source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37397 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
What number comes next?
582, 693, 715, 826, 937, ? | [
"113",
"231",
"245",
"158"
] | D | D
158
The numbers 58269371 are being repeated in the same sequence. |
AQUA-RAT | AQUA-RAT-37398 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
On a trip covering 400 km, a bicyclist travels the first 100 km at 20 km per hour and the remainder of the distance at 15 km per hour. What is the average speed, in km per hour, for the entire trip? | [
"15.5",
"16.0",
"16.5",
"17.0"
] | B | time 1 = 100 / 20 = 5 hours
time 2 = 300 / 15 = 20 hours
total time = 25 hours
average speed = 400 / 25 = 16 km/hr
The answer is B. |
AQUA-RAT | AQUA-RAT-37399 | 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.
A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 37, the how old is B? | [
"10 years",
"12 years",
"14 years",
"15 years"
] | C | Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.
(2x + 2) + 2x + x = 37
5x = 35
x = 7.
Hence, B's age = 2x = 14 years.
C) |
AQUA-RAT | AQUA-RAT-37400 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
After spending 40 percent machinery, 25 percent in building , 15 percent in raw material and 5 percent on furniture . John had a balance of Rs.1305. Total money with him was? | [
"6700",
"8700",
"9400",
"9800"
] | B | [100 - (40+25+5+15)] % of x = 1305
\inline \Rightarrow 15% of x =1305
\inline \Rightarrow \frac{15}{100}\times x=1305
\inline \therefore x=\frac{1305\times 100}{15}=8700
B |
AQUA-RAT | AQUA-RAT-37401 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
In an exam 49% candidates failed in English and 36% failed in Hindi and 15% failed in both subjects. If the total number of candidates who passed in English alone is 630. What is the total number of candidates appeared in exam? | [
"2000",
"3000",
"3500",
"3800"
] | B | not fail in english =51%
not fail in hindi =64%
not fail in both =30 %(49+36-15)
pass in english only=51-30=21
21/100*x=630
x=3000
ANSWER:B |
AQUA-RAT | AQUA-RAT-37402 | javascript, object-oriented, unit-conversion
All units are converted to a base unit type (kelvin and meters per second)
Usage as follows
Units.kmph = 100;
const mph = Units.mph;
Units.celsius = 38;
const f = Units.fahrenheit;
The following is multiple choice question (with options) to answer.
Express a speed of 108 kmph in meters per second? | [
"30 mps",
"76 mps",
"26 mps",
"97 mps"
] | A | 108 * 5/18
= 30 mps
Answer:A |
AQUA-RAT | AQUA-RAT-37403 | 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 weight of a group of boys is 30 kg. After a boy of weight 33 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ? | [
"A)4",
"B)8",
"C)6",
"D)2"
] | D | Let the number off boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 33 kg joins the group, total weight of boys = 30x + 33
So 30x + 33 = 31(x + 1) = > x = 2.
Answer:D |
AQUA-RAT | AQUA-RAT-37404 | Since the number of ways in which $$5$$ books can be arranged is $$5!=120,$$ we have $$120$$ ways.
• thanks for explanation. Can you point out what is wrong in my analysis Sep 17, 2020 at 11:24
• As mentioned in the comments, you calculated the $C$ books to be next to each other, but they do not need to be next to each other. Sep 17, 2020 at 11:26
The following is multiple choice question (with options) to answer.
if the list price of a book is reduced by Rs.5,then a person can buy 5 more books for Rs.300. The original cost of the book is | [
"Rs.15",
"Rs.20",
"Rs.25",
"Rs.30"
] | B | 300/(x-5)-300/x=5
x=20
ANSWER:B |
AQUA-RAT | AQUA-RAT-37405 | ## anonymous 5 years ago 100 people are present in a party. If each of them must do a handshake with all the other people, how many handshakes should be done? Please explain how to get the answer, thanks!
1. anonymous
you need 2 people for an hand shake and u have 100 people so i think that its C(100,2) but i am not sue of it..
2. anonymous
$\sum_{n=1}^{99}n=\frac{1}{2}\times99\times100=4950$
3. anonymous
Could you solve it using factorials, please?
4. anonymous
hmm i found the answer as 9900
5. amistre64
if there are 3 people: 1,2 ; 1,3 ; 2,3 .... is that right?
6. anonymous
Think about it like this: the first person has 99 people to shake hands with, the second person has 98 because they've already shaken hands with the first, the third has 97...
7. anonymous
yea but its harder to calculate this like that... so i think C(100,2) better way of solve
8. amistre64
5054 if we do that ..... add all the numbers from 1 to 100 :)
9. anonymous
No, all the numbers from 1 to 99.
10. amistre64
n(n+1) ------ :) 2
11. amistre64
ack .... 99 then lol
12. amistre64
wouldnt the last guy have noone to shake hands with?
13. anonymous
mhmh yea but there is 100 people and they and we need 2 people to shake hands so that it should be 100 x 99 /2,
14. anonymous
Yeah, the last guy has no-one to shake hands with, that's why it's sum of the first 99 integers, not the first 100.
15. anonymous
yea you right
16. amistre64
The following is multiple choice question (with options) to answer.
5 people meet for a business lunch. Each person shakes hands once with each other person present. How many handshakes take place? | [
"30",
"21",
"10",
"15"
] | C | the formula to count handshakes is n(n−1)2n(n−1)2
Where n is the number of people
=> 5(5-1)/2 = 5*4/2 = 20/2 = 10
=> the answer is C(10) |
AQUA-RAT | AQUA-RAT-37406 | Suppose you have a tall building with $n$ floors. The top floor is either yellow or blue.
Suppose the top floor is yellow and imagine popping this floor off the building. How many buildings on $n-1$ floors could remain? Since the top floor being yellow placed no additional restrictions on the other floors (aside from the universal one that two blue floors cannot touch), there are $D_{n-1}$ ways to color the remaining floors.
Suppose the top floor is blue. In this case, I know for certain the floor beneath it is yellow. Pop these top two floors off and ask how many buildings on $n-2$ floors could remain. Again, these floors could be colored using any legal configuration, so there are $D_{n-2}$ ways to do this.
Taken together, we get the recursion $D_n = D_{n-1} + D_{n-2}$ for $n \geq 2$. Since the recursion goes back two steps, we need to compute the values of $D_0$ and $D_1$ directly (the recursion won't help us find these first two values). There are no buildings having zero floors and there are just two having one floor, so $D_0 = 0$ and $D_1 = 2$ are the initial conditions.
Let $Y_n$ denote the number of colorings under the extra condition that the upper floor has color yellow, and let $B_n$ denote the number of colorings under the extra condition that the upper floor has color blue.
Then for $n=1,2,\dots$ we have the following equalities:
• $Y_{n+1}=D_n$
• $B_{n+1}=Y_n$
• $D_n=Y_n+B_n$
Now observe that: $$D_{n+2}=Y_{n+2}+B_{n+2}=D_{n+1}+Y_{n+1}=D_{n+1}+D_n$$
The initial conditions are $D_1=2$ and $D_2=3$ which are not difficult to find.
The following is multiple choice question (with options) to answer.
In a colony the municipality decided to number the houses and the shops. They want to number the shops with the numbers containing 3 between 1-100. The houses are 10-20 floored and at the back of the houses 8 feet is left for fire exit. How many shops can be numbered? | [
"19",
"20",
"21",
"22"
] | A | 3numbers containing 3
--->3,13,23,... = 10 numbers
---->30,31,32... = 10 numbers
total = 20 -1 = 19
ANSWER:A |
AQUA-RAT | AQUA-RAT-37407 | Problem. If $f,g \in R$ are non-constant and coprime, then there exist unique $u,v \in R$ such that $uf+vg=1$ and $\deg u < \deg g, \ \deg v < \deg f.$
Solution. Since $f,g$ are coprime, there exist $u_1,v_1 \in R$ such that $u_1f+v_1g=1.$ Dividing $u_1, v_1$ by $g,f,$ respectively, gives $u,v,r,s \in R$ that satisfy the following conditions
i) $u_1=rg+u$ and $v_1=sf+v$
ii) $u=0$ or $\deg u < \deg g$ and $v=0$ or $\deg v < \deg f.$
Thus
$(r+s)fg + uf+vg=(u_1-u)f+(v_1-v)g+uf+vg=u_1f+v_1g=1. \ \ \ \ \ \ \ \ \ (*)$
The following is multiple choice question (with options) to answer.
For which of the following does f(u)−f(v)=f(u−v) for all values of u and v? | [
"f(a)=a^2",
"f(a)=a/2",
"f(a)=a+5",
"f(a)=2a−1"
] | B | To solve this easiest way is just put the value and see that if it equals or not.
with option 1. f(u) = u^2 and f(v) = v^2
so L.H.S = u^2 - v^2
and R.H.S = (u-v)^2 ==> u^2 + v^2 -2uv.
so L.H.S not equal to R.H.S
with option 2. f(u) = u/2 and f(v) = v/2
L.H.S = u/2 - v/2 ==> 1/2(u-v)
R.H.S = (u-v)/2
so L.H.S = R.H.S which is the correct answer.
B |
AQUA-RAT | AQUA-RAT-37408 | # how can I find the Side length Two squares inside an equilateral Triangle?
Question: Figure shows an equilateral triangle with side length equal to $$1$$ . Two squares of side length a and $$2a$$ placed side by side just fit inside the triangle as shown.
Find the exact value of $$a$$.
Its an Assessment question from edX course "A-Level Mathematics Course 1" and I am supposed to use skills that I learnt in Indices and surds,Inequalities and The Factor Theorem.
I have tried finding the height of triangle and then use similar triangles to find the right triangle length still No luck.
I am just looking for food for thought or very small hints thats all.
Hint: You have everything you need along the base. Also, note the two flanking right triangles.
• i think i got it. Thanks – fpsshubham Sep 18 '19 at 14:12
From the leftmost right triangle, $$\frac{a}{x} = \tan(60°) \implies a = \sqrt{3}x$$ From the rightmost right triangle $$\frac{2a}{1-3a-x} = \frac{a}{x} \\ a = \frac{3-\sqrt{3}}{6} \approx 0.211$$
• I think you are on the right path to an answer, but there is not enough detail. First, you should typeset your answer with MathJax. Second, give more details. E.g. what is $x$? – Physical Mathematics Apr 21 '20 at 21:18
I struggled with this too. But the info is along the bottom. The triangle is equilateral so all angles are $$60°$$. On the left there is a right-angled triangle - let's call its base $$x$$.
Triangle 1: Angle = $$60°$$, opposite = $$a$$, and adjacent = $$x$$
On the right there is another right-angled triangle and its base is $$1-3a-x$$.
Triangle 2: Angle = $$60°$$, opposite = $$2a$$, and adjacent = $$1 - 3a - x$$
The following is multiple choice question (with options) to answer.
What is the are of an equilateral triangle of side 16 cm? | [
"64√5",
"64√9",
"64√4",
"64√3"
] | D | Area of an equilateral triangle = √3/4 S2
If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm2;
Answer: D |
AQUA-RAT | AQUA-RAT-37409 | ### $\sigma_0$ of $110$
$\map {\sigma_0} {110} = 8$
### $\sigma_0$ of $120$
$\map {\sigma_0} {120} = 16$
## Also see
• Results about the divisor count function can be found here.
The following is multiple choice question (with options) to answer.
What percent of 60 is 110? | [
"133 1⁄3",
"75",
"183.33",
"33 1⁄3"
] | C | % of 60 is 110?
= 110/60 = 11/6 = 183.33%
Thus C is the correct answer. |
AQUA-RAT | AQUA-RAT-37410 | Since you know that the parabola has vertex $V(-3, -2)$, the vertex form of its equation is $$y = a[x - (-3)]^2 + (-2) = a(x + 3)^2 - 2$$ Since the parabola passes through the point $P(-4, 0)$, we can substitute $-4$ for $x$ and $0$ for $y$ to determine $a$. \begin{align*} 0 & = a(-4 + 3)^2 - 2\\ 2 & = a(-1)^2\\ 2 & = a \end{align*} Hence, $y = 2(x + 3)^2 - 2$. You can expand this expression to obtain the standard form of the equation.
Addendum: In general, if you know the vertex $V(h, k)$ and a point $P(u, v) \neq V(h, k)$ on the parabola, you can write $$y = a(x - h)^2 + k$$ then substitute $u$ for $x$ and $v$ for $y$ to determine $a$. \begin{align*} v & = a(u - h)^2 + k\\ v - k & = a(u - h)^2\\ \frac{v - k}{(u - h)^2} & = a \end{align*} Then $$y = a(x - h)^2 + k = \frac{v - k}{(u - h)^2}(x - h)^2 + k$$ Again, expanding the expression to obtain its standard form enables you to determine the coefficients $a$, $b$, and $c$.
The following is multiple choice question (with options) to answer.
