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Q:

The surface area of a cube is 1734 sq. cm. Find its volume

A) 2334 cubic.cm	B) 3356 cubic.cm
C) 4913 cubic.cm	D) 3478 cubic.cm

Answer & Explanation Answer: C) 4913 cubic.cm

Explanation:

Let the edge of the cube bea. Then,

6 $a^{2}$ = 1734

=> a = 17 cm.

Volume = $a^{3}$ = $17^{3}$ = 4193 cu.cm

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Filed Under: Volume and Surface Area

Q:

The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 cu. meters. Find the breadth of the wall.

A) 40cm	B) 30cm
C) 20cm	D) 10cm

Answer & Explanation Answer: A) 40cm

Explanation:

Let the breadth of the wall be x metres.

Then, Height = 5x metres and Length = 40x metres.

x * 5x * 40x = 12.8

=> $x^{3} = 12.8 / 200 = 128 / 2000 = 64 / 1000$

=> x = 4/10 m

=> x = (4/10)*100 = 40 cm

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Filed Under: Volume and Surface Area

Q:

In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

A) 0.25	B) 0.5
C) 0.75	D) 0.40

Answer & Explanation Answer: D) 0.40

Explanation:

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40

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Filed Under: Probability

Q:

The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

A) 1/4	B) 1/2
C) 1/6	D) 1/3

Answer & Explanation Answer: C) 1/6

Explanation:

P(X) = $\frac{1}{5}$ , P(Y) = $\frac{1}{4}$ , P(Z) = $\frac{1}{3}$

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

= $(\frac{1}{4} * \frac{1}{3} * \frac{4}{5}) + (\frac{3}{4} * \frac{1}{3} * \frac{1}{5}) + (\frac{2}{3} * \frac{1}{4} * \frac{1}{5}) + (\frac{1}{4} * \frac{1}{3} * \frac{1}{5})$

= $\frac{4}{60} + \frac{3}{60} + \frac{2}{60} + \frac{1}{60}$ = $\frac{10}{60}$ = $\frac{1}{6}$

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Filed Under: Probability

Q:

In a race, the odd favour of cars P,Q,R,S are 1:3, 1:4, 1:5 and 1:6 respectively. Find the probability that one of them wins the race.

A) 319/420	B) 27/111
C) 114/121	D) 231/420

Answer & Explanation Answer: A) 319/420

Explanation:

$P (P) = \frac{1}{4}, P (Q) = \frac{1}{5}, P (R) = \frac{1}{6}, P (S) = \frac{1}{7}$
All the events are mutually exclusive hence,

Required probability = P(P)+P(Q)+P(R)+P(S)

$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}$ = $\frac{319}{420}$

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Filed Under: Probability

Q:

What is a drawback of MVT?

Answer

1) ability to support multiple processors
2) virtual storage
3) source level debugging

Workspace

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Subject: Operating Systems

Q:

What is pre-emptive and non-preemptive scheduling?

Answer

Preemptive scheduling: The preemptive scheduling is prioritized. The highest priority process should always be the process that is currently utilized.

Non-Preemptive scheduling: When a process enters the state of running, the state of that process is not deleted from the scheduler until it finishes its service time.

Workspace

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Subject: Operating Systems

Q:

If x is chosen at random from the set {1,2,3,4} and y is to be chosen at random from the set {5,6,7}, what is the probability that xy will be even?

A) 1/2	B) 2/3
C) 3/4	D) 4/3

Answer & Explanation Answer: B) 2/3

Explanation:

S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)}
Total element n(S)=12

xy will be even when even x or y or both will be even.
Events of x, y being even is E.
E ={(1,6),(2,5),(2,6),(2,7),(3,6),(4,5),(4,6),(4,7)}
n(E) = 8

P(E)=n(E)n(S)=812 = 4/3

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Filed Under: Probability

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