Computer Science Study Materials for Competitive Exams

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Friday, 11 May 2018

UGC NET Computer Science Paper III Re-test September 2013 - Part 5

41.       Given a binary search trees for a set of n=5 keys with the following probabilities :
The expected optimal cost of the search is
(A) 2.65
(B) 2.70
(C) 2.75
(D) 2.80
Answer: C
42.       Given 0-1 knapsack problem and fractional knapsack problem and the following statements :
S1 : 0-1 knapsack is efficiently solved using Greedy algorithm.
S2 : Fractional knapsack is efficiently solved using Dynamic programming.
Which of the following is true?
(A) S1 is correct and S2 is not correct.
(B) Both S1 and S2 are correct.
(C) Both S1 and S2 are not correct.
(D) S1 is not correct and S2 is correct.
Answer: C
43.       Equivalence class partitioning approach is used to divide the input domain into a set of equivalence classes, so that if a program works correctly for a value, then it will work correctly for all the other values in that class. This is used .................
(A) to partition the program in the form of classes.
(B) to reduce the number of test cases required.
(C) for designing test cases in white box testing.
(D) all of the above.
Answer: B
44.       The failure intensity for a basic model as a function of failures experienced is given as λ(μ)-λ0[1 – (μ)/(V0)] where λ0 is the initial failure intensity at the start of the execution, μ is the average or expected number of failures at a given point in time, the quantity V0 is the total number of failures that would occur in infinite time.
Assume that a program will experience 100 failures in infinite time, the initial failure intensity was 10 failures/CPU hr. Then the decrement of failures intensity per failure will be
(A) 10 per CPU hr.
(B) 0.1 per CPU hr.
(C) –0.1 per CPU hr.
(D) 90 per CPU hr.
Answer: C
45.       Improving processing efficiency or performance or restructuring of software to improve changeability is known as
(A) Corrective maintenance
(B) Perfective maintenance
(C) Adaptive maintenance
(D) Code maintenance
Answer: B

46.       In ...............,modules A and B make use of a common data type, but perhaps perform different operations on it.
(A) Data coupling
(B) Stamp coupling
(C) Control coupling
(D) Content coupling
Answer: B
47.       Consider the following schemas :
Branch_Schema = (branch_name, assets, city)
Customer_Schema = (cutstomer_name, street, city)
Deposit_Schema = (branch_name, account_number, customer_name, balance)
Borrow_Schema = (branch_name, loan_number, customer_name, amount)
Which of the following tuple relational calculus finds all customers who have loan amount more than Rs 12,000?
(A) {t(customer_name) | tborrow[?] t[amount]>12000}
(B) {t |t(customer_name) | tborrow[?] t[amount]>12000}
(C) {t |[?] borrow (t(customer_name=s(customer_ name))[?] [amount]>12000}
(D) {t |[?] borrow (t(customer_name)[?] s[amount] >12000}
Answer: C
48.       Match the following :
(a) Create          (i) The E-R Model
(b) Select           (ii) Relationship Model
(c) Rectangle    (iii) DDL
(d) Record         (iv) DML
Codes :
      (a)   (b)  (c)  (d)
(A) (iii)  (iv)  (i)   (ii)
(B) (iv)  (iii)  (ii)  (i)
(C) (iv)  (iii)  (i)  (ii)
(D) (iii)  (iv)  (ii)  (i)
Answer: A
49.       Match the following :
Codes :
      (a)   (b)  (c)  (d)
(A) (iii)  (iv)  (ii)   (i)
(B) (iv)  (iii)  (ii)   (i)
(C) (ii)  (iii)  (iv)   (i)
(D) (iii)  (iv)  (i)   (ii)
Answer: C
50.    Sixty (60) reusable components were available for an application. If only 70% of these components can be used, rest 30% would have to be developed from scratch. If average component is 100 LOC and cost of each LOC is Rs 14, what will be the risk exposure if risk probability is 80%?
(A) Rs 25,200
(B) Rs 20,160
(C) Rs 25,160
(D) Rs 20,400
Answer: B

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