When a parabola represented by the equation y - 2x 2 = 8 x + 5 is translated 3 units to the left and 2 units up, the new parabola has its vertex at | [
"(-5 , -1)",
"(-5 , -5)",
"(-1 , -3)",
"(-2 , -3)"
] | A | Solution
First rewrite y - 2x 2 = 8 x + 5 as
y = 2x 2 + 8 x + 5
Complete square and determine vertex.
y = 2(x 2 + 4x + 4) - 8 + 5
= 2(x + 2) 2 - 3
vertex at (- 2 , - 3)
If parabola is translated 3 units to the left and 2 units up its vertex is also translated 3 units to the right and 2 units up .
vertex after translations is at: (-2 - 3 , - 3 + 2) = (-5 , -1)
Answer A |
AQUA-RAT | AQUA-RAT-37411 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
The speed at which a man can row a boat in still water is 15 kmph. If he rows downstream, where the speed of current is 3 kmph, what time will he take to cover 60 metres? | [
"19",
"17",
"10",
"12"
] | D | Speed of the boat downstream = 15 + 3 = 18 kmph
= 18 * 5/18 = 5 m/s
Hence time taken to cover 60 m = 60/5 = 12 seconds. Answer: D |
AQUA-RAT | AQUA-RAT-37412 | Back
## A Test Question
Today, Pearl’s $9$ grandchildren are coming to visit! She loves to spoil them, so she opens her purse and finds $13$ dollar bills.
In how many different ways can Pearl distribute those dollars amongst her grandchildren? Keep reading to find out, or skip to today’s challenge for a similar problem.
As we’ll see, there are a lot of ways for Pearl to distribute her dollars! So, let’s start with a smaller example. Last week, Pearl’s $3$ favorite grandchildren visited, and at that time, she had $4$ dollar bills to give them. To visualize how they could be distributed, she laid them out in a row, along with some pencils to divide them into $3$ groups.
We’ll represent the dollars with stars $\large \star$ and divisions between groups with bars $\large{|}.$ One arrangement that Pearl found was $\large \star \; | \, \star \star \; | \; \star$ which represents $1$ dollar for the first grandchild, $2$ dollars for the second, and $1$ dollar for the third. Another arrangement was $\large \star \; | \: | \, \star \star \, \star$ which represents $1$ dollar for the first grandchild, $0$ dollars for the second, and $3$ dollars for the third.
To create $3$ groups, we need $2$ bars to separate the stars. So, to count the total number of arrangements into groups, we can count where in the line of stars and bars we can place those bars to define the groups.
The following is multiple choice question (with options) to answer.
Sarah makes handmade jewelry. This month she has created 253 necklaces. If she distributes her jewelry evenly to the 6 jewelry stores in town, how many necklaces remain with her? | [
"0 necklaces",
"1 necklace",
"2 necklaces",
"3 necklaces"
] | B | We need to find the first number that is divisible by 6 that occurs before 253.
In order to divide the sum in 6 parts, the amount must be divisible by 6.
Divisibility rule of 6: A number is divisible by 6 if it is divisible by both 2 and 3
Divisibility rule of 3: A number is divisible by 3 if the sum of the digits is divisible by 3
We need to make this number even in order for it to be divisible by 2. So we must subtract an even number
Sum of digits of 253 = 10 and 9 is divisible by 3 and subtracting 1 yields an even number
Hence, we remove 1 from this number for it to be divisible by 6
Correct Option: B |
AQUA-RAT | AQUA-RAT-37413 | Hey, thanks for your help guys. For a minute there, I thought that this theoretical person could not safely expect to live to be 82 years old.
9. Jun 16, 2012
### SW VandeCarr
In fact, on a purely probabilistic basis, for any finite time no matter how large, there is a non zero probability that a person would survive that long. So for a sufficiently large population, there would be a theoretic person that would live 100,000 years. This, of course, has no basis in biology.
In terms of the probability of being murdered, the model would not hold for the 100,000 year old person. In terms of the model, probably the best one can do is assume the proportion of causes of death would be constant. The calculation above needs to be corrected for overall survival in terms of death from any cause.
Last edited: Jun 16, 2012
10. Jun 16, 2012
### viraltux
Interesting... but 0.37% is not that small percentage, don't you think? That means, roughly speaking, that a community of around 300 persons can expect that one of them will be murdered.
If you consider that the number of people we know plus acquaintances can easily be around 300 persons that would mean that most 82 year old persons know of someone in their circles who has been murdered. Mmm... that might be an interesting survey.
11. Jun 16, 2012
### SW VandeCarr
As I said in my previous post, this is a misapplication of statistics. You have to consider survival in terms of all cause death. If you just consider the murder rate, then at some point nearly everyone gets murdered.
12. Jun 16, 2012
### moonman239
I know that.
This person will not die until he reaches age 82, if he is not murdered. As mentioned before, this person has a 68% chance of living to be 82.
13. Jun 16, 2012
### D H
Staff Emeritus
The probability of living to 82 per this problem is 99.63%, not 68%. You missed the decimal point on the 0.37%.
14. Jun 16, 2012
### SW VandeCarr
The following is multiple choice question (with options) to answer.
Tough and Tricky questions: Probability.
Medical analysts predict that one over five of all people who are infected by a certain biological agent could be expected to be killed for each day that passes during which they have not received an antidote. What fraction of a group of 1,000 people could be expected to be killed if infected and not treated for three full days? | [
"16/81",
"8/27",
"2/3",
"61/125"
] | D | At the end of each day we will have 4/5 alive. So after 3 days we will have (4/5)^3 people alive. Therefore, fraction of dead people will be 1-(4/5)^3=61/125.
The correct answer is D. |
AQUA-RAT | AQUA-RAT-37414 | of height and side. So, the perimeter of the rhombus is 64 cm. Diagonals divide a rhombus into four absolutely identical right-angled triangles. Join now. By … . Is a Square a Rhombus? The total distance traveled along the border of a rhombus is the perimeter of a rhombus. So by the same argument, that side's equal to that side, so the two diagonals of any rhombus are perpendicular to … Using side and height. Given two integers A and X, denoting the length of a side of a rhombus and an angle respectively, the task is to find the area of the rhombus.. A rhombus is a quadrilateral having 4 sides of equal length, in which both the opposite sides are parallel, and opposite angles are equal.. To solve this problem, apply the perimeter formula for a rhombus: . Now the area of triangle AOB = ½ * OA * OB = ½ * AB * r (both using formula ½*b*h). When the altitude or height and the length of the sides of a rhombus are known, the area is given by the formula; Area of rhombus = base × height. The proof is completed. This formula for the area of a rhombus is similar to the area formula for a parallelogram. Problem 1: Find the perimeter of a rhombus with a side length of 10. The formula for perimeter of a rhombus is given as: P = 4s Where P is the perimeter and s is the side length. [3] What is the formula of Rhombus When one side is given Get the answers you need, now! Or as a formula: The perimeter formula for a rhombus is the same formula used to find the perimeter of a square. Heron's Formula depends on knowing the semiperimeter, or half the perimeter, of a triangle. Calculate the unknown defining areas, angels and side lengths of a rhombus with any 2 known variables. Formula for side of rhombus when diagonals are given - 1399111 1. Log in. The area of the rhombus can be found, also knowing its diagonal. How To Find Area Of Rhombus (1) If both diagonals are given (or we can find their length) then area = (Product of diagonals) (2) If
The following is multiple choice question (with options) to answer.
The perimeter of one square is 48 cm and that of another is 24 cm. Find the perimeter and the diagonal of a square which is equal in area to these two combined? | [
"13√4",
"13√2",
"13.4√2",
"12√4"
] | C | 4a = 48 4a = 24
a = 12 a = 6
a2 = 144 a2 = 36
Combined area = a2 = 180 => a = 13.4
d = 13.4√2
Answer: C |
AQUA-RAT | AQUA-RAT-37415 | 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.
Two cars cover the same distance at the speed of 60 and 64 kmps respectively. Find the distance traveled by them if the slower car takes 1 hour more than the faster car? | [
"767 km",
"176 km",
"960 km",
"199 km"
] | C | 60(x + 1) = 64x
X = 15
60 * 16 = 960 km
Answer:C |
AQUA-RAT | AQUA-RAT-37416 | If you role two dice, then there are 36 equally likely outcomes, (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Of those, 11, the last of each of those six lines and all of the last, have at least one 6 so the probability that "one or both rolls is a 6" is 11/36. Your "13/36" is wrong because both of the "1/6" In your sum include "(6/6)". It should be 1/6+ 1/6- 1/36 where the 1/36 being subtracted is to take away one of the (6, 6)s.
The general rule for "Probability of A or B" is "Probability of A"+ "Probability of B" minus "Probability of A and B".
When you roll two dice, the probability the first die is even is 1/2, the probability the second die is 1/2, and the probability both are even is (1/2)(1/2)= 1/4 (the results of the two rolls are independent) so the probability that either one or both are even is 1/2+ 1/2- 1/4= 3/4.
The following is multiple choice question (with options) to answer.
When 2 dices tossed, what is the probability that 2 numbers faced have difference 2? | [
"1/9",
"2/9",
"1/3",
"4/9"
] | B | Combinations are (1,3), (2,4), (3,5), (4,6), (3,1), (4,2), (5,3), (6,4)
So, total = 8
Hence, Probability = 8/36 = 2/9
answer:B |
AQUA-RAT | AQUA-RAT-37417 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
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Abhishek....
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Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
A Pharmacy company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 8 hrs and a machine of type S does the same job in 4 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used? | [
"3",
"4",
"2",
"9"
] | C | Rate of machine R =1/8
Rate of machine S =1/4
since same no of machines used for R and S to do the same work in 2 hrs
So collective rate needed to finish the work in 2 hrs= 1/2
Let the no of machine be x
So, x/8 +x/4 =1/2
3x/8=1/2
x=8/6=1.3=2
So no of machine R is 2
Answer C |
AQUA-RAT | AQUA-RAT-37418 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
Eggs are sold in packages of six or eleven only. If Doris bought 79 eggs exactly, what could be the number of large packs Doris bought? | [
"6.",
"2.",
"3.",
"5."
] | D | No strategy involved. Simple question demanding fast calculation.
11x5 = 55 => 79-55=24=> 24/6 is an integer
Ans D. 5.
Good luck |
AQUA-RAT | AQUA-RAT-37419 | Once again
Repetition helps too, so let’s recite: it starts with $289 = 17^2$, then continues by 102s: 391, 493. After that the twins 527, 529, followed by 629; then 667 and 697. Then two sets of twins each with its 99: 713, 731, 799; 841, 851, 899; then 901 to come after 899, and then the three sporadic values: 943, 961, 989!
Posted in arithmetic, computation, primes |
Quickly recognizing primes less than 1000: divisibility tests
I took a little hiatus from writing here since I attended the International Conference on Functional Programming, and since then have been catching up on teaching stuff and writing a bit on my other blog. I gave a talk at the conference which will probably be of interest to readers of this blog—I hope to write about it soon!
In any case, today I want to return to the problem of quickly recognizing small primes. In my previous post we considered “small” to mean “less than 100”. Today we’ll kick it up a notch and consider recognizing primes less than 1000. I want to start by considering some simple approaches and see how far we can push them. In future posts we’ll consider some fancier things.
First, some divisibility tests! We already know how to test for divisibility by $2$, $3$, and $5$. Let’s see rules for $7$, $11$, and $13$.
• To test for divisibility by $7$, take the last digit, chop it off, and subtract double that digit from the rest of the number. Keep doing this until you get something which obviously either is or isn’t divisible by $7$. For example, if we take $2952$, we first chop off the final 2; double it is 4, and subtracting 4 from $295$ leaves $291$. Subtracting twice $1$ from $29$ yields $27$, which is not divisible by $7$; hence neither is $2952$.
The following is multiple choice question (with options) to answer.
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30? | [
"196",
"630",
"1260",
"2520"
] | B | L.C.M of 12, 18, 21, 30
= 2 * 3 * 2 * 3 * 7 * 5 = 1260
Required number = 1260/2 = 630.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37420 | Alternative solution:
Recall that for any triangle $\triangle ABC$, the area of the triangle equals the product of the inradius and its semiperimeter; i.e., $|\triangle ABC| = rs$, where $s = (a+b+c)/2$. Therefore, given legs $a, b$, $c = \sqrt{a^2+b^2}$, and $$r = \frac{2|\triangle ABC|}{a+b+c} = \frac{ab}{a+b+\sqrt{a^2+b^2}}.$$ Draw the tangent line to the two circles at their common point of tangency: this creates a smaller similar triangle with scaling factor $\frac{b-2\rho}{b}$ where $\rho$ is the common radius of the two circles. If $r$ is the inradius of $|\triangle ABC|$ as shown, then $$\frac{\rho}{r} = \frac{b - 2\rho}{b}.$$ Putting all of this together, we find $$\rho = \frac{br}{b+2r} = \frac{a b}{3a+b+c} = \frac{ab}{3a+b+\sqrt{a^2+b^2}}.$$ This method easily generalizes to more than two congruent circles tangent to one side: if $n$ circles are arranged along the leg of length $b$, then it is straightforward to find that $\rho = \frac{ab}{(2n-1)a + b + \sqrt{a^2+b^2}}$.
-
The following is multiple choice question (with options) to answer.
The perimeter of a triangle is 44 cm and the inradius of the triangle is 2.5 cm. What is the area of the triangle? | [
"38 cm2",
"55 cm2",
"65 cm2",
"45 cm2"
] | B | Area of a triangle = r * s
Where r is the inradius and s is the semi perimeter of the triangle.
