The number of stamps that Kaye and Alberto had were in the

Given: \(\frac{K}{A}=\frac{5x}{3x}\) (where \(x\) is a positive integer) and \(\frac{5x-10}{3x+10}=\frac{7}{5}\) --> \(x=30\) --> \(K=150\) and \(A=90\).

Question: \((K-10)-(A+10)=?\) --> \((K-10)-(A+10)=40\).

Answer: C.
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Another method is to use ratios:

Note that the total number of stamps is the same in both the cases.
On ratio scale, 5:3 gives 8 total stamps while 7:5 gives 12 total stamps. So total stamps must be at least LCM of 8 and 12 i.e. 24.
So in first case, Kaye had 15 stamps while Alberto had 9. After the exchange, Kaye had 14 stamps while Alberto had 10. So Kaye gave 1 stamp to Alberto. But actually, Kaye gave 10 stamps to Alberto. So actual total number of stamps is 240.
So the multiplier of 7:5 is 20 (because 12*20 = 240). Kaye had (7-5)*20 = 40 more stamps than Alberto.

Answer (C)
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so before exchange ratio is 5x:3x (x basically cancels out but they might have 10:6, 15:9 etc. which reduces to 5:3) so after exchange we have 5x-10:3x+10 = 7:5 You can represent this in fractions like:
(5x-10)/(3x+10) = 7/5 and cross multiply to get
25x-50 = 21x + 70 -> 4x = 120 x = 30
NOW 30 IS NOT THE ANSWER - that is the multiplier so you need to find the number of stamps in possession so...

Kaye 7*30 =210
Alberto 5*30 = 150
210-150 = 60 = D

Answer to this question is C (40), provided OA is wrong.

When we write \(\frac{K}{A}=\frac{5x}{3x}\) it means that \(x\) is the multiple of the initial ratio of 5/3 and not 7/5.

We found that \(x=30\): initially K had \(5*30=150\) stamps and now K has \(150-10=140\) stamps. Initially A had \(3*30=90\) stamps and now A has \(90+10=100\) stamps. Current difference is \(140-100=40\).

Hope it's clear.
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1. We know the initial ratio of the number of stamps. We can represent the initial number of stamps that Kaye and Alberto had, as 5x and 3x resp
2. We can represent the final number of stamps that Kaye and Alberto had as 5x-10 and 3x + 10 resp. The final ratio is 5x-10/3x+10
3. The final ratio is also given as 7:5
4. As (2) and (3) both represent the final ratio of number of stamps and so as both are equivalent we can equate the two. 5x-10/ 3x+10 = 7/5
5. x=30
6. We now need to find out how many stamps Kaye and Alberto had after the gift.
7. Kaye had 5x-10 = 140 stamps. Alberto had 3x-10 = 100 stamps
8. So Kaye has 40 more stamps than Alberto after the gift.

Correct choice is c.
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5+3 = 8
7+5= 12
LCM of 12 and 8 = 24
8*3 = 24. Thus (5*3) / (3*3) = 15/9
12*2 = 24. Thus (7*2) / (5*2) = 14/10
10 – 9 = 1unit
What we need is 14-10= 4
Therefore 10*4 = 40

Initially their counts is
5x:3x

After K gave 10 stamps to A their new ratio is
7y:5y

i.e. 5x-10=7y-----Eq 1
3x+10=5y ----- eq 2

Solving eq 1 qnd 2 we get y=20

so from 2nd ratio

7*20 - 5 *20 = 40

Once x = 30 is calculated, we require to calculate (5x-10) - (3x+10)
= 2x-20
= 60-20
= 40 = Answer = C

so you can think about it this way. For some number (x) Kaye had 5x stamps and Alberto had 3x
then kaye gave 10 stamps and the ration became 7:5, we can turn this into an equation like so:

(5x-10)/(3x+10)=7/5
cross multiplying gives us: 21x+70=25x-50 or 120=4x x=30
but remember, we are trying to figure out how many more stamps kaye had than alberto after she gave him ten.
originally, kaye had 150 stamps and alberto had 90
then she gave alberto 10, so she had 140 and he had 100
the difference is 40 stamps (C)

One option is to solve the question using TWO VARIABLES.
Let K = # of stamps K had after the exchange
Let A = # of stamps A had after the exchange
This means that K+10 = # of stamps K had before the exchange
This means that A-10 = # of stamps A had before the exchange

Note: Our goal is to find the value of K-A

The number of stamps that K and A (originally) had were in the ratio 5:3
So, (K+10)/(A-10) = 5/3
We want a prettier equation, so let's cross multiply to get 3(K+10) = 5(A-10)
Expand: 3K + 30 = 5A - 50
Rearrange: 3K - 5A = -80

After K gave A 10 of her stamps, the ratio of the number K had to the number A had was 7:5
So, K/A = 7/5
We want a prettier equation, so let's cross multiply to get 5K = 7A
Rearrange to get: 5K - 7A = 0

At this point we have two equations:
5K - 7A = 0
3K - 5A = -80

Our goal is to find the value of K - A.
IMPORTANT: We need not solve for the individual values of K and A. This is great, because something nice happens when we subtract the blue equation from the red equation.
We get: 2K - 2A = 80
Now divide both sides by 2 to get: K - A = 40

Answer:

Cheers,
Brent

Lets work -




So, \(\frac{(5x-10)}{(3x+10)} = \frac{7}{5}\)

