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Q:

The dimensions of an open box are 50 cm, 40 cm and 23 cm. Its thickness is 2 cm. If 1 cubic cm of metal used in the box weighs 0.5 gms, find the weight of the box.

A) 8.04kg B) 8.14kg
C) 8.24kg D) 9.04kg
 
Answer & Explanation Answer: A) 8.04kg

Explanation:

Volume of the metal used in the box = External Volume - Internal Volume 

                                                    = [(50 x 40 x 23) - (44 x 34 x 20)] cu.cm  

                                                    = 16080 cu.cm 

 

Weight of the metal =[(16080 x 0.5)/1000] kg = 8.04 kg.

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Q:

Find the number of bricks, each measuring 24 cm x 12 cm x 8 cm, required to construct a wall 24 m long, 8m high and 60 cm thick, if 10% of the wall is filled with mortar?

A) 35000 B) 45000
C) 55000 D) 65000
 
Answer & Explanation Answer: B) 45000

Explanation:

Volume of the wall = (2400 x 800 x 60)  cu.cm  

 

Volume of bricks   = 90% of the volume of the wall 

= [(90/100) x 2400 x 800 x 60] cu.cm  

 

Volume of 1 brick = (24 x 12 x 8)  cu.cm 

 

Number of bricks = [(90/100) x (2400 x 800 x 60) ]/ (24 x 12 x 8) = 45000

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Q:

I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?

A) 1/999 B) 1/1001
C) 1/1000 D) 4/1000
 
Answer & Explanation Answer: C) 1/1000

Explanation:

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.
Hence required probability =1103 = 1/1000

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Q:

A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:

A) 123/897 B) 23/67
C) 7/44 D) 12/45
 
Answer & Explanation Answer: C) 7/44

Explanation:

Here, n(E) = 7C1×5C1×5C1

 

And,  n(S) = 12C1*11C1*10C1

P(S) = 7*6*512*11*10 = 7/44

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Q:

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

A) 1/6 B) 1/3
C) 1/2 D) 1/4
 
Answer & Explanation Answer: C) 1/2

Explanation:

P(odd) = P (even) =12 1(because there are 50 odd and 50 even numbers)

 

Sum or the three numbers can be odd only under the following 4 scenarios:

 

Odd + Odd + Odd = 12*12*1218

 

Odd + Even + Even = 12*12*12=18

 

Even + Odd + Even = 12*12*12=18

 

Even + Even + Odd = 12*12*12 = 18

 

Other combinations of odd and even will give even numbers. 

 

Adding up the 4 scenarios above:

 

1818+1818 = 48 = 12

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Q:

Differentiate between Complier and Interpreter?

Answer

An interpreter reads one instruction at a time and carries out the actions implied by that instruction. It does not perform any translation. But a compiler translates the entire instructions

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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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Q:

There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

A) 1/2 B) 3/4
C) 4/7 D) 3/8
 
Answer & Explanation Answer: D) 3/8

Explanation:

Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.

Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.

Required probability =24/64  = 3/8

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Filed Under: Probability