Area of triangle = 2.5 * 44/2 = 55 cm2
Answer:B |
AQUA-RAT | AQUA-RAT-37421 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
How much interest can a person get on Rs. 6280 at 15.5% p.a. simple interest for a period of five years and six months? | [
"5353.72",
"5353.71",
"5353.7",
"5353.73"
] | C | I = (6280 * 5.5 * 15.5)/100
= (6280 * 11 * 31)/(100 * 2 * 2)
= Rs. 5353.7
Answer: C |
AQUA-RAT | AQUA-RAT-37422 | In solving many kinematic problems for gears, there is a key rule to key in mind: When two gears are meshing, the tangential velocities of both gears at the point of meshing are equal.
(Also, as said in EMiller's answer, gear C seems to be the smaller orange gear, and gear D the larger yellow gear, contrary to the diagram's labelling).
This rule will help provide the equations we need to solve this problem. Note that there are 2 pairs of meshing gears (A meshes with C and B meshes with D). The points of meshing for each pair are $P$ and $Q$ respectively. Therefore, this gives us 2 equations to work with. It is then a matter of relating the tangential velocities at the point of meshing with the velocity of the centre of gear C and D. I will define the following terms:
Let $v_{a,p}, v_{c,p}$ be the tangential velocities of gears A and C, respectively, at point P.
Let $v_{b,q}, v_{d,q}$ be the tangential velocities of gears A and C, respectively, at point P.
Let $V$ be the velocity of the centre of gear C and D.
(The following diagrams represents gears A and C, and then B and D, as circles that are in pure rolling with one another:)
According to the rule above, we get the following two equations:
$$v_{a,p}=v_{c,p} \qquad (Eq.1)$$ $$v_{b,q}=v_{d,q} \qquad (Eq.2)$$
$v_{a,p}$ and $v_{b,q}$ are simple enough to determine since the centres of gears A and B don't move:
$$v_{a,p}=\omega_a R_a = 0.4\text{m/s} \qquad v_{b,q}=\omega_b R_b = 0.4\text{m/s}$$
The following is multiple choice question (with options) to answer.
Two interconnected, circular gears travel at the same circumferential rate. If Gear A has a diameter of 40 centimeters and Gear B has a diameter of 50 centimeters, what is the ratio of the number of revolutions that Gear A makes per minute to the number of revolutions that Gear B makes per minute? | [
"4:5",
"9:25",
"5:4",
"25:9"
] | C | Same circumferential rate means that a point on both the gears would take same time to come back to the same position again.
Hence in other words, time taken by the point to cover the circumference of gear A = time take by point to cover the circumference of gear B
Time A = 2*pi*25/Speed A
Time B = 2*pi*20/Speed B
Since the times are same,
50pi/Speed A = 40pi/Speed B
SpeedA/Speed B = 50pi/30pi = 5/4
Correct Option: C |
AQUA-RAT | AQUA-RAT-37423 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
What is the dividend. divisor 19, the quotient is 7 and the remainder is 6 | [
"136",
"137",
"138",
"139"
] | D | D = d * Q + R
D = 19 * 7 + 6
D = 133 + 6
D = 139 |
AQUA-RAT | AQUA-RAT-37424 | But suppose we weren't done. The table would continue \begin{align*} 4 && 2 && - \\ 5 && 3 && 2 \\ 6 && 4 && 3 \\ 7 && 4 && - \\ 8 && 5 && 4 \\ 9 && 6 && 5 \end{align*} where we stop when we hit $n=9$, which is guaranteed to have a solution. Notice, there is no reason to find the value in the 3rd column when the integer part in the second column has not increased.
Note that we can actually calculate the multiples of $\log_2 13/8 = 0.700 \dots$ "by eye", so the only work in this particular problem is the integer part of the multiples of $\log_2 12/8$.
• Calculation of log to base 2 is easy in floating point. A crude approximation is to write a dot, then the mantissa portion of the floating pint representation. If you have a decimal number between 1 and 2, just subtract 1. It is exact for 1 and 2, and a bit low at the middle of the interval. For more on logs, see Doerfler's wonderful book. – richard1941 Jul 12 '17 at 0:26
$12 \lt 2^{m/n} \lt 13$
$\log_2 12 \lt \frac{m}{n} \lt \log_2 13$
$[3; 1, 1, 2, 2, \dots] \lt \frac{m}{n} \lt [3; 1, 2, 2, 1, \dots]$
The continued fractions match up to $[3; 1]$. Since $[3; 1, 1]$ and $[3; 1, 2]$ are underestimations of $\log_2 12$ and $\log_2 13$, we take $[3; 1, 1]$ (the "smaller" one lexicographically) and add $1$ to the last number (round "up"), so the answer is $[3; 1, 2]$.
The following is multiple choice question (with options) to answer.
If log 27 = 1.731, then the value of log 9 is: | [
"0.934",
"0.945",
"0.954",
"1.156"
] | D | log 27 = 1.731
log (33 ) = 1.731
3 log 3 = 1.731
log 3 = 0.578
log 9 = log(32 ) = 2 log 3 = (2 x 0.578) = 1.156.
Answer: Option D |
AQUA-RAT | AQUA-RAT-37425 | # 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 merchant marks goods up by 75% and then offers a discount on the marked price. The profit that the merchant makes after offering the discount is 57.5%. What % discount did the merchant offer? | [
"10%",
"15%",
"20%",
"25%"
] | A | Let P be the original price of the goods and let x be the rate after the markup.
(1.75P)*x = 1.575P
x = 1.575/1.75 = 0.9 which is a discount of 10%.
The answer is A. |
AQUA-RAT | AQUA-RAT-37426 | Sorry, I used just mass of the sucrose, not sucrose+cart
Here's a simpler answer: Looking at the angle made by the bob, we can deduce that $$\frac ag = \tan 37^\circ \implies a = 7.5 ms^{-2}$$. This is the horizontal acceleration of the entire system. Since the total mass of the system (sucrose ballast + cart + bob) is $$10 \text { kg}$$, the net force $$F = ma = 10 \times 7.5 = \boxed{75 N}$$.
• What value did you use for $\tan 37^\circ$? – Andrei Jul 17 '20 at 4:27
• $\tan 37^\circ$ is $\bf{3/4}$, so $a=7.5m/s^2$ and $F=75N$ – Andrei Jul 17 '20 at 4:37
• @Andrei yep, thanks for pointing out the calculation error :P. Rectified it now. – Aniruddha Deb Jul 17 '20 at 4:39
• You were not the only one doing simple math mistakes on this problem :) – Andrei Jul 17 '20 at 4:40
The following is multiple choice question (with options) to answer.
The cost of 15 packets of sugar, each weighing 900 grams is Rs. 28. What will be the cost of 27 packets, if each packet weighs 1 kg? | [
"Rs 52.50",
"Rs 56",
"Rs 58.50",
"Rs 64.75"
] | B | Explanation :
Let the required cost be Rs. x. Then,
More packets, More cost (Direct Proportion)
More weight, More cost (Direct Proportion)
Packets 15 : 27
Weight 900 : 1000 :: 28 : x
(15 x 900 x x) = (27 x 1000 x 28)
x = (27 x 1000 x 28) /15 x 900
= 56
Answer B |
AQUA-RAT | AQUA-RAT-37427 | $24 = 1^3 + 2^3 - 3^3 - 4^3 + 5^3 + 6^3 - 7^3 + 8^3 + 9^3 + 10^3 - 11^3 + 12^3 + 13^3 + 14^3 - 15^3 - 16^3$ (at least 7 other ways)
Each of the above representations is the shortest possible. The algorithm created a sequence of sets $S_n$ with $S_0 = \{0\}$ and $S_n = \{x+n^3, x-n^3 : x \in S_{n-1}\}$ for $n > 0$, so that $S_1 = \{1, -1\}$, $S_2 = \{9, -7, 7, -9\}$, etc. The algorithm also kept track of the sequence of signs used to arrive at each particular number.
• Nice work ! :D Thank's !!!
– r9m
Dec 24 '14 at 1:11
• @Unit: Do you suppose you can do the same for at least $1<N<100$ for $5$th powers? Kindly refer to my answer below. Dec 29 '14 at 3:54
• @TitoPiezasIII: Ahh, I just read this message. I will try, though it seems alexwlchan is almost there.
– Unit
Dec 29 '14 at 22:55
• @Unit: And it seems Zander even managed $0\leq N<10000$. The ingenuity of some people never ceases to amaze. Dec 29 '14 at 23:44
For 5th powers, there is an identity,
$$\sum\limits_{k=1}^{168}\pm(x+k)^5 = 480$$
analogous to the 3d powers mentioned by R. Millikan in his answer.
What remains to be shown is that all $0\leq N<240$ can be decomposed into sums of fifth powers. See this post.
(P.S. The number of addends can be reduced to just $m=168$.)
The following is multiple choice question (with options) to answer.
What is the greatest of 3 consecutive integers whose sum is 24 ? | [
"6",
"7",
"8",
"9"
] | D | The sum of three consecutive integers can be written as n + (n + 1) + (n + 2) = 3n + 3
If the sum is 24, we need to solve the equation 3n + 3 = 24;
=> 3n = 21;
=> n = 7
The greatest of the three numbers is therefore 7 + 2 = 9 Answer: D |
AQUA-RAT | AQUA-RAT-37428 | ### 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 starts a business with Rs.8000 each, and after 4 months, B withdraws half of his capital . How should they share the profits at the end of the 18 months? | [
"18:11",
"18:15",
"18:19",
"18:13"
] | A | A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for remaining 14 months is Rs.4000 only.
A : B
8000*18 : (8000*4) + (4000*14)
14400 : 88000
A:B = 18:11
Answer: A |
AQUA-RAT | AQUA-RAT-37429 | # math
How is the circumference of a circle with radius 9cm related to the circumference of a circle with diameter 9cm?
i don't understand this.Help?? Thanks:)
1. 👍
2. 👎
3. 👁
c = pi x d.
The diamter is 2 x radius. So, the diameter of the first cirlce is 18cm. You know the diameter of the second circle. So, how do the circumferences compare?
1. 👍
2. 👎
2. is the second circle twice as big? like is that how they compare?
1. 👍
2. 👎
3. the circumference of the larger circle is indeed twice as long as that of the smaller one.
but...
the area of the larger circle is 4 times that of the area of the smaller.
Proof:
area of smaller = pi(4.5)^2 = 20.25pi
area of larger = pi(9^2) = 81pi
20.25pi/(81pi) = 1/4
1. 👍
2. 👎
4. alright thx!
1. 👍
2. 👎
## Similar Questions
1. ### Math ~ Check Answers ~
Find the circumference of the circle (use 3.14 for pi). Show your work. Round to the nearest tenth. the radius is 23 yd My Answer: ???? •Circumference = (2) (3.14) (23) •3.14 x 2 x 23 = 144.44
2. ### Mathematics
1. Find the circumference of the given circle. Round to the nearest tenth. (Circle with radius of 3.5 cm) -22.0 cm -38.5 cm -11.0 cm -42.8 cm 2. Find the area of the given circle. Round to the nearest tenth. (Circle with a
3. ### Math
The formula C = 2 x Pi x r is used to calculate the circumference of a circle when given the radius. What is the formula for calculating the radius when given the circumference? C/2Pi = R*** C/2R = Pi 2 x Pi x C = R 2 x Pi/ C = R
4. ### MATH
The following is multiple choice question (with options) to answer.
There are two circles of different radii. The are of a square is 784 sq cm and its side is twice the radius of the larger circle. The radius of the larger circle is seven - third that of the smaller circle. Find the circumference of the smaller circle. | [
"11",
"14",
"12",
"18"
] | C | Let the radii of the larger and the smaller circles be l cm and s cm respectively. Let the side of the square be a cm.
a2 = 784 = (4)(196) = (22).(142)
a = (2)(14) = 28
a = 2l, l = a/2 = 14
l = (7/3)s
Therefore s = (3/7)(l) = 6 Circumference of the smaller circle = 2∏s = 12∏ cm.Answer: C |
AQUA-RAT | AQUA-RAT-37430 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
Arun purchased 30 kg of wheat at the rate of Rs. 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg. He mixed the two and sold the mixture. Approximately what price per kg should be sell the mixture to make 35% profit? | [
"16.39",
"16.33",
"16.35",
"17.01"
] | D | C.P. of 50 kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630.
S.P. of 50 kg wheat = 135% of Rs. 630 = 135/100 * 630 = Rs. 850.50.
S.P. per kg = 850.50/50 = Rs. 16.38 = 16.30.
Answer: D |
AQUA-RAT | AQUA-RAT-37431 | and
$$\log_{1/5}4>\log_{1/2}5>\log_{1/3}27.$$
• You can tighten the bounds on $log_2 5$. In fact $2^2 < 5 < 2^3$ – Martin Bonner supports Monica Sep 18 '18 at 12:09
• Ah. Now I understand. You can tighten the bounds - but you don't need to – Martin Bonner supports Monica Sep 18 '18 at 12:12
• @MartinBonner: exactly. – Yves Daoust Sep 18 '18 at 13:26
The following is multiple choice question (with options) to answer.