Or, \(25x - 50 = 21x + 70\)

Or, \(4x = 120\)

Or, \(x = 30\)

We are asked to find -


Or, (5x-10) - (3x+10) => 2x - 20

Or, 2(30) - 20 =40

Hence answer is (C) 40
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We are given that the number of stamps that Kaye and Alberto had were in the ratio 5 : 3. We can represent this as:

K : A = 5x : 3x

We are next given that after Kaye gave Alberto 10 of her stamps, the ratio of the number Kaye had to the number Alberto had was 7 : 5. Using this information we can create the following equation:

(5x – 10)/(3x + 10)= 7/5

5(5x – 10) = 7(3x + 10)

25x – 50 = 21x + 70

4x = 120

x = 30

Kaye now has 5(30) – 10 = 140 stamps and Alberto has 3(30) + 10 = 100 stamps. So Kaye has 140 – 100 = 40 more stamps than Alberto.

Answer: C
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We are given that the number of stamps that Kaye and Alberto had was in the ratio 5 : 3. We can represent this as:

K : A = 5x : 3x

We are next given that after Kaye gave Alberto 10 of her stamps, the ratio of the number Kaye had to the number Alberto had was 7 : 5. Using this information, we can create the following equation:

(5x – 10)/(3x + 10) = 7/5

5(5x – 10) = 7(3x + 10)

25x – 50 = 21x + 70

4x = 120

x = 30

Kaye now has 5(30) – 10 = 140 stamps, and Alberto has 3(30) + 10 = 100 stamps. So Kaye has 140 – 100 = 40 more stamps than Alberto has.

Answer: C
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This is a fairly easy question on ratios, but with a big trap in terms of language. The most common answer that a lot of students get on this question is 60 i.e. option D. However, this is the trap answer.

Questions like these on ratios test your knowledge of mutliples of numbers. As such, although the conventional approach of assuming the stamps as 5x and 3x and then building an equation followed by solving the equation yields you the answer, it’s always good to approach ratio questions using numbers.

We know that the number of stamps that Kaye and Alberto had are multiples of 5 and 3 respectively. So, what do we do? Do we start with 5 and 3? The answer is a bigg no. Observe the options, all the options are bigger numbers than 5 and 3. And remember that we are trying to find the difference between the number of stamps. We need to take bigger multiples of 5 and 3, which are multiples of 10 as well (preferably).

Let’s try 50 and 30, then. After gifting 10 stamps to Alberto, the respective numbers with Kaye and Alberto would be 40 and 40. This is not in the ratio of 7:5. So, 50 and 30 are not the number of stamps.

If we try 100 and 60, we obtain the numbers 90 and 70 after the gifting process. This is in the ratio of 9:7, NOT 7:5.

When we try 150 and 90, we see that the resultant numbers after Kaye gifts alberto would be 140 and 100 which are in the ratio of 7:5. Therefore, Kaye must have started off with 150 stamps and Alberto with 90 stamps. So, Kaye had 60 more stamps than Alberto?? No, that’s not correct.

Observe that the question mentions “As a result of the gift…”. This means, we need to find out how much more Kaye had AFTER gifting 10 stamps to Alberto. AFTER gifting 10 stamps to Alberto, Kaye had 140 and Alberto had 100, so Kaye had 40 more stamps.

The correct answer option is C.

Every word, every phrase and every sentence in a GMAT question is valuable. If you rush through the questions without paying due attention, you run the risk of falling for the trap answers. If required, after you compute an answer, re-read the question statement and satisfy yourselves that you have indeed answered the actual question and NOT what you thought the question was.

Hope that helps!
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Hi VeritasKarishma

Two quick questions

Q1) Where can one learn more about this ratio methodology as i would like to inculcate this / practice some more problems using this methodology ?


Q2) Also -- can one assume, this ratio methodology can be done ONLY IF THE TOTAL remains the same ?

Reason is, In this problem also involving ratios -- the total has changed and thus ONLY ALGEBRA works in this problem

https://gmatclub.com/forum/the-present- ... 29648.html

Here are some videos to understand the "ratio methodology"

Ratios: https://youtu.be/5ODENGG5dvc
Weighted Averages: https://www.youtube.com/watch?v=_GOAU7moZ2Q
Mixtures: https://www.youtube.com/watch?v=VdBl9Hw0HBg
TSD: https://youtu.be/7ASEIvxYPCM
Work - Rate: https://youtu.be/88NFTttkJmA

and here are some posts:
https://anaprep.com/arithmetic-ratios-t ... ll-starts/
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
https://anaprep.com/arithmetic-applying ... -concepts/
https://anaprep.com/arithmetic-work-rat ... s-and-all/
https://anaprep.com/arithmetic-avoiding ... work-rate/
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K : A
before sharing 5 : 3 =8
after sharing 7 : 5 =12

Even after quantities are exchange overall quantity remain same so LCM of 8 and 12 is 24

Multiply before share quantity by 3 and after sharing by 2
Then apply unit law

5 : 3 => 15 : 9 =24
14 : 10=> 14 : 10=24

now 1 unit =10 (15-14=1 also 10-9=1)

Ans=(14-10)x10=40(14 and 10 are after sharing ratio)

Ans:C.40

Hi Bunuel ,

Why we cannot follow the above approach by user shaselai

Regards