GATE Questions

Q:

5 C in an OF?

A) 5 Cards in an Old Fan B) 5 Circles in an Olympic Flag
C) 5 Colors in an Old Football D) 5 Colors in an Old Flag
 
Answer & Explanation Answer: B) 5 Circles in an Olympic Flag

Explanation:

5 C in an OF denotes 5 Circles in an Olympic Flag.

These are riddles of type

12 S of the Z

52 W in a Y.

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192 29263
Q:

If E = 5 and READ is coded as 7, then what is the code of 'DEAR' ?

A) 6 B) 7
C) 8 D) 9
 
Answer & Explanation Answer: B) 7

Explanation:

Here E = 5 = 5/1 = 5

=> READ = 18 + 5 + 1 + 4 = 28/4 = 7

=> DEAR = 4 + 5 + 1 + 18 = 28/4 = 7

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Filed Under: Coding and Decoding
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110 28975
Q:

Find the appropriate relation for quantity 1 and quantity 2 in the following question:

Quantity I: the unit digit in (6817)754

Quantity II: the unit digit in (365×659×771)

A) Quantity I > Quantity II B) Quantity I < Quantity II
C) Quantity I ≥ Quantity II D) Quantity I ≤ Quantity II
 
Answer & Explanation Answer: A) Quantity I > Quantity II

Explanation:
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1 28875
Q:

The First Mechanical Computer Designed by Charles Babbage was called?

A) Super Computer B) Abacus
C) Calculator D) Analytical Engine
 
Answer & Explanation Answer: D) Analytical Engine

Explanation:

Computer was invented by Charles Babbage.

The First Mechanical Computer Designed by Charles Babbage, a British mathematician was called as Analytical Engine which resembles today's modern machines  between 1833 and 1871.

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145 28807
Q:

Artists are generally whimsical. Some of them are frustrated. Frustrated people are prone to be drug addicts.

Based on these statements which of the following conclusions is true?

A) All frustrated people are drug addicts B) Frustrated people are whimsical
C) All drug addicts are artists D) Some artists may be drug addicts
 
Answer & Explanation Answer: D) Some artists may be drug addicts

Explanation:
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44 28798
Q:

Find the appropriate relation for quantity1 and quantity2 in the following question:

Quantity 1: In an examination, Ankita scored 35 marks less than Puneeta. Puneeta scored 65 more marks than Meenakshi. Rakhi scored 115 marks, which is 20 marks more than Meenakshi's. Simpy scored 108 marks less than the maximum marks of the test. What approximate percentage of marks did Simpy score in the examination, if she got 67 marks more than Ankita?

Quantity 2: The length of a rectangle is increased by 60%. By what percent would the width have to be decreased to maintain the same area?

A) Quantity1 < Quantity2 B) Quantity1 ≤ Quantity2
C) Quantity1 ≥ Quantity2 D) Quantity1 > Quantity2
 
Answer & Explanation Answer: D) Quantity1 > Quantity2

Explanation:

Quantity1-
Rakhi’s marks= 115
Meenakshi’s marks= 115 - 20 = 95
Puneeta’s marks= 95 + 65= 160
Ankita’s marks=160 - 35= 125
Simpy’s marks= 125+ 67= 192
Total maximum marks= 192 + 108= 300
Required percentage marks of Simpy
192300×100=64%
Quantity2-
Let length and breadth be 100.

After increase in length it become 160, then
reduction in breadth be ‘x’
Now, 160*x= 100*100
Hence, x = 10000160=62.5

100-62.5=37.5%

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2 28672
Q:

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

A) 23/42 B) 19/42
C) 7/32 D) 16/39
 
Answer & Explanation Answer: B) 19/42

Explanation:

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

 

(ii) Selecting bag II and then drawing a red ball from it.

 

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7

  PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) = PE1×PAE1+PE2×PAE2 

 

                          = 12×47+12×26=1942

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Filed Under: Probability
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64 28508
Q:

In a queue, Amrita is 10th from the front while Mukul is 25th from behind and Mamta is just in the middle of the two. If there be 50 persons in the queue. What position does Mamta occupy from the front ?

A) 14th B) 16th
C) 18th D) 20th
 
Answer & Explanation Answer: C) 18th

Explanation:

Number of persons between Amrita and Mukul = 50 - (10 + 25) = 15. Since Mamta lies in middle of these 15 persons, so Mamta`s position is 8th from Amrita i.e. 18th from the front.

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102 28507