)^T$
$(A^3+4A^2-2A+7I)^T=0^T=0$
$(A^3)^T+(4A^2)^T-(2A)^T+(7I)^T=0$
$(A^T)^3+4(A^T)^2-2(A^T)+7I=0$
so $A^T$ satisfies the equation as well.
3. ## Re: show for a if a square matrix A...it will hold for A^T
The following is multiple choice question (with options) to answer.
If t > 0, which of the following could be true?
I. t^3 > t^2
II. t^2 = t
III. t^2 > t^3 | [
"I only",
"III",
"IIIII",
"All of the above"
] | D | 1. t^3 > t^2 Well this is true for all t > +1 (but not for values 0 < t < 1)
2. t^2 = t This is true for only one positive number, 1
3. t^2 > t^3 Again this is true only for values of t such that 0 < t < 1
So the answer is D, since there exists at least one value of t which satisfies the equation/ in equation(s) |
AQUA-RAT | AQUA-RAT-34878 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
Marts income is 60 percent more than Tims income and Tims income is 50 percent less than Juans income. What percentage of Juans income is Marts income | [
"124%",
"b) 120 %",
"c) 96 %",
"d) 80 %"
] | D | M = (160/100)T
T = (50/100)J
=> M = (80/100)J
Answer D. |
AQUA-RAT | AQUA-RAT-34879 | & = \frac{13^4}{\dfrac{52!}{4!48!}}\\[2mm] & = \frac{13^4}{\dbinom{52}{4}} \end{align*}
The following is multiple choice question (with options) to answer.
5358 x 53 = ? | [
"283974",
"283758",
"273298",
"273258"
] | A | A
5358 x 53 = 5358 x (50 + 3)
= 5358 x 50 + 5358 x 3
= 267900 + 16074
= 283974. |
AQUA-RAT | AQUA-RAT-34880 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
A dishonest milkman wants to make a profit on the selling of milk. He would like to mix water (costing nothing) with milk costing Rs.33 per litre so as to make a profit of 20% on cost when he sells the resulting milk and water mixture for Rs.36 In what ratio should he mix the water and milk? | [
"1:20",
"1:10",
"1:8",
"1:4"
] | B | cost needed to net a 20% profit:
(36-x)/x = .2
x=30
actual cost:
33
solution (x=liters of water needed to be added to the 1liter of milk):
33/(1+x)=30
x = 1/10
so to get the cost down to 30
Milk:Water
1:1/10
or (10/10):(1/10)
ANSWER:B |
AQUA-RAT | AQUA-RAT-34881 | 1021 671 1175 1021 11 25 1021 = 1021
1027 1008 1045 1027 18 19 1027 = 13 * 79
1027 128 1085 1027 2 31 1027 = 13 * 79
1033 928 1113 1033 16 21 1033 = 1033
1039 885 1144 1039 15 22 1039 = 1039
1051 384 1189 1051 6 29 1051 = 1051
1057 65 1088 1057 1 32 1057 = 7 * 151
1057 793 1200 1057 13 24 1057 = 7 * 151
1063 195 1147 1063 3 31 1063 = 1063
1069 744 1225 1069 12 25 1069 = 1069
1087 1003 1155 1087 17 21 1087 = 1087
1093 455 1247 1093 7 29 1093 = 1093
1099 640 1269 1099 10 27 1099 = 7 * 157
1099 915 1219 1099 15 23 1099 = 7 * 157
1117 585 1288 1117 9 28 1117 = 1117
1123 67 1155 1123 1 33 1123 = 1123
1129 201 1216 1129 3 32 1129 = 1129
1141 1121 1160 1141 19 20 1141 = 7 * 163
1141 335 1271 1141 5 31 1141 = 7 * 163
1147 1037 1232 1147 17 22 1147 = 31 * 37
1147 715 1323 1147 11 27 1147 = 31 * 37
1153 992 1265 1153 16 23 1153 = 1153
1159 136 1221 1159 2 33 1159 = 19 * 61
1159 469 1320 1159 7 30 1159 = 19 * 61
1171 896 1325 1171 14 25 1171 = 1171
1183 408 1333 1183 6 31 1183 = 7 * 13^2
1183 603 1363 1183 9 29 1183 = 7 * 13^2
1201 1159 1239 1201 19 21 1201 = 1201
1213 737 1400 1213 11 28 1213 = 1213
1231 680 1421 1231 10 29 1231 = 1231
1237 280 1353 1237 4 33 1237 = 1237
1249 871 1431 1249 13 27 1249 = 1249
The following is multiple choice question (with options) to answer.
