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Raj played three rounds of a game of chance. In each round

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Raj played three rounds of a game of chance. In each round [#permalink] New post 15 Nov 2012, 22:02
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Raj played three rounds of a game of chance. In each round he doubled the amount with himself and then gave a certain amount to his friend. The amount he gave his friend in each round was a quarter of his gain in the first round. If he had instead given to his friend in each round an amount equal to half of his gain in the first round he would have finally remained with rs 70 less than he actually did. Find the amount he started with in rs?

A. 80
B. 160
C. 40
D. 60
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Nov 2012, 03:02, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Raj played three rounds of a game of chance. In each round [#permalink] New post 16 Nov 2012, 03:02
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siddharthamaridi wrote:
Raj played three rounds of a game of chance. In each round he doubled the amount with himself and then gave a certain amount to his friend. The amount he gave his friend in each round was a quarter of his gain in the first round. If he had instead given to his friend in each round an amount equal to half of his gain in the first round he would have finally remained with rs 70 less than he actually did. Find the amount he started with in rs?

A. 80
B. 160
C. 40
D. 60


Say Raj has x rs.

Case when he gives to his friend in each round a quarter of his gain in the first round:
The amount he has after the first round: \(2x-\frac{x}{4}=\frac{7x}{4}\);
The amount he has after the second round: \(2*\frac{7x}{4}-\frac{x}{4}=\frac{13x}{4}\);
The amount he has after the third round: \(2*\frac{13x}{4}-\frac{x}{4}=\frac{25x}{4}\).

Case when he gives to his friend in each round half of his gain in the first round:
The amount he has after the first round: \(2x-\frac{x}{2}=\frac{3x}{2}\);
The amount he has after the second round: \(2*\frac{3x}{2}-\frac{x}{2}=\frac{5x}{2}\);
The amount he has after the third round: \(2*\frac{5x}{2}-\frac{x}{2}=\frac{9x}{2}\).

We are told that the difference is 70, thus \(\frac{25x}{4}-\frac{9x}{2}=70\) --> \(x=40\).

Answer: C.
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Re: Raj played three rounds of a game of chance. In each round [#permalink] New post 08 Dec 2012, 04:54
sometimes the wording of the questions confuse me

:cry:
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Re: Raj played three rounds of a game of chance. In each round [#permalink] New post 14 Aug 2013, 01:31
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Re: Raj played three rounds of a game of chance. In each round [#permalink] New post 29 Sep 2014, 05:20
Bunuel wrote:
Bumping for review and further discussion.



this will be relatively easy if we chose 4x as the amount he starts with as we know that we want to give the quarter of the amount he won in the first round to his friend.

so
at start he has 4x

we doubled that in first round so now has 8x ( so he won 4x ) and quarter of that is x , this x is waht he gives to his friend ...so he is left with ( 4x + 4x -x) = 7x

after second round 14 x after third round 28 x ....and he gives x+x+x = 3x to his frnd so left with 25 x

second case when he gives double of what he won to his frnd

at start 4x

he doubles it to 8x ...so half of what he won he gives to his frnd i.e 2x so end of first round he is left with ........8x - 2x =6x

after second round he doubles to 12 x and then aftewr third 24 x ..but he gives 2x+2x+2x to his frnd so he is left with 24-6 = 18 x


now 25x-18 x = 70

x = 10

so initial amount = 40
Re: Raj played three rounds of a game of chance. In each round   [#permalink] 29 Sep 2014, 05:20
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