It is currently 16 Jan 2018, 13:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Director
Joined: 29 Nov 2012
Posts: 861

Kudos [?]: 1535 [4], given: 543

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

### Show Tags

07 Jan 2013, 04:27
4
KUDOS
10
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

61% (01:17) correct 39% (01:39) wrong based on 379 sessions

### HideShow timer Statistics

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

_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Kudos [?]: 1535 [4], given: 543

Math Expert
Joined: 02 Sep 2009
Posts: 43296

Kudos [?]: 139205 [5], given: 12779

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

### Show Tags

07 Jan 2013, 04:52
5
KUDOS
Expert's post
8
This post was
BOOKMARKED
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 $$x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5$$ --> $$x=26$$.

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.
_________________

Kudos [?]: 139205 [5], given: 12779

Board of Directors
Joined: 01 Sep 2010
Posts: 3451

Kudos [?]: 9750 [0], given: 1237

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

### Show Tags

07 Jan 2013, 04:52
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
_________________

Kudos [?]: 9750 [0], given: 1237

Math Expert
Joined: 02 Sep 2009
Posts: 43296

Kudos [?]: 139205 [1], given: 12779

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

### Show Tags

07 Jan 2013, 04:58
1
KUDOS
Expert's post
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.
_________________

Kudos [?]: 139205 [1], given: 12779

Board of Directors
Joined: 01 Sep 2010
Posts: 3451

Kudos [?]: 9750 [0], given: 1237

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

### Show Tags

07 Jan 2013, 05:04
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
_________________

Kudos [?]: 9750 [0], given: 1237

Director
Joined: 29 Nov 2012
Posts: 861

Kudos [?]: 1535 [0], given: 543

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

### Show Tags

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 $$x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5$$ --> $$x=26$$.

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?
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Kudos [?]: 1535 [0], given: 543

Math Expert
Joined: 02 Sep 2009
Posts: 43296

Kudos [?]: 139205 [0], given: 12779

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

### Show Tags

07 Jan 2013, 05:34
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 $$x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5$$ --> $$x=26$$.

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.
_________________

Kudos [?]: 139205 [0], given: 12779

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7862

Kudos [?]: 18450 [7], given: 237

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

### Show Tags

07 Jan 2013, 09:12
7
KUDOS
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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 18450 [7], given: 237

Intern
Joined: 03 Jul 2016
Posts: 1

Kudos [?]: [0], given: 24

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

### Show Tags

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

Kudos [?]: [0], given: 24

Intern
Joined: 24 Apr 2016
Posts: 11

Kudos [?]: [0], given: 203

Location: Germany
Concentration: Finance
WE: Analyst (Other)
Kay began a certain game with x chips. On each of the next [#permalink]

### Show Tags

06 May 2017, 10:23
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

Kudos [?]: [0], given: 203

Intern
Joined: 03 Dec 2016
Posts: 16

Kudos [?]: 1 [0], given: 19

Location: United States
Concentration: Strategy, Finance
GPA: 3.97
WE: Business Development (Consulting)
Kay began a certain game with x chips. On each of the next [#permalink]

### Show Tags

23 Nov 2017, 10:27
x - Beginning amount of chips

After playing 1st game: x*(1/2)-1

After playing 2nd game: (x*(1/2)-1)*(1/2)-1

Then we have to solve for x, given that we know that "she had 5 chips remaining after her two plays".

(x*(1/2)-1)*(1/2)-1=5
x*(1/2)-1=12
x=26

Ans: D

Kudos [?]: 1 [0], given: 19

Kay began a certain game with x chips. On each of the next   [#permalink] 23 Nov 2017, 10:27
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.