What is the unit digit in 7105 ? | [
"1",
"5",
"7",
"9"
] | C | Unit digit in 7105 = unit digit in [(74)26 * 7]
But, unit digit in (74)26 = 1
unit digit in 7105 = (1 * 7) = 7
ANSWER:C |
AQUA-RAT | AQUA-RAT-34882 | % Example: Concession stand profit
% revenue = 3*soda + 5*popcorn + 2*candy;
% cost = 1*soda + 2*popcorn + 0.75*candy;
% objective = revenue - cost; % profit
% Edit the lines below with your calculations.
objective = a*(y - x^2)^2 + (1 - x)^2;
end
The following is multiple choice question (with options) to answer.
The manager of a theater noted that for every 10 admissions tickets sold, the theater sells 3 bags of popcorn at 2.40 $ each, 4 sodas at 1.50 $ each and 2 candy bars at 1.00$ each. To the nearest cent, what is the average (arithmetic mean) amount of these snacks sales per ticket sold? | [
"1.48$",
"1.52$",
"1.60$",
"1.64$"
] | B | For every 10 tickets amount of snacks sold is 3*2.40+4*1.5+2*1=$15.2, hence amount of the snacks sales per ticket is $15.2/10=~$1.52.
Answer: B. |
AQUA-RAT | AQUA-RAT-34883 | ## Exercise 21
An individual is to select from among n alternatives in an attempt to obtain a particular one. This might be selection from answers on a multiple choice question, when only one is correct. Let A be the event he makes a correct selection, and B be the event he knows which is correct before making the selection. We suppose P(B)=pP(B)=p and P(A|Bc)=1/nP(A|Bc)=1/n. Determine P(B|A)P(B|A); show that P(B|A)P(B)P(B|A)P(B) and P(B|A)P(B|A) increases with n for fixed p.
### Solution
P(A|B)=1P(A|B)=1, P(A|Bc)=1/nP(A|Bc)=1/n, P(B)=pP(B)=p
P ( B | A ) = P ( A | B ) P ( B ) P A | B ) P ( B ) + P ( A | B c ) P ( B c ) = p p + 1 n ( 1 - p ) = n p ( n - 1 ) p + 1 P ( B | A ) = P ( A | B ) P ( B ) P A | B ) P ( B ) + P ( A | B c ) P ( B c ) = p p + 1 n ( 1 - p ) = n p ( n - 1 ) p + 1
(22)
P ( B | A ) P ( B ) = n n p + 1 - p increases from 1 to 1 / p as n P ( B | A ) P ( B ) = n n p + 1 - p increases from 1 to 1 / p as n
(23)
## Exercise 22
The following is multiple choice question (with options) to answer.
There were two candidates in an election. Winner candidate received 52% of votes and won the election by 288 votes. Find the number of votes casted to the winning candidate? | [
"776",
"3744",
"299",
"257"
] | B | W = 52% L = 48%
52% - 48% = 4%
4% -------- 288
52% -------- ? => 3744
Answer:B |
AQUA-RAT | AQUA-RAT-34884 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
Three containers A,B and C are having mixtures of milk and water in the ratio of 1:5 and 3:5 and 5:7 respectively. If the capacities of the containers are in the ratio of all the three containers are in the ratio 5:4:5, find the ratio of milk to water, if the mixtures of all the three containers are mixed together. | [
"53:115",
"55:115",
"115:55",
"5:115"
] | A | Assume that there are 500,400 and 500 liters respectively in the 3 containers.
Then ,we have, 83.33, 150 and 208.33 liters of milk in each of the three containers.
Thus, the total milk is 441.66 liters.
Hence, the amount of water in the mixture is 1400-441.66=958.33liters.
Hence,
the ratio of milk to water is 441.66:958.33 => 53:115(using division by .3333)
The calculation thought process should be
(441*2+2):(958*3+1)=1325:2875
Dividing by 25 => 53:115.
ANSWER A 53:115 |
AQUA-RAT | AQUA-RAT-34885 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
For 600 km journey it will take 8hrs, if 120 km is done by bus and the rest by car. It takes 20 minutes more, if 200 km is done by busand the rest by car. The ratio of the speed of the train to that of the cars is? | [
"1 : 3",
"3 : 4",
"2 : 3",
"2 : 5"
] | B | Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120 + 480 = 8 1 + 4 = 1 ....(i)
x y x y 15
And, 200 + 400 = 25 1 + 2 = 1 ....(ii)
x y 3 x y 24
Solving (i) and (ii), we get: x = 60 and y = 80.
Ratio of speeds = 60 : 80 = 3 : 4.
