**Question:** In a contest of intelligence, three problems A, B and C were posed.

* Among the contestants there were 25 who solved at least one problem each.

* Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C.

* The number of participants who solved only problem A was one more than the number who solved problem A and at least one other problem.

* Of all students who solved just one problem, half did not solve problem A.

How many students solved only problem B?

**Answer:**

6 students solved only problem B

X => Students who solved only problem A

Y => Students who solved only problem B

Z => Students who solved only problem C

P => Students who solved both problem B and problem C

From 4 :

Students who solved only problem A = Students who solved only problem B + Students who solved only problem C

X = Y + Z

From 3 :

Students who solved problem A and at least one other = X - 1

From 2 :

(Y + P) = 2 * (Z + P)

Y + P = 2 * Z + 2 * P

Z = (Y - P) / 2

From 1 and Figure:

X + X - 1 + Y + Z + P = 25

2*X + Y + Z + P = 26

2*(Y + Z) + Y + Z + P = 26 (from 4)

3*Y + 3*Z + P = 26

3*Y + 3* (Y - P) / 2 + P = 26 (from 2)

6*Y + 3*Y - 3*P + 2*P = 52

9*Y - P = 52

Y = (52 + P) / 9

Now, it is obvious that all values are integer. Hence, P must be 2 and Y must be 6.

So 6 students solved only problem B.