Question

A three digit number has its tens digit as twice its hundred digit which in turn is one third its units digit. The sum of its extreme digits is 8. Find its middle digit?Select one:a. 8b. 4c. 6d. 2

Answer 1

A is 100a+10b+c

B is 100c+10b+a

Let's see the statements

1) B-A is div by 8

So (100c+10b+a)-(100a+10b+c)=99(c-a) is div by 8...

Since C and a are digits, ways the equation is satisfied

a) c is 9 and a is 1 , number becomes 9b1 and 1b9

Even you take b as max possible 9, sum becomes 991+199=1190, <1200

b) c is 8 and a is 0, but then A becomes two digit number so not possible..

So number can be only 9b1 and 1b9, and answer will be NO irrespective of value of B

Sufficient

2) A is a multiple of 5..

So C can be either 5 or 0 but 0 is not possible as the other number B becomes two digit number.

Hence ab5 and 5ba where 5ba>ab5 but the larger number has to be > 1200/2 or 600.. hence Ans is NO

Sufficient