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From the total amount available, a man keeps 25,000$ for

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From the total amount available, a man keeps 25,000$ for [#permalink] New post 16 Sep 2012, 01:54
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From the total amount available, a man keeps 25,000$ for himself and then distributes the remaining between two of his sons in the ratio of 3:2 (3 parts for the elder and 2 parts for the younger son). Later, he decides to give the dollar 25,000 (which he had initially kept for himself) to his younger son. This makes the ratio of amount with elder son to ratio of amount with younger son 2:3. Find the amount received by the elder brother.


A. 30,000
B. 100,000
C. 25,000
D. 40,000
E. 500
[Reveal] Spoiler: OA

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Last edited by Bunuel on 16 Sep 2012, 01:56, edited 1 time in total.
Renamed the topic.
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Re: From the total amount available, a man keeps 25,000$ for [#permalink] New post 16 Sep 2012, 02:08
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Pansi wrote:
From the total amount available, a man keeps 25,000$ for himself and then distributes the remaining between two of his sons in the ratio of 3:2 (3 parts for the elder and 2 parts for the younger son). Later, he decides to give the dollar 25,000 (which he had initially kept for himself) to his younger son. This makes the ratio of amount with elder son to ratio of amount with younger son 2:3. Find the amount received by the elder brother.
A. 30,000
B. 100,000
C. 25,000
D. 40,000
E. 500


There are two conventional algebraic ways to solve these types of problems. In the first, we just introduce an unknown for the amounts each brother received. We then use the fact that a ratio is just a fraction in order to translate each statement in the question into algebra:

If the elder brother initially got $e, and the younger brother initially got $y, then from the ratio given, we know that e/y = 3/2, or 2e = 3y. Further, if the younger brother is given $25,000, he will then have y + 25000 dollars. We know the ratio of e to y+25000 is 2 to 3, so e/(y + 25000) = 2/3, or 3e = 2y + 50000. We now have two equations in two unknowns:

3e = 2y + 50,000
2e = 3y

If we multiply the first equation by 3 and the second equation by 2, we can then subtract the second from the first:

9e = 6y + 150,000
4e = 6y
5e = 150,000

So e = 30,000.

It's faster to still to use a multiplier. If the ratio of the amounts given to the elder and younger brothers is 3 to 2, then for some number x, the elder brother got $3x and the younger brother got $2x. We want to find $3x. Since the ratio of 3x to 2x+25,000 is 2 to 3, we have

3x/(2x + 25,000) = 2/3
9x = 4x + 50,000
5x = 50,000
x = 10,000

And since we wanted to find 3x, the answer is 30,000.

Finally, you can solve this in a kind of conceptual way. If we rewrite each ratio so that the elder brother's amount is the same in each we have:

* before any money is transferred, ratio of elder's $ to younger's $: 6 to 4
* after the money is transferred, ratio of elder's $ to younger's $: 6 to 9

So the $25,000 transfer is equivalent to 5 parts in the ratio (the difference between 9 and 4). Since the amount the elder brother has is equivalent to 6 parts, his amount is $30,000.

And I suppose you could also backsolve the question fairly easily, though if the numbers were different, that could be a very bad approach.
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Re: From the total amount available, a man keeps 25,000$ for [#permalink] New post 16 Sep 2012, 07:09
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IanStewart wrote:
Pansi wrote:
From the total amount available, a man keeps 25,000$ for himself and then distributes the remaining between two of his sons in the ratio of 3:2 (3 parts for the elder and 2 parts for the younger son). Later, he decides to give the dollar 25,000 (which he had initially kept for himself) to his younger son. This makes the ratio of amount with elder son to ratio of amount with younger son 2:3. Find the amount received by the elder brother.
A. 30,000
B. 100,000
C. 25,000
D. 40,000
E. 500


There are two conventional algebraic ways to solve these types of problems. In the first, we just introduce an unknown for the amounts each brother received. We then use the fact that a ratio is just a fraction in order to translate each statement in the question into algebra:

If the elder brother initially got $e, and the younger brother initially got $y, then from the ratio given, we know that e/y = 3/2, or 2e = 3y. Further, if the younger brother is given $25,000, he will then have y + 25000 dollars. We know the ratio of e to y+25000 is 2 to 3, so e/(y + 25000) = 2/3, or 3e = 2y + 50000. We now have two equations in two unknowns:

3e = 2y + 50,000
2e = 3y

If we multiply the first equation by 3 and the second equation by 2, we can then subtract the second from the first:

9e = 6y + 150,000
4e = 6y
5e = 150,000

So e = 30,000.

