Lex has $\$ 2.65$. He has only dimes (worth $\$ 0.10$ each) and quarters (worth $\$ 0.25$ each). If Lex has more quarters than dimes, how many coins does he have in total?
(A) 12
(B) 13
(C) 16
(D) 19
(E) 22
Lex has 265 cents in total.

Since a quarter is worth 25 cents, the total value in cents of the quarters that Lex has is a multiple of 25 , and so must end in $00,25,50$ or 75 .

Since the remaining part of the 265 cents is made up of dimes only, the remaining part is a multiple of 10 , so ends in a 0 .

Thus, the value of the quarters must end with a 5 , so ends with 25 or 75 .

Since Lex has more quarters than dimes, we start by trying to determine the largest possible number of quarters that he could have.

The largest possible value of his quarters is thus 225 cents, which would be $225 \div 25=9$ quarters, leaving $265-225=40$ cents in dimes, or 4 dimes.

Thus, Lex has $9+4=13$ coins in total.

(The next possible largest value of his quarters is 175 cents, which would come from 7 quarters and so 90 cents in dimes or 9 dimes. This does not satisfy the condition that the number of quarters is larger than the number of dimes.)

ANSWER: (B)

#