source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-39697 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
A girl was riding her scooter. Her father wanted to calculate her speed. He knew that the street was 600 meters from end to end. He saw his daughter make the roundtrip 6 times. It took her 20 minutes to complete those 6 trips. How fast was the girl riding her scooter in kilometers per hour? | [
"2",
"2.4",
"1.2",
"2.1"
] | B | The distance of the street was 600 meters.
The made the roundtrip 6 times, which would be 2 trips for each roundtrip.
Multiply 12 times 600 to get the total distance. 600 * 12 = 7200 meters of 7.2 km.
The question asked for the speed in km/hour and 20 minutes is one third of an hour.
Multiply 7200 by 1/3 to get approximately 2400 meters.
Correct answer is B |
AQUA-RAT | AQUA-RAT-39698 | Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:37
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:53
Bunuel wrote:
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
I though about the same but but when i see that TOO as three things to be arranged in 3! ways then i also thought that OO ARE TWO DIGITS AND THEY ARE TWO DIFFERENT PRIME NOS SO WHY DIVIDE BY 2!
this might clear my entire probability confusion i hope...
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Posts: 43322
Kudos [?]: 139450 [1], given: 12790
The following is multiple choice question (with options) to answer.
An staff identification code consists of a vowel followed by a 3-digit number greater than 200. Exactly 2 of the 3 digits in the code should be identical. How many different codes is it possible to form? | [
"211",
"216",
"1075",
"1080"
] | C | An staff identification code will be of the type -,-,-,-
first digit can be selected from any of the 5 vowels in 5C1 ways
now for the remaining three digit lets consider the following two cases
case 1: when the number is greater than 200 but less than 300
number will be of the type 2,_,_. now suppose repeating number is same as first digit number i.e. 2. and the third number is one of the remaining 9 numbers (we are rejecting 2 here, because it will result in 222, which is not acceptable as per the given condition). thus these two number can arrange themselves in two blank spaces in 2! ways. hence total number of numbers in which repeating digit is same as the first digit = 1.9.2! =18
now, suppose that repeating number is different than first digit. thus possible case in this case are 8 as listed below:
211
233
244
255
266
277
288
299
here again we have rejected 200( because number must be greater than 200) and 222 ( exactly two repeating digits are allowed)
thus total possible cases are 18 + 8 =26
case 2: number ranging from 300 to 999
here for first digit we have 7 cases (3,4,5,6,7,8,9)
now if the repeating number is same as the first number then we will have 18 cases ( same reasoning as mentioned in the previous case)
if the repeating number is different than first digit number then we will have 9 cases ( because here number ending with two zeros are allowed)
hence total number of ways = 7(18+9) = 189
thus different number of codes = 5(189+26) = 1075
hence C |
AQUA-RAT | AQUA-RAT-39699 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
A thief is spotted by a policeman from a distance of 100metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 15km/hr and that of the policeman 20km/hr, how far the thief will have run before he is overtaken ? | [
"A)1km",
"B)300m",
"C)650m",
"D)750m"
] | B | Relative speed of the policeman = 20-15 = 5 km/hr
time taken by policeman to cover 100m = (100/1000)*(1/5) = 1/50 hr
In 1/50 hr the thief covers a distance of 15/50 km = 3/10 km = 300m
Answer is B |
AQUA-RAT | AQUA-RAT-39700 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
The total of the ages of Jayant, Prem and Saransh is 93 years. Ten years ago, the ratio of their ages was 2 : 3 : 4. What is the present age of Jayant? | [
"24 years",
"32 years",
"34 years",
"38 years"
] | A | Solution
Let the ages of Jayant, Prem and Saransh 10 years ago be 2x, 3x and 4x years respectively. Then,
(2x + 10)+(3x + 10)+(4x + 10) = 93.
‹=›9x = 63
x = 7.
∴ Jayant's present age = 2x + 10 = 24 years.
Answer A |
AQUA-RAT | AQUA-RAT-39701 | ### 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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Posts: 9558
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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.
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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.
Cereal A is 8% sugar by weight, whereas healthier but less delicious Cereal B is 2% sugar by weight. To make a delicious and healthy mixture that is 4% sugar, what should be the ratio of Cereal A to Cereal B, by weight? | [
" 2:9",
" 2:7",
" 1:2",
" 1:4"
] | C | (8/100)A+(2/100)B = (4/100)(A+B)
4A =2B
=> A/B = 1/2
Answer is C. |
AQUA-RAT | AQUA-RAT-39702 | Question 9
Let $f(x) = \sqrt {x^2-2x-3}$.
a) Write $x^2-2x-3$ in the form $a(x-h)^2 + k$
b) Find the inverse of function $g$ defined by $g(x) = \sqrt {x^2-2x-3} \quad , \quad x \ge 3$
c) Find the range of the inverse of $g$ defined above.
Solution
a)
Complete the square of the given expression
$x^2-2x-3 = (x-1)^2 - 1 - 3 = (x-1)^2 - 4$
b)
Let $y = \sqrt {x^2-2x-3}$
Substitute the radicand $x^2-2x-3$ by $(x-1)^2 - 4$
$y = \sqrt {(x-1)^2 - 4}$
Square both sides
$y^2 = (x-1)^2 - 4$
Solve the above for $x$ to find the two solutions
$x = 1 \pm \sqrt {y^2 + 4}$
Since $x \ge 3$ (see definition of $g$ above), we select the solution
$x = 1 + \sqrt {y^2 + 4}$
The inverse of $g$ is given by
$g^{-1}(x) = 1 + \sqrt {x^2 + 4}$
c)
One of the properties of inverse functions is that range of $g^{-1}$ is the domain of $g$ which is given by the interval
$[3 , \infty)$
The following is multiple choice question (with options) to answer.
What is the range of all the roots of |x^2 - 3| = x ? | [
"4",
"3",
"2",
"1"
] | C | we get 2 quadratic equations here ..
1) x^2-x-3=0 ....... roots 2 , -1
2) x^2+x-3=0 ........ roots -2, 1
Inserting each root in given equation , it can be seen that -1 and -2 do not satisfy the equations .
So value of x for given equation .... x=3 or x=1
I guess range is 3-1 =2
C |
AQUA-RAT | AQUA-RAT-39703 | 2. How many ways can one choose three distinct characters to be placed into the forms from above.
3. Watch out for any duplication or overcounting.
There are two options (using Joe's suggestion in the comments):
• $\{a,b,c,c,c\}$. We can choose $3$ of the $5$ positions for the $c$'s, then there are two options for the $a$ and $b$ ($a$ first and then $b$ or the other way around). This gives $$2\binom{5}{3}=20$$ ways to write $a$, $b$, and $c$.
• $\{a,b,b,c,c\}$. We can choose $1$ of the $5$ positions for the $a$, and then $2$ of the remaining $4$ positions for the $b$'s. This gives $$5\binom{4}{2}=30$$ ways to write $a$, $b$, and $c$.
Now, there are $10\cdot 9\cdot 8=720$ ways to choose three digits for the positions $a$, $b$, and $c$. So, one might expect $$720\cdot (20+30)=36000$$ arrangements. The problem with this is that there is some overcounting since $abbcc$ and $accbb$ result in the same arrangement when the numbers in $b$ and $c$ are also flipped. This occurs for $a\leftrightarrow b$ in the first option and $b\leftrightarrow c$ in the second form. Therefore, we've overcounted by a factor of $2$ and there are $18000$ total arrangements. (Note that $720\cdot 30/2=10800$ matching @Green's approach).
The following is multiple choice question (with options) to answer.
Will must choose a 3-character computer password, consisting of 3 distinct digits, in any order.From how many different passwords can Will choose? | [
"390",
"2,340",
"4,320",
"7,020"
] | C | 10C1*9C1*8C1 = 720.
=>720 * 3 ! = 4320.
Option E is correct answer...but OA is C. |
AQUA-RAT | AQUA-RAT-39704 | 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 moves past a post and a platform 264 metre long in 8 seconds and 20 seconds respectively. What is the speed of the train? | [
"79.2 km/hr",
"82.4 km/hr",
"69.7 km/hr",
"49.8 km/hr"
] | A | The train can cover a distance equal to its length in 8 seconds. Therefore, additional
(20−8)=12 seconds was taken to cross the platform.
Hence, speed of the train=264/12=22 m/s
=22 × 18/5=79.2 km/hr
ANSWER:A |
AQUA-RAT | AQUA-RAT-39705 | ### Show Tags
23 Oct 2009, 23:42
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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
Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.
In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.
Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).
Consider:
$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:
$$ttttt|||$$
Means that first nephew will get all the tickets,
$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1
And so on.
How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.
The following is multiple choice question (with options) to answer.
A person buys 18 local tickets for Rs 110. Each first class ticket costs Rs 10 and each second class ticket costs Rs 3. What will another lot of 18 tickets in which the numbers of first class and second class tickets are interchanged cost? | [
"112",
"118",
"121",
"124"
] | D | Explanation :
Let, there are x first class ticket and (18-x) second class tickets.
Then,
110 = 10x + 3(18−x).
=>110=10x+54−3x.
=>7x=56.
=>x=8.
If the number of the first class and second class tickets are interchanged, then the total cost would be 10×10 + 3 × 8=124.
Answer : D |
AQUA-RAT | AQUA-RAT-39706 | # 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.
In traveling from a dormitory to a certain city, a student went 1/5 of the way by foot, 2/3 of the way by bus, and the remaining 6 kilometers by car. What is the distance, in kilometers, from the dormitory to the city? | [
"30",
"58",
"60",
"90"
] | B | I believe there is a better way to do it.
Basically one of the options should satisfy the given criteria.
60 did
1/5 *60 =12
2/3*60=40
so total distance
52 + remaining 6 =58
Answer B |
AQUA-RAT | AQUA-RAT-39707 | 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.
The price of a microchip declines by 67 percent every 6 months. At this rate, approximately how many years will it take for the price of an $81 microchip to reach $1? | [
"1.5 years",
"2 years",
"3 years",
"13 years"
] | B | I think the fastest way is
81 *(1/3)^n = 1
n =4
So 2 yrs
ANSWER:B |
AQUA-RAT | AQUA-RAT-39708 | python, beginner, algorithm, strings, regex
Test 8: 1e is an invalid number.
--------------------------------------------------
Test 9: e3 is an invalid number.
--------------------------------------------------
Test 10: 6e-1 is a valid number.
--------------------------------------------------
Test 11: 99e2.5 is an invalid number.
--------------------------------------------------
Test 12: 53.5e93 is a valid number.
--------------------------------------------------
Test 13: --6 is an invalid number.
--------------------------------------------------
Test 14: -+3 is an invalid number.
--------------------------------------------------
Test 15: 95a54e53 is an invalid number.
The following is multiple choice question (with options) to answer.
Find the invalid no.from the following series 13, 18, 25, 30, 37, 40 | [
"23",
"26",
"40",
"42"
] | C | The differences between two successive terms from the beginning are 7, 5, 7, 5, 7, 5. So, 40 is wrong.
C |
AQUA-RAT | AQUA-RAT-39709 | # Finding the probability using a normal distrubtion.
