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

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

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.

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

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

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

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:

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

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

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!

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.

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!

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.