It's a problem based on the binomial distributions of Probability.
Let us consider a normal test where the performance of the last question won't effect the next question.
If the probability of correctness of one problem is p, then the probability of correctness of r problems out of n problems is
P(r) = nCr p^r (1-p)^n-r
To find this one we need total number of problems (n) and number of problems to be correct in order to succeed in the test (r).
Now come to the present scenario, where performance to the last problem is applied on the next-to-be-given problem. If we consider statement(2), it is given that the probability that he answers successive questions correctly is p, which is not different from the probability that he answers first question correctly. So, this scenario is not different from the scenario where the performance on a problem won't effect the next problem. To say it simply, the given scenario is no different from a normal scenario where probability of correctly answering any problem is same and is 'p'.
So statement(2) together with statement(1), which gives total number of questions, is sufficient.
Answer "C"
Note: the value of p, which is not given is taken granted as p and the final answer will be in terms of p.
P(7)+P(8)+P(9)+P(10) = 10C7 p^7 (1-p)^3 + 10C8 p^8 (1-p)^2 + 10C9 p^9 (1-p)^1 + 10C10 p^10