I have a stats question that says, "An airline flies airplanes that hold 100 passengers. Typically, some 10% of the passengers with reservations do not show up for the flight. The airline generally overbooks fights in an attempt to fill them."
And the question is asking 4 parts, the first being "Find the probability that a flight booked for 100 people flies full." The second being, "Find the probability that for a flight booked for 100 passengers, between 90 and 100 passengers inclusively will show up for the flight" Third being, "Find the probability that for a flight with 105 reservations, everyone will get a seat." and fourth "If the flight is booked for 106 passengers, can the airline be at least 95% sure that everyone will get a seat?"
The only hints that I have about this question are that it is a normal distribution because that is what the homework assignment is about, but I cannot even think of how to start this problem because the mean and standard deviation do not seem to be given.
• The problem really concerns the binomial distribution. However you can solve it using a Normal approximation to the binomial. In particular $X \sim B(n,p)$ can be approximated by $X \sim N(np,np(1-p))$ Does this help? – Karl Apr 17 '15 at 12:27
• It does a bit, so using sqrt(np*(1-p)) that should give me my standard deviation for this problem right? – Ramirez77 Apr 17 '15 at 12:36
• Yes, then you can use the standard transformation into $Z$. – Karl Apr 17 '15 at 12:38
• Oh okay that makes a lot of sense then. – Ramirez77 Apr 17 '15 at 12:39
• Ok, good luck. For future reference I'd have a look at the conditions by which the approximation is a good one. – Karl Apr 17 '15 at 12:45
I'm pretty sure you are supposed to use the normal approximation to the normal, as suggested by @Karl. I hope you used continuity corrections to give that the best chance to work. Here are exact binomial answers from R statistical software to compare with your approximate answers:
The following is multiple choice question (with options) to answer.
20% of major airline companies equip their planes with wireless internet access. 70% of major airlines offer passengers free on-board snacks. What is the greatest possible percentage of major airline companies that offer both wireless internet and free on-board snacks? | [
"20%",
"30%",
"40%",
"70%"
] | A | To maximize the percentage of companies offering both, let's assume that all 20% of companies which offer wireless internet also offer snacks.
The answer is A. |
AQUA-RAT | AQUA-RAT-39710 | 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
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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.
A reduction of 40% in the price of bananas would enable a man to obtain 64 more for Rs.40, what is reduced price per dozen? | [
"5",
"6",
"2",
"3"
] | D | 40*(40/100) = 16 --- 64
? --- 12 => Rs.3
Answer: D |
AQUA-RAT | AQUA-RAT-39711 | # 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 trader purchased two colour televisions for a total of Rs. 35000. He sold one colour television at 30% profit and the other 40% profit. Find the difference in the cost prices of the two televisions if he made an overall profit of 32%? | [
"Rs. 21000",
"Rs. 17500",
"Rs. 19000",
"Rs. 24500"
] | A | Let the cost prices of the colour television sold at 30% profit and 40% profit be Rs. x and Rs. (35000 - x) respectively.
Total selling price of televisions = x + 30/100 x + (35000 - x) + 40/100 (35000 - x)
=> 130/100 x + 140/100 (35000 - x) = 35000 + 32/100 (35000)
x = 28000
35000 - x = 7000
Difference in the cost prices of televisions = Rs. 21000
ANSWER:A |
AQUA-RAT | AQUA-RAT-39712 | units
(Note that $1\ \mathrm{Pa}=1\ \mathrm{kg\ m^{-1}\ s^{-2}}$ and that $1\ \mathrm J=1\ \mathrm{kg\ m^2\ s^{-2}}$.)
The following is multiple choice question (with options) to answer.
Mother,her daughter and her grand child weighs 140 kg. daughter and her daughter(child) weighs 60 kg. child is 1/5th of her grand mother. What is the age of the daughter? | [
"44",
"47",
"48",
"49"
] | A | mother + daughter + child = 140kg
daughter + child = 60kg
mother = 140 - 60 = 80kg
child = 1/5th of mother = (1/5)*80 = 16kg
So now daughter = 140 -(mother + child) = 140 - (80+16) = 44kg
ANSWER:A |
AQUA-RAT | AQUA-RAT-39713 | If 2 fair dice are rolled, what is the probability that the sum is 6, given both dice show odd numbers?
When the sum is 6, there combinations are: (1,5),(2,4),(3,3),(4,2),(5,1). Probability = 5/36
Probability of 2 odd numbers is 1/2.
What I'm not sure about is when we work out Pr(sum is 6|both numbers odd). I initially thought that the top line was Pr(sum is 6), but then the result would be 5/2. This can't be right.
So, does that mean that we only choose combinations which have odd pairs? So, the answer would be 3/2?
Still makes no sense as the number is greater than 1???
Can somebody shed some light on this for me?
-
What you're trying to compute is the probability that two dice sum to $6$, given that both dice have odd values. Note that there are three ways for pairs of dice with odd values to sum to six, namely $(1,5),(3,3),(5,1)$. There are three odd values each die can take, so there are nine ways for pairs of dice to both have odd values. Thus the probability you are looking for is $3/9=1/3$.
The following is multiple choice question (with options) to answer.
If two dice are thrown simultaneously, then find the probability that the sum of numbers appeared on the dice is 6 or 7? | [
"5/6",
"7/36",
"5/36",
"11/36"
] | D | The sum of numbers appeared is 6 or 7. Therefore, the required sums are 6 or 7, i.e., the required events are (1,5), (5,1), (2,4), (4,2), (3,3), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)
i.e., for 6, n(E) = 5 and for 7, n(E) = 6
Therefore, the required probability = n(E)/n(S) = 11/36.
ANSWER:D |
AQUA-RAT | AQUA-RAT-39714 | 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.
The average age of 36 students in a group is 14 years. When teacher's age is included to it, the average increases by one. What is the teacher's age in years ? | [
"66",
"77",
"51",
"98"
] | C | Explanation:
Age of the teacher = (37 * 15 - 36 * 14) years = 51 years.
Answer: C) 51 years |
AQUA-RAT | AQUA-RAT-39715 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
If a car went the first third of the distance at 80 kmh, the second third at 18 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip? | [
"34 kmh",
"40 kmh",
"42 kmh",
"44 kmh"
] | A | Assume D/3 = 720 (this number is convenient because it is divisible by 80, 18 and 48)
So:
720 = 80*T1 = 9 hrs
720 = 18*T2 = 40 hrs
720 = 48*T3 = 15 hrs
T = T1 + T2 + T3 = 64 hrs
D = RT
(720*3) = R*64
R = 33.75
ANSWER: A |
AQUA-RAT | AQUA-RAT-39716 | # 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.
Cindy has her eye on a sundress but thinks it is too expensive. It goes on sale for 15% less than the original price. Before Cindy can buy the dress, however, the store raises the new price by 25%. If the dress cost $68 after it went on sale for 15% off, what is the difference between the original price and the final price? | [
"$0.00",
"$1.00",
"$3.40",
"$5.00"
] | D | 0.85*{Original Price} = $68 --> {Original Price} = $80.
{Final Price} = $68*1.25 = $85.
The difference = $85 - $85 = $5.
Answer: D. |
AQUA-RAT | AQUA-RAT-39717 | 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 360 m long is running at a speed of 54 km/hr. In what time will it pass a bridge 140 m long? | [
"40 sec",
"11 sec",
"88 sec",
"33.33 sec"
] | D | Speed = 54 * 5/18 = 15 m/sec
Total distance covered = 360 + 140 = 500 m
Required time = 500 * 1/15 = 33.33 sec
Answer: D |
AQUA-RAT | AQUA-RAT-39718 | # Math Help - Calculus Help Please
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
2. Originally Posted by Luke007
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
a)
$\frac {dN}{dt} = -0.25N$
$\Rightarrow \frac {dN}{N} = -0.25 dt$
$\Rightarrow \ln N = -0.25t + C$
$\Rightarrow N = e^{-0.25t + C}$
$\Rightarrow N = e^C e^{-0.25t}$
$\Rightarrow N = Ae^{-0.25t}$ ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book
when $t = 0$, $N = 1000$
$\Rightarrow N(0) = Ae^0 = 1000$
$\Rightarrow A = 1000$
$\Rightarrow N(t) = 1000e^{-0.25t}$
b)
$\frac {dN}{dt} = -0.25N$
when $N$ is half it's size:
$\frac {dN}{dt} = -0.25 (0.5N)$
$\frac {dN}{dt} = -0.125N$
So the population is decreasing at a rate of -0.125
The following is multiple choice question (with options) to answer.
After 10% of the inhabitants of a village disappeared, a panic set in during which 25% of the remaining inhabitants left the village. At that time, the population was reduced to 5535. What was the number of original inhabitants? | [
"7900",
"8200",
"8500",
"8800"
] | B | Let the total number of original inhabitants be x.
(75/100) * (90/100) * x = 5535
(27/40) * x = 5535
x = 5535 * 40 / 27 = 8200
The answer is B. |
AQUA-RAT | AQUA-RAT-39719 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
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The following is multiple choice question (with options) to answer.
A particular parking garage is increasing its rates by 30 percent per month. Bob decides to reduce the number of days he uses the garage per month so that the amount he spends at the garage per month remains unchanged. Which of the following is closest to Bob’s percentage reduction in the number of days he uses the garage each month? | [
"12%",
"16%",
"20%",
"23%"
] | D | Let x be the original number of days that Bob used the garage.
Let P be the original cost per day.
Let N*x be the new number of days that Bob uses the garage.
N*x*1.3P = x*P
N*x = x/1.3 which is about 0.77x, thus a 23% reduction.
The answer is D. |
AQUA-RAT | AQUA-RAT-39720 | # Is this a simple answer to the classic problem "A certain city has 10 bus routes..."
A certain city has 10 bus routes. Is it possible to arrange the routes and the bus stops so that if one route is closed, it is still possible to get from anyone stop to any other (possibly changing along the way), but if any two routes are closed, there are at least two stops such that it is impossible to get from one to the other?
Given the solution above, why did the authors feel compelled to answer: Yes. Consider 10 straight lines in the plane, no 2 are parallel & no 3 are concurrent. Let lines be bus routes & let points of intersection be stops. We get from anyone stop to any other (if the stops lie on 1 line, w/o changing; & if not, then with just 1 change). If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once. However, if we discard 2 lines, then 1 stop-their point of intersection-will have no bus routes passing thru it, & it'll be impossible to get from this stop to any other.
Source: A. M. Yaglom and l. M. Yaglom CHALLENGING MATHEMATICAL PROBLEMS WITH ELEMENTARY SOLUTIONS Volume II Problems From Various Branches of Mathematics Translated by James McCawley, Jr. Revised and edited by Basil Gordon DOVER PUBLICATIONS, INC. NEW YORK
• Yes, the answer is correct. Jun 9, 2019 at 6:12
• My guess, Joe, is that when they wrote "changing", they meant "changing once". Jun 9, 2019 at 13:04
• Your answer is correct and fine. It is possible the authors just didn't think of this solution and thus put their solution into the book. Everybody overlooks something every now and then. Jun 9, 2019 at 15:30
• My solution fails to meet the requirement, "If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once." Jun 10, 2019 at 0:55
• Gerry, your interpretation of the wording is what makes the problem sufficiently complex. Thanks. Jun 10, 2019 at 0:58
Let’s go back to the roots.