It's faster to still to use a multiplier. If the ratio of the amounts given to the elder and younger brothers is 3 to 2, then for some number x, the elder brother got $3x and the younger brother got $2x. We want to find $3x. Since the ratio of 3x to 2x+25,000 is 2 to 3, we have

3x/(2x + 25,000) = 2/3
9x = 4x + 50,000
5x = 50,000
x = 10,000

And since we wanted to find 3x, the answer is 30,000.

Finally, you can solve this in a kind of conceptual way. If we rewrite each ratio so that the elder brother's amount is the same in each we have:

* before any money is transferred, ratio of elder's $ to younger's $: 6 to 4
* after the money is transferred, ratio of elder's $ to younger's $: 6 to 9

So the $25,000 transfer is equivalent to 5 parts in the ratio (the difference between 9 and 4). Since the amount the elder brother has is equivalent to 6 parts, his amount is $30,000.

And I suppose you could also backsolve the question fairly easily, though if the numbers were different, that could be a very bad approach.


For the conceptual solution, the attached drawing can be helpful.
Initially, elder son gets 6 parts and younger son gets 4 parts. Then, the $25,000 received by the younger son is represented by the 5 short line segments.
One segment represents 25,000/5 = 5,000. So, elder son gets 6*5,000 = 30,000.
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Ratios-Concept(IanStewart).jpg
Ratios-Concept(IanStewart).jpg [ 13.84 KiB | Viewed 1484 times ]


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Re: From the total amount available, a man keeps 25,000$ for [#permalink] New post 26 Feb 2013, 20:25
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Work back from options,
A. 30000 implies , younger stands at 20000,
so when gifted another 25000, ratio stands at 30000:45000::2:3

Hence A.
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Re: From the total amount available, a man keeps 25,000$ for [#permalink] New post 16 Sep 2012, 02:10
Pansi wrote:
From the total amount available, a man keeps 25,000$ for himself and then distributes the remaining between two of his sons in the ratio of 3:2 (3 parts for the elder and 2 parts for the younger son). Later, he decides to give the dollar 25,000 (which he had initially kept for himself) to his younger son. This makes the ratio of amount with elder son to ratio of amount with younger son 2:3. Find the amount received by the elder brother.


A. 30,000
B. 100,000
C. 25,000
D. 40,000
E. 500


Let us say Elder got 3x and younger 2x. So after addition of 25,000 the ratio changes to 2:3. we can write that as follows:

3x/(2x+25000) = 2/3
=> 9x = 4x + 50000
=> 5x = 50000
=>x=10000

Therefore initial amount of Elder ie 3x = 30000

Hence Correct Choice is A.
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Re: From the total amount available, a man keeps 25,000$ for [#permalink] New post 26 Feb 2013, 20:07
I spend around 20 mins trying to solve the problem and the answer is 20,000 then I realized that I was solving for the youngest son. LOL :( I Keep doing these careless mistakes
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Re: From the total amount available, a man keeps 25,000$ for [#permalink] New post 24 May 2013, 18:30
EvaJager wrote:

For the conceptual solution, the attached drawing can be helpful.
Initially, elder son gets 6 parts and younger son gets 4 parts. Then, the $25,000 received by the younger son is represented by the 5 short line segments.
One segment represents 25,000/5 = 5,000. So, elder son gets 6*5,000 = 30,000.


Awesome conceptual solution. Thanks!
Re: From the total amount available, a man keeps 25,000$ for   [#permalink] 24 May 2013, 18:30
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