A bank teller has $255 in $5 bills and $10 bills. If the number of $10

Question

A bank teller has $255 in $5 bills and $10 bills. If the number of $10 bills is greater than the number of $5 bills, what is the maximum possible number of bills that the bank teller has?

(A) 35
(B) 33
(C) 31
(D) 29
(E) 26

Regor60 · Accepted Answer

To maximize the number of bills you need to maximize the number of $5 bills, since that's the lowest denomination.

If the number of $10 and $5 bills were equal:

15X = 255 and X = 17 for a

total of 34 bills.

However, the number of $10 bills is constrained to >$5 bills and we need to minimize their number.

If we increase the number of $10 bills by 1 to 18 then we need to reduce the number of $5 bills by 2 to 15 to preserve the total of $255.

18+15 = 33

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gmatophobia · Answer

Assume that -

The number of $10 bills =  x  
 The number of $5 bills =  y

10x + 5y = 255

Dividing the equation by 5 on both sides we get

2x + y = 51

x = \frac{50 + 1 - y}{2}

x = 25 + \frac{1 - y}{2}

For y = 1

x = 25 + 0 = 25

Hence, (x,y) = (25,1) is one possible solution.

Other values of (x,y), that satisfy the equation are

(x,y) = (25,1) ⇒ Sum = 26
(x,y) = (24,3) ⇒ Sum = 27
(x,y) = (23,5) ⇒ Sum = 28
(x,y) = (22,7) ⇒ Sum = 29

Inference: For every one-point decrease in the value of x, the value of y increases by 2 points. The sum increases by 1 point.

Let's write a few more terms based on this logic

(x,y) = (25,1) ⇒ Sum = 26
(x,y) = (24,3) ⇒ Sum = 27
(x,y) = (23,5) ⇒ Sum = 28
(x,y) = (22,7) ⇒ Sum = 29
(x,y) = (21,9) ⇒ Sum = 30
(x,y) = (20,11) ⇒ Sum = 31
(x,y) = (19,13) ⇒ Sum = 32
(x,y) = (18,15) ⇒ Sum = 33
(x,y) = (17,17) ⇒ Sum = 34 ⇒ We can stop here, because x must be greater than y.

Maximum Sum = 18 + 15 = 33

Option B

GMATGuruNY · Answer

Let x = the number of $5 bills and y = the number of $10 bills.
 A bank teller has $255. 
Thus:
5x+10y = 255
x+2y = 51

We can PLUG IN THE ANSWERS, which represent the value of x+y: the total number of bills.
Since the question stem asks for the greatest possible number of bills, start with the greatest answer choice.
When the correct answer is plugged in, the number of $10 bills will be greater than the number of $5 bills:
y > x

A: x+y = 35
Subtracting x+y=35 from x+2y=51, we get:
(x+2y) - (x+y) = 51-35
y=16
Since x+y=35, the implication is that x=19, with the result that y < x.
Eliminate A.

B: x+y = 33
Subtracting x+y=33 from x+2y=51, we get:
(x+2y) - (x+y) = 51-33
y=18
Since x+y=33, the implication is that x=15, with the result that y > x.
Success!

B

Dheeraj_079 · Answer

Let the number of 5$ bils be a and 10$ bills be b.
5a + 10b = 255 and b>a
from this equation we can gather that a is odd.

To maximised the number of notes, a and b must be nearby with the caveat that b > a
It is possible for value a = 15 and b = 8.

Hence answer is B

GMATinsight · Answer

$255 in $5 bills and $10 bills

Let, there are x $5 bill and y $10 bills

so, 5x + 10y = 255
i.e. x + 2y = 51

Given,  the number of $10 bills is greater than the number of $5 bills,  i.e. y > x

Let's write all solutions possible
(x, y) = (1, 25) OR (3, 24) OR (5, 23) OR (7, 22) OR (9, 21) OR (11, 20) OR (13, 19) OR   (15, 18)   OR (17, 17)

i.e.  (x+y)_{max} = 15+18 = 33

Answer: Option B

KarishmaB · Answer

Responding to a pm: This is an application of integer solutions to equations. Check here Integer Solutions to equation in 2 variables: https://anaprep.com/algebra-integer-sol ... iables/Say we have F $5 bills and T $10 bills. 5F + 10T = 255 F + 2T = 51 We know that there are many solutions possible starting from T = 0, F = 51. More the $10 bills, fewer the total number of bills. We want $10 bills to be more than $5 bills but just about. So let's try to make them equal. T + 2T = 3T = 51 T = 17 Hence if F = T = 17, then $5 bills are 17, $10 bills are 17. But we need more $10 bills so let's make $10 bills 18. This means that F goes down to 15 (For every 1 extra bill of $10, we reduce two bills of $5 to keep the sum same) So we have 18 + 15 = 33 total bills at max. Answer (B) If we had to minimize the number of bills with more $10 bills than $5 bills, all we need to do is get all $10 bills as much as possible. To make $255, I can get $250 by using 25 $10 bills and only 1 bill of $5. So minimum number of bills would be 25 + 1 = 26 If we had to minimize the number of bills with more $5 bills, then we do the same as original question but when we reach the point of F = T = 17, then we increase F by 2 and reduce T by 1 to get 19 + 16 = 35 bills.

SRIVISHUDDHA22 · Answer

https://gmatclub.com/forum/download/file.php?id=86802 GMAT-Club-Forum-s4lwsv83.png

A bank teller has $255 in $5 bills and $10 bills. If the number of $10

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A bank teller has $255 in $5 bills and $10 bills. If the number of $10

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