Practice MCQ Questions and Answer on Arithmetic Reasoning

1.

A total of 324 coins of 20 paise and 25 paise make a sum of Rs. 71. The number of 25-paise coins is

• (A) 120
• (B) 124
• (C) 144
• (D) 200

Show Answer

 Let the number of 20-paise coins be x. Then, number of 25-paise coins = (324 - x). Therefore 0.20 x x + 0.25 (324 - x) = 71 ⇔ 20x + 25 (324 - x) = 7100 ⇔ 5x= 1000 ⇔ x = 200. Hence, number of 25-paise coins = (324 - x) - 124.

2.

A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?

• (A) 45
• (B) 60
• (C) 75
• (D) 90

Show Answer

 Let number of notes of each denomination be x. Then, x + 5x + 10x = 480 ⇔ 16x = 480 ⇔ x = 30. Hence, total number of notes = 3x = 90.

3.

Mