Kay began a certain game with x chips

[Sinta]

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

{all <> signs are inclusive : < >}

While solving this I could not understand why is there more than one solution for x, and the gmatprep explanation was no help either...
any one?

Source: Gmat Prep Q Pack1
OA: D. 25<x<30
Screen shot + OA :http://uploadpic.net/ZFKE.png

[jnelson0612]

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

{all <> signs are inclusive : < >}

While solving this I could not understand why is there more than one solution for x, and the gmatprep explanation was no help either...
any one?

Source: Gmat Prep Q Pack1
OA: D. 25<x<30
Screen shot + OA :http://uploadpic.net/ZFKE.png

I would work backward on this. I would think okay, I'm left with five chips after subtracting half the chips and one additional chip. Let's add back the one additional chip. That makes six. Six must be half of what I had before this play, so I started this play with twelve. Double check: start with 12, subtract half and then one more. Left with 5--cool!

Okay, so now we're left with 12. Let's work backward again. Add the one that was left, which makes 13. 13 is half of what I started with, so I must have started with 26.

Notice that D is the only answer choice that contains this number, so I would pick it and move on. All the question is saying is that the correct answer is *contained* in one of these intervals; the question does not say that every integer in these intervals is a possible value.

I hope that this helps! :-)

[elle_piri]

x-[(1/2x)+1] - [1/2 (1/2x+1) + 1] = 5
simplify:
x - (x+2)/2 - (x+6)/4 = 5
Simplify more:
(4x-2x-4-x-6)/4 = 5 ===> x-10=20 ===> x=30

Is it correct?

[RonPurewal]

Is it correct?

it can't be, since you got an answer that disagrees with the correct value of 26 (see above).

your mistake is here:

x-[(1/2x)+1] - [1/2 (1/2x+1) + 1] = 5

the red part is the number of chips you lost the first time, but that's not what it's supposed to be; it should represent the number of chips you have remaining after the first loss.

by the way, when you are writing expressions like these on a forum, please write x/2, not "1/2x". (the latter makes it seem as though x is in the denominator.)

[geezer0305]

Is it correct?

it can't be, since you got an answer that disagrees with the correct value of 26 (see above).

your mistake is here:

x-[(1/2x)+1] - [1/2 (1/2x+1) + 1] = 5

the red part is the number of chips you lost the first time, but that's not what it's supposed to be; it should represent the number of chips you have remaining after the first loss.

by the way, when you are writing expressions like these on a forum, please write x/2, not "1/2x". (the latter makes it seem as though x is in the denominator.)

Hi,

I used the same method, however, I am not clear on why it is incorrect.

Number of chips remaining after 2 plays = (Total number of chips at the beginning of the game) - (total chips lost in the first play) - (total chips lost in the second play)

Number of chips remaining after 2 plays = 5
Total number of chips at the beginning of the game = x
total chips lost in the first play = (x/2 + 1)
total chips lost in the second play = (1/2*(x/2 +1) + 1)

on simplifying we get the answer as 30. Option D is inclusive of 30.

Am i slipping somewhere?

[RonPurewal]

you're missing a step. i'll show you where.

Number of chips remaining after 2 plays = 5
Total number of chips at the beginning of the game = x
total chips lost in the first play = (x/2 + 1)

that's the total chips LOST in the first play.

what you haven't done yet -- and need to do -- is calculate how many chips you still HAVE after that play.
that would be x - (x/2 + 1), which simplifies to x/2 - 1.

total chips lost in the second play = (1/2*(x/2 +1) + 1)

the plus sign that's green, there, should therefore be a minus sign instead.

then, you have to subtract this value from the number you started this step with (= x/2 - 1), to find the number of chips you HAVE after the second loss.

if you set that result equal to 5, you can solve for x = 26.

[RonPurewal]

in any case, whenever you have a problem with particularly obnoxious algebra -- like this one -- you should try to find OTHER, non-algebraic ways to solve the problem.

in this case, even though the correct answers are stated as intervals instead of numbers, it should be clear that there's going to be a single x that solves the problem (because increasing the value of x will always increase the number of chips left over).

so, you can just try a random even number for x, and then adjust up or down depending on how that turns out.
(you know that x has to be even, because one more than half of x has to be a whole number of chips. if x is odd, then you have to lose a half-chip, which isn't reasonable.)

so ... well, looking at the choices, let's take x = 22, because that looks like it's somewhere in the middle.
then, in the first try, i lose 11 + 1 = 12 chips, leaving me with 10.
then, in the second try, i lose 5 + 1 = 6 chips, leaving me with 4.
that's not enough, so 22 is too small.
try a bigger number (not too much bigger, as 4 is not that far away from 5).
if you try 24, you'll lose a half-chip on the second turn. if you try 26, you'll get the right answer.