The following is multiple choice question (with options) to answer.
20 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus? | [
"None of these",
"380",
"312",
"324"
] | B | Explanation :
He can go in any bus out of the 20 buses.
Hence He can go in 20 ways.
Since he can not come back in the same bus that he used for travelling,
He can return in 19 ways.
Total number of ways = 20 x 19 = 380. Answer : Option B |
AQUA-RAT | AQUA-RAT-39721 | 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 man sitting in a train which is traveling at 30 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.? | [
"50 kmph",
"82 kmph",
"62 kmph",
"65 kmph"
] | B | Relative speed = 280/9 m / sec = ((280/9)*(18/5)) kmph = 112 kmph.
Speed of goods train = (112 - 30) kmph = 82 kmph.
Answer : B. |
AQUA-RAT | AQUA-RAT-39722 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
The average age of 3 boys is 30 years and their ages are in the proportion 1:2:3. The age of the youngest boy is? | [
"15years",
"20years",
"22years",
"25years"
] | A | Total age of 3 boys = 30*3 = 90
ratio of their ages = 1:2:3
Age of the youngest = 90*1/6 = 15 years
Answer is A |
AQUA-RAT | AQUA-RAT-39723 | # 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.
In the series 0, 3, 8, 15,__ What is the next number? | [
"22",
"23",
"24",
"25"
] | C | 3-0=3
8-3=5
15-8=7
difference increasing by 2 every time
next difference 9
15+9=24
ANSWER:C |
AQUA-RAT | AQUA-RAT-39724 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
Machine–A produces 40% of the total output and Machine-B produces 60% of the total output. An average of nine units out of a thousand goods manufactured by Machine-A and one unit of 50 units produced by Machine-B prove to be defective. What is the probability that a unit chosen at random from the total daily output of the factory is defective? | [
"a. 0.156",
"b. 0.01506",
"c. 0.0156",
"d. 0.0001566"
] | C | Let total Production be 10000 units....
A produces 4000 units and 36 units are defective
B produces 6000 units and 120 units are defective
So, Out of total 10,000 units 156 units are defective...
So the required probability = 156/10000 => 0.0156
Answer will be (C) |
AQUA-RAT | AQUA-RAT-39725 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
In the first hour of a two-hour trip, a car traveled d kilometers, and in the second hour of the trip, the car traveled one that distance. What is the average rate at which the car traveled during the trip, in kilometers per hour? | [
" d",
" 1/3*d",
" 1/2*d",
" 3/4*d"
] | A | Solution:
Lets look at the information given in the question...
Total time travelled = 2 hrs
Distance travelled first hour = d
Distance travelled second hour = d
The question is asking for the avg speed at which the car travels for 2 hrs
We know that avg speed = total distance/total time
Lets place the information given in the question to the formula above = (d+d)/2
=d
We find that option A is the answer as it gives the same expression as found in our calculation above. Answer option A. |
AQUA-RAT | AQUA-RAT-39726 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
City A and City B are 140 miles apart. Train C departs City A, heading towards City B, at 4:00 and travels at 40 miles per hour. Train D departs City B, heading towards City A, at 4:20 and travels at 20 miles per hour. The trains travel on parallel tracks. At what time do the two trains meet? | [
"5:00",
"5:30",
"6:00",
"6:20"
] | D | Train C has traveled 20 mi in the half hour before Train D has started its journey.
140-20=120
40+20 =60 mph
120 mi/ 60 mph = 2 hrs
4:20pm + 2 hrs = 6:20pm
Answer:
D. 6:20 |
AQUA-RAT | AQUA-RAT-39727 | of 69” which states the doubling period as 0.35 + 69/Interest The time value of money is a concept integral to all parts of business. How to Calculate Present Value, and Why Investors Need to Know It, Understanding the Compound Annual Growth Rate – CAGR. We could put the equation more concisely and use the$10,000 as FV. The above future value equation can be rewritten as follows: PV=FV(1+i)n\begin{aligned} &\text{PV} = \frac{ \text{FV} }{ ( 1 + i )^ n } \\ \end{aligned}PV=(1+i)nFV, PV=FV×(1+i)−nwhere:PV=Present value (original amount of money)FV=Future valuei=Interest rate per periodn=Number of periods\begin{aligned} &\text{PV} = \text{FV} \times ( 1 + i )^{-n} \\ &\textbf{where:} \\ &\text{PV} = \text{Present value (original amount of money)} \\ &\text{FV} = \text{Future value} \\ &i = \text{Interest rate per period} \\ &n = \text{Number of periods} \\ \end{aligned}PV=FV×(1+i)−nwhere:PV=Present value (original amount of money)FV=Future valuei=Interest rate per periodn=Number of periods. … The welfare of the owners would be maximized when net worth or net value is created from making a financial decision. It is underlying theme embodies in financial concepts such as:eval(ez_write_tag([[580,400],'xplaind_com-box-4','ezslot_5',134,'0','0'])); It is the basis used to work out the intrinsic value of a firm, a share of common stock, a bond or any other financial instrument. The time value of money implies that: 1. a person will have to pay in future more, for a rupee received today and 2. a person may accept less today, for a rupee to be
The following is multiple choice question (with options) to answer.
The rate at which a sum becomes four times of itself in 15 years at S.I., will be : | [
"10 %",
"15 %",
"20 %",
"40 %"
] | C | Sol.
Let sum = x. Then, S.I. = 3x
∴ Rate = [100 * S.I. / P*T] = [100 * 3x / x * 15]% = 20%.
Answer C |
AQUA-RAT | AQUA-RAT-39728 | Using the law of sines, one can get
$$s=\frac{1}{2}bc \sin(\alpha)=\frac{1}{2}bc \frac{a}{2R}=\frac{abc}{4R}$$
where $R$ is the radius of the circumscribed circle.
• note that : Dr. Sonnhard Graubner says this first of all – Khosrotash Jul 26 '15 at 13:28
• @Khosrotash comments should not be used to supply answers. This is a misuse of the comment feature. – miracle173 Apr 24 '17 at 5:03
The following is multiple choice question (with options) to answer.
Consider a quarter of a circle of radius 25. Let r be the radius of the circle inscribed in this quarter of a circle. Find r. | [
"16*(sqr2 -1)",
"8*(sqr3 -1)",
"25*(sqr2 - 1)",
"12* (sqr7 -1)"
] | C | I got 16/(sqr2 +1) and just forgot to multiply by (sqr2 -1).
Answer is C |
AQUA-RAT | AQUA-RAT-39729 | 4,8),(3,5,8),(2,6,8),(1,7,8),(1,9,8),(9,0,9),(8,1,9),(7,2,9),(6,3,9),(5,4,9),(4,5,9),(3,6,9),(2,7,9),(1,8,9)]
The following is multiple choice question (with options) to answer.
78, 64, 48, 30, 10, (...) | [
"-12 number",
"-14",
"2",
"8"
] | A | Explanation :
78 - 14 = 64
64 - 16 = 48
48 - 18 = 30
30 - 20 = 10
10 - 22 = -12
Answer : Option A |
AQUA-RAT | AQUA-RAT-39730 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
Find large number from below question The difference of two numbers is 1000. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder | [
"1345",
"1197",
"1540",
"1600"
] | B | Let the smaller number be x. Then larger number = (x + 1365).
x + 1000 = 6x + 15
5x = 985
x = 197
Large number = 197 +1365 = 1197
B |
AQUA-RAT | AQUA-RAT-39731 | Related Rate Prob - Ships sailing in different directions
• November 5th 2010, 09:15 PM
dbakeg00
Related Rate Prob - Ships sailing in different directions
Ship A is 15 miles east of point O and moving west at 20mi/h. Ship B is 60 miles south of O and moving north at 15mi/h. a)Are they approaching or seperating after 1 hour and at what rate? b)after 3hrs?
Let D=distance between the ships at time t
$D^2=(60-15t)^2+(15-20t)^2$
$2D*\frac{dx}{dt}=(2)(-15)(60-15t)+(2)(-20)(15-20t)$
$2D*\frac{dx}{dt}=(-30)(60-15t)+(-40)(15-20t)$
$2D*\frac{dx}{dt}= 1250t-2400$
$\frac{dx}{dt}=\frac{1250t-2400}{2D}$
$D=\sqrt{5^2+45^2}=5\sqrt{82}$
$\frac{dx}{dt}=\frac{1250t-2400}{10\sqrt{82}}$
a) $\frac{dx}{dt}=\frac{1250*1-2400}{10\sqrt{82}}=approaching\ at \frac{-115}{\sqrt{82}}$ mi/h
b) $\frac{dx}{dt}=\frac{1250*3-2400}{10\sqrt{82}}=seperating\ at \frac{135}{\sqrt{82}}$ mi/h
The book agrees with me on part A, but on part B it says the answer should be $seperating\ at\ \frac{9\sqrt{10}}{2}$ mi/h. Am I missing something obvious or is the book wrong here?
• November 6th 2010, 02:24 AM
Ackbeet
The following is multiple choice question (with options) to answer.
From the starting point in a boat race, one competitor started to sail north at a speed of 1.3 Km/h, the other competitor started to sail west at a speed of 1.2 Km/h. What is the distance in Km between the two competitors after 5 hours? | [
"8.8",
"12.",
"12.5.",
"14."
] | A | Both competitors are sailing making angle of 90 degrees.
After 5 hrs one competitor will cover a distance of = 1.3 *5= 6.5 KM
And, other competitor will cover a distance of = 1.2*5= 6KM
distance between them after 5 hrs = (6.5^2+ 6^2)^1/2= 8.8KM
A is the answer |
AQUA-RAT | AQUA-RAT-39732 | i.e. we need to check that $$-1-[-x]\leq x<-[-x].$$
The inequality on the right, $[-x]<-x$, is the first inequality of the definition of $[-x]$, i.e. $[-x]\leq-x<[-x]+1$, and therefore it is true. Notice the equality doesn't hold on the left, because it is not an integer.
The inequality on the left is $-x\leq [-x]+1$, which is equivalent to $-x<[-x]+1$, again because it is not an integer. And this is the other part of the definition of $[-x]$, and therefore it is true.
This proves that $$-1-[-x]=[x].$$
-
Thanks for the effort, but I think I'm getting confused following your answer. Could you please label the inequalities as (1), (2), etc., and rewrite the answer? Here are a few examples of how I'm getting confused: "The equality on the right, $[−x] < −x$" Did you mean inequality? "The inequality on the right is $−x \leq [−x]+1$" Where does this come from? Also, is $x$ is not an integer, how can we aim to establish that $-1-[-x] \leq x$, as the left part is an integer? – dotslash Jul 19 '13 at 13:14
The following is multiple choice question (with options) to answer.
What inequality represents the condition 1<x<5? | [
"|x|<3",
"|x+5|<4",
"|x-3|<2",
"|-5+x|<4"
] | C | 1<x<5 represents segment with
length: 5-1 = 4
center: (5+1)/2 = 3
For any |x-a|<b a is a center and b is a half of length. So, |x-3|<4/2 represents our inequality (answer C) |
AQUA-RAT | AQUA-RAT-39733 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The ratio of two quantities is 5 to 6. If each of the quantities is increased by 3, what is the ratio of these two new quantities? | [
"2/15",
"7/5",
"9/5",
"It cannot be determined from the information given."
