Andy has a certain number of $1 bills, $2 bills and $10 bills. If the

Let the number of 2$ bill be X,
Number of 1$ bill is 6X
Total value of these two becomes 8X
As the total amount is 160, and the denomination left is 10$ bill, so 8x should be a multiple of 10,
So that we can have an integer value for number of 10$ bills,
Minimum Value X can take is 5 , so that 8x = 40
So 10y= 120
Y= 12
So A , IMO is the answer

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Solution:

If we let x = the number of $2 bills Andy has, then 6x = the number of $1 bills he has. Furthermore, since we are looking for the maximum number of $10 bills Andy could have, let’s check the answer choices starting with the largest number given, which is 12. If he has 12 $10 bills, then he has $40 left for the $1 bills and $2 bills. We can create the equation:

1(6x) + 2(x) = 40

8x = 40

x = 5

Since x is an integer, we see that it is possible for Andy to have 12 $10 bills (note: the only way we can reject 12 as the correct answer is if we can show x, the number of $2 bills, is not a positive integer).

Alternate Solution:

If we let x = the number of $2 bills Andy has, then 6x = the number of $1 bills he has. The dollar value of the $1 bills is 6x dollars and the dollar value of the $2 bills is 2x dollars. Together, the $1 and $2 bills make up 6x + 2x = 8x dollars. Thus, the $10 bills are worth 160 - 8x dollars and it follows that the number of $10 bills is (160 - 8x)/10. Since the number of $10 bills must be an integer, it must be true that (160 - 8x)/10 is an integer. To maximize the expression (160 - 8x)/10, we must minimize the negative term, 8x. The smallest non-zero value of x for which 160 - 8x is divisible by 10 is x = 5. Thus, the greatest value of $10 bills is (160 - 8x)/10 = (160 - 8(5))/10 = (160 - 40)/10 = 120/10 = 12.

Answer: A

If there is 1 bill of $2 then there will be 6 bills for $1.

Total: $1(6) + $2(1) = 8.

This pair can give us a $40 amount as a minimum so that the remaining amount $160- $40 = $120 is divisible by $10.

So, 12 * 10 = 120

Answer A

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