The number A is a two-digit positive integer; the number B is the two-digit positive integer formed by reversing the digits of A. if Q=10B-A, what is the value of Q?
(1) The tens digit of A is 7
(2) The tens digit of B is 6
I am able to get the value of Q using (1) which is contrary to the provided answer.
A = 7/10 + x
B = x/10 + 7
Q=10(x/10 + 7) - (7/10 + x)
Q= x + 70 - 7/10 - x
Q= 10x + 700 - 7 - 10x
Q= 700 - 7
Does this mean that the answer provided by Manhattan is wrong?
Let A= 10x + y and B =10y + x
it is stated that the tens digit is 7. so x=7 which makes A= 70 + y and B=10y +7
now on solving: Q= 10B-A= 10 (10y+7) - 70-y
Q=99y, we don't know value of y, so insuff.
tens digit of b is 6.
A= 10x+6 and B = 60+x
Q= 10 (60+x) - 10x-6
Q=594 so (2) is sufficient ALONE