B |
AQUA-RAT | AQUA-RAT-34886 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
Robin is traveling from one end of a forest to the other. In order to find her way back, she is leaving morsels of bread in the following pattern: 2 morsels of GRAIN, 3 morsels of white, and 1 morsel of rye. The pattern repeats after she leaves the morsel of rye. If Robin drops 2,000 morsels of bread, what are the last 3 morsels of bread that she drops? | [
"rye − wheat − wheat",
"wheat − wheat − white",
"white − rye − wheat",
"RYE-GRAIN-GRAIN"
] | D | Q is nothing BUT asking about remainder..
total morsels thrown before the pattern is repeated = 2+3+1 = 6..
so REMAINDER when 2000 is div by 6..
1998 is EVEN and also div by 3, so remainder is 2..
Last 3 of morsels are 1998-1999-2000
so he uses the last of morsel of pattern in 1998 two of the first morsels of the pattern in 1999 and 2000, and the pattern is G, G, W, W, W, R..
ans R-G-G
D |
AQUA-RAT | AQUA-RAT-34887 | ### Show Tags
22 Mar 2017, 01:52
If n is the product of 3 consecutive integers, which of the following must be true?
I. a multiple of 2 II. a multiple of 3 III. a multiple of 4
A. I only
B. II only
C. III only
D. I and II
E. II and III
_________________
The following is multiple choice question (with options) to answer.
Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is : | [
"9",
"11",
"13",
"15"
] | D | Solution
Let the three integers be x, x +2 and x + 4. Then, 3x = 2 (x + 4) + 3 ⇔ x = 11.
∴ Third integer = x + 4 = 15.
Answer D |
AQUA-RAT | AQUA-RAT-34888 | Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Originally posted by EgmatQuantExpert on 13 Jul 2018, 06:28.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:04, edited 2 times in total.
Senior Manager
Joined: 04 Aug 2010
Posts: 322
Schools: Dartmouth College
Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
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Updated on: 18 Jul 2018, 12:09
3
EgmatQuantExpert wrote:
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
Let the tank = 60 liters.
Since P takes 3 hours to fill the 60-liter tank, P's rate $$= \frac{60}{3} = 20$$ liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate $$= \frac{60}{4} = 15$$ liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate $$= \frac{60}{-5} = -12$$ liters per hour.
Since R works to EMPTY the tank, R's rate is negative.
The following is multiple choice question (with options) to answer.
A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely? | [
"3 hrs 45 min",
"4 hrs",
"2 hrs",
"3 hrs 15 min"
] | A | time taken by one tap to fill half = 3 hrs
part filled by four taps in 1 hr (4*1/6)=2/3
remain =(1-1/2)=1/2
=>2/3:1/2=>1:x
tot time = 3 hrs 45 min
ANSWER A |
AQUA-RAT | AQUA-RAT-34889 | algorithms
Title: How many times can you pour milk among 3 buckets? This is a problem from usaco, which I solved, but I don't know how to estimate the number of times you can pour milk among the jugs, my solutions just uses a big enough number.
Here's the problem statement:
Farmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk. Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.
Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.
For example, if the input is 8, 9, 10 then the output is 1, 2, 8, 9, 10.
Another example: if the input is 2, 5, 10 then the output is 5, 6, 7, 8, 9, 10.
My approach is to recursively try all possible move sequences, keeping track of the contents of bucket C each time bucket A is empty. Currently I stop the recursion at the arbitrary depth 200. What is the correct depth to stop the recursion at? Since the total amount of water is C liters (C being the capacity of bucket C), the number of possible states is at most the number of non-negative integer solutions of $x+y+z = C$, which is $\binom{C+2}{2}$. Any sequence of moves longer than that must repeat a state. Therefore it is enough to recurse up to level $\binom{C+2}{2}$.
The following is multiple choice question (with options) to answer.
Total 30 cows 10 cow gives each 2 liter milk 10 cow gives each 3/4 liter milk 10 cow gives each 1/4 liter milk this is split into 3 son per each 10 cows & 10 liter milk how? | [
"10",
"12",
"15",
"16"
] | A | 10 cow 2 liter each =20 liter
10cow 3/4 liter each= 3/4=0.758*10=7.5
10 cow 1/4 liter each = 1/4=0.25*10=2.5
add 20+7.5+2.5=30
milk split into 3 son each 10 liter then 30/3=10
ANSWER:A |
AQUA-RAT | AQUA-RAT-34890 | # 1988 AIME Problems/Problem 15
## Problem
In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order $1, 2, 3, 4, 5, 6, 7, 8, 9$.
While leaving for lunch, the secretary tells a colleague that letter $8$ has already been typed but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based on the above information, how many such after-lunch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)
## Solution 1
Re-stating the problem for clarity, let $S$ be a set arranged in increasing order. At any time an element can be appended to the end of $S$, or the last element of $S$ can be removed. The question asks for the number of different orders in which all of the remaining elements of $S$ can be removed, given that $8$ had been removed already.