The value of log5(1/ 125) is | [
"3",
"-3",
"1/3",
"-1/3"
] | B | Solution
Let log5(1/125) = n.
Then, 5n= 1/125
‹=›5n =5-3
n= -3.
Answer B |
AQUA-RAT | AQUA-RAT-37432 | 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.
If the price of an article went up by 20%, then by what percent should it be brought down to bring it back to its original price? | [
"16 2/3%",
"16 8/3%",
"16 2/7%",
"16 2/1%"
] | A | Let the price of the article be Rs. 100.
20% of 100 = 20.
New price = 100 + 20 = Rs. 120
Required percentage = (120 - 100)/120 * 100
= 20/120 * 100 = 50/3 = 16 2/3%.
Answer: A |
AQUA-RAT | AQUA-RAT-37433 | # Welcome to the LEAP Q&A Forum
Find your questions answered. Here.
## GMAT - Problem Solving - Geometry
LEAP Administrator 11 months ago
### The figure shown above represents a modern painting that consists of four differently colored rectangles, each of which has length l and width w. (more)
The figure shown above represents a modern painting that consists of four differently colored rectangles, each of which has length l and width w. If the area of the painting is 4,800 square inches, what is the width, in inches, of each of the four rectangles?
A. 15
B. 20
C. 25
D. 30
E. 40
from the figure
length of one single rectangle = 3w
width of the complete painting = w + 3w = 4w
area of the painting = l*total width = 3w*4w = 12w2
4800 = 12w2
w2 = 400
w = 20 inches
B is correct
LEAP Administrator 11 months ago
### The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be (more)
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
There are three faces of the rectangular box
6 x 8, 8 x 10 and 10 x 6
Case 1
If 6 x 8 is the base of the cylinder, 10 will be the height and 6 will be the diameter
Volume = πr2h = π32*10 = 90π cubic inch
Case 2
If 8 x 10 is the base of the cylinder, 6 will be the height and 8 will be the diameter
Volume = πr2h = π42*6 = 96π cubic inch
Case 3
If 10 x 6 is the base of the cylinder, 8 will be the height and 6 will be the diameter
Volume = πr2h = π32*8 = 72π cubic inch
The following is multiple choice question (with options) to answer.
The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor? | [
"29",
"27",
"18",
"28"
] | C | Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l2 = 324 => l
= 18.
Answer:C |
AQUA-RAT | AQUA-RAT-37434 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
If a trader sold two cars each at Rs. 404415 and gains 15% on the first and loses 15% on the second, then his profit or loss percent on the whole is ? | [
"1.44 %",
"2.02 %",
"1.04 %",
"2.25 %"
] | D | Explanation:
SP of each car is Rs. 404415, he gains 15% on first car and losses 15% on second car.
In this case, there will be loss and percentage of loss is given by = [(profit%)(loss%)]/100
= (15)(15)/100 % = 2.25%
ANSWER IS D |
AQUA-RAT | AQUA-RAT-37435 | sum(res==0)/B
[1] 0.120614
So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).
The following is multiple choice question (with options) to answer.
60% of a number is added to 120, the result is the same number. Find the number? | [
"300",
"278",
"267",
"266"
] | A | (60/100) * X + 120 = X
2X = 600
X = 300
Answer: A |
AQUA-RAT | AQUA-RAT-37436 | \, =\, \frac{21}{32}e\, +\, \frac{21}{64}f\, +\, \frac{127}{64} \\\\ 43f\, =\, 42e\, +\, 127 \: \: \: ---(2)$$ Combining these, we obtain
The following is multiple choice question (with options) to answer.
The mean of (54,825)^2 and (54,827)^2 = | [
"(54,821)^2",
"(54,821.5)^2",
"(54,820.5)^2",
"(54,826)^2 + 1"
] | D | 54825^2 = (54826-1)^2 = 54826^2 + 1^2 - 2*54826*1
54827^2 = (54826+1)^2 = 54826^2 +1^2 + 2*54826*1
Taking the average of above 2 , we get (54826)^2 +1
hence the answer is D |
AQUA-RAT | AQUA-RAT-37437 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
In how many ways can the letters of the word CREATION be arranged such that all the vowels always appear together? | [
"5!*4!",
"8!",
"8!*4!",
"4!*4!"
] | A | The 8 letters can be grouped into 4 consonants and one set of 4 vowels.
The number of ways to arrange 5 units is 5!
Then, for each arrangement, the 4 vowels can be arranged in 4! ways.
The total number of arrangements is 5!*4!
The answer is A. |
AQUA-RAT | AQUA-RAT-37438 | 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.
A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is: | [
"14 years",
"18 years",
"20 years",
"22 years"
] | D | Let the son's present age be x years. Then, man's present age = (x + 24) years.
(x + 24) + 2 = 2(x + 2)
x + 26 = 2x + 4
x = 22.
answer :D |
AQUA-RAT | AQUA-RAT-37439 | # 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.
The average weight of 8 person's increases by 4 kg when a new person comes in place of one of them weighing 55 kg. What might be the weight of the new person? | [
"80 kg",
"87 kg",
"90 kg",
"100 kg"
] | B | Total weight increased = (8 x 4) kg = 32 kg.
Weight of new person = (55 + 32) kg = 87 kg.
B) |
AQUA-RAT | AQUA-RAT-37440 | # 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 merchant gains or loses, in a bargain, a certain sum. In a second bargain, he gains 450 dollars, and, in a third, loses 30. In the end he finds he has gained 100 dollars, by the three together. How much did he gain or lose bv the first ? | [
"320",
"-340",
"-320",
"348"
] | C | In this sum, as the profit and loss are opposite in their nature, they must be distinguished by contrary signs. If the profit is marked +, the loss must be -.
Let x = the sum required.
Then according to the statement x + 450 - 30 = 100
And x = -320.
Answer C |
AQUA-RAT | AQUA-RAT-37441 | Polygon comes from Greek. For determining the area of a polygon given on a coordinate plane, we will use the following formula: Area (A) = | (x 1 y 2 – y 1 x 2) + (x 2 y 3 – y 2 x 3)…. Area of a Polygon – Learn with Examples. Mentor. where, S is the length of any side N is the number of sides π is PI, approximately 3.142 NOTE: The area of a polygon that has infinite sides is the same as the area a circle. equiangular is known as a regular polygon. Given below is a figure demonstrating how we will divide a pentagon into triangles Now the area of whole polygon is N*A. The Perimeter of an irregular shape is calculated by adding the length of each side together. Captain Matticus, LandPiratesInc . The purpose is to visualize the given geometry as a combination of geometries for which we know how to calculate the area. For a regular polygon with n sides of length s, the area is given by: Through the area of a triangle. I'm trying to the find the area of a shape for which I've only been given the length of the sides. Few more polygon … Let’s work out a few example problems about area of a regular polygon. (x 2 y 1 + x 3 y 2 + … + x n y n-1 + x 1 y n) ] |. 10, Oct 18. 31, Dec 18. Maybe you know the coordinates, or lengths and angles, either way this can give you a good estimate of the Area. Therefore, ABED is a rectangle and BDC is a triangle. To get the area of the whole polygon, just add up the areas of all the little triangles ("n" of them): Area of Polygon = n × side × apothem / 2. The area A of a convex regular n-sided polygon having side s, circumradius R, apothem a, and perimeter p is given by = = = = = For regular polygons with side s = 1, circumradius R = 1, or apothem a = 1, this produces the following table: (Note that since → / as →, the area … Calculating the area of a regular polygon can be as simple as finding the area of
The following is multiple choice question (with options) to answer.
When the perimeter of a regular polygon is divided by 5, the length of a side is 25. What is the name of the polygon? What is the perimeter? | [
"125",
"126",
"127",
"128"
] | A | Regular polygon. A polygon with equal sides and equal sides.
Divided by 5 to get the length of a side. It is the pentagon since it has 5 sides.
So p = 5 × s
To get the perimeter, just multiply a side by 5.
Since 25 × 5 = 125, the perimeter is 125.
Answer A |
AQUA-RAT | AQUA-RAT-37442 | # Find the number of ways so that each boy is adjacent to at most one girl.
There are $7$ boys and $3$ girls who need to be lined up in a row. Find the number of ways so that each boy is adjacent to at most one girl.
In simple terms the situation demands that any distribution of the type $$...GBG...$$ must not come into play.
First of all the total number of arrangements are $10!$ and we can actually find a complement of those situations which we don't want.
In order to calculate the number of ways in which the wrong position can be true, I considered $GBG$ to be kind of a single package. The number of ways to make this package are:$${7 \choose 1} \cdot {3 \choose 2} \cdot 2!$$, now considering this package and $6$ boys plus the $1$ girl left, we can permute them all in 8! ways [$6$ boys, $1$ girl and our "package"], thus making the total to be $${7 \choose 1} \cdot {3 \choose 2} \cdot 2! \cdot 8! \tag{1}$$
Things seem to be tractable hithero, but as I was writing this questions I saw one problem in my argument: The cases containing the configurations $GBGBG$ have been possible counted several times thus $(1)$ is not giving the correct number of ways to be subtracted.
Can we anyhow make some changes in this approach and find the solution?
• Well, inclusion-exclusion lets you first subtract the cases in which at least one boy is alongside two girls so long as you add back the cases in which at least two boys are alongside two girls. – lulu Apr 11 '18 at 16:59
• I suppose the boys are distinguishable? That would be a fair supposition I guess... – Fimpellizieri Apr 11 '18 at 17:10
• @Fimpellizieri Certainly! – Harsh Sharma Apr 11 '18 at 17:12
• Do they self-identify in accordance with the genders assigned to them? – Strawberry Apr 12 '18 at 9:09
## 7 Answers
The following is multiple choice question (with options) to answer.
A total of 320 chocolates were distributed among 100 boys and girls such that each boy received 2 chocolates and each girl received 3 chocolates. Find the respective number of boys and girls? | [
"10,10",
"30,40",
"70, 50",
"20,20"
] | D | Let the number of boys be x.
Number of girls is 100 - x.
Total number of chocolates received by boys and girls = 2x + 3(100 - x) = 320
=> 300 - x = 320 => x = 20.
So, the number of boys or girls is 20.
ANSWER:D |
AQUA-RAT | AQUA-RAT-37443 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 85% like cherries. What is the minimum percentage of people who like all three? | [
"15%",
"20%",
"25%",
"30%"
] | D | It seems that something is wrong with your answer options.
1) minimum applesbananas:
[XXXXXXXXXXXXXX------] - apples
[-----XXXXXXXXXXXXXXX] - bananas
mimimum ab = 70 - (100-75) = 45%
2) minimum (applesbananas)cherries:
[XXXXXXXXX-----------] - applesbananas
[----XXXXXXXXXXXXXXXX] - cherries
mimimum ab = 45 - (100-85) =30%
D |
AQUA-RAT | AQUA-RAT-37444 | 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$$.Physics
The following is multiple choice question (with options) to answer.
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 sec. The speed of the train is? | [
"25 km/hr",
"50 km/hr",
"12 km/hr",
"40 km/hr"
] | B | Answer: Option B
Speed of the train relative to man = 125/10 = 25/2 m/sec.
= 25/2 * 18/5 = 45 km/hr
Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
x - 5 = 45 => x = 50 km/hr. |
AQUA-RAT | AQUA-RAT-37445 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 26 m * 2 m * 0.75 m? | [
"26000",
"27782",
"27891",
"25000"
] | A | 26 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
26 = 1/1000 * x => x = 26000
Answer: A |
AQUA-RAT | AQUA-RAT-37446 | => 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.
Rahul can do a work in 3 days while Rajesh can do the same work in 2days. Both of them finish the work together and get $355. What is the share of Rahul? | [
"$50",
"$142",
"$60",
"$100"
] | B | Rahul's wages: Rajesh's wages = 1/3 : 1/2 = 2:3
Rahul's share = 355*2/5 = $142
Answer is B |
AQUA-RAT | AQUA-RAT-37447 | Thanks.
• Binomial Theorem, do you know? – IAmNoOne Apr 29 '14 at 2:58
• I've heard of it, and I know it via Pascal's Triangle for low $n$, such as $n=2,3,4$, but I'm not familiar with its general formula. – Sujaan Kunalan Apr 29 '14 at 2:59
• I will post an answer then. – IAmNoOne Apr 29 '14 at 2:59
• You can find many posts about this here. For example, this question and this question and other posts linked there. – Martin Sleziak Sep 27 '15 at 20:35
There is an easy way to prove the formula.
Suppose you want to buy a burger and you have a choice of 6 different condiments for it - mustard, mayonnaise, lettuce, tomatoes, pickles, and cheese. How many ways can you choose a combination of these condiments for your burger?
Of course, you can choose either 6 different condiments, 5 different condiments, 4 different condiments, etc. So the obvious way to solve the problem is:
\begin{align}{{6}\choose{6}} + {{6}\choose{5}} + {{6}\choose{4}} + {{6}\choose{3}} + {{6}\choose{2}} + {{6}\choose{1}} = \boxed{64}\end{align}
But there is a better way. Imagine 6 spaces for 6 condiments:
_____ _____ _____ _____ _____ _____
For every space, there are $2$ possible outcomes: Yes or No, meaning the condiment was chosen or the condiment was not chosen. With $2$ possible outcomes for each space, there are $2^6 = \boxed{64}$ possible ways.
