Practice MCQ Questions and Answer on Arithmetic Reasoning

51.

In a garden, there are 10 rows and 12 columns of mango trees. The distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden. The length of the garden is

• (A) 20 m
• (B) 22 m
• (C) 24 m
• (D) 26 m

Show Answer

 Each row contains 12 plants. There are 11 gapes between the two corner trees (11 x 2) metres and 1 metre on each side is left. Therefore Length = (22 + 2) m = 24 m.

52.

In a class, 3/5 of the students are girls and rest are boys. If 2/9 of the girls and 1/4 of the boys are absent, what part of the total number of students is present ?

• (A) 17/25
• (B) 18/49
• (C) 23/30
• (D) 23/36

Show Answer

53.

A number consists of two digits whose sum is 11. If 27 is added to the number, then the digits change their places. What is the number ?

• (A) 47
• (B) 65
• (C) 83
• (D) 92

Show Answer

 Let the ten's digit be x. Then, unit's digit = (11 - x). So, number = 10x + (11 - x) = 9x + 11. Therefore (9x + 11) + 27 = 10 (11 - x) + x ⇔ 9x + 38 = 110 - 9x ⇔ 18x = 72 ⇔ x = 4. Thus, ten's digit = 4 and unit's digit = 7. Hence, required number = 47.

54.

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

Show Answer

 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.

55.

Nitin's age was equal to square of some number last year and the following year it would be cube of a number. If again Nitin's age has to be equal to the cube of some number, then for how long he will have to wait?

• (A) 10 years
• (B) 38 years
• (C) 39 years
• (D) 64 years

Show Answer

 Clearly, we have to first find two numbers whose difference is 2 and of which the smaller one is a perfect square and the bigger one a perfect cube. Such numbers are 25 and 27. Thus, Nitin is now 26 years old. Since the next perfect cube after 27 is 64, so required time period = (64 - 26) years = 38 years.

56.

The number of boys in a class is three times the number of girls. Which one of the following numbers cannot represent the total number of children in the class ?

• (A) 48
• (B) 44
• (C) 42
• (D) 40

Show Answer

 Let number of girls = x and number of boys = 3x. Then, 3x + x = 4x = total number of students. Thus, to find exact value of x, the total number of students must be divisible by 4.

57.

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

Show Answer

 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.

58.

The 30 members of a club decided to play a badminton singles tournament. Every time a member loses a game he is out of the tournament. There are no ties. What is the minimum number of matches that must be played to determine the winner ?

• (A) 15
• (B) 29
• (C) 61
• (D) None of these

Show Answer

 Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.

59.

A, B, C, D and E play a game of cards. A says to B, "If you give me 3 cards, you will have as many as I have at this moment while if D takes 5 cards from you, he will have as many as E has." A and C together have twice as many cards as E has. B and D together also have the same number of cards as A and C taken together. If together they have 150 cards, how many cards has C got ?

• (A) 28
• (B) 29
• (C) 31
• (D) 35

Show Answer

 Clearly, we have : A = B - 3 ...(i) D + 5 = E ...(ii) A+C = 2E ...(iii) B + D = A+C = 2E ...(iv) A+B + C + D + E=150 ...(v) From (iii), (iv) and (v), we get: 5E = 150 or E = 30. Putting E = 30 in (ii), we get: D = 25. Putting E = 30 and D = 25 in (iv), we get: B = 35. Putting B = 35 in (i), we get: A = 32. Putting A = 32 and E = 30 in (iii), we get: C = 28.

60.

Five bells begin to toll together and toll respectively at intervals of 6, 5, 7, 10 and 12 seconds. How many times will they toll together in one hour excluding the one at the start ?

• (A) 7 times
• (B) 8 times
• (C) 9 times
• (D) 11 times

Show Answer

 L.C.M. of 6, 5, 7, 10 and 12 is 420. So, the bells will toll together after every 420 seconds i.e. 7 minutes. Now, 7 x 8 = 56 and 7 x 9 = 63. Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.