Can anybody explain how to solve this question?

70% of students answered question A correctly, 55% of students answered question B correctly, 20% of students answered neither question correctly. What is the percentage of students answered both question A and B correctly?

45%

Ok, I guess you would like an explanation too.

Here goes. Lets say that X is the number of all people who are in this test (answering questions). Now let A be the number of people who answered to question A correctly, so A = 0.7*X, which means 70% of X. In the same manner, B = 0.55*X. Let W be the number of people who answered both given questions wrong. So W = 0.2*X. Now X consists of people who answered to A correctly, those who answered B correctly, and those who answered nothing correctly. But beware: Those that answered A correctly MIGHT HAVE answered B correctly too. So it's not a simple addition. Let's call A&B those who answered A and B questions correctly. Then we can say that X consists of people who answered ONLY A correctly, those who only answered B correctly, those who answered BOTH A AND B correctly, and those who know nothing. So X = A + B + W - A&B. Think about it a bit, it makes perfect sense. So, with that in mind we have:
A&B = A + B + W - X = 0.7*X + 0.55*X + 0.2*X - 1*X = 0.45*X
And that means that 45% of people answered both A and B questions correctly.

P(A) = 0.70
P(B) = 0.55
P(A' & B') = 0.20

P(A & B) =??

--------------------

P(A & B) = 1 - P ((A & B)') = 1 - P(A' U B') (De Morgan)
= 1 - [ P(A') + P(B') - P(A' & B') ]
= 1 - [ 0.30 + 0.45 - 0.20]
= 0.45

Yeah, 45%

P(A) + P(A') = 1.

P(A) = 0.70 = 1 - P(A')
P(A') = 0.30

P(B) = 0.55 = 1- P(B')
P(B') = 0.45

P(A' & B') = 0.20

1 - (0.30 + 0.45 - 0.20) = 1 - 0.55 = 0.45
note: you are subtracting the 0.20 because it is in both P(A') and P(B'), because it is overlapped.

if you have done venn diagrams:
let a venn diagram of 4 sections explain.
section E = all A'
section F = all B'
section G = intersection of A' with B'
section H = all within the universe outside the two circles A', B'
circle A' = section E,G
circle B' = section F,G
so to determine the area of the portion not in circle E or circle F.
subtract circle E and circle F from the universe, subtracting section G from the area of circle F before subtracting from the universe, so that area of section G is not counted twice (because it is in both circle E and in circle F).