We know that both ways have foolproof logic and will both give identical answers no matter how many condiments there are. So this means we have proven:
\begin{align}\sum_{k = 0}^{n} \binom{n}{k} = 2^n\end{align}
The following is multiple choice question (with options) to answer.
During one month at a particular restaurant, 1/6 of the burgers sold were veggie burgers and 1/3 of the rest of the burgers sold were double-meat. If x of the burgers sold were double-meat, how many were veggie burgers? | [
"x/8",
"x/2",
"2x/3",
"3x/5"
] | D | Let y be the number of total burgers.
veggie = y/6 , non veggie = 5y/6
1/4 of the rest of the burgers sold were double-meat
=> 5y/6 * 1/3 = double meat = x
=> y/6 = 3x/5 = veggie
Hence D |
AQUA-RAT | AQUA-RAT-37448 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
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The following is multiple choice question (with options) to answer.
A pump can fill a tank with a water in 2 hours. Because of a leak, it took 6 hours to fill the tank. The leak can drain all the water of the full tank in how many hours? | [
"2",
"3",
"4",
"5"
] | B | The rate of the pump + leak = 1/6
1/2 - leak's rate = 1/6
leak's rate = 1/2 - 1/6 = 1/3
The leak will empty the tank in 3 hours.
The answer is B. |
AQUA-RAT | AQUA-RAT-37449 | • Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15
• From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24
Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.
In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.
The following is multiple choice question (with options) to answer.
A toy store sells small, medium, large, and super-large toy trucks in each of the colors red, blue, black, green, orange, pink, and yellow. The store has an equal number of trucks of each possible color-size combination. If Paul wants a medium, red truck and his mother will randomly select one the trucks in the store, what is the probability that the truck she selects will have at least one of the two features Paul wants? | [
"3/7",
"4/7",
"3/14",
"5/14"
] | D | Probability of NOT selecting medium out of 4 sizes = 3/4
Probability of NOT selecting red out of 7 colours = 6/7
Total probability of NOT selecting red and medium = (3/4)*(6/7) = 9/14
Required probability = 1 - 9/14 (this will select at least one of red and medium)
= 5/14
The answer is D. |
AQUA-RAT | AQUA-RAT-37450 | I'm assuming that the four fingers are given, e.g., the fingers of one hand without the thumb. An arrangement of five distinguishable rings on the four labeled fingers amounts to a linear arrangement of $1$, $2$, $3$, $4$, $5$, and three indistinguishable zeros as separators. There are ${8!\over3!}=6720$ such arrangements. Note that a green ring and a blue ring on the index finger can be worn in $2$ ways.
• If we let $0$ denote the lack of a ring and if given a linear arrangement, the first pair go on the first finger (in the specified order), etc then the sequence $10234500$ and $01234500$ should be considered equivalent but I don't see how your solution accounts for that? – benguin Apr 9 '17 at 4:07
• @benguin : The zeros serve as separators. – Christian Blatter Apr 9 '17 at 7:05
• Thanks for the clarification, that makes more sense :) I was considering an argument that first orders the rings ($5!$ ways) then uses stars-and-bars to place the rings onto the fingers ($C(5+4-1,4-1)$ ways) for a total of $5!C(8,3)$ which gives the same result. – benguin Apr 9 '17 at 8:43
Suppose you had only one ring. It then would be $4^1=4$ different arrangements/placements. Now if we add one more ring, it will be $4\cdot4$ because we combine the four possible arrangements with yet another four arrangements which makes it $16$. This means you can have two rings on one finger. By the same token we would have $4^5=1024$ arrangements for five rings. And the person can have five rings on one finger or no ring at all on the finger.
If you had $5^4$ arrangements, it would mean $5^1=5$ different arrangements for one ring on four fingers. And that doesn’t make sense.
And finger $a$ can have six possible arrangements: either of five rings or no ring at all.
Hope my explanation was not confusing.
The following is multiple choice question (with options) to answer.
In how many ways can nine different rings be worn on four fingers of one hand? | [
"10",
"129",
"135",
"126"
] | D | Required number of ways = ways of selecting 4 objects out of 9 given objects
= 9C4 = 9*8*7*6/24= 126
Answer D |
AQUA-RAT | AQUA-RAT-37451 | # System of three equations
$$3x+2y-z=4$$ $$4y-2z+6x=8$$ $$x-2y=5$$
I found that the first two equations were the same. Then, looking at the second and third equations, I thought the answer was No Solution, because I couldn't find a way to solve it, but instead the answer was Infinitely Many Solutions. Can someone explain this to me?
• "I couldn't find a way to solve it": there are zillion equations that have solutions though we don't know how to address them ! – Yves Daoust Sep 20 '16 at 12:49
• Usually I can solve such problems though – thunderbolt Sep 21 '16 at 2:59
## 4 Answers
As one of the equations is a duplicate, you can drop it and solve.
$$\begin{cases}3x+2y-z=4\\ x-2y=5\end{cases}$$
Let us move $z$ to the RHS and consider it as known to obtain a system of two equations in two unknowns,
$$\begin{cases}3x+2y=z+4\\ x-2y=5.\end{cases}$$
Then by elimination
$$\begin{cases}4x=z+9\\ 8y=z-11.\end{cases}$$
We can assign $z$ any value, yet get values of $x,y,z$ that satisfy all equations. For instance $(3,-1,3)$ or $(1,-2,-5)$. There are infinitely many others. In the solution space, $\mathbb R^3$, the solutions describe a straight line.
This is the same as$$\begin{bmatrix}3&2&-1\\6&4&-2\\4&-2&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\8\\5\end{bmatrix}$$
The following is multiple choice question (with options) to answer.
If 5x+2y=3x-4y,then what equation will arise? | [
"x=-y",
"y=-x",
"A and B",
"x=y"
] | D | 5x+2y=3x-4y
5x-3x=4y-2y
2x=2y
x=y
Option:D |
AQUA-RAT | AQUA-RAT-37452 | They're compounding this much every day, so if I were to write this as a decimal ... Let me just write that as a decimal. 0.06274%. As a decimal this is the same thing as 0.0006274. These are the same thing, right? 1% is .01, so .06% is .0006 as a decimal. This is how much they're charging every day. If you watch the compounding interest video, you know that if you wanted to figure out how much total interest you would be paying over a total year, you would take this number, add it to 1, so we have 1., this thing over here, .0006274. Instead of just taking this and multiplying it by 365, you take this number and you take it to the 365th power. You multiply it by itself 365 times. That's because if I have$1 in my balance, on day 2, I'm going to have to pay this much x $1. 1.0006274 x$1. On day 2, I'm going to have to pay this much x this number again x $1. Let me write that down. On day 1, maybe I have$1 that I owe them. On day 2, it'll be $1 x this thing, 1.0006274. On day 3, I'm going to have to pay 1.00 - Actually I forgot a 0. 06274 x this whole thing. On day 3, it'll be$1, which is the initial amount I borrowed, x 1.000, this number, 6274, that's just that there and then I'm going to have to pay that much interest on this whole thing again. I'm compounding 1.0006274. As you can see, we've kept the balance for two days. I'm raising this to the second power, by multiplying it by itself. I'm squaring it. If I keep that balance for 365 days, I have to raise it to the 365th power and this is counting any kind of extra penalties or fees, so let's figure out - This right here, this number, whatever it is, this is - Once I get this and I subtract 1 from it, that is the mathematically
The following is multiple choice question (with options) to answer.
United Telephone charges a base rate of $11.00 for service, plus an additional charge of $0.25 per minute. Atlantic Call charges a base rate of $12.00 for service, plus an additional charge of $0.20 per minute. For what number of minutes would the bills for each telephone company be the same? | [
"2 minutes",
"10 minutes",
"20 minutes",
"40 minutes"
] | C | Lets take number of minutesx.
Given that, 11+0.25x=12+0.2x ->0.05x=2 -> x=20minutes
ANS C |
AQUA-RAT | AQUA-RAT-37453 | => 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.
In a colony of 70 residents, the ratio of the number of men and women is 3 : 4. Among the women, the ratio of the educated to the uneducated is 1 : 4. If the ratio of the number of educated to uneducated persons is 8 : 27, then find the ratio of the number of educated to uneducated men in the colony? | [
"A)1:6",
"B)1:2",
"C)1:8",
"D)1:9"
] | B | Number of men in the colony = 3/7 * 70 = 30.
Number of women in the colony = 47 * 70 = 40.
Number educated women in the colony = 1/5 * 30 = 6.
Number of uneducated women in the colony = 4/5 * 50 = 24.
Number of educated persons in the colony = 8 /35 * 70 = 16.
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 30 - 10 = 20.
Number of educated men and uneducated men are in the ratio 10 : 20 i.e., 1:2.
Answer:B |
AQUA-RAT | AQUA-RAT-37454 | 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.
Of the 400 members at a health club, 280 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between: | [
"200 to 260",
"240 to 280",
"220 to 260",
"220 to 300"
] | B | W=280
P=300
ALL=400
N(NEITHER)=AT LEAST 60
BOTH=?
W+P-BOTH+N=ALL
280+300-BOTH+60=400
BOTH=240
now pay attention to the fact, that both at most can be 280, since W=280
B |
AQUA-RAT | AQUA-RAT-37455 | 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.
A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is: | [
"14 years",
"18 years",
"20years",
"22 years"
] | D | Let the son's present age be x years. Then, man's present age = (x + 24) years.
(x + 24) + 2 = 2(x + 2)
x + 26 = 2x + 4
x = 22.
ANSWER:D |
AQUA-RAT | AQUA-RAT-37456 | # 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.
Two train 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 45 kmph respectively. In how much time will they cross each other, if they are running in the same direction? | [
"80 sec",
"252 sec",
"320 sec",
"330 sec"
] | B | Sol.
Relative speed = (45 - 40) kmph = 5 kmph = [5 * 5/18] m/sec = 25/18 m/sec.
Total distance covered = Sum of lengths of trains = 350 m.
∴ Time taken = [350 * 18/25] sec = 252 sec.
Answer B |
AQUA-RAT | AQUA-RAT-37457 | At most, 90% of the private homes are brick. So that's 9/10 x 3/8 = 27/80.
100% of the apartments are brick. So that's 5/8 = 50/80.
Add those two together and you get 77/80 not 15/16 (=75/80 as the answer suggests)...
Look at it this way.
Say there were 80 houses in total.
The ration tells us that 30 would be private homes and 50 would be apartments.
10% of the private homes are wooden which makes 3.
All 50 apartments are brick and the remaining 27 private houses are unknown which could mean that all 27 are brick.
Hence a maximum of 77 brick houses out of 80.
5. Hmmm....I'm not a maths expert but let's see what we get if we reverse the ratios. Perhaps there was a communication error.
$p:a=3:5$
$\frac {1}{10}$of a=wooden
$\frac{1}{10}\times\frac{5}{8}=\frac{5}{80}$
= number of wooden.
number of brick = 1-number of wooden
$=\frac{75}{80} = \frac{15}{16}$
Edit: Are you sure you wrote the ratio the correct way around? The way you wrote it suggests that the person above has the correct answer.
I don't know how it got 15/16. I actually can't figure out why that is.
At most, 90% of the private homes are brick. So that's 9/10 x 3/8 = 27/80.
100% of the apartments are brick. So that's 5/8 = 50/80.
Add those two together and you get 77/80 not 15/16 (=75/80 as the answer suggests)...
Look at it this way.
Say there were 80 houses in total.
The ration tells us that 30 would be private homes and 50 would be apartments.
10% of the private homes are wooden which makes 3.
All 50 apartments are brick and the remaining 27 private houses are unknown which could mean that all 27 are brick.
Hence a maximum of 77 brick houses out of 80.
The following is multiple choice question (with options) to answer.
85% of the population of a city is 85000. The total population of the city is? | [
"95000",
"97500",
"100000",
"105000"
] | C | X * (85/100) = 85000
X = 1000 * 100
X = 100000
Answer: C |
AQUA-RAT | AQUA-RAT-37458 | # Is $\frac{200!}{(10!)^{20} \cdot 19!}$ an integer or not?
A friend of mine asked me to prove that $$\frac{200!}{(10!)^{20}}$$ is an integer
I used a basic example in which I assumed that there are $200$ objects places in $20$ boxes (which means that effectively there are $10$ objects in one box). One more condition that I adopted was that the boxes are distinguishable but the items within each box are not. Now the number of permutations possible for such an arrangement are : $$\frac{200!}{\underbrace{10! \cdot 10! \cdot 10!\cdots 10!}_{\text{20 times}}}$$ $$\Rightarrow \frac{200!}{(10!) ^{20}}$$
Since these are just ways of arranging, we can be pretty sure that this number is an integer.
Then he made the problem more complex by adding a $19!$ in the denominator, thus making the problem: Is $$\frac{200!}{(10!)^{20} \cdot 19!}$$ an integer or not?
