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Kay began a certain game with x chips. On each of the next [#permalink]

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07 Jan 2013, 04:27

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Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:

A. 7<x<12 B. 13<x<18 C. 19<x<24 D. 25<x<30 E. 31<x<35

Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:

A. 7<x<12 B. 13<x<18 C. 19<x<24 D. 25<x<30 E. 31<x<35

On the first play she lost \(\frac{x}{2}+1\) chips and she was left with \(x-(\frac{x}{2}+1)=\frac{x-2}{2}\) chips.

On the second play she lost \(\frac{x-2}{4}+1\) chips.

So, we have that \(x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5\) --> \(x=26\).

Answer: D.

We can also solve this question backward: At the end Kay had 5 chips, so before that she had (5+1)*2=12 chips: 12-(12/2+1)=5. The same way, she had 12 chips, so before that she had (12+1)*2=26 chips: 26-(26/2+1)=12.

Notice that x cannot be odd, because she lost one more than half the number of chips, which means that the number of chips must be even.
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Re: Kay began a certain game with x chips. On each of the next [#permalink]

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07 Jan 2013, 05:27

Bunuel wrote:

fozzzy wrote:

Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:

A. 7<x<12 B. 13<x<18 C. 19<x<24 D. 25<x<30 E. 31<x<35

On the first play she lost \(\frac{x}{2}+1\) chips and she was left with \(x-(\frac{x}{2}+1)=\frac{x-2}{2}\) chips.

On the second play she lost \(\frac{x-2}{4}+1\) chips.

So, we have that \(x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5\) --> \(x=26\).

Answer: D.

We can also solve this question backward: At the end Kay had 5 chips, so before that she had (5+1)*2=12 chips: 12-(12/2+1)=5. The same way, she had 12 chips, so before that she had (12+1)*2=26 chips: 26-(26/2+1)=12.

Hope it's clear.

I didn't understand this step did you divide by 2?
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Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:

A. 7<x<12 B. 13<x<18 C. 19<x<24 D. 25<x<30 E. 31<x<35

On the first play she lost \(\frac{x}{2}+1\) chips and she was left with \(x-(\frac{x}{2}+1)=\frac{x-2}{2}\) chips.

On the second play she lost \(\frac{x-2}{4}+1\) chips.

So, we have that \(x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5\) --> \(x=26\).

Answer: D.

We can also solve this question backward: At the end Kay had 5 chips, so before that she had (5+1)*2=12 chips: 12-(12/2+1)=5. The same way, she had 12 chips, so before that she had (12+1)*2=26 chips: 26-(26/2+1)=12.

Hope it's clear.

I didn't understand this step did you divide by 2?

On the second play she also lost one more than half the number of chips she had at the beginning of that play. Since at the begging of the second play she had \(\frac{x-2}{2}\) chips, then she lost \(\frac{(\frac{x-2}{2})}{2}+1=\frac{x-2}{4}+1\) chips.

Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:

A. 7<x<12 B. 13<x<18 C. 19<x<24 D. 25<x<30 E. 31<x<35

The most important thing to understand here is this: she loses one more than half the number she has. This implies that at the end of a play, she is left with one less than half the number she has at the beginning. After two plays, she is left with 5 (which is 1 less than half of what she had at the beginning of the second play). So she had 6*2 = 12 at the end of the first play. Since 12 is one less than half of what she had at the beginning of the first play, she must have had 13*2 = 26 at the beginning of the first play. Hence, x = 26
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Re: Kay began a certain game with x chips. On each of the next [#permalink]

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05 Dec 2014, 10:19

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Re: Kay began a certain game with x chips. On each of the next [#permalink]

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16 Dec 2015, 07:33

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Kay began a certain game with x chips. On each of the next [#permalink]

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02 Nov 2016, 18:18

fozzzy wrote:

Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:

A. 7<x<12 B. 13<x<18 C. 19<x<24 D. 25<x<30 E. 31<x<35

Here's how I solved it: At first she had x chips After the first play, she had x/2 -1 chips After the second play, she had 1/2 (x/2 -1)-1 which equals 5. Solving for x, we get: 1/2 (x/2 -1)-1 =5 1/2 (x/2 -1) = 6 x/2 -1 =12 x/2 =13 x =26

gmatclubot

Re: Kay began a certain game with x chips. On each of the next
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02 Nov 2016, 18:18

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