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Kay began a certain game with x chips. On each of the next

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Kay began a certain game with x chips. On each of the next [#permalink] New post 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
[Reveal] Spoiler: OA

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Re: Kay began a certain game with x chips. On each of the next [#permalink] New post 07 Jan 2013, 04:52
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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.

Hope it's clear.
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Re: Kay began a certain game with x chips. On each of the next [#permalink] New post 07 Jan 2013, 04:52
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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




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Re: Kay began a certain game with x chips. On each of the next [#permalink] New post 07 Jan 2013, 04:58
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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


At the beginning she had 26 chips not 23. Check here: kay-began-a-certain-game-with-x-chips-on-each-of-the-next-145373.html#p1165407

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] New post 07 Jan 2013, 05:04
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Bunuel wrote:
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


At the beginning she had 26 chips not 23. Check here: kay-began-a-certain-game-with-x-chips-on-each-of-the-next-145373.html#p1165407

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.



Got it :) N+1.........

But here is a risk to be a bit careless or depends on the question posed ?? I mean the logic was correct, after all

Thanks for suggesting
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Re: Kay began a certain game with x chips. On each of the next [#permalink] New post 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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Re: Kay began a certain game with x chips. On each of the next [#permalink] New post 07 Jan 2013, 05:34
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.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Kay began a certain game with x chips. On each of the next [#permalink] New post 07 Jan 2013, 09:12
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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


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] 07 Jan 2013, 09:12
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