# Aptitude and Reasoning Questions

• #### Verbal Reasoning - Mental Ability

Q:

The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8748, its purchase price was :

 A) 10000 B) 12000 C) 14000 D) 16000

Explanation:

Purchase price = $Rs.87481-101003$ = Rs. [8748 * (10/9) * (10/9 )* (10/9)] = Rs.12000

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Q:

A grocer has a sale of Rs 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs, 6500 ?

 A) 4991 B) 5467 C) 5987 D) 6453

Explanation:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs.[(6500 x 6) - 34009]

= Rs. (39000 - 34009)

= Rs.  4991.

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Q:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

 A) 1/2 B) 3/5 C) 9/20 D) 8/15

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

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Q:

The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4 ?

 A) 76 B) 79 C) 85 D) 87

Explanation:

Average = total runs / no.of innings = 32

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs

Total runs = new avg x new no. of innings = 36 x 11 = 396

Runs made in the 11th inning = 396 - 320 = 76

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Q:

If each edge of a cube is increased by 50%, find the percentage increase in Its surface area

 A) 125% B) 150% C) 175% D) 110%

Explanation:

Let the edge = a cm

So increase by 50 % = a + a/2 = 3a/2

Total surface Area of original cube = $6a2$

TSA of new cube = $63a22$ =$69a24$=  $13.5a2$

Increase in area = $13.5a2-6a2$ =$7.5a2$

$7.5a2$ Increase % =$7.5a26a2×100$ = 125%

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Q:

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

 A) 40 B) 80 C) 120 D) 200

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

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Q:

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

 A) 52/221 B) 55/190 C) 55/221 D) 19/221

Explanation:

We have n(s) =$52C2$ 52 = 52*51/2*1= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = $52*512*1$ = $26C2$ = 325, n(B)= $26*252*1$= 4*3/2*1= 6  and  n(A∩B) = $4C2$ = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

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Q:

8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?

 A) 18 litres B) 24 litres C) 32 litres D) 42 litres

Explanation:

Let the quantity of the wine in the cask originally be x litres

Then, quantity of wine left in cask after 4 operations =$x1-8x4$litres

$⇒1-8x4=234$

$⇒x=24$