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Q:

What is silly window syndrome?

Answer

It is a problem that can ruin TCP performance. This problem occurs when data are passed to the sending TCP entity in large blocks, but an interactive application on the receiving side reads 1 byte at a time.

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Subject: Networking
Job Role: Database Administration

Q:

What are Digrams and Trigrams?

Answer

The most common two letter combinations are called as digrams. e.g. th, in, er, re and an. The most common three letter combinations are called as trigrams. e.g. the, ing,and, and ion.

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Subject: Networking
Job Role: Network Engineer

Q:

What is wide-mouth frog?

Answer

Wide-mouth frog is the simplest known key distribution center (KDC) authentication protocol

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Subject: Networking
Job Role: Database Administration

Q:

What is Mail Gateway

Answer

It is a system that performs a protocol translation between different electronic mail delivery protocols

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Subject: Networking
Job Role: Network Engineer

Q:

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

A) 2/91	B) 1/22
C) 3/22	D) 2/77

Answer & Explanation Answer: A) 2/91

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = $15 C_{3}$ = $\frac{15 * 14 * 13}{3 * 2 * 1}$ = 455.

Let E = event of getting all the 3 red balls.

n(E) = $5 C_{3}$ = $\frac{5 * 4}{2 * 1}$ = 10.

=> P(E) = n(E)/n(S) = 10/455 = 2/91.

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Filed Under: Probability

Q:

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

A) 3/20	B) 29/34
C) 47/100	D) 13/102

Answer & Explanation Answer: D) 13/102

Explanation:

Let S be the sample space.

Then, n(S) = $52 C_{2}$ =(52 x 51)/(2 x 1) = 1326.

Let E = event of getting 1 spade and 1 heart.

n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = $13 C_{1} * 13 C_{1}$ = 169.

P(E) = n(E)/n(S) = 169/1326 = 13/102.

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Filed Under: Probability

Q:

In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by:

A) 21 m	B) 26 m
C) 28m	D) 29m

Answer & Explanation Answer: C) 28m

Explanation:

A : B = 100 : 75

B : C = 100 : 96.

A : C = (A/B*B/C) = (100/75*100/96) = 100/72 =100:72

A beats C by (100 - 72) m = 28 m.

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Filed Under: Races and Games

Q:

A runs $1 \frac{2}{3}$ times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

A) 200 m	B) 300m
C) 270m	D) 160m

Answer & Explanation Answer: A) 200 m

Explanation:

Ratio of the speeds of A and B =5 : 1 = 5 :3

Thus, in race of 5 m, A gains 2 m over B.

2 m are gained by A in a race of 5 m.

80 m will be gained by A in race of $(\frac{5}{2} * 80)$ m =200m

=> Winning post is 200 m away from the starting point.

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Filed Under: Races and Games

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