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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, 05:27

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55% (hard)

Question Stats:

57% (02:20) correct
43% (01:47) wrong based on 180 sessions

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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

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

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07 Jan 2013, 05:52

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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.

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. _________________

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

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07 Jan 2013, 06:27

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? _________________

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

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07 Jan 2013, 06:34

Expert's post

fozzzy wrote:

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?

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.

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

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07 Jan 2013, 10:12

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Expert's post

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

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 _________________

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

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05 Dec 2014, 11:19

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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, 08:33

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