The $19!$ in the denominator seemed to be pretty odd and hence I couldn't find any intuitive way to determine the thing. Can anybody please help me with the problem?
The following is multiple choice question (with options) to answer.
2,200 has how many positive divisors? | [
"12",
"16",
"20",
"24"
] | D | By factorization, we can write 2200 as 2200=2^3*5^2*11.
The number of factors is (3+1)(2+1)(1+1) = 24
The answer is D. |
AQUA-RAT | AQUA-RAT-37459 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
At exactly what time past 6:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time? | [
"20 13/21 minutes past 7:00",
"20 13/17 minutes past 7:00",
"21 9/11 minutes past 6:00",
"21 9/11 minutes past 7:00"
] | C | 5.5 is the angle between minute n hour, this is what I was taught...so shouldn't it be solve by dividing 90 with 5.5?
That would have been the case if your initial difference between the hour and the minute hand was = 0 degrees or in other words, both minute and hour hands were at the same location. But as per the question, you are asked for time AFTER 6:00. At 6:00, the angle between the hour and the minute hand is 210 degrees. you need to take this into account as well.
So in order for the difference to decrease to 90 degrees, the minute hand must eat away this difference of 210-90 = 120 degree at the rate of 5.5 degrees per minute ---> 120/5.5 = 21 9/11 minutes.
Thus, C is the correct answer. |
AQUA-RAT | AQUA-RAT-37460 | 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 two trains of lengths 400 m, 600 m respectively, running at same directions. The faster train can cross the slower train in 180 sec, the speed of the slower train is 48 km. then find the speed of the faster train? | [
"87 Kmph",
"68 Kmph",
"67 Kmph",
"57 Kmph"
] | B | Length of the two trains = 600m + 400m
Speed of the first train = X
Speed of the second train= 48 Kmph
1000/X - 48 = 180
1000/x - 48 * 5/18 = 180
50 = 9X - 120
X = 68 Kmph
Answer: B |
AQUA-RAT | AQUA-RAT-37461 | In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
The following is multiple choice question (with options) to answer.
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 5? | [
"897",
"98,910",
"64,749",
"49,700"
] | B | Find the number , Upon Sum of 3 Digits of a number Gives a Reminder 2 when it is Divided by 5
Seeing the Options After Dividing and Finding the Reminder of 2
My Answer was B |
AQUA-RAT | AQUA-RAT-37462 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Sam invested $15000 @10% per annum for one year. If the interest is compounded half yearly, then the amount received by Sam at the end of the year will be? | [
"$16537.50",
"$24512.56",
"$18475.89",
"$15478.56"
] | A | P = $15000
R = 10%
p.a. = 5%
T = 2 half years
Amount = 15000 * (1 + 5/100)^2 = 15000* 21/20 * 21/20 = $16537.50
Answer is A |
AQUA-RAT | AQUA-RAT-37463 | 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 of equal lengths take 10 sec and 12 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction? | [
"11 sec",
"12 sec",
"17 sec",
"21 sec"
] | A | Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/12 = 10 m/sec.
Relative speed = 12 + 10 = 22 m/sec.
Required time = (120 + 120)/22 = 11 sec.
Answer: A |
AQUA-RAT | AQUA-RAT-37464 | # Math Help - Divisibility
1. ## Divisibility
Hi all.
I'm trying to figure out the following problem:
Find the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4.
Help will be appreciated. Looking for a simple/elementary proof.
Thanks.
2. Originally Posted by pollardrho06
Hi all.
I'm trying to figure out the following problem:
Find the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4.
Help will be appreciated. Looking for a simple/elementary proof.
Thanks.
Hint:
$|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3 \text{ but not by } 4\}|$
$= |\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3\}|-|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 12\}|$
This is a consequence of $A\backslash B=A\backslash(A\cap B)$, and $|X\backslash Y|=|X|-|Y|$, if $Y\subseteq X$.
3. Originally Posted by Failure
Hint:
$|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3 \text{ but not by } 4\}|$
$= |\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3\}|-|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 12\}|$
This is a consequence of $A\backslash B=A\backslash(A\cap B)$, and $|X\backslash Y|=|X|-|Y|$, if $Y\subseteq X$.
I don't see it...
4. ## Hint
Look for cycles.
5. Hello, pollardrho06!
Find the number of positive integers not exceeding 1000
that are divisible by 3 but not by 4.
The following is multiple choice question (with options) to answer.
Find how many positive integers less than 10,000 are there suchthat the sum of the digits of the no. is divisible by 3? | [
"1256",
"2354",
"2678",
"3334"
] | D | If sum of the digits is divisible by 3, the number is divisible by 3.
Therefore, required number of non-negative integers is equal to count of numbers less than 10,000 which are divisible by 3.
Such numbers are (3, 6, 9, ... , 9999) (arithmetic progression with first term=3, last term=9999, common difference=3).
Count of such numbers =
9999
3
=
3333
99993=3333
But zero is also divisible by 3.
This makes our total count 3334
D |
AQUA-RAT | AQUA-RAT-37465 | The final number of outcomes is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8 - 20 \cdot 3^8 + 15 \cdot 2^8 -6\cdot 1^8 = 191520$$
And the probability of rolling 8 dice and getting at least one of each number is $$\frac{191520}{6^8} = \frac{665}{5832} \approx 0.114$$
Thus the counterintuitive result is that you are likely to be missing at least one number when you roll 8 dice. In fact, in order to have better than a 50% chance of getting one of each number, you would have to roll 13 dice!
• Of course, the main problem with the question at hand, is that the user asked for the number of combinations, not for the probability of getting them. And as such, the answer depends on whether or not identical combinations are considered the same. For the sake of calculating probability, they are obviously not considered the same. But since the question is not about probability, it remains slightly unclear. I would guess that identical combinations are not considered the same, which makes your answer correct. – barak manos Mar 4 '15 at 21:43
To have all numbers appear at least once, you can have two pairs of the same number or one triplet. The number of ways to get a triplet is $6$ (choose the repeated number) $8 \choose 3$ (choose the positions for the triplet) $5!$ (order the remaining numbers)$=40320$ Can you do the two pairs?
The following is multiple choice question (with options) to answer.
On eight consecutive flips of a coin, what is the probability that all eight produce the same result? | [
"1/16",
"1/128",
"1/8",
"1/4"
] | B | TTTTTTTT case --> P = (1/2)^8 = 1/256
HHHHHHHH case --> P = (1/2)^8 = 1/256
P(TTTTTTTT or HHHHHHHH) = 1/8 + 1/8 = 1/128
Answer: B |
AQUA-RAT | AQUA-RAT-37466 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
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Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
Pradeep has to obtain 20% of the total marks to pass. He got 390 marks and failed by 25 marks. The maximum marks are | [
"300",
"2075",
"800",
"1000"
] | B | Explanation :
Let their maximum marks be x.
Then, 20% of x = 390 + 25
=>20/100x= 415
x = (41500/20)
x= 2075.
Answer : B |
AQUA-RAT | AQUA-RAT-37467 | # Identifying Prime Numbers from 507 to 10647
#### ErikHall
##### New member
So lets say you have the Numbers (2677; 6191 and 1091), how would you go about the check that 2677 and 1091 are Prime ?
The conditions are:
- You dont have a calculator
- You just have a Pen
Is there a clever way to do it except looking at the Last Digit ?
Thanks for the Help !
Also small fun fact, if you use the Google Calculator and divide by Zero, it says "Infinity". Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
#### lookagain
##### Senior Member
Look at 1,091, for instance. Look at the prime numbers whose squares are
less than or equal to 1,091. If none of those divide 1,091, then 1,091 is prime.
#### lev888
##### Full Member
Is there a clever way to do it except looking at the Last Digit ?
Could you explain the last digit method?
#### ErikHall
##### New member
Could you explain the last digit method?
So if you see a Number like 546342 you know its not Prime since you can divide it by 2. Same with 4 and 8. Those numbers are, as far as i know, never Prime. So this is a way to see non Primes. But it dosnt help you do see if a Number is Prime. It only shows the ones who are not. And not even that many of them
#### ErikHall
##### New member
Look at 1,091, for instance. Look at the prime numbers whose squares are
less than or equal to 1,091. If none of those divide 1,091, then 1,091 is prime.
Interessting way, but that would mean you have to know the Primes up to like 10.000 right ? Or at least the ones, that are near to 10.000 Squared.
#### Subhotosh Khan
##### Super Moderator
Staff member
Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
Incorrect.
The following is multiple choice question (with options) to answer.
Which of the following is a prime number ? | [
"65",
"35",
"97",
"35"
] | C | C
97
Explanation: Clearly, 97 is a prime number. |
AQUA-RAT | AQUA-RAT-37468 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
The population of a town increased from 1,33,400 to 1,93,500 in a decade. The average percent increase of population per year is: | [
"4.5%",
"5.5%",
"6.5%",
"8.75%"
] | A | Explanation: Increase in 10 years = (193500 - 133400) = 60100.
Increase% = (60100/133400 x 100)% = 45%.
Required average = (45/10)% = 4.5%.
Answer: Option A |
AQUA-RAT | AQUA-RAT-37469 | - 2 years, 3 months ago
- 2 years, 3 months ago
I was getting the answer as 36.
My cases were similar to that of Deeparaj.
Case 1: When (4,8) is one of the selected pair.
Among the remaining 6 numbers only (2,6) have GCD=2. We can select any 3 pairs from the remaining 6 numbers in ((6C2)(4C2)(2C2)/3!)=15 ways( Note that we have to only select the pairs, hence the factor of 3! in the denominator). From this we need to subtract the ways where (2,6) is one of the pairs. Hence the answer of case 1 is 15-3=12.
Case 2: When (4,8) is not of the pairs.
In this case we can show that in each of the 4 pairs we must have one odd number and one even. Therefore total number ways of selecting 4 pairs in this case is simply 4!=24.
- 2 years, 6 months ago
Ah...
I forgot to divide by 2! in my first case to remove the ordering. Thanks for the clarification.
- 2 years, 6 months ago
Can a number be repeated in the pairs?
- 2 years, 6 months ago
No.
- 2 years, 6 months ago
Case 1: One of the pairs is (4,8): $$4\times {4\choose2}$$
Still working on Case 2.
- 2 years, 6 months ago
Case 2:$$4!$$.
So, on the whole $$\boxed{ 48 }$$ ways. Am I right?
- 2 years, 6 months ago
The following is multiple choice question (with options) to answer.
Ratio between Rahul and Deepak is 4:3, After 22 Years Rahul age will be 26 years. What is Deepak present age | [
"10",
"3",
"5",
"7"
] | B | Explanation:
Present age is 4x and 3x,
=> 4x + 22 = 26 => x = 1
So Deepak age is = 3(1) = 3
Answer: Option B |
AQUA-RAT | AQUA-RAT-37470 | We have 8 members and a 3 slots. So total number of ways is 8*7*6, divided by 3! for double counts. 8*7*6/3!=56 .
Now what are the number of ways can we have two teams members on the committee?
We can pick any of the 8 for the first slot. The second slot is reserved for the team member of the first slot, so 1 ways. And the last slot can be any of the remaining 6.
So 8*1*6. But we have to account for double counts. Do we divide by 3! or 2!? This is were I am stuck. If we divide by 2!, we get 8*1*6/2=24
56-24=32, our answer. But I don't understand why we divide by 2!??
I hope I was a little more clearer. Thank you again for responding Bunuel!
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Re: Committee Combination Tough Problem [#permalink]
### Show Tags
02 Oct 2012, 04:36
Bunuel wrote:
alphabeta1234 wrote:
A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?
A)20
B)22
C)26
D)30
E)32
I'm not sure what are you exactly dong in your last approach.
Anyway, if you want to solve with slot method the simpler solution would be: 8*6*4/3!=32, 8 ways to choose for the first slot, 6 ways to choose for the second slot, 4 ways to choose for the third slot and dividing by 3! to get rid of duplication.
The following is multiple choice question (with options) to answer.
75 people are being divided into committees. However, the project calls for 9 teams of 9 people each. How many more people are needed to reach this quota? | [
"6",
"8",
"19",
"10"
] | A | 9 x 9 = 81
81 - 75 = 6
The answer is A. |
AQUA-RAT | AQUA-RAT-37471 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
Due to construction, the speed limit along an 7-mile section of highway is reduced from 55 miles per hour to 35 miles per hour. Approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | [
" 2",
" 3",
" 4",
" 5"
] | C | 7/35 - 7/55 = 7/5 * ( 11 - 7)/77
= 7/5 * 4/77 * 60 min
= 7 * 12 * 4/77
= 336/77 ~ 4.3
Answer - C |
AQUA-RAT | AQUA-RAT-37472 | ### 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.
The ratio of investments of two partners P and Q is 7:5 and the ratio of their profits is 7:10. If P invested the money for 2 months, find for how much time did Q invest the money? | [
"8",
"10",
"18",
"4"
] | D | 7*5: 2*x = 7:10
x = 4
Answer:4
Answer is D |
AQUA-RAT | AQUA-RAT-37473 | # 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.
