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by daretodream » Fri Feb 19, 2010 3:27 am
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
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by harsh.champ » Fri Feb 19, 2010 3:58 am
daretodream 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
Let p be 10a + b (ab being the two digit no.)
Then,q=10b + a
Cash left with Ken would be $200-(11a + 11b)
=200 - 11(a+b)
Now,we are subtracting a multiple of 11 from 200.

Plugging no.s
79:- 11(a+b) =200-79 = 121
90:- 11(a+b) =200-90 = 110
113:-11(a+b) =200-113 = 87(which is not a multiple of 11)

So, the answer would be C.
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by ajith » Fri Feb 19, 2010 4:03 am
daretodream 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
p = 10a+b
q= 10b+a

p+q = 11a+11b
= 11(a+b)

Now The numbers given are 200 -( p+q)

A p+q = 121 = 11*11
B p+q = 110
C p+q = 87 Not possible

Hence C
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