] | D | let the actual ratio be 5k/6k, increase numerator and denominator by 4=> 5k+3/6k+3
Until the value of k is unknown we can not determine the actual ratio.
OR
let the actual be 3/4. After increasing 3 => 6/7
let the actual be 6/8. After increasing 3 => 9/11
Therefore without actual quantities we cannot determine the actual ratio.
Answer : D |
AQUA-RAT | AQUA-RAT-39734 | What we're looking for: How many subsets have $6$ as an element but not $7$?
We must choose to include $6$, that gives $1$ choice. We may choose to to include $1$ or not, that gives $2$ choices. We may choose to include $2$ or not, that gives $2$ choices. We may choose to include $3$ or not, that gives $2$ choices. We may choose to include $4$ or not, that gives $2$ choices. We may choose to include $5$ or not, that gives $2$ choices. We must choose not to include $7$, that gives $1$ choice. By the multiplication principle the number subsets with $6$ but not $7$ is:
$$1•2•2•2•2•2•1=2^5$$
The following is multiple choice question (with options) to answer.
How many different subsets of the set {0, 1, 2, 3, 4, 5,6,7} do not contain 0? | [
"A.16",
"B.72",
"C.93",
"D.32"
] | C | Number of subset
Since we have 7 digits other than 0, we can take any numbers from the set of 7 to make a subset. Also it is a matter of selection and not arrangement.So we will consider combinations.
7c1+7c2+7c3+7c4+7c5+7c6+7c7=92
And one set is the NULL set having no elements in it so
92+1=93.
ANSWER C. |
AQUA-RAT | AQUA-RAT-39735 | 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.
The dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of white washing the four walls of the room at Rs. 5 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each? | [
"4538",
"4529",
"4528",
"4530"
] | D | Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls
= 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5
= Rs. 4530
Answer: D |
AQUA-RAT | AQUA-RAT-39736 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
A train sets off at 2 p.m. at the speed of 70 kmph. Another train starts at 3:30 p.m. in the same direction at the rate of 85 kmph. At what time the trains will meet? | [
"09:20p.m",
"04:40p.m",
"10.30p.m",
"02.32p.m"
] | C | D = 70 * 1 ½ = 105 km
RS = 85 – 70 = 15
T = 105/15 = 7 h
3.30 + 7 h = 10.30 p.m.
Answer:C |
AQUA-RAT | AQUA-RAT-39737 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
Two positive integers differ by 4, and sum of their reciprocals is 4. Then one of the numbers is | [
"a) 3",
"b) 1",
"2",
"d) 21"
] | C | Algebraic approach:
Let n be the smaller integer => 1/n + 1/(n+4) = 4
or ((n+4)+n)/n(n+4) =4 or (n^2+4n)*4 =2n+4 or n=2 as n cannot be -negative
Solve for n => n=2. Hence,
C |
AQUA-RAT | AQUA-RAT-39738 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
C and D started a business investing Rs. 49,000 and Rs. 35,000 respectively. In what ratio the profit earned after 4 years be divided between C and D respectively? | [
"7:4",
"7:5",
"6:4",
"5:5"
] | B | C:D= 49000 : 35000 = 7:5.
ANSWER:B |
AQUA-RAT | AQUA-RAT-39739 | 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 and B can do a piece of work in 3 days, B and C in 4 days, C and A in 6 days. How long will C take to do it? | [
"18 days",
"20 days",
"24 days",
"30 days"
] | C | Explanation:
2c = ¼ + 1/6 – 1/3 = 1/12
c = 1/24 => 24 days
Answer: C |
AQUA-RAT | AQUA-RAT-39740 | Seven-game series: . $\_\:\_\:\_\:\_\:\_\:\_\:W$
The first six games must have three W's and three L's in some order.
There are: . ${6\choose3,3} = {\color{red}20}$ ways.
Therefore, there are: . $1 + 4 + 10 + 20 \:=\:{\color{blue}35}$ ways.
The following is multiple choice question (with options) to answer.
Look at this series: 97, 97, 97, 87, 87, 87, 77, 77, 77 ... What number should come next? | [
"66",
"67",
"68",
"69"
] | B | Explanation:
In this series, each number is repeated, then 10 is subtracted to arrive at the next number.
Answer: Option B |
AQUA-RAT | AQUA-RAT-39741 | As others have said, one can ask that the girls' chairs, and the block of chairs for the boys, be chosen together (in this way, we treat the group of boys as "another girl", for a total of 5 "girls" choosing among 5 "seats"); thus, we can have the boys choose one of the 5 "seats" (in reality, it is their block of chairs), the first girl chooses one of the 4 remaining chairs, the second girl chooses one of the 3 remaining chairs, etc. making for $$5!=5\times 4\times 3\times 2\times 1=120$$ ways of seating the girls and choosing the block of chairs for the boys. Then, all that remains is for the boys to arrange themselves within the block, which there are $$3!=3\times 2\times 1=6$$ ways to do, making (again) a total of $$(5!)\times (3!)=120\times 6=720$$ ways of arranging them.
-
+1 for the graphics! :-) – joriki Jul 8 '12 at 19:53
Thanks!$\text{}$ – Zev Chonoles Jul 8 '12 at 20:00
Treat the group of three boys as an additional girl. That makes five girls, which can be arranged in $5!$ ways. Then you can replace the additional girl by any permutation of the three boys, of which there are $3!$.
-
thanks joriki, but not quiet clear about what you say – user1419170 Jul 8 '12 at 19:49
@user1419170: Could you be more specific? What part isn't clear to you? – joriki Jul 8 '12 at 19:49
this part Treat the group of three boys as an additional girl – user1419170 Jul 8 '12 at 19:51
@user1419170: See Brian's and Saurabh's answers, which use the same idea but phrase it differently. – joriki Jul 8 '12 at 19:53
The following is multiple choice question (with options) to answer.
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? | [
"80",
"100",
"25",
"206"
] | D | We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number
of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
= (24 + 90 + 80 + 15)
= 209.
ANSWER D |
AQUA-RAT | AQUA-RAT-39742 | This is an A.P. The below workout with step by step calculation shows how to find what is the sum of natural numbers or positive integers from 1 to 250 by applying arithmetic progression. is 56. If n is an integer, then n, n+1, and n+2 would be consecutive integers. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are . OPtion 1) 312750 2) 2502 62500 is a sum of number series by applying the values of input parameters in the formula. On each iteration, we add the number num to sum, which gives the total sum in the end. Write a program in java to print the sum of all numbers from 50 to 250(inclusive of 50 and 250) that are multiples of 3 and not divisible by 9. i.e. asked Jan 14 in Binomial theorem by Ritik01 (48.1k points) The sum of numbers from 250 to 1000 which are divisible by 3 is (A) 135657 (B) 136557 (C) 161575 (D) 156375. binomial theorem; jee; jee mains; Share It On Facebook Twitter Email. Here, we will not only tell you what the sum of integers from 1 to 300 is, but also show you how to calculate it fast. , 249, 250.The first term a = 1The common difference d = 1Total number of terms n = 250 Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … The number series 1, 2, 3, 4, . The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. + 249 + 250 = 31375 Therefore, 31375 is the sum of positive integers upto 250. I. DIVISIBILITY OF NUMBERS 1. (1). Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275. Next, it’s going to add those numbers
The following is multiple choice question (with options) to answer.
The sum of all the integers d such that -26 < d < 24 is | [
"0",
"-2",
"-25",
"-49"
] | D | Easy one -
-25, -24, -23,-22,...... -1,0, 1, 2...., 22, 23
Cancel everyhitng and we're left with - -25 and -24 d= -49.
D is the answer. |
AQUA-RAT | AQUA-RAT-39743 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Of the four numbers, whose average is 80, the first is one-fourth of the sum of the last three. The first number is ? | [
"28",
"64",
"11",
"48"
] | B | Explanation:
Let the first number be x,
Then, sum of the four numbers = x + 4x = 5x.
so, 5x/4 = 80 or x = (80 * 4) / 5 = 64
Answer: B |
AQUA-RAT | AQUA-RAT-39744 | 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.
The dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of white washing the four walls of the room at Rs. 5 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each? | [
"Rs.4532",
"Rs.4526",
"Rs.4598",
"Rs.4530"
] | D | Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls
= 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5
= Rs.4530
Answer:D |
AQUA-RAT | AQUA-RAT-39745 | # another probability: number of ways for 4 girls and 4 boys to seat in a row
4 posts / 0 new
Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row
In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
KMST
Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.
For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.
There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).
With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.
Ej-lp ACayabyab
thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?
Jhun Vert
As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.
First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)
Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)
Total number of ways = First Case + Second Case
The following is multiple choice question (with options) to answer.
In how many different number of ways 4 boys and 2 girls can sit on a bench? | [
"700",
"720",
"780",
"800"
] | B | npn = n!
6p6 = 6 × 5 × 4 × 3 × 2 × 1 = 720
B |
AQUA-RAT | AQUA-RAT-39746 | # How to calculate the area covered by any spherical rectangle?
Is there any analytic or generalized formula to calculate area covered by any rectangle having length $l$ & width $b$ each as a great circle arc on a sphere with a radius $R$?
Note: Spherical rectangle is a quadrilateral having equal opposite sides but non-parallel & all the interior angles are equal in magnitude & each one is greater than $90^\circ$.
• What do you mean by "rectangle"? Do you just want a quadrilateral, or some stronger condition? (Note that you cannot have four right angles in a quadrilateral on the sphere.) – Nick Matteo Mar 25 '15 at 13:16
• Yes, a quadrilateral having equal opposite sides (but not parallel) each as a great circle arc on a sphere. – Harish Chandra Rajpoot Mar 25 '15 at 13:21
• What do you mean by length and width? As the sides are not parallel, we don't have the usual idea of width and length. Do you simply mean the lengths of the sides? – robjohn Apr 4 '15 at 14:12
• Yes, length & width are simply the sides of the rectangle as great circles arcs on the spherical surface. – Harish Chandra Rajpoot May 28 '15 at 23:56
Assume we are working on a sphere of radius $1$, or consider the lengths in radians and the areas in steradians.
Extend the sides of length $l$ until they meet. This results in a triangle with sides $$w,\quad\frac\pi2-\frac l2,\quad\frac\pi2-\frac l2$$
The following is multiple choice question (with options) to answer.
The length of a rectangle is three - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 2500 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 15 units? | [
"440 sq.units",
"550 sq.units",
"455 sq.units",
"450 sq.units"
] | D | Given that the area of the square = 2500 sq.units
=> Side of square = √2500 = 50 units
The radius of the circle = side of the square = 50units.
Length of the rectangle = 3/5 * 50 = 30 units
Given that breadth = 15 units
Area of the rectangle = lb = 30 * 15= 450 sq.units
Answer:D |
AQUA-RAT | AQUA-RAT-39747 | # 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.
