Total Cash = 200 $
p = xy = 10 x + y
q = yx = 10y + x

p + q = 11 ( x + y)

Remaining cash = 200 - 11 (x + y) where x+y = 1, 2 .....

200 - 11( 3 ) = 167
200 - 11( 55) = 145
200 - 11( 10) = 90
200 - 11( 11) = 79

So our Ans . C

IMO C

Pluggin in is the best way to solve these kinds of problems.

We know that (p+q) is the amount spent and 200 is the total balance before spending it.

200 - (p+q) = x .... (1)

x is the remaining balance.

p+q can be written as (10a+b)+(10b+a) since we are told that p and q are two-digit integers that have the same digits but in reverse order.

(10a+b)+(10b+a) = 11a + 11b = 11 (a+b) ....... (2)

Plugin numbers now.

Lets start with E,

x = 167. This is the remaining amount. Therefore, 33 is the amount spent i.e. (p+q)
Now, p+q = 11 (a+b) from (2). So, 33 is divisible by 11. This could be the amount. Eliminate.

Lets try C now,

x = 113. Therefore, 87 is the amount spent i.e. (p+q)
87 is NOT divisible by 11. So, This could not be the amount.

Hence, C.

armaan700+ wrote:Ken withdrew $200 in cash from the bank. Using the cash, he bought a book for exactly p dollars and a hat for exactly q dollars, where p and q are two-digit integers that have the same digits but in reverse order. Assuming no sales tax, each of the following could be the amount of cash, in dollars, that Ken has remaining EXCEPT:

(A) 79
(B) 90
(C) 113
(D) 145
(E) 167

OA will be posted later.

Two 2-digit numbers containing same digits but in reverse order would add up to be a multiple of 11, say 11 x; where x is a positive integer. The remaining money subtracted from 200 must be a multiple of 11, therefore. (A) 200 - 79 = 121 = 11*11 (yes); (B) 200 - 90 = 110 = 11*10 (yes); (C) 200 - 113 = 87 = 11*(NO positive integer); hence [spoiler]C[/spoiler]