Practice MCQ Questions and Answer on Arithmetic Reasoning

11.

The total of the ages of Amar, Akbar and Anthony is 80 years. What was the total of their ages three years ago ?

(A) 71 years
(B) 72 years
(C) 74 years
(D) 77 years

Solution:

 Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.

12.

If you write down all the numbers from 1 to 100, then how many times do you write 3 ?

(A) 11
(B) 18
(C) 20
(D) 21

Solution:

 Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39. So, required number = 10 + 10 = 20.

13.

A shepherd had 17 sheep. All but nine died. How many was he left with ?

(A) Nil
(B) 8
(C) 9
(D) 17

Solution:

 'All but nine died' means 'All except nine died' i.e. 9 sheep remained alive.

14.

In three coloured boxes - Red, Green and Blue, 108 balls are placed. There are twice as many balls in the green and red boxes combined as there are in the blue box and twice as many in the blue box as there are in the red box. How many balls are there in the green box ?

(A) 18
(B) 36
(C) 45
(D) None of these

Solution:

 Let R, G and B represent the number of balls in red, green and blue boxes respectively. Then, . R + G + B = 108 ...(i), G + R = 2B ...(ii) B = 2R ...(iii) From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R. Putting G = 3R and B = 2R in (i), we get: R + 3R + 2R = 108 ⇔ 6R = 108 ⇔ R = 18. Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

15.

A student got twice as many sums wrong as he got right. If he attempted 48 sums in all, how many did he solve correctly ?

(A) 12
(B) 16
(C) 18
(D) 24

Solution:

 Suppose the boy got x sums right and 2x sums wrong. Then, x + 2x = 48 ⇔ 3x = 48 ⇔ x = 16.

16.

In a cricket match, five batsmen A, B, C, D and E scored an average of 36 runs. D Scored 5 more than E; E scored 8 fewer than A; B scored as many as D and E combined; and B and C scored 107 between them. How many runs did E score ?

(A) 62
(B) 45
(C) 28
(D) 20

Solution:

 Total runs scored = (36 x 5) = 180. Let the runs scored by E be x. Then, runs scored by D = x + 5; runs scored by A = x + 8; runs scored by B = x + x + 5 = 2x + 5; runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x. So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120. Therefore 3x + 120 =180 ⇔ 3X = 60 ⇔ x = 20.

17.

A bird shooter was asked how many birds he had in the bag. He replied that there were all sparrows but six, all pigeons but six, and all ducks but six. How many birds he had in the bag in all?

(A) 9
(B) 18
(C) 27
(D) 36

Solution:

 There were all sparrows but six' means that six birds were not sparrows but only pigeons and ducks. Similarly, number of sparrows + number of ducks = 6 and number of sparrows + number of pigeons = 6. This is possible when there are 3 sparrows, 3 pigeons and 3 ducks i.e. 9 birds in all.

18.

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) 85
(B) 80
(C) 75
(D) 70

Solution:

 Clearly, out of every 16 persons, there is one captain. So, number of captains (1200/16) = 75.

19.

Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77 but three tickets from city A to B and two tickets from city A to C cost Rs. 73. What are the fares for cities B and C from A ?

(A) Rs. 4, Rs. 23
(B) Rs. 13, Rs. 17
(C) Rs. 15, Rs. 14
(D) Rs. 17, Rs. 13

Solution:

 Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A. Then, 2x + 3y = 77 ...(i) and 3x + 2y = 73 ...(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17. Putting y = 17 in (i), we get: x = 13.

20.

A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed

(A) 30 birds
(B) 60 birds
(C) 72 birds
(D) 90 birds

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Practice MCQ Questions and Answer on Arithmetic Reasoning

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