Ramu bought an old car for Rs. 34000. He spent Rs. 12000 on repairs and sold it for Rs. 65000. What is his profit percent? | [
"16%",
"17%",
"18%",
"41%"
] | D | Total CP = Rs. 34000 + Rs. 12000
= Rs. 46000 and SP = Rs. 65000
Profit(%)
= (65000 - 46000)/46000 * 100
= 41%
Answer: D |
AQUA-RAT | AQUA-RAT-39748 | # 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.
When a plot is sold Rs. 18,700, the owner loses 15%. At what price must the plot be sold in order to gain 15%? | [
"Rs. 21,000",
"Rs. 22,500",
"Rs. 25,300",
"Rs. 25,800"
] | C | Solution
85 : 18700 = 115 : x or x = (18700 x 115 / 85)
= 25300.
Hence, S.P. = Rs. 25300.
Answer C |
AQUA-RAT | AQUA-RAT-39749 | newtonian-mechanics, rotational-dynamics, estimation, stability
Title: Human Lean Equation For a medical experiment I am doing, I need an equation to find the angle at which someone will lean before falling. I am not mathematically inclined in terms of advanced stuff, I am more so of a trigonometry person.
I assume factors that will be needed are BMI, including height and weight, but that is really all I have. For example, I am a 5'11" (180.3cm) tall female. I weigh 165 pounds (74.8kg). I want to find out, computationally (because I could measure myself), how far I could lean before I fall.
Any ideas of how I could go upon this? Even though your body is not a simple, homogeneous, rigid object, we can calculate a little... With a lot of assumptions.
I assume, your feet stand next to each other, not in a step-forward-position. And I assume, the question is about leaning forward/backward, not sideways.
And I assume that the arms have to be aligned at the side of the body.
First step, we need to locate your center of gravity (CG). Even if you feel light/heavy hearted, we assume you to be symmetric. We also assume you to be front-back-symmetric, except for your feet.
Now we need the hight-coordinate of CG. You can find it via plancking across a bar. Let's assume your result is that CG is 99cm from floor up.
Next up, we need your shoe size. I guess you at an american women's size of 11.5, which equates to 27cm foot-length.
If you ever took dance classes, you may remember that you were instructed to "put your weight over your heel", or over the center of your feet. Let's assume that this little shift can be done without "leaning".
As you are "more of a trigonometry person", here comes the fun part:
We think of you (having "weight over heel") as a L-shape, with your feet being 27cm, and the stem of the "L" being 99cm (height of CG). Now you gradually lean forward, and when CG passes over your toes, you fall.
The following is multiple choice question (with options) to answer.
In a certificate by mistake a candidate gave his height as 25% more than actual height. In the interview panel, he clarified that his height was 5 feet 8 nches. Find the % correction made by the candidate from his stated height to his actual height? | [
"10",
"20",
"30",
"17"
] | D | His height was = 5 feet 8 inch = 8 + 60 = 68 inch. Required % correction =68*(1.25-1) = 17
D |
AQUA-RAT | AQUA-RAT-39750 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
Three candidates contested an election and received 1136, 7636 and 13628 votes respectively. What percentage of the total votes did the winning candidate got? | [
"55%",
"56%",
"57%",
"60.8%"
] | D | Total number of votes polled = (1136 + 7636 + 13628) = 22400
So, Required percentage = 13628/22400 * 100 = 60.8%
ANSWER : D |
AQUA-RAT | AQUA-RAT-39751 | When morning comes and all $N$ pirates wake up, they see the greatly diminished pile and a happy fat monkey.
The question is: how many bananas could there have been in the original pile for this to occur?
• Is this for all sets of possible bananas? – awesomepi May 22 '14 at 15:58
• ...The number of possible bananas is limited depending on N, N can be any natural number (except 1). – kaine May 22 '14 at 16:27
• The way I remember it, concludes with the pirates waking up, finding a greatly diminished pile of bananas, each taking their share of $x$ bananas, and feeding the final banana to the monkey. That would give you a starting pint, but I'm not sure it's correct. – SQB May 23 '14 at 7:01
• @SQB I did not remember it ending that way but that could be my bad memory. Mathematically though, I am pretty sure that would be the same as N+1 pirates so Ross's answer still applies. – kaine May 23 '14 at 12:52
• @kaine it is not the same, since N+1 pirates would always divide by (N+1) here it is always a division by N, only it happens N+1 times with the waking up part. – Falco Aug 6 '14 at 11:22
The basic idea is to work backwards.
The last pirate must have found $N+1$ bananas, because he had to find enough bananas remaining for at there to be least 1 banana in the pile left for each pirate. He took 1, fed 1 to the monkey, and left $N-1$. That $N+1$ means that the next to last pirate found $\frac N{N-1}\cdot (N+1)+1$ and so on.
A cute trick is to recognize that there could have been
$-(N-1)$ bananas. Each pirate gives the monkey $1$, takes $-1$, and leaves the pile the same size as before. Since we need to divide it by $N$ for each pirate, the next solution is higher by $N^N$, so the minimal positive solution is $N^N-N+1$ bananas.
The following is multiple choice question (with options) to answer.
Suppose 8 monkeys take 8 minutes to eat 8 banana's.How many minutes would it take 3 monkeys to eat 3 banana's? | [
"5 min",
"6 min",
"7 min",
"8 min"
] | D | there are equal no. of monkeys and equal no. of bananas and they take equal time and the time is 8 mint to eat a banana so each monkey take 8 mints to eat a banana's so 3 monkeys will take 8 mints to eat 3 banana's.
ANSWER:D |
AQUA-RAT | AQUA-RAT-39752 | ## 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.
The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 6360, what is the length of the plot in metres? | [
"333",
"200",
"240",
"276"
] | C | Let length of plot = L meters, then breadth = L - 20 meters
and perimeter = 2[L + L - 20] = [4L - 40] meters
[4L - 40] * 26.50 = 6360
[4L - 40] = 6360 / 26.50 = 240
4L = 280
L = 280/4= 70 meters.
Answer: C |
AQUA-RAT | AQUA-RAT-39753 | 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.
A charitable association sold an average of 66 raffle tickets per member. Among the female members, the average was 70 raffle tickets. The male to female ratio of the association is 1:2. What was the average number T of tickets sold by the male members of the association | [
"50",
"56",
"58",
"62"
] | C | Given that, Total average T sold is 66, Male/Female = 1/2 and Female average is 70. Average of Male members isX.
(70*F+X*M)/(M+F) = 66 -> Solving this equation after substituting 2M=F, X = 58. ANS C. |
AQUA-RAT | AQUA-RAT-39754 | • Perfect thank you very much, just the guidance I needed – Darragh O'Flaherty Apr 20 '18 at 11:28
You have $v= \alpha u + w$. Now, dot product by $u$ gives $$v \cdot u = \alpha u \cdot u + w \cdot u$$ We know $w \cdot u=0$ so the equation gives $$3 =\alpha 9$$ so $$\alpha={ 1 \over 3}$$. Knowing $\alpha$ it is straightforward that $$w=\left ({7 \over 3}, {5 \over 3}, -{4 \over 3} \right )$$
The following is multiple choice question (with options) to answer.
What is the simplified result of following the steps below in order?
(1) add 5y to 2w
(2) multiply the sum by 3
(3) subtract w + y from the product | [
"5w + 14y",
"5x + 16y",
"5x + 5y",
"6x + 4y"
] | A | 3(5Y+2W) -W-Y= 14Y+5W
'A' is the answer |
AQUA-RAT | AQUA-RAT-39755 | 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.
The length of the bridge, which a train 120 m long and traveling at 45 km/hr can cross in 30 sec is? | [
"377",
"367",
"237",
"255"
] | D | Speed = 45 * 5/18 = 25/2 m/sec.
Time = 30 sec
Let the length of bridge be x meters.
Then, (120 + x)/30 = 25/2
x = 255 m.
Answer: D |
AQUA-RAT | AQUA-RAT-39756 | => 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.
Paul's income is 20% less than Rex's income, Quentin's income is 20% less than Paul's income, and Sam's income is 40% less than Paul's income. If Rex gave 60% of his income to Sam and 40% of his income to Quentin, Quentin's new income would be what fraction of Sam's new income? | [
"\t26/27",
"\t13/17",
"\t13/19",
"\t12/19"
] | A | MAKE R = 10
P = 0.8R = 8
Q = 0.8P = 6.4
S= 0.6P = 4.8
FOR THAT WE GET S = 10.8
AND Q 10.4
SO 10.4/10.8= 2.6/2.7
Ans:A |
AQUA-RAT | AQUA-RAT-39757 | general-relativity, astrophysics, orbital-motion, perturbation-theory, differential-equations
$${\rm{perihelion}} \, {\rm{advance}}\,{\rm{relation:}}\,6\pi M{u_0} \to \frac{{6\pi GM}}{{{r_0}{c^2}}},$$
in which $r_0$ is the mean orbital radius (the mean distance from Sun). Considering Sun with the mass $1.989 \times 10^{30}$ kg and the Mercury with the mean distance from Sun about $5.79 \times 10^7$ km, one finds the famous result of $43$ $\frac{\rm{arc-seconds}}{\rm{century}}$. (data from Solar System Data)
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
References
The following is multiple choice question (with options) to answer.
Mercury travels around the Sun at an approximate speed of 28.8 miles per second. This speed is how many kilometers per hour? (1 km = 0.6 miles) | [
"172,800",
"124,600",
"93,400",
"72,200"
] | A | The speed is 28.8 miles/s. Then 28.8/0.6 = 48 km/s
48*3600 = 172,800 kph
The answer is A. |
AQUA-RAT | AQUA-RAT-39758 | # 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? | [
"75%",
"25%",
"28%",
"85%"
] | B | 800 --- 200
100 --- ? => 25%
Answer: B |
AQUA-RAT | AQUA-RAT-39759 | 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 man can do a job in 15 days. His father takes 10 days and his son finishes it in 25 days. How long will they take to complete the job if they all work together? | [
"6.3",
"6.9",
"7.1",
"6.1"
] | C | 1 day work of the three persons = (1/15 + 1/10 + 1/25) = 21/150
So, all three together will complete the work in 150/21 = 7.1 days.
Answer:C |
AQUA-RAT | AQUA-RAT-39760 | ### 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.
A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. What is the share of B in the profit. | [
"2660",
"1000",
"2300",
"4000"
] | A | Explanation :
A is a working member and for that, he receive 5% of the profit
= 5% of 7400 = 5*7400/100 = 370
Remaining amount = 7400-370 = 7030
Ratio of their investments = 6500*6 : 8400*5 : 10000*3
= 65*6 : 84*5 : 100*3 = 13*6 : 84 : 20*3 = 13*2 : 28 : 20
= 13 : 14 : 10
Share of B in the profit = 7030 * (14/37) = 190*14 = 2660. Answer : Option A |
AQUA-RAT | AQUA-RAT-39761 | 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.
Two trains are traveling on a collision course. If train A is traveling at a speed of 350 mph and train B is traveling 28% slower, how much time will it take the trains to collide if the initial distance between the two is 3010 miles? | [
"Five hours and 0 minutes.",
"One hour and 10 minutes.",
"Two hours and 25 minutes.",
"Three hours and 15 minutes."