Ram sold two bicycles, each for Rs.990. If he made 10% profit on the first and 10% loss on the second, what is the total cost of both bicycles? | [
"Rs.2000",
"Rs.1980",
"Rs.1891",
"Rs.1750"
] | A | Explanation:
(10*10)/100 = 1%loss
100 --- 99
? --- 1980 => Rs.2000
ANSWER IS A |
AQUA-RAT | AQUA-RAT-37474 | Since 1000 = 999 + 1, 100 = 99 + 1, and 10 = 9 + 1, this equation can be rewritten
$$999 d_{1}+d_{1}+99 d_{2}+d_{2}+9 d_{3}+d_{3}+d_{4}=3 k$$
Rearranging gives
$$d_{1}+d_{2}+d_{3}+d_{4}=3 k-999 d_{1}-99 d_{2}-9 d_{3}$$
$$=3 k-3\left(333 d_{1}\right)-3\left(33 d_{2}\right)-3\left(3 d_{3}\right)$$
We can now factor a 3 from the right side to get
$$d_{1}+d_{2}+d_{3}+d_{4}=3\left(k-333 d_{1}-33 d_{2}-d_{3}\right)$$
since $$\left(k-333 d_{1}-33 d_{2}-d_{3}\right)$$ is an integer, we have shown that $$d_{1}+d_{2}+d_{4}+d_{4}$$ is divisible by 3.
(2) Assume $$d_{1}+d_{2}+d_{3}+d_{4}$$ is divisible by $$3 .$$ Consider the number $$d_{1} d_{2} d_{3} d_{4} .$$ As remarked above,
$$d_{1} d_{2} d_{3} d_{4}=d_{1} \times 1000+d_{2} \times 100+d_{3} \times 10+d_{4}$$
so
The following is multiple choice question (with options) to answer.
The value of 99^(93/99) x 99 is: | [
"9989",
"9896",
"9894",
"9809"
] | C | (100 – 6/99) x 99 = 9900 – 6
= 9894.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37475 | Let one woman complete the job in $$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 piece of work in 6 days, but with the help of his son, he can do it in 3 days. In what time can the son do it alone? | [
"5",
"7",
"6",
"8"
] | C | Son's 1day's work = (1/3)-(1/6) = 1/6
The son alone can do the work in 6 days
Answer is C |
AQUA-RAT | AQUA-RAT-37476 | For some reason I find it easier to think in terms of letters of a word being rearranged, and your problem is equivalent to asking how many permutations there are of the word YYYYBBBBB.
The formula for counting permutations of words with repeated letters (whose reasoning has been described by Noldorin) gives us the correct answer of 9!/(4!5!) = 126.
The following is multiple choice question (with options) to answer.
In how many ways can the letters of the word REPEATED be arranged? | [
"1206",
"1348",
"6720",
"3000"
] | C | REPEATED has 8 words, but E is repeated three times. Thus ways = 8! / 3! = 6720 ways
C |
AQUA-RAT | AQUA-RAT-37477 | algorithms
Title: How to calculate least common multiple (LCM) for two numbers with constants (shifts)? What's the most effective algorithm to calculate LCM for 2 numbers where each of them has their own "shift"?
Example:
We want to find LCM for 25 + 5 as c and 30 + 10 as c
Then for the first case, the sequence will look like this:
25*1+5, 25*2+5, 25*3+5, 25*4+5...
And the second one:
30*1+10, 30*2+10, 30*3+10, 30*4+10...
In this case, the result should be 130 - 5th member of the first sequence and 4th of the second one. Given integers $a,b \geq 1$ and $c,d \geq 0$, you are looking for the set of solutions of $ax + c = by + d$ over integers $x,y \geq 0$. We can assume without loss of generality that $d = 0$ (replace $c,d$ with $c-d,0$ or $0,d-c$) and that $(a,b,c) = 1$ (otherwise, divide everything by the GCD).
I claim that if there is any solution then necessarily $(a,b) = 1$. Indeed, suppose that $(a,b) = g > 1$. If $ax + c = by$ then $g \mid ax,by$ implies $g \mid c$, and so $(a,b,c) = g > 1$, contrary to assumption. Therefore $(a,b) = 1$.
If $ax + c = by$ then $ax + c \equiv 0 \pmod{b}$ and so $x \equiv -c/a \pmod{b}$ (note that we can divide by $a$ since $(a,b)=1$). Similarly, $y \equiv c/b \pmod{a}$. These equations essentially give us both the minimal solution and all solutions.
The following is multiple choice question (with options) to answer.
If the LCM of two integers I, J (where J> I and I>1) is I*J, then which of the following can not be true? | [
"All Prime factors of I must be Prime factors of J.",
"Both I and J can be consecutive integers.",
"Both I and J can be Prime Numbers.",
"I and J do not share any Prime factors."
] | A | If LCM of two numbers I and J is the product IJ then both numbers are Co- prime.
(Co-prime numbers are any two numbers which have an HCF of 1 i.e. when two numbers have no common prime factor apart from the number 1).
Numbers that can be Co-Prime are :
1) Two consecutive natural numbers.
2) Two consecutive odd numbers.
3) Two prime numbers.
4) One prime number and the other a composite number such that the composite number is not a multiple of the prime number.
With above four statements it can be concluded that Options B,C,D and E are all true.
Only Option A cannot be true.
Hence answer is A |
AQUA-RAT | AQUA-RAT-37478 | 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.
Total of ages of A,B and C at present is 90 years. If ratio of this ages is 2 : 3 : 4 what will be the age of C after 6 years? | [
"45 years",
"42 years",
"46 years",
"37 years"
] | C | 2+3+4=9
4*(90/9)=40yrs
after 6yrs=46
ANSWER:C |
AQUA-RAT | AQUA-RAT-37479 | => 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.
In a colony of 70 residents, the ratio of the number of men and women is 5 : 2. Among the women, the ratio of the educated to the uneducated is 1 : 4. If the ratio of the number of educated to uneducated persons is 8 : 27, then find the ratio of the number of educated to uneducated men in the colony? | [
"A)1:4",
"B)1:1",
"C)1:8",
"D)1:9"
] | A | Number of men in the colony = 5/7 * 70 = 50.
Number of women in the colony = 2/7 * 70 = 40.
Number educated women in the colony = 1/5 * 30 = 6.
Number of uneducated women in the colony = 4/5 * 50 = 24.
Number of educated persons in the colony = 8 /35 * 70 = 16.
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 50 - 10 = 40.
Number of educated men and uneducated men are in the ratio 10 : 40 i.e., 1:4.
Answer:A |
AQUA-RAT | AQUA-RAT-37480 | Since the the percentage of tagged fish in the second catch approximates the percentage of tagged fish in the pond, the approximate number of fish in the pond is:
0.04(total fish) = 50
This equation assumes that there are 50 fish tagged in the total population. We do not know that. The only thing we know is that that the percentage of tagged fish in the second catch is 4%. The question says that 4% approximates the number of tagged fish in the pond. So this is the true equation we have:
Quote:
.04 (total) = tagged fish
We are missing two variables. We don't know the total fish and we don't know the tagged fish.
If there are 16 tagged fish, then choice A is correct.
If there are 26 tagged fish, then choice B is correct, and etc.
If we assume that the number of fish in the second catch (50) is the number of fish tagged, then yes the total fish would be 1250. However, that's not what the question provides. I think this question is written poorly.
So the question states that "In a certain pond, 50 fish were caught, tagged, and returned to the pond."
From this sentence, we can deduce that there are indeed a total of 50 tagged fish in the pond. The only way to have some other number of tagged fish in the pond is if there were already some number of tagged fish in the pond (in which case, the question would have told us so) or if either more fish were tagged afterward or some of the tagged fish were removed from the pond (again, we would have been told). Since we have no such information, we cannot assume that there might be some other number of tagged fish in the pond.
Perhaps you are missing the fact that 50 fish are caught TWICE: first all of them are tagged, and the second time, the tagged fish are counted.
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Of the goose eggs laid at a certain pond, 2/3 hatched and 3/4 of the geese that hatched from those eggs survived the first month. Of the geese that survived the first month, 3/5 did not survive the first year. If 100 geese survived the first year and if no more than one goose hatched from each egg, how many goose eggs were laid at the pond? | [
" 280",
" 400",
" 540",
" 500"
] | D | Of the goose eggs laid at a certain pond, 2/3 hatched and 3/4 of the geese that hatched from those eggs survived the first month:
2/3*3/4 = 1/2 survived the first month.
Of the geese that survived the first month, 3/5 did not survive the first year:
(1-3/5)*1/2 = 1/5 survived the first year.
100 geese survived the first year:
1/5*(total) = 100 --> (total) = 500.
Answer: D. |
AQUA-RAT | AQUA-RAT-37481 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and
second half at the rate of 24 km/hr. Find the total journey in km | [
"200 Km",
"222 Km",
"224 Km",
"248 Km"
] | C | Explanation:
Let time taken to travel the first half = x hr
Then time taken to travel the second half = (10 - x) hr
Distance covered in the the first half = 21x [because, distance = time*speed]
Distance covered in the the second half = 24(10 - x)
Distance covered in the the first half = Distance covered in the the second half
So,
21x = 24(10 - x)
=> 45x = 240
=> x = 16/3
Total Distance = 2*21(16/3) = 224 Km [multiplied by 2 as 21x was distance of half way]
Answer: C |
AQUA-RAT | AQUA-RAT-37482 | of the half of the and! = 18π: PM as the perimeter of a circle ) ] or\ c.: 6169ec2ffad00696 • Your IP: 185.2.4.37 • Performance & security by cloudflare, Please complete the security check access. ) r units that area of the window is 20 ft, find the perimeter and area 7!, y }, { rx, ry } ] gives an ellipse! Two congruent triangles d or πr, and area 1470 perimeter ( circumference ) of a semicircle has a is! You temporary access to the circumference of a semicircle including its diameter arc depends on. Gives a circle ) of that circle and the 2r because those diameters are part of semicircle.
The following is multiple choice question (with options) to answer.
A semicircle has a radius of 9. What is the approximate perimeter of the semicircle? | [
"16",
"25",
"33",
"46"
] | D | The perimeter of a circle is 2*pi*r.
The perimeter of a semicircle is 2*pi*r/2 + 2r = pi*r + 2r
The perimeter is pi*9 + 2*9 which is about 46.
The answer is D. |
AQUA-RAT | AQUA-RAT-37483 | Hint: Find the probability of $A$ only plus the probability of $B$ only.
Let $x= P(A~ \text{only})$, $y=P(B ~\text{only})$. Then $x +y + 0.2 =0.4$. Thus $x+y= 0.2$
-
THe probability that at least one of them occurs is $0.4$ and the probability that both occur is $0.2$. Thus, the probability that exactly one occurs is $0.4-0.2=0.2$.
-
$P((A \cup B)\setminus(A \cap B)) = P(A\setminus B) + P(A\setminus B) = P(A) + P(B) − 2P(A \cap B)$ you will get the answer!
-
what is $P((A \cup B)(A \cap B))$? – Surb May 3 at 13:40
@timmbob: There was a backslash in the source text; yagmur intended and tried to write $(A\cup B)\setminus (A\cap B)$ as one would expect based on what followed. – Jonas Meyer May 3 at 17:21
The following is multiple choice question (with options) to answer.
The probability of two events A and B are 0.20 and 0.40 respectively. The probability that both A and B occur is 0.15. The probability that neither A nor B occur is _________ | [
"0.45",
"0.4",
"0.5",
"0.55"
] | D | we are apply that formula..............
P(AorB)=P(A)+P(B)-P(A AND B)
=.20+.40-.15
=.45
but the probability of neither A nor B=1-.45
=0.55
ANSWER:D |
AQUA-RAT | AQUA-RAT-37484 | 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 work four times as fast as B. Together they can finish a job in 12 days. In how many days can A complete the job alone? | [
"14",
"15",
"16",
"17"
] | B | B's rate is x and A's rate is 4x.
x + 4x = 1/12
x = 1/60
A's rate is 4x = 1/15
A can complete the job in 15 days.
The answer is B. |
AQUA-RAT | AQUA-RAT-37485 | comes up heads. The student will count the coins and write their answer to the right of each problem. The probability of all three tosses is heads: P ( three heads) = 1 × 1 + 99 × 1 8 100. ' 'The coin is just as likely to land heads as tails. Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. problems as “if you tossed a coin 6 times, what is the probability of getting two heads?” Let p denote the probability of the outcome of interest, Hence, the probability of the other outcome must be (1 − p). Practice Problem. A coin has a 50% chance of landing on heads the each time it is thrown. Determine the probability of each event: a) an odd number appears in a toss of a fair die; b) one or more heads appear in the toss of four fair. If it isn’t a trick coin, the probability of each simple outcome is the same. An experiment could be rolling a fair 6-sided die, or. Subjective Probability. Coin Probability Problems Coin is a currency token which has two faces, one is head and other is tail. For a fair coin, what is the probability that the longest run of heads or tails in a sequence of 30 tosses is less than or equal to 5? (pg 107) Because the coin toss is the simplest random event you can imagine, many questions about coin tossing can be asked and answered in great depth. (15 – 20 min) Homework Students flip a coin. Describe the sample space. Think of it this way: What is the probability of tossing 2 heads in a row if you toss a fair coin 7 times? Multiplication would lead you to think the probability is 6*1/4=1. 125 Stacy and George are playing the heads or tails game with a fair coin. Well, that is unless you failed to spin the coin, there is probability involved there too. Assume that the probability a girl is born is the same as the probability a boy is born. Jack has coins C_1, C_2,. This page continues to illustrate probability facts using the flip-a-coin-4-times-and-count-the-number-of-heads problem. What is the probability that you’ll toss a coin and get heads? What about twice in a row? Three times? Probability questions ask you determine the likelihood that an
The following is multiple choice question (with options) to answer.
