CS GATE 2014 PAPER 03 - Online Test
Q1. 
The above synchronous sequential circuit built using JK flip-flops is initialized with
Q2Q1Q0 = 000. The state sequence for this circuit for the next 3 clock cycles is
Answer : Option C
Explaination / Solution:
Q2. Let ⊕ denote the Exclusive OR (XOR) operation. Let ‘1’ and ‘0’ denote the binary
constants. Consider the following Boolean expression for F over two variables P and Q.
F(P,Q) = ((1⊕P)⊕(P⊕Q))⊕((P⊕Q)⊕(Q⊕0))
The equivalent expression for F is
Answer : Option D
Explaination / Solution:
Q3. Choose the word that is opposite in meaning to the word “coherent”.
Answer : Option C
Explaination / Solution:
No Explaination.
Q4. Consider the following processors (ns stands for nanoseconds).
Assume that the pipeline registers have zero latency.
P1: Four-stage pipeline with stage latencies 1 ns, 2 ns, 2 ns, 1 ns.
P2: Four-stage pipeline with stage latencies 1 ns, 1.5 ns, 1.5 ns, 1.5 ns.
P3: Five-stage pipeline with stage latencies 0.5 ns, 1 ns, 1 ns, 0.6 ns, 1 ns.
P4: Five-stage pipeline with stage latencies 0.5 ns, 0.5 ns, 1 ns, 1 ns, 1.1 ns.
Which processor has the highest peak clock frequency?
Answer : Option C
Explaination / Solution:
Clock period (CP) = max stage delay + overhead
Q5. The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012-2013. For
international comparison, the GDP is compared in US Dollars (USD) after conversion based
on the market exchange rate. During the period 2012-2013 the exchange rate for the USD
increased from Rs. 50/ USD to Rs. 60/ USD. India’s GDP in USD during the period 2012-
2013
Answer : Option D
Explaination / Solution:
Per 100 Rs final value 107 Rs
⇒ Per 100/50 Dollars final value 107/60
for 100 dollars____?
= ((100 × 50)/100) × (107/60) = 89.16
Decreased by 11%.
Q6. Which of the following statements are CORRECT?
1) Static allocation of all data areas by a compiler makes it impossible to implement
recursion.
2) Automatic garbage collection is essential to implement recursion.
3) Dynamic allocation of activation records is essential to implement recursion.
4) Both heap and stack are essential to implement recursion.
Answer : Option D
Explaination / Solution:
To implement recursion, activation record should be implemented by providing dynamic
memory allocation. This dynamic allocation is done from runtime stack. Heap is essential to
allocate memory for data structures at run-time, not for recursion.
So statement 1 and 3 are correction.
Q7. What is the optimized version of the relation algebra expression 
where
A1, A2 are sets of attributes in with A1 ⊂ A2 and F1, F2 are Boolean expressions based on
where
A1, A2 are sets of attributes in with A1 ⊂ A2 and F1, F2 are Boolean expressions based on
Answer : Option A
Explaination / Solution:
No Explaination.
Q8. In the context of modular software design, which one of the following combinations is
desirable?
Answer : Option B
Explaination / Solution:
Cohesion is a measure of internal strength within a module, whereas coupling is a measure of
inter dependency among the modules. So in the context of modular software design there
should be high cohesion and low coupling.
Q9. A prime attribute of a relation scheme R is an attribute that appears
Answer : Option B
Explaination / Solution:
A prime attribute or key attribute of a relation scheme R is an attribute that appears in any of
the candidate key of R, remaining attributes are known as non-prime or non-key tribute
Q10. Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below.
T1 : r1 (X) ; r1 (z) ; w1 (X) ; w1 (z)
T2 : r2 (X) ; r2 (z) ; w2 (z)
T3 : r3 (X) ; r3 (X) ; w3 (Y)
S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z)
S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z)
Which one of the following statements about the schedules is TRUE?
Answer : Option A
Explaination / Solution:
Precedence graph for S1 & S2 are as follows
∴Only S1 is conflict serializable.