] | A | Answer is A. Five hours and 0 minutes.
Since they are racing towards each other, the speed is accumulative, ie. 350 mph + 350*0.72 = 602 (0.72 because the second train is 28% slower)
So time before collision is total distance / total speed = 3010/602 = 5 hours |
AQUA-RAT | AQUA-RAT-39762 | 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.
A and B go around a circular track of length 600 m on a cycle at speeds of 36 kmph and 54 kmph. After how much time will they meet for the first time at the starting point? | [
"120 sec",
"176 sec",
"178 sec",
"187 sec"
] | A | Time taken to meet for the first time at the starting point
= LCM { length of the track / speed of A , length of the track / speed of B}
= LCM { 600/ (36 * 5/18) , 600/ (54 * 5 /18) }
= LCM (60, 40) = 120 sec.
Answer: A |
AQUA-RAT | AQUA-RAT-39763 | 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.
When all the students in a school are made to stand in rows of 50, 25 such rows are formed. If the students are made to stand in rows of 30, how many such rows will be formed ? | [
"41",
"36",
"38",
"19"
] | A | Explanation:
Total number of students = 50 x 25
When arranged in rows of 30, number of rows formed are,
= 41.
Answer: A |
AQUA-RAT | AQUA-RAT-39764 | materials, density
densitytotal = masstotal / volume =
volume = masstotal / densitytotal = 0.2 kg / 2300 kg/m3 = 0.00008696 m3
masssand = 2700 kg/m3 * 0.0000869 m3 = 0.2348 kg
massplastic = 1400 kg/m3 * 0.0000869 m3 = 0.1217 kg
But I don't think this step was the right choice since the mass of the sand came out to be larger than the mass of the sample. Was I correct in trying to figure out the value of the mass of sand and plastic or was that the wrong approach to determine the material's composition? If it's the first case, how should I find out the mass values? Your mistake is that you have calculated the mass of a sample with the same volume as ours consisting entirely of sand, and a sample with the same volume as ours consisting entirely of sand.
volume = mass_total / density_total = 0.2 kg / 2300 kg/m^3 = 0.00008696 m^3
gives the initial volume of the sample before the salt is removed, however by multiplying this by the densities of sand and plastic, we calculate the mass of a 0.0000869 m^3 sample of sand and a 0.0000869 m^3 sample of plastic.
First we need to find the new volume and density of the sample after the salt has been removed. This can be done by calculating the volume of salt removed:
volume_salt=0.03kg_salt/2100 kg/m^3=0.00001429m^3
By subtracting this volume from the initial volume you already calculated:
0.00008696m^3-0.00001429m^3=0.00007267m^3
dividing the new mass by this gives:
0.170kg/0.00007267m^3=2,339kg/m^3
The following is multiple choice question (with options) to answer.
A cement mixture is composed of 3 elements. By weight, 1/3 of the mixture is sand, 1/4 of the mixture is water, and the remaining 10 pounds of the mixture is gravel. What is the weight of the entire mixture in pounds? | [
"24",
"26",
"28",
"30"
] | A | Let the total weight be x.
Sand content= (1/3)x
Water content= (1/4)x
Gravel=x-(1/3)x-(1/4)x=(5/12)x=10
x=24
Then answer will be A=24 |
AQUA-RAT | AQUA-RAT-39765 | Question
# The number of positive numbers less than $$1000$$ and divisible by $$5$$ ( no digits being replaced) is
A
150
B
154
C
166
D
None of these
Solution
## The correct option is B $$154$$Here, the available digits are $$0,1,2,3,4,5,6,7,8,9.$$The numbers can be of one, two or three digits and in each of them unit's place must have $$0$$ or $$5$$ as they must be divisible by $$5$$.The number of numbers of one digit $$=1$$$$(\because$$ is the only number$$)$$.The number of numbers of two digits divisible by $$5=$$ number of all the numbers of two digits divisible by $$5-$$ number of numbers of two digits divisible by $$5$$ and having $$0$$ in ten's place $$\displaystyle =^{ 2 }{ { P }_{ 1 } }\times ^{ 9 }{ { P }_{ 1 } }-1$$,$$(\because$$unit's place can be filled by either $$0$$ or $$5$$ in first category and only by $$5$$ in the second category$$)$$$$=2\times 9-1=17$$.The number of numbers of numbers of three digits divisible by $$5=$$ number of all the numbers of three digits divisible by $$5=$$ number of numbers of three digits divisible by $$5=$$ number of numbers of three digits divisible by $$5$$ and having $$0$$ in hundred's place $$\displaystyle =^{ 2 }{ { P }_{ 1 } }\times ^{ 9 }{ { P }_{ 2 } }\times ^{ 8 }{ { P }_{ 1 } }\times 1=2\times 9\times 8-8=136.$$$$\therefore$$ required number of numbers $$=1+17+136=154$$ Mathematics
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The following is multiple choice question (with options) to answer.
Find the least number of 5 digits, which is exactly divisible by 89. | [
"27",
"37",
"932",
"12"
] | A | ANSWER:A |
AQUA-RAT | AQUA-RAT-39766 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
If a sequence of 8 consecutive odd integers with increasing values has 5 as its 7th term, what is the sum of the terms of the sequence? | [
"-8",
"-4",
"0",
"4"
] | C | Let x be the first term.
Then x + 12 = 5 and x = -7
The sum is: x + (x+2) +...+ (x+14) =
8x + 2(1+2+...+7) = 8x + 2(7)(8) / 2 = 8(-7) +56 = 0
The answer is C. |
AQUA-RAT | AQUA-RAT-39767 | # 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 clock store sold a certain clock to a collector for 25 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 55 percent of what the collector had paid. The shop then sold the clock again at a profit of 70 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $125, for how much did the shop sell the clock the second time? | [
"$500",
"$467.5",
"$935",
"$525.5"
] | B | Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 25% profit.
This means the clocks' selling price was C (1.25) and this becomes cost price for the collector.
Now, when the collector tries to sell the same clock to the store, the store buys it for 55% the price at which the collector bought it.
Thus, you get = 1.25*0.55*C = 0.6875 C
Furthermore, the store sells the clock for the second time for 70% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.7 = 1.7 * 0.6875 C
Finally given that C - 0.6875 C = 125 ----> C = 400$
Thus, the cost of the clock the second time around = 1.7*0.6875 C = 1.7 * 0.6875 * 400 = 467.5$. Hence B is the correct answer. |
AQUA-RAT | AQUA-RAT-39768 | +0
# 3.00 x 10^8 / 6.50 x 10^-7 the result is 4.61 x 10^14 my question is how 10^8 / 10^-7 = 10^14? thank u
0
330
2
3.00 x 10^8 / 6.50 x 10^-7
4.61 x 10^14
my question is how 10^8 / 10^-7 = 10^14?
thank u
Guest Nov 21, 2014
#1
+91465
+5
3.00 x 10^8 / 6.50 x 10^-7
4.61 x 10^14
my question is how 10^8 / 10^-7 = 10^14? IT DOES NOT!
$$\\(3\times 10^8)/(6.5\times 10^{-7})\\\\ =(3/6.5)\times 10^8/10^{-7}\\\\ =0.4615\times 10^{8--7}\\\\ =0.4615\times 10^{15}\\\\$$
BUT this answer is to be in scientific notation so there mus be just one non-zero number in front of the point
$$\\=4.4615\times 10^{-1} \times 10^{15}\\\\ =4.4615\times 10^{14}\qquad \mbox{Correct to 4 significant figures}\\\\$$
So does that all make sense now?
Melody Nov 21, 2014
Sort:
#1
+91465
+5
3.00 x 10^8 / 6.50 x 10^-7
4.61 x 10^14
my question is how 10^8 / 10^-7 = 10^14? IT DOES NOT!
$$\\(3\times 10^8)/(6.5\times 10^{-7})\\\\ =(3/6.5)\times 10^8/10^{-7}\\\\ =0.4615\times 10^{8--7}\\\\ =0.4615\times 10^{15}\\\\$$
The following is multiple choice question (with options) to answer.
Which one of the following is equal to 3.14×10(6)? | [
"314",
"3140",
"3140000",
"None of these"
] | C | Solution
Given, 3.14 ×106 =3.140000×1000000
= 3140000.
Answer C |
AQUA-RAT | AQUA-RAT-39769 | ### 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 40 percent less than Juans income. What percentage of Juans income is Marts income | [
"124%",
"120 %",
"96 %",
"80 %"
] | C | Tim=100
Mary's income is 60 percent more than tims income;
Mary's Income = T+0.6T = 1.6T = 1.6*100=160
Tims income is 40 percent less than juans income .
100 is 40% less than Juan's Income
100=J-0.4J
100=0.6J
J = 100/0.6=1000/6
What percentage of juans income is marts income
(160/(1000/6))*100 = (160*6*100)/1000 = 96%
ANSWER:C |
AQUA-RAT | AQUA-RAT-39770 | 4,8),(3,5,8),(2,6,8),(1,7,8),(1,9,8),(9,0,9),(8,1,9),(7,2,9),(6,3,9),(5,4,9),(4,5,9),(3,6,9),(2,7,9),(1,8,9)]
The following is multiple choice question (with options) to answer.
4, 9, 13, 22, 35, ? | [
"57",
"70",
"63",
"75"
] | A | Sum of two consecutive numbers of the series gives the next number.
Answer : A. |
AQUA-RAT | AQUA-RAT-39771 | 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 does 4/5th of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work? | [
"23 days",
"37 days",
"37 ½ days",
"40 days"
] | C | Explanation:
A can finish the whole work in 20 × 5/4 days = 25 days
A and B together finish the whole work in 5 × 3 days = 15 days
Therefore, B can finish the whole work in 25 B/ 25 + B = 15
25 B = 15 ( 25 + B)= 375 + 15B
10B = 375 and B = 375/10 = 37 ½ days.
Answer: Option C |
AQUA-RAT | AQUA-RAT-39772 | ### Show Tags
27 Oct 2016, 16:26
3
GMATPrepNow wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
We are given that Bill and Ted each competed in a 240-mile bike race. We are also given that Bill’s average speed was 5 miles per hour slower than Ted’s average speed.
Both Bill and Ted had distance of 240 miles. If we let Ted’s speed = r, we can let Bill’s speed = r - 5. We can use the above information to determine the time of Bill and Ted in terms of variable r.
Since time = distance/rate, Ted’s time = 240/r and Bill’s time = 240/(r - 5).
Since Ted completed the race 4 hours sooner than Bill did, we can create the following equation:
240/r + 4 = 240/(r - 5)
To eliminate the denominators of the fractions we can multiply the entire equation by r(r-5) and we have:
240(r - 5) + 4[r(r - 5)] = 240r
240r - 1,200 + 4r^2 - 20r = 240r
4r^2 - 20r - 1,200 = 0
r^2 - 5r - 300 = 0
(r - 20)(r + 15) = 0
r = 20 or r = -15
Since r must be positive, r = 20.
Thus, Bill’s rate = 20 - 5 = 15 mph.
_________________
# Scott Woodbury-Stewart
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Scott@TargetTestPrep.com
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Re: Bill and Ted each competed in a 240-mile bike race. [#permalink]
The following is multiple choice question (with options) to answer.