Since $8$ had already been added to the pile, the numbers $1 \ldots 7$ had already been added at some time to the pile; $9$ might or might not have been added yet. So currently $S$ is a subset of $\{1, 2, \ldots 7\}$, possibly with $9$ at the end. Given that $S$ has $k$ elements, there are $k+1$ intervals for $9$ to be inserted, or $9$ might have already been placed, giving $k+2$ different possibilities.
Thus, the answer is $\sum_{k=0}^{7} {7 \choose k}(k+2)$ $= 1 \cdot 2 + 7 \cdot 3 + 21 \cdot 4 + 35 \cdot 5 + 35 \cdot 6 + 21 \cdot 7 + 7 \cdot 8 + 1 \cdot 9$ $= \boxed{704}$.
The following is multiple choice question (with options) to answer.
In an office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. Secretary takes the top letter and types it. Boss delivers in the order 1, 2, 3, 4, 5 which cannot be the order in which secretary types? | [
"4, 5, 2, 3, 9",
"4, 5, 2, 3, 1",
"4, 5, 2, 3, 0",
"4, 5, 2, 3, 2"
] | B | 4, 5, 2, 3, 1
Answer: B |
AQUA-RAT | AQUA-RAT-34891 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon? | [
"267",
"180",
"287",
"177"
] | B | Angle traced by the hour hand in 6 hours
=(360/12)*6
Answer: B |
AQUA-RAT | AQUA-RAT-34892 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
A factory produces x widgets per day. The factory's fixed costs are $7000 per day. The price per widget is $80 and the variable costs are $20 per widget. How many widgets need to be produced for profits of $6200 a day? | [
"42.33",
"90.33",
"168",
"220"
] | D | profits=6200=60x-8000 --> x=220
Answer D. |
AQUA-RAT | AQUA-RAT-34893 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
Mani buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share must be: | [
"10",
"13",
"15",
"20"
] | C | shares = 20%
==> 9%
interest of 12%==>15
ANSWER C |
AQUA-RAT | AQUA-RAT-34894 | # One-zero dividend
## Challenge description
For every positive integer n there exists a number having the form of 111...10...000 that is divisible by n i.e. a decimal number that starts with all 1's and ends with all 0's. This is very easy to prove: if we take a set of n+1 different numbers in the form of 111...111 (all 1's), then at least two of them will give the same remainder after division by n (as per pigeonhole principle). The difference of these two numbers will be divisible by n and will have the desired form. Your aim is to write a program that finds this number.
## Input description
A positive integer.
## Output description
A number p in the form of 111...10...000, such that p ≡ 0 (mod n). If you find more than one - display any of them (doesn't need to be the smallest one).
## Notes
Your program has to give the answer in a reasonable amount of time. Which means brute-forcing is not permited:
p = 0
while (p != 11..10.00 and p % n != 0)
p++
Neither is this:
do
p = random_int()
while (p != 11..10.00 and p % n != 0)
Iterating through the numbers in the form of 11..10..00 is allowed.
Your program doesn't need to handle an arbitrarily large input - the upper bound is whatever your language's upper bound is.
## Sample outputs
2: 10
3: 1110
12: 11100
49: 1111111111111111111111111111111111111111110
102: 1111111111111111111111111111111111111111111111110
The following is multiple choice question (with options) to answer.
The sum of the first 100 numbers, 1 to 100 is always divisible by | [
"2",
"2 and 4",
"2, 4 and 8",
"2, 4 and 9"
] | A | Explanation:
The sum of the first 100 natural numbers = (100 x 101)/2
= (50 x 101)
As 101 is an odd number and 50 is divisible by 2, so the sum is always divisible by 2.
Answer: Option A |
AQUA-RAT | AQUA-RAT-34895 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train when moves at an average speed of 40 kmph, reaches its destination on time. When its average speed becomes 35 kmph, then it reaches its destination 15 minutes late. Find the length of journey. | [
"70 km",
"80 km",
"40 km",
"20 km"
] | A | Sol.
Difference between timings = 15 min = 1/4 hr.
Let the length of journey be x km.
Then, x/35 - x/40 = 1/4
⇔ 8x - 7x = 70
⇔ x = 70 km.
Answer A |
AQUA-RAT | AQUA-RAT-34896 | Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.
The following is multiple choice question (with options) to answer.
Two positive integers differ by 4, and sum of their reciprocals is 6. Then one of the numbers is | [
"a) 3",
"b) 1",
"c) 5",
"4"
] | D | Algebraic approach:
Let n be the smaller integer => 1/n + 1/(n+4) = 6
or ((n+4)+n)/n(n+4) =6 or (n^2+4n)*6 =2n+4 or n=2 as n cannot be -negative
Solve for n => n=4. Hence,
D |
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