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Ratio Problem

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Ratio Problem [#permalink] New post 31 Mar 2007, 05:47
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Can anyone help with this one?

The number of stamps that Kaye and Alberto had respectively were in the ratio 5:3. After Kaye gave Alberto 10 of her stamps, the ratio went to 7:5. As a result of this gift, Kaye had how many more stamps than Alberto?
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Data Sufficiency Ratio problem [#permalink] New post 31 Mar 2007, 07:24
If m, r, x, and y are positive, is the ratio of m:r = the ratio of x:y?

(1) The ratio of m to y is equal to the ratio of x to r
(2) The ratio of m + x to r + y is equal to the ratio of x to y
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 [#permalink] New post 31 Mar 2007, 10:59
The number of stamps that Kaye and Alberto had respectively were in the ratio 5:3. After Kaye gave Alberto 10 of her stamps, the ratio went to 7:5. As a result of this gift, Kaye had how many more stamps than Alberto?

Equation 1: K/A = 5/3
Equation 2 (after the gift): (K-10)/(A+10) = 7/5

You have 2 equations with 2 variables, so you can solve for K and A.
A=90 and K=150. But the question asks for the number of stamps as a result of the gift:
A+10 = 100
K-10 = 140
-> Kaye had 40 more stamps than Alberto
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 [#permalink] New post 31 Mar 2007, 11:21
If m, r, x, and y are positive, is the ratio of m:r = the ratio of x:y?

(1) The ratio of m to y is equal to the ratio of x to r
(2) The ratio of m + x to r + y is equal to the ratio of x to y

Solution:

(1) if m=1, y=2, x=1, r=2 (complies to the condition m/y=x/r) then
m/r=x/y
BUT if m=5, y=15, x=1, r=3 (also complies to the condition m/y=x/r)
then m/r<>x/y ---> Not sufficient

(2) if x/y = (m+x)/(r+y) then x/y = m/r
You can try any selection of values for x,y,m, and r
e.g. x=5, y=20, m=1, r=4
---> Sufficient

My answer, B
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[#permalink] New post 31 Mar 2007, 11:24
If m, r, x, and y are positive, is the ratio of m:r = the ratio of x:y?

(1) The ratio of m to y is equal to the ratio of x to r
(2) The ratio of m + x to r + y is equal to the ratio of x to y


(1) m/y = x/r
Cross multiplying we get m*r = x*y
NOT SUFF

(2) (m+x)/(r+y) = x/y

Cross multiplying we get m*y +x*y = r*x + x*y

Cancelling out x*y from both sides m*y = r*x OR m/r = x/y

SUFF

Hence (B)
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 [#permalink] New post 01 Apr 2007, 01:22
Thank you for the answers, I have always had to think a while about these questions, now that I know how to set them up they are quite trivial :)
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 [#permalink] New post 01 Apr 2007, 04:07
Can anyone help with this one?

The number of stamps that Kaye and Alberto had respectively were in the ratio 5:3. After Kaye gave Alberto 10 of her stamps, the ratio went to 7:5. As a result of this gift, Kaye had how many more stamps than Alberto?

Assuming initial ratio co-efficient as x and it is y afterwards so initially kaye and alberto had 5X and 3X stamps respectively. After the exchange, they had 7Y and 5Y stamps respectively. since kaye gave 10 to Alberto,it can be derived that 5X-7Y=10 & 5Y-3X =10

from the above equations Y =20 and X=30 so Kaye had 60 more stampsearlier and 40 more stamps after exchange

This can be solved by other methods also.
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 [#permalink] New post 01 Apr 2007, 04:12
If m, r, x, and y are positive, is the ratio of m:r = the ratio of x:y?

(1) The ratio of m to y is equal to the ratio of x to r
(2) The ratio of m + x to r + y is equal to the ratio of x to y

from 1 => m/y =x/r which doesn't give the answer of question
from 2 => m+x/r+y =x/y =>my+xy = rx +xy => my =rx so m/r =x/y

and it answers the question so i would go with B

there is a standard theorem also that if a/b =c/d then both are equal to a+c/b+d here it is given that a/b =a+b/c+d which is only possible when a/b =c/d
  [#permalink] 01 Apr 2007, 04:12
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