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 <equal x <equal 12
B) 13<equal x <equal 18
C) 19<equal x <equal 24
D) 25<equal x <equal 30
E) 31<equal x <equal 35
OAD
Please explain.
Many thanks in advance.
Kavin
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Kay began a certain game with x chips
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DavidG@VeritasPrep
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You could test the answer choices. If she loses one more than 1/2 of the original amount, we know that she must have started with an even number of chips, otherwise, we'd be dealing with non-integer values.Needgmat 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 <equal x <equal 12
B) 13<equal x <equal 18
C) 19<equal x <equal 24
D) 25<equal x <equal 30
E) 31<equal x <equal 35
OAD
Please explain.
Many thanks in advance.
Kavin
Try C. Pick an even number in the interval. Say x = 22. If she loses one more than 1/2 of her chips, she'll lose 11 + 1 = 12 chips, leaving her with 10 chips. If she has 10 chips to start round 2, and again, she loses one more than 1/2 of her chips, she'll lose 5 + 1 = 6 chips, leaving her with 4. But we know she has 5. (Notice that if you were to test another value in this interval and set x = 24, you'd end up with a non-integer value after the second round.)
So Try D. Say x = 26. If she loses one more than 1/2 of her chips, she'll lose 13 + 1 = 14 chips, leaving her with 12 chips. If she has 12 chips to start round 2, and again, she loses one more than 1/2 of her chips, she'll lose 6 + 1 = 7 chips, leaving her with 5. And so D is our answer.
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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
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
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DavidG@VeritasPrep
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And there's always good old-fashioned algebra.Needgmat 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 <equal x <equal 12
B) 13<equal x <equal 18
C) 19<equal x <equal 24
D) 25<equal x <equal 30
E) 31<equal x <equal 35
OAD
Please explain.
Many thanks in advance.
Kavin
Round 1: If we start with x, and we lose 1 more than 1/2 of those chips, we'll lose (1/2)x + 1 chips, leaving us with x - [(1/2)x + 1] or (1/2)x - 1 chips.
Round 2: If we were to lose 1 more than 1/2 of those chips, we'd lose [(1/2)x - 1]/2 + 1 chips. Simplifying, this expression, we're losing (1/4)x - (1/2) + 1 or (1/4)x + 1/2 chips. Because we started this round with (1/2)x - 1 chips, we'd be left with (1/2)x - 1 - [(1/4)x + 1/2] chips. Simplify to get (1/4)x - 3/2.
So (1/4)x - 3/2 = 5, or (1/4)x = 13/2, and x = 52/2 = 26.
