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by yumi2012 » Sun Aug 18, 2013 8:26 pm
An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?

A. $31
B. $29
C. $26
D. $23
E. $21
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Source: — Problem Solving |

by Brent@GMATPrepNow » Sun Aug 18, 2013 9:06 pm
yumi2012 wrote:An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?

A. $31
B. $29
C. $26
D. $23
E. $21
To determine the amount that CANNOT be paid out, we'll eliminate the amounts that CAN be paid out.

When we scan the 5 answer choices, we can see that we can eliminate 3 of them almost immediately.
We can pay out $21 using three $7 chips, so ELIMINATE E.
Since $26 is $5 greater than $21, we can pay out $26 using three $7 chips and one $5 chip. ELIMINATE C.
Since $31 is $10 greater than $21, we can pay out $31 using three $7 chips and two $5 chips. ELIMINATE A.

This leaves us with B and D.
We can pay out $29 using two $7 chips, and three $5 chips so ELIMINATE B.

By the process of elimination (POE), the correct answer must be D

Cheers,
Brent
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by ganeshrkamath » Sun Aug 18, 2013 11:55 pm
yumi2012 wrote:An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?

A. $31
B. $29
C. $26
D. $23
E. $21
So every amount can be represented by N = 5a + 7b where a and b are whole numbers.

31 = 10 + 21
29 = 15 + 14
26 = 5 + 21
21 = 21

Choose D
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by GMATGuruNY » Mon Aug 19, 2013 3:17 am
yumi2012 wrote:An eccentric casino owner decides that his casino should only use chips in $5 and $7 denominations. Which of the following amount cannot be paid out using these chips?

A. $31
B. $29
C. $26
D. $23
E. $21
Make a list of the MULTIPLES OF 5 and the MULTIPLES OF 7 less than 30:
5, 10, 15, 20, 25, 30
7, 14, 21, 28.

Determine the four answers answer choices that CAN be yielded by the list of values above.
A: 31 = 10+21
B: 29 = 15+14
C: 26 = 5+21
E: 21 = 21.

The correct answer is D.
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by tarik » Mon Aug 19, 2013 1:22 pm
Got it on time by elimination.
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