100 identical coins each with probability 'pp' showing up heads & tossed. | [
"13/101",
"11/101",
"51/101",
"53/102"
] | C | Let aa be the number of coins showing heads
Then, P(A=50)=P(A=51)P(A=50)=P(A=51)
⇒100C50×P50×(1−P)50⇒100C51×P51×(1−P)49⇒100C50×P51×(1−P)50=100C51×P⇒51(1âˆ′P)=50P⇒P=51/101
C |
AQUA-RAT | AQUA-RAT-37486 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
If a#b = ab – b + b^2, then 3#4 = | [
"2",
"8",
"15",
"24"
] | D | Solution -
Simply substitute 3 and 4 in equation in the place of a and b respectively.
3#4 = 3*4 - 4 + 4^2 = 12 - 4 + 16 = 24. ANS D |
AQUA-RAT | AQUA-RAT-37487 | 5C2 * 4!(5C2 for selecting 2 out of 5 places for E & rest can be arranged in 4! ways.)
Ans with this approach 240 which is correct in this case.
Now the question discussed above:
2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?
Going as per the above approach it comes
8C4 * (7!/4!2!) (8C4 for filling 4 I's in 8 blank spaces and arranging rest 7 alphabets in 7!/4!2!)
But in this case answer doesn't match.
Please if you could explain the flaw in my approach will be very helpful.
Cheers.
_________________
Sailing through rough waters. Stars & few kudos will steer me past.
Re: In how many ways can the letters of the word PERMUTATIONS be &nbs [#permalink] 09 May 2018, 01:12
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R? | [
"59",
"77",
"36",
"99"
] | A | Explanation:
The first letter is E and the last one is R.
Therefore, one has to find two more letters from the remaining 11 letters.
Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.
The second and third positions can either have two different letters or have both the letters to be the same.
Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.
Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.
Total number of possibilities = 56 + 3 = 59
Answer: A) 59 |
AQUA-RAT | AQUA-RAT-37488 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. Let $m=k-2l$
$38m + 7l = 34$.
$(3+ 5*7)m + 7l = 34$
$3m + 7(5m + l) = 34$. Let $n = 5m+l$.
$3m + 7n = 34$
$3m + 2*3n + n = 34$
$3(m + 2n) + n = 34$. Let $a = m+2n$.
$3a + n = 34$.
Let $a=11; n = 1$. So $a=m+2n; 11=m + 2; m = 9$. and $n= 5m + l; 1=5*9 +l; l = -44$. and $m = k - 2l; 9=k +88; k = -79$.
So $25 - 79*38 = -2977$ and $59 - 44*69=-2977$.
So $-2977\equiv 25 \mod 38$ and $-2977 \equiv 59 \mod 69$.
That's of course not positive but.
$25 - 79*38 = 59- 44*69 \iff$
$25 - 79*38 + 69*38 = 59 - 44*69 + 69*38 \iff$
$25 - 10*38 = 59 - 6*69 \iff$
$25 - 10*38 + 69*38 = 59 - 6*69 + 69*38 \iff$
$25 + 59*38 = 59 + 32*69$
And $25+59*38 = 59 + 32 * 69 =2267$.
$2267\equiv 25\mod 38$ and $2267\equiv 59\mod 69$ and as the lowest common multiple of $38$ and $69$ is $2622$ this is the smallest positive such number.
The following is multiple choice question (with options) to answer.
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: | [
"101",
"107",
"111",
"185"
] | C | Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Answer: Option C |
AQUA-RAT | AQUA-RAT-37489 | valuePV=Present value (original amount of money)i=Interest rate per periodn=Number of periods\begin{aligned} &\text{FV} = \text{PV} \times ( 1 + i )^ n \\ &\textbf{where:} \\ &\text{FV} = \text{Future value} \\ &\text{PV} = \text{Present value (original amount of money)} \\ &i = \text{Interest rate per period} \\ &n = \text{Number of periods} \\ \end{aligned}FV=PV×(1+i)nwhere:FV=Future valuePV=Present value (original amount of money)i=Interest rate per periodn=Number of periods. We hope you like the work that has been done, and if you have any suggestions, your feedback is highly valuable. Conversely, the time value of money (TVM) also includes the concepts of future value (compounding) and present value … What is compound interest? Time Value of Money is a concept that recognizes the relevant worth of future cash flows arising as a result of financial decisions by considering the opportunity cost of funds. Of course, because of the rule of exponents, we don't have to calculate the future value of the investment every year counting back from the10,000 investment in the third year. What is simple interest? There are time value of money concepts that are designed to calculate the future value of money. In essence, all you are doing is rearranging the future value equation above so that you may solve for present value (PV). In other words, to find the present value of the future $10,000, we need to find out how much we would have to invest today in order to receive that$10,000 in one year. Suppose you are one of the lucky people to win the lottery. So, here is how you can calculate today's present value of the $10,000 expected from a three-year investment earning 4.5%: $8,762.97=10,000×(1+.045)−3\begin{aligned} &\8,762.97 = \10,000 \times ( 1 + .045 )^{-3} \\
The following is multiple choice question (with options) to answer.
Find the value of 85P3 . | [
"565350",
"595650",
"535950",
"565350"
] | B | 85P3 = 85!/(85-3)!
= 85!/82!
= 85*84*83*82!/82!
= 85*84*83
= 595650
ANSWER:B |
AQUA-RAT | AQUA-RAT-37490 | 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$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer’s suggested retail price. If during a sale, the store discounts an additional 20 percent from the discount price, what would be the lowest possible price of a container of pet food that had a manufacturer’s suggested retail price o f $ 30.00? | [
"$ 10.00",
"$ 11.20",
"$ 14.40",
"$ 16.80"
] | D | For retail price = $30
First maximum discounted price = 30 - 30% of 30 = 30-9 = 21
Price after additional discount of 20% = 21 - 20% of 21 = 21 - 4.2 = 16.8
Answer: Option D |
AQUA-RAT | AQUA-RAT-37491 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
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Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
If two different solutions of alcohol with a respective proportion of water to alcohol of 3:1 and 2:3 were combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts? | [
"30.0%",
"36.6%",
"42.5%",
"44.4%"
] | C | Alcohol: Water
First mixture -- 3:1 -- 15:5
Second mixture -- 2:3 -- 8:12
Equal amounts of the mixture were combined, hence the final mixture will be in the ratio -- (15+8): (5+12) -- 23:17
% of alcohol -- 17/(23+17) * 100
= (17/40) * 100
= 42.5
ANSWER:C |
AQUA-RAT | AQUA-RAT-37492 | We have to decide among $\text{(A)}$ and $\text{(B)}$. Note that the $26$th prime is $101$. This implies that if $p_{n}$ denotes the $n$ th prime, then $$\sum_{n=1}^{168}p_{n} = \sum_{n=1}^{25}p_{n}+\sum_{n=26}^{168} p_{n} > \sum_{n=26}^{168} 101 =101 \times 143=14443 >\text{(A)}=11555$$
The answer is thus $\text{(B)}$, $76127$. The answer can be confirmed through direct calculation or can be verfied here.
• Why did you choose the 26th prime here? Or is it just arbitrary? Jan 30, 2017 at 14:22
• It is somewhat commonly known piece of a trivia - there are 25 primes below 100 and the primes below 25 sum to 100. Jan 30, 2017 at 14:31
• @MarioDS It is as Jon Claus said; many people memorized it, so I was able to answer it quickly. Jan 30, 2017 at 15:00
There are $168$ primes with the first one equal to $2$ the rest $\ge 2k-1$ for $k=2,3,4,...,168$. So their sum is at least $168^2+1=28 225$.
• (a) would mean that the average prime is $<70$, which is horrendously implausible, for me good enough to pick (b) instead. - But this answer is the formal reason why it's implausible Jan 30, 2017 at 20:22
I just wanted to carry forward your observation about "Every prime can be written in of the form $(6n-1),(6n+1)$ except $2$ & $3$".
The following is multiple choice question (with options) to answer.
How many prime numbers are between 25/8 and 213/6? | [
"6",
"7",
"8",
"9"
] | D | 25/8 = 4-
213/6 = 36-
Prime numbers between 4 and 36 are 5, 7, 11, 13, 17, 19, 23, 29, and 31
- sign signifies that the number is marginally less .
Answer D |
AQUA-RAT | AQUA-RAT-37493 | This series of cash flows will yield exactly 10 % is $56.07 scenario... Stan also wants his son to be paid or received in the future amount that you expect receive! Is extremely important in many financial calculations 510.68 ; discount rate. discount rate is investment! Money received in the discussion above, we looked at one investment over the course of year! Now knows all three variables for the first offer suggests the value of$ 100 today or can... For the four discount rates trend, the amount $100 being$... Would not have realized a future sum the applicable discount rate. a present value formula shown! Costs, inflation will cause the price they pay for an investment might earn example, future... ; discount rate or the interest rate ” is used in the future cashflows expected from an investment might.... That with an initial investment of exactly $100 today or I can pay you back 100! Periods interest rates rise and the CoStar product suite often used as the present value provides a basis assessing... Today, you can buy goods at today 's prices, i.e must... Earned on the funds over the next five years time refers to future value and a... If you receive money today, you can buy goods at today 's prices, i.e from now 1 4... Aone-Size-Fits-All approach to determining the appropriate discount rate is used when referring to present. Discount lost business profits to a present value, the future value of cash flows will yield exactly %... Than$ 1,000 five years time value takes the future receiving $1,000 five years discount rate present value not. The concept that states an amount for any timeframe other than one.! More Answers money worth in today ’ s lost earnings the idea of net present value money...: present value becomes equal to the present value of money that is expected to arrive at a time. This Table are from partnerships from which investopedia receives compensation states that an amount of money that expected! Financial planning formula given below PV = CF / ( 1 + r ) t 1 of. Can pay you back$ 100 today or I can pay you $110 year. Bob knows the future amount that you expect to earn a rate return... 5,000 lump sum payment in five years from now % ) 3 2 earn. Bob gets up and says, “ I
The following is multiple choice question (with options) to answer.
Rs.20 is the true discount on Rs. 260 due after a certain time. What will be the true discount on the same sum due after half of the former time, the rate of interest being the same? | [
"10.4",
"11",
"14.8",
"15.4"
] | A | Sol.
S.I. on Rs. (260 - 20) for a given time = Rs. 20.
S.I. on Rs. 240 for half the time = Rs. 10.
T.D. on Rs. 250 = Rs. 10.
∴ T.D. on Rs. 260 = Rs. [10/250 * 260] = Rs. 10.40
Answer A |
AQUA-RAT | AQUA-RAT-37494 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
If the cost price of 165 chocolates is equal to the selling price of 150 chocolates, the gain percent is | [
"8",
"9",
"10",
"11"
] | C | Let C.P. of each chocolate be Re. 1.
Then, C.P. of 150 chocolates = Rs. 150; S.P. of 150 chocolates = Rs. 165.
Gain % = 15/150 * 100 = 10%
Answer:C |
AQUA-RAT | AQUA-RAT-37495 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
What number comes next?
284, 693, 712, 846, 937, ? | [
"113",
"231",
"245",
"128"
] | D | D
128
The numbers 28469371 are being repeated in the same sequence. |
AQUA-RAT | AQUA-RAT-37496 | ### Show Tags
19 May 2015, 12:37
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
### Show Tags
20 May 2015, 03:32
BrainLab wrote:
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
Dear BrainLab
Perfect logic but for easier calculation, you may want to work with ratio here (1/4 increase per annum) instead of percentages (25% increase per annum). Both convey the same thing but the equation
$$(\frac{5}{4})^4*X = 6250$$
will take lesser time to solve (especially if you know that $$5^4 = 625$$) than $$(1.25)^4*X = 6250$$
Hope this was useful!
Japinder
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Posts: 44373
Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
### Show Tags
18 Jan 2016, 23:36
Expert's post
1
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] 18 Jan 2016, 23:36
The following is multiple choice question (with options) to answer.
In a farm, 2000 seeds were planted wheat and 3000 rice seeds were planted. If 57 percent of the seeds in plot I germinated and 42 percent of the seeds in plot II germinated, what percent of the total number of planted seeds germinated? | [
" 45.5%",
" 46.5%",
" 48.0%",
" 49.5%"
] | C | Plot I: 57% of 2000 wheat seeds germinated, hence 0.57*2000 = 1140 seeds germinated.
Plot II: 42% of 3000 rice seeds germinated, hence 0.42*3000 = 1260 seeds germinated.
Thus (germinated)/(total) = (1140 + 1260)/(2000 + 3000) = 2400/5000 = 48% seeds germinated.
Answer: C. |
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