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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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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, 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?
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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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02 Nov 2016, 19:18

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

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

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06 May 2017, 11:23

carcass wrote:

well here is difficult to attack the question upfront, but reading carefully the stem you can do.

she lost one more than half the number of chips she had at the beginning of that play and she ends with 5 chips.

Work in reverse engineering. If she finishes with 5 chips and after each game she lost the half of chip + 1. 5*2= 10 + 1 =11 then 11*2=22 +1 =23

At the beginning she has 23 chips. So answer must be D

Regards

You also have to do the calculation steps in the reversed order.

To get to the 5 chips at the end of the second play, you must divide the number of chips after the first play by 2 and subtract 1 from it. Therefore to get the number of chips after the first play, just do everything in reverse, e.g. add 1 to the 5 chips and multiple the result by 2 to get 12. Therefore, 12 chips were left after the first play. Just repeat this process to get to the answer: (12+1)*2=26