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

Question

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

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

PS04475

Bunuel · Accepted Answer

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.

KarishmaB · Answer

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

Carcass · Answer

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

Bunuel · Answer

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.

Carcass · Answer

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

fozzzy · Answer

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

Bunuel · Answer

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.

ninjawannabe · Answer

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

ms17 · Answer

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

dimson · Answer

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

EMPOWERgmatRichC · Answer

Hi Kavin,

This question can be solved by TESTing THE ANSWERS.

The prompt tells us that Kay lost 1 MORE than HALF of her chips each 'play' and ended up with 5 chips after 2 'plays', so the number of starting chips has to be more than 20 (since half of 20 is 10 and half of 10 is 5). In addition, the starting number of chips CANNOT be an odd number, since you would end up with a 'fraction of a chip' at some point, which is not possible.

Let's TEST Answer C. The upper-end of that range is 24...

IF...Kay started with 24 chips....

Half+1 of 24 = 12+1 = 13 which leaves 11 chips 
Half+1 of 11 = 5.5 + 1 = 6.5 which leaves 4.5 chips. This is TOO SMALL (it's supposed to be 5 chips).

This example is pretty close to what we're looking for though, so let's try a number just a little bigger...

Answer D.

IF...Kay started with 26 chips....

Half+1 of 26 = 13+1 = 14 which leaves 12 chips 
Half+1 of 12 = 6 + 1 = 7 which leaves 5 chips. This is an exact MATCH for what we were told, so this MUST be the answer.

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

lucajava · Answer

x - (1+\frac{x}{2}) - (1 + \frac{x}{2} - \frac{1}{2} - \frac{x}{4}) = 5

x - x - 2 + \frac{1}{2} + \frac{x}{4} = 5

\frac{x}{4} = 7 - \frac{1}{2}

x = 26

Gmat20201 · Answer

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

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

Soln:

Lets make it simple and quick to avoid mistake and save time. Imp part is to understand question stem well .
 
Question says that 
Kay began a game with- x chips ( start value)

Played two games- Lost one more than half chips in each game 
-So game 1 started with  x chips  : lost x/2+ 1 and was left with L1 chips ( after losing half plus one chips)
 and game 2 started with  L1 chips : lost L1/2+1 and was left with L2 chips
 
 Given-5 chips remaining after her two plays , which means our  L2=5

Lets put all together to get value of x ( start value).We go in reverse order since we know value of L2 we count L1 first and than count X

L1 were the chips when game 2 was started.

L1= chips losts + chips left after game 2
 = L1/2+1+L2 
 = L1/2+1+5 ( since L2=5)
2L1= L1+12

solving this we get L1=12 
 
 x = x/2+1+L1 
 = x/2+1+ 12
 2x = x+26

So x=26

Ans choice D

Hope this helps!

Basshead · Answer

She lost  one more  than  half the number  of chips.

Since she ended with 5, we have to add one and then double this number. 5 + 1 = 6; 6 * 2 = 12

Since there were two plays, we must do this process again: 12 + 1 = 13 * 2 = 26

She started with 26, which is Answer D.

Rebaz · Answer

This how i reverse engineer this question:

At the end of her 2nd play she ends up with 5 chips.

5 is one less than half of what she had, meaning she had  12 chips .

50% of 12 chips is 6 chips and she had one fewer than 6, which is 5.

Now we reverse engineer how many chips she had at beginning of her first game which (X) numbers of chips as indicated by the question stem.

12 chips is one less that half of what she had, meaning she had  26  chips and here is why:

50% of 26 chips is 13 chips and she had one fewer than 13, which is 12.

Thus X=26

At the beginning of the game: X=26
At the end of the first play, she has 12 chips 
At the end of her second play, she has 5 chips.

26-------12 (12 is one fewer than half of 26) --------5  (5 is one fewer than half of 12)

The answer on this question is  D

I hope it is clear!

DanTheGMATMan · Answer

Best way is just to build it in reverse starting from 5. Also have the algebraic setup on here for fun:

https://www.youtube.com/watch?v=Tq_13bQ9YCQ

bumpbot · Answer

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

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

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