Two dogsled teams raced across a 500-mile course in Alaska. Team A finished the course in 5 fewer hours than team B. If team A’s average speed was 5 miles per hour greater than that of team B, what was team B’s average speed, in miles per hour? | [
"12",
"15",
"18",
"20"
] | D | Let v be team B's average speed.
Time = Distance/Rate and the time difference was 5 hours.
500/v - 500/(v+5)= 5
500v + 2500 - 500v = 5(v)(v+5)
500 = (v)(v+5)
(20)(25) = (v)(v+5)
v = 20 mph
The answer is D. |
AQUA-RAT | AQUA-RAT-39773 | # 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.
Robert is travelling on his cycle andhas calculated to reach point A at 2 PM. if he travels at 10 kmph, he will reach there at 12Pm if he travels at 15kmph. At what speed must he travel to reach A at 1 PM? | [
"10 kmph",
"11 kmph",
"12 kmph",
"14 kmph"
] | C | Let the distance travelled by x km.
Then, x - x = 2
10 15
3x - 2x = 60
x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60 hrs = 6 hrs.
10
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = 60 kmph. = 12 kmph.
5
C |
AQUA-RAT | AQUA-RAT-39774 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is: | [
"152600m2",
"153500m2",
"153600m2",
"153800m2"
] | C | Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
As per question length is 3x and width is 2x
We know perimeter of rectangle is 2(L+B)
So, 2(3x+2x) = 1600
=> x = 160
So Length = 160*3 = 480 meter
and Width = 160*2 = 320 meter
Finally, Area = length * breadth
= 480 * 320 = 153600
ANSWER IS C |
AQUA-RAT | AQUA-RAT-39775 | # another probability: number of ways for 4 girls and 4 boys to seat in a row
4 posts / 0 new
Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row
In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
KMST
Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.
For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.
There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).
With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.
Ej-lp ACayabyab
thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?
Jhun Vert
As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.
First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)
Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)
Total number of ways = First Case + Second Case
The following is multiple choice question (with options) to answer.
In how many different number of ways 4 boys and 4 girls can sit on a bench? | [
"40320",
"42345",
"23456",
"40567"
] | A | npn = n!
8p8 = 8 x 7 x 6 × 5 × 4 × 3 × 2 × 1 = 40320
A) |
AQUA-RAT | AQUA-RAT-39776 | 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 finish a work in 18 days and B can do same work in half the time taken by A. then working together, what part of same work they can finish in a day | [
"1/6",
"1/9",
"1/11",
"1/3"
] | A | A's 1 day work = 1/18
B's 1 day work = 1/9
(A+B)'s one day work = (1/18+1/9)=1/6
Answer: A |
AQUA-RAT | AQUA-RAT-39777 | $y = x$ if $x \geq -1$.
It also means
$y = -(x + 1) - 1$ if $x + 1 < 0$, i.e. $x < -1$
$y = -x - 2$ if $x < -1$.
So all the points which satisfy these two linear equations will be your solution.
The following is multiple choice question (with options) to answer.
If x ≠ 0 and (−y^2 + xy)/x=y+x, which of the following expressions is equal to its opposite? | [
"x^2",
"x^2 + y^2",
"y^2",
"(y + x)^2"
] | B | (−y^2 + xy)/x=y+x
=> −y^2 + xy = xy +x^2
=> x^2 + y^2 = 0
Answer B |
AQUA-RAT | AQUA-RAT-39778 | # 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 marks up the price of his goods by 20% and gives a discount of 10% to the customer. He also uses a 900 gram weight instead of a 1 kilogram weight. Find his percentage profit due to these maneuvers? | [
"8%",
"12%",
"20%",
"16%"
] | C | Solution: He sells only 900 grams when he takes the money for 1 kg.
and
he sells at a 8% profit (20% markup, 10% discount).
Hence, his selling price is Rs. 108 for 900 grams.
% profit = (18/90)*100 = 20%.
Answer: Option C |
AQUA-RAT | AQUA-RAT-39779 | +0
# At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
0
2788
3
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Feb 25, 2015
#3
+99377
+5
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$
so
$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$
$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$
Obviously there is not a neg number of people so there must be 12 people.
Feb 25, 2015
#1
0
can you even solve that
Feb 25, 2015
#2
+98196
+5
We can solve this by this "formula"
n(n-1)/ 2 = 66 multiply by 2 on each side
n(n-1) = 132 simplify and rearrange
n^2 - n - 132 = 0 factor
The following is multiple choice question (with options) to answer.
Number of handshakes in a party is 28.Find the number of people present in the party altogether? | [
"5",
"6",
"7",
"8"
] | D | no. of handshakes (ie) n(n-1)/2= 28
therefore no. of people present
n(n-1)=56
n2-n-56=0
n=8
No. of people present = 8
ANSWER:D |
AQUA-RAT | AQUA-RAT-39780 | Shanonhaliwell April 8th, 2018 03:31 PM
Quote:
Originally Posted by romsek (Post 591335) outstanding, you seem to be getting the hang of things.
Thanks to you, I was able to do it.
All times are GMT -8. The time now is 12:30 AM.
The following is multiple choice question (with options) to answer.
If the time is currently 1:30pm, what time will it be in exactly 239 hours? | [
"12:30am",
"1:30am",
"12:30pm",
"2:30pm"
] | C | Every 24 hours will bring us to the same 1:30pm. 240 is divisible by 24 (10*24=240), so if it were 240 hours we would have 1:30pm, but since it's 239 hours then it will be 1 hour less, so 12:30pm.
Answer: C. |
AQUA-RAT | AQUA-RAT-39781 | 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. 375",
"Rs. 400",
"Rs. 600",
"Rs. 800"
] | B | Explanation:
C's 1 day's work =1/3-(1/6+1/8)=1/3-7/24=1/24
A's wages : B's wages : C's wages =1/6:1/8:1/24=4:3:1
C's share (for 3 days) = Rs.(3X1/24X3200)=Rs. 400
ANSWER IS B |
AQUA-RAT | AQUA-RAT-39782 | ### Show Tags
18 Feb 2015, 07:51
1
Using smart numbers:
Jennifer Purchases watches for $5: B=5 Markup of 100%; now selling for$10: x=10
She sells 5 watches: N=5
Her total profit will be (N*x)-(B*N)
(5*10)-(5*5)= 25
T=25
Now using the variables we must look at the answer choices that will solve for the % markup of the watches, which we chose to be 100.
B=5
T=25
N=5
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Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
### Show Tags
18 Feb 2015, 21:51
Cost price per watch = b
Selling price per watch = $$\frac{t}{n}$$
Let x = percent of the markup from her buy price to her sell price
$$x = 100 *\frac{t}{n} * \frac{1}{b} = \frac{100t}{nb}$$
_________________
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Joined: 02 Sep 2009
Posts: 50544
Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
### Show Tags
22 Feb 2015, 11:17
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?
A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
The following is multiple choice question (with options) to answer.
An agent buys a T.V. set listed at Rs. 10000 and gets 10% and 20% successive discounts. He spends 10% of his C.P. on transport. At what price (in rupees) should he sell the T.V. set to earn a profit of 10%? | [
"8692",
"8702",
"8712",
"8722"
] | C | Net price after two discounts=80% of 90% of Rs. 10000=Rs. 7200.
Net C.P.=(7200+10% of Rs. 7200)=7920.
Therefore, S.P. =110% of Rs. 7920=Rs. 8712.
ANSWER:C |
AQUA-RAT | AQUA-RAT-39783 | 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 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 the measurements on the S scale of 30 and 60 respectively. What measurement on the R scale corresponds to a measurement of 90 on the S scale? | [
"32",
"36",
"42",
"48"
] | C | A change of +18 on the R scale is a change of +30 on the S scale.
R = (18/30)*S + K = 3S/5 + K
6 = 3(30)/5 + K
K = -12
R = 3S/5 - 12
R = 3(90)/5 - 12 = 54-12 = 42
The answer is C. |
AQUA-RAT | AQUA-RAT-39784 | ### 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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Posts: 9558
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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.
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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.
In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 14 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can? | [
"40",
"77",
"48",
"52"
] | B | Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T - 14)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 - 14 = 4/9(T - 14)
10T = 1386 - 616 => T = 77.
ANSWER:B |
AQUA-RAT | AQUA-RAT-39785 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
The sum of three consecutive even numbers is 87. Find the middle number of the three? | [
"31",
"21",
"29",
"22"
] | C | Middle number =87/3
=29
Ans C |
AQUA-RAT | AQUA-RAT-39786 | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
The following is multiple choice question (with options) to answer.
Rs.460 was divided among 41 boys and girls such that each boy Rs.12 and each girl got Rs.8. What is
the number of boys? | [
"33",
"30",
"36",
"28"
] | A | Explanation :
-------------------------------------------------------------------
Solution 1
-------------------------------------------------------------------
Assume that the number of boys = b and number of girls is g
number of boys + number of girls = 41
=> b + g = 41 ------------ (Equation 1)
Given that each boy got Rs.12 and each girl got Rs.8 and Total amount = Rs.460
=> 12b + 8g = 460 -------- (Equation 2)
Now we need solve Equation 1 and Equation 2 to get b and g
(Equation1)×8=>8b+8g=8×41=328−−−−−−−−(Equation3)
(Equation 2) - (Equation 3) = 4b = 460 - 328 = 132
=>b=1324=33
-------------------------------------------------------------------
Solution 2
-------------------------------------------------------------------
Given that Amount received by a boy = Rs.12 and
Amount received by a girl =Rs.8
Total amount = 460
Given that number of boys + Number of girls = 41
Hence mean amount = 460/41
By the rule of alligation, we have
Amount received by a boy Amount received by a girl
12 8
Mean amount
460/41
460/41 - 8 =132/41 12 - 460/41 = 32/41
Number of boys : Number of girls = 132/41 : 32/41 = 132 : 32 = 66 : 16 = 33 : 8
Given that number of boys + Number of girls = 41
Hence number of boys = 41×3341=33
Answer : Option A |
AQUA-RAT | AQUA-RAT-39787 | # 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.
Vijay bought 160 shirts at the rate of Rs. 215 per shirt. The transport expenditure was Rs. 1400. He paid an octroi at the rate of Rs. 1.75 per shirt and labour charges were Rs. 320. What should be the selling price of one shirt, if he wants a profit of 20%? | [
"Rs. 258",
"Rs. 270",
"Rs. 273",
"Rs. 285"
] | C | Total CP per shirt = 215 + 1400/160 + 1.75 + 320/160 = Rs. 227.5
SP = CP[(100 + profit%)/100]
= 227.5 * [(100 + 20)/100] = Rs. 273.
ANSWER:C |
AQUA-RAT | AQUA-RAT-39788 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
Three candidates contested an election and received 45990, 20008 and 34884 votes respectively. What percentage of the total votes did the winning candidate get ? | [
"57.25%",
"45.54%",
"75.20%",
"45.59%"
] | D | Explanation :
Total number of votes polled = (45990 + 20008 + 34884 )
= 61200.
Required percentage = (45990/ 100882) x 100%
= 45.59%.
