50+ Arithmetic Reasoning Questions and Answers Pdf - 1

Question: 1

A farmer built a fence around his square plot. He used 27 fence poles on each side of the square. How many poles did he need altogether?

(A) 100

(B) 101

(C) 102

(D) 104

Ans: D

Since each pole at the corner of the plot is common to its two sides, so we have :

Total number of poles needed = 27 x 4 – 4 = 108 – 4 = 104.

Question: 2

Between two book ends in your study are displayed your five favourite puzzle books. If you decide to arrange the five books in every possible combination and moved juts one book every minute, how long would it take you?

(A) 4 hours

(B) 3 hours

(C) 2 hours

(D) 1 hour

Ans: C

Clearly, number of ways of arranging 5 books = 5! = 5 x 4 x 3 x 2 x 1 = 120.

So, total time taken = 120 minutes = 2 hours.

Question: 3

Ayush was born two years after his father’s marriage. His mother is five years younger than his father but 20 years older than Ayush who is 10 years old. At what age did the father get married?

(A) 21 years

(B) 23 years

(C) 33 years

(D) 37 years

Ans: B

Ayush’s present age = 10 years.

His mother’s present age = (10 + 20) years = 30 years.

Ayush’s father’s present age = (30 + 5) years = 35 years.

Ayush’s father’s age at the time of Ayush’s birth = (35 – 10) = 25 years.

∴ Ayush’s father’s age at the time of marriage = (25 – 2) years = 23 years.

Question: 4

A group of 1200 persons consisting of captains and soldiers is travelling in a train. For every 15 soldiers there is one captain. The number of captains in the group is

(A) 70

(B) 75

(C) 80

(D) 85

Ans: B

Clearly, out of every 16 persons, there is one captain.

So, number of captains = 1200÷16 = 75.

Question: 5

A man has a certain number of small boxes to pack into parcels. If he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcel, none is left over. What is the number of boxes, he may have to pack?

(A) 300

(B) 301

(C) 303

(D) 309

Ans: B

Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5, 6 and no remainder when divided by 7.

Thus, the number must be of the form (L.C.M. of 3, 4, 5, 6) x + 1 i.e. (60x + 1) and a multiply of 7. Clearly, for x = 5, the number is multiply of 7. So, the number is 301.

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