A SPEAK TRUTH IN 60% OF THE CASE AND B IN 90% OF THE CASE IN WHAT PERCENTAGE OF CASES ARE THEY LIKELY TO CONTRADICT EACH OTHER IN STATING THE SAME FACT

let P(A) be the probability that A speaks truth.
then P(A') be the probability that A does not speak truth.
let P(B) be the probability that B speak truth.
then P(B') be the probability that B does not speak truth.
P(A) = 60 / 100 = 0.6
P(A') = 1-0.6 = 0.4
P(B) = 90/100 = 0.9
P(B') = 1- 0.9 = 0.1
the probability that A speak truth and B does not speak truth = P(A)*P(B')
= 0.6*0.1
=0.06
the probability that B speak truth and A does not = P(B)*P(A')
=0.9*0.4
=0.36
therefore the probability that contradiction arise
= 0.06+0.36
=0.42
thus the required percentage = 0.42 *100 = 42%
hope this helps you.