Answer : D |
AQUA-RAT | AQUA-RAT-39789 | +0
# URGENT! I NEED HELP RIGHT AWAY!
0
261
1
+755
Larry has 4-cent stamps and 9-cent stamps, which he can combine to produce various amounts of postage. For example, he can make 40 cents by using four 9-cent stamps and a 4-cent stamp, or by using ten 4-cent stamps. However, there are some amounts of postage he can't make exactly, such as 10 cents.
What is the largest number of cents that Larry CANNOT make exactly from a combination of 4- and/or 9-cent stamps?
Explain how you know your answer is correct. (You should explain two things: why Larry can't make the amount of your answer, and why he CAN make any bigger amount.)
Jan 26, 2018
#1
+21199
+1
Larry has 4-cent stamps and 9-cent stamps, which he can combine to produce various amounts of postage.
For example, he can make 40 cents by using four 9-cent stamps and a 4-cent stamp, or by using ten 4-cent stamps.
However, there are some amounts of postage he can't make exactly, such as 10 cents.
What is the largest number of cents that Larry CANNOT make exactly from a combination of 4- and/or 9-cent stamps?
Explain how you know your answer is correct. (You should explain two things:
why he CAN make any bigger amount.)
The following is multiple choice question (with options) to answer.
The number of stamps that P and Q had were in the ratio of 5:3 respectively. After P gave Q 10 stamps, the ratio of the number of P's stamps to the number of Q's stamps was 9:7. As a result of the gift, P had how many more stamps than Q? | [
"20",
"30",
"40",
"60"
] | A | P started with 5k stamps and Q started with 3k stamps.
(5k-10)/(3k+10) = 9/7
35k - 27k = 160
k = 20
P has 5(20) - 10 = 90 stamps and Q has 3(20)+10 = 70 stamps.
The answer is A. |
AQUA-RAT | AQUA-RAT-39790 | • $P_2$ will fly $\big[1-(d+r+y)\big]$ distance away from the airport in the counter-clockwise direction to meet up with $P_3$.
• At this point, $P_2$ will donate $z$ fuel to $P_3$.
• $P_2$ and $P_3$ will then both fly back $z$ distance, arriving at a distance of $1-d-r-y-z$ from the airport with no fuel.
• After refuelling at the airport, $P_1$ will fly the distance towards $P_2$ and $P_3$ and refund each of them for that much fuel. All three planes will then head back to the airport together.
From this, we must have
• $0 \leqslant s\leqslant d/3$: $P_1$ can fly $s$ distance forward and backwards, and refund $P_2$ for $s$ distance
• $z\geqslant 0$: cannot donate negative fuel
• $2x + 1-d-r-y \leqslant d+r+y$: $P_3$ must not run out of fuel before $P_2$ can reach it again
• $1-d-r-y - z \leqslant d/4$: $P_1$ can reach $P_2$ and $P_3$, refund them both, and the three of them will have enough fuel to head back to the airport
• $2x + 2s + 1-d-r-y - z\leqslant d+r+y + z$: $P_2$ and $P_3$ must not run out of fuel before $P_1$ can reach them again
Putting these together:
The following is multiple choice question (with options) to answer.
the first flight out of Phoenix airport had a late departure. If the next three flights departed on-time, how many subsequent flights need to depart from Phoenix on-time, for the airport's on-time departure rate to be higher than 90%? | [
"6",
"7",
"9",
"10"
] | B | We need on-time departure rate to be higher than 9/10, so it should be at least 10/11 which means that 10 out of 11 flights must depart on time. Since for now 3 out of 4 flights departed on time then 10−3=7 subsequent flights need to depart on-time.
Answer: B |
AQUA-RAT | AQUA-RAT-39791 | $\therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130$
Which is the same answer as yours, however the answer in the book says $2097170$. They must of added the 20 instead of subtracted it. OK thanks
4. Originally Posted by chancey
Right, but thats not the answer in the back of the book...
I did it this way:
$\sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20$
$\sum_{n=1}^{20} (2^n) = \frac{1-2^{21}}{1-2} - 1 = 2097150$
$\because \sum_{n=0}^{k} (ar^n) = \frac{a(1-r^{k+1})}{1-r}$
$\therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130$
Which is the same answer as yours, however the answer in the book says $2097170$. They must of added the 20 instead of subtracted it. OK thanks
Then the book is inncorect.
The following is multiple choice question (with options) to answer.
If the sum of a number and its square is 20, what is the number? | [
"15",
"26",
"28",
"4"
] | D | Let the number be x. Then,
x + x2 = 20
(x + 5)(x - 4) = 0
x = 4
ANSWER:D |
AQUA-RAT | AQUA-RAT-39792 | $P(-5) = 100,000 * (1.0594631)^{-5} \approx 74,915$.
=====
Alternatively
$(1 + rate)^{12} = 2$. What is $(1 + rate)^5$?(1+rate)^{12} = 2 \implies rate = .0594631$so$(1+rate)^5 = 1.594631^5 = 1.3348398541700343648308318811845$So from$1995$to$2000$the population grew by a factor of$1.3348398541700343648308318811845$. Or$x*1.3348398541700343648308318811845 = 100,000$so$x= \frac {100,000}{1.3348398541700343648308318811845} = 74,915\$
• Thank you very much. I'm sorry but I don't have enough reputation to declare your answer as useful but of course, it is. – Bachir Messaouri Dec 11 '17 at 22:16
The following is multiple choice question (with options) to answer.
If population of certain city increases at the rate of 5%. If population in 1981 was 370440, then population in 1978 was? | [
"320000",
"120029",
"187899",
"178968"
] | A | X * (105/100) * (105/100) * (105/100) = 370440
X = 370440/1.157625
X = 320000
Answer:A |
AQUA-RAT | AQUA-RAT-39793 | 3 , 10 , 12 , 5 , 18 , 6 = { x1 , x2 , x3 , x4 , x5 , x6 }
n = 6 , μ = 9
σ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{3^{2}+10^{2}+12^{2}+5^{2}+18^{2}+6^{2}}{6}\:&space;-\:&space;9^{2}}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{638}{6}\:&space;-\:&space;81}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{23}$ = 5.03
----------------------
With the Standard Deviation being an indicator of how far away the values are from the mean/average. Let's use it with the list of values from example (1.1).
For 5 , 7 , 3 , 5 , 6 , 4, the mean was 5.
The Standard Deviation was 1.29.
We can work out how far way one Standard Deviation is from the mean in both a positive and negative direction.
51.29 = 3.71 , 5 + 1.29 = 6.29
This means we would expect that the majority of the values will be between 3.71 and 6.29.
Looking at the list, this does turn out to be the case.
With only 3 and 7 lying out side this range.
Now looking at the list from example (1.2).
3 , 10 , 12 , 5 , 18 , 6, the mean was 9.
The Standard Deviation was 5.03.
95.03 = 3.97 , 9 + 5.03 = 14.03
This means we would expect that the majority of the values will be between 3.97 and 14.03.
Similar to the list from (1.1), this again is the case.
The following is multiple choice question (with options) to answer.
Set Q consists of the following five numbers: Q = {5, 8, 13, 21, 32}. Which of the following sets has the same standard deviation as Set Q?
I. {35, 38, 43, 51, 64}
II. {10, 16, 26, 42, 68}
III. {46, 59, 67, 72, 75} | [
" I only",
" III",
" IIII",
" IIIII"
] | D | (1) Multiplying all the numbers change Standard Deviation
(2) Adding and subtracting all the numbers with the same number keeps the standard deviation same.
If you observe Set I and III are added versions of Set Q .
Set I: 5 has been added to the Set Q
Set III: Subtract each element from 80 and you would find a number there in the Set Q.
Set II: elements are multiplied by 2 and standard deviation changes.
Hence the answer is D. |
AQUA-RAT | AQUA-RAT-39794 | Now
$$y+5 = 2k+5$$
Now we just need to show that $$2k+5$$ is 2 times an integer plus 1
$$2k+5 = 2(k+2)+1$$
So $$2k+5$$ is odd because it can be written in the form 2*integer +1 where the integer here is $$k+2$$. So $$y+5$$ is odd since $$y+5 = 2k+5$$
So if any number is even. Then that number plus 5 is odd. It doesn't matter if the original number is 3x or 8z or 3x^2-5x+x^3 etc...
The following is multiple choice question (with options) to answer.
If x and y are integers and x=50y + 2, which of the following must be odd? | [
"none of the following",
"x + y",
"x + 2y",
"3x-1"
] | A | x = 50y + 2, since y is integer, 50y will always be even hence 50y + 2 will be even hence x = even
you don't know whether Y is even or odd, so you need to try.
E: even * even = even ; ->not true
B: even + even = even ; > not true
C: even + 2*even = even ;
A |
AQUA-RAT | AQUA-RAT-39795 | ## Example 2: Surgical implants
Surgeries involving implants sometimes result in the patient's body rejecting the implant. A certain surgery has a rejection rate of 11, percent. The rest of the patients successfully accept the implant.
Assume that the results for each patient are independent.
Example 2
In a random sample of 8 of these surgeries, find the probability that at least one patient rejects the implant.
P, left parenthesis, start text, a, t, space, l, e, a, s, t, space, o, n, e, space, r, e, j, e, c, t, s, end text, right parenthesis, equals
## Example 3: Free-throws
Esther is a basketball player who makes 75, percent of the free-throws she attempts. Assume that the results of each shot are independent.
Example 3
If Esther attempts 3 free-throws, find the probability that she misses at least one free-throw.
P, left parenthesis, start text, a, t, space, l, e, a, s, t, space, o, n, e, space, m, i, s, s, end text, right parenthesis, equals
## Generalizing the strategy
In general, we can use these strategies:
P, left parenthesis, start text, a, t, space, l, e, a, s, t, space, 1, space, s, u, c, c, e, s, s, end text, right parenthesis, equals, 1, minus, P, left parenthesis, start text, a, l, l, space, f, a, i, l, u, r, e, s, end text, right parenthesis
or similarly,
P, left parenthesis, start text, a, t, space, l, e, a, s, t, space, 1, space, f, a, i, l, u, r, e, end text, right parenthesis, equals, 1, minus, P, left parenthesis, start text, a, l, l, space, s, u, c, c, e, s, s, e, s, end text, right parenthesis
## Want to join the conversation?
• So, P(at least 1 success)=1−P(all failures)
The following is multiple choice question (with options) to answer.
During the past week, a local medical clinic tested N individuals for two infections. If 1/2 of those tested had infection A and, of those with infection A, 1/6 also had infection B, how many individuals did not have both infection A and B? | [
"N/12",
"4N/15",
"14N/12",
"11N/12"
] | D | 1/3 of tested had infection A, thus N/2 had infection A;
Of those with infection A, 1/6 also had infection B, thus 1/2*N/6 = N/12 had both infections A and B.
Therefore, N - N/12 = 11N/12 did not have both infection A and B.
Answer: D. |
AQUA-RAT | AQUA-RAT-39796 | ### 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.
A and B invests Rs.8000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year? | [
"32:45",
"32:37",
"32:41",
"32:47"
] | A | A : B
(8000*4)+(4000*8) : (9000*6)+(6000*6)
64000 : 90000
32 : 45
Answer: A |
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