number of stamps

[pbathia]

from GMAT Prep test 1

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

A) 20
b) 30
c) 40
d) 60
e) 90

I was able to come up with the correct answer (c), but it took me the better part of 3.5 minutes, can anyone suggest attacking this problem a better way?

[RonPurewal]

two ways to do the problem:
(1) use an unknown multiplier: let the original numbers of stamps be 5x for kaye and 3x for alberto. then after the exchange, kaye has 5x - 10 stamps, and alberto has 3x + 10.

therefore:
(5x - 10) / (3x + 10) = 7 / 5

cross multiply:
25x - 50 = 21x + 70
4x = 120
x = 30

plug back in to get
kaye's new total = 5x - 10 = 140
alberto's new total = 3x + 10 = 100
answer = c

--

you could also solve the problem by starting with the answer choices, and running them through the problem backwards. for instance, starting from the correct answer (c), you'd need to find two #s of stamps that are 40 apart, and are also in the ratio 7:5. those numbers are 140 and 100; you could then undo the exchange, finding that there were originally 150 and 90 stamps. since these numbers are indeed in the ratio 5:3 - which won't happen with any of the other answer choices - you must have started with the correct choice.

[Guest]

I was confused by this question as well. Ron - I like your answer as it is a much more efficient way of doing it then what I learned from the course.

This is what I learned from the course, that in these ratio problems you should setup 2 equations:

1). 3K = 5A
2). 5K - 50 = 7A + 70

Then when you solve this system of equations you get K = 150, and A = 90.

So a couple of questions here:
1). Should I abolish this way of thinking when it comes to ratio problems? (setting up 2 equations as opposed to using the unknown multiplier - or are there specific cases when to use both?)
2). I initially got the answer wrong because I subtracted 150 - 90 = 60. When in fact I should have gone back to the questions and subtracted 10 from K (150 - 10 = 140) and added 10 to A (90 + 10 = 100) then subtracted those 2 amounts to get 40. So my question is when setting up these equations do you always have to go back to the second equation and perform those addition/subtractions to the answer found ( I thought that once the equations were properly setup these calcs would already have been taken care.)

Thanks!

[rfernandez]

I'll chime in here in Ron's stead. Responding to your questions:

1) The nice thing about using the unknown multiplier technique is that you're able to create one equation in one variable. That's a big plus. One small minus about the strategy is that you're solving only for the multiplier -- you need to go back and plug that multiplier back in to find the amount you're interested in.

2) It's possible to set up the question differently from the start so that you don't have to adjust the result. It's all in how you define your variables:

7x is the number of stamps Kaye has after the transaction (so she had 7x + 10 beforehand)
5x is the number of stamps Alberto has after the transaction (so he had 5x - 10 beforehand)

Now, set up the equation as (7x + 10) / (5x - 10) = 5/3
3(7x + 10) = 5(5x - 10)
21x + 30 = 25x - 50
x = 20

So, Kaye has 140 stamps after the transaction and Alberto has 100 stamps after the transaction. 140 - 100 = 40.

Generally speaking, it's always a good idea to make sure you're answering the question that's being asked. You can ensure this by re-reading the question at the end of your work to make sure you've answered it.

Rey

[sumit_pune]

Hello, my $.02 for this problem

first ratio = 5:3
2nd = 7:5

now number of stamps will be same even after the exchange , i multiply first by 12 ( why bec 2nd one 7:5 = 12) and 2nd by 8 ( 5:3=8)

1st = 5:3*12 = 60:36
2nd = 7:5*8 = 56:40

these ratio represent exchange of 4 stamp ( can u see !!)
and the K has 16 more than the A; in this case , hence after 10 stamps exchange, K will have 16*2.5 = 40 more !!

this will give answer within minute.

[RonPurewal]

Hello, my $.02 for this problem

first ratio = 5:3
2nd = 7:5

now number of stamps will be same even after the exchange , i multiply first by 12 ( why bec 2nd one 7:5 = 12) and 2nd by 8 ( 5:3=8)

1st = 5:3*12 = 60:36
2nd = 7:5*8 = 56:40

these ratio represent exchange of 4 stamp ( can u see !!)
and the K has 16 more than the A; in this case , hence after 10 stamps exchange, K will have 16*2.5 = 40 more !!

this will give answer within minute.

that's an unconventional approach, but i think it's legit.

[danielle.wipperfurth]

Hi - Thanks for all this insight. Can someone please explain to me why you don't use x as a variable attached to both ratios - only to one?

i.e. why isn't the equation set up as follows?:

(5x - 10) / (3x + 10) = 7x / 5x

I think it has to do with the multiplier being different once the exchange is made but don't have perfect clarity.

[tim]

Ask yourself what x represents and you'll probably see the answer. If you still have a question, let us know!

[eric.d.giles]

Why can't you plug x = 30 into 7x/5x rather than the 5x/3x ratio? Since this is the new ratio, shouldn't we get the right answer from it?

[RonPurewal]

Why can't you plug x = 30 into 7x/5x rather than the 5x/3x ratio? Since this is the new ratio, shouldn't we get the right answer from it?

no. since that's a whole new ratio, you can't use the same multiplier -- i.e., you can't use the letter "x" anymore.

the correct numbers are a few posts up:
* before the exchange, kaye had 150 and alberto had 90.
* after the exchange, kaye had 140 and alberto had 100.

... so, if you write the pre-exchange (5:3 ratio) numbers as "5x" and "3x", then x = 30.
if the post-exchange numbers are written as "7y" and "5y", then y is 20, not 30.

the point here is that you can't use the same letter as a multiplier unless EVERYTHING is in one ratio.
i.e.,
* if you have four things in the ratio 3:5:5:7, then you can go ahead and call those 3x, 5x, 5x, and 7x.
* on the other hand, if you have two things in the ratio 3:5, and two totally different things in the ratio 5:7, then you would need two different multipliers: 3x and 5x, and then 5y and 7y.

(you can avoid having to use a second variable by expressing the second ratio as a proportion, since you already have expressions for those quantities. see above.
in general, you should basically never need more than 1 variable for a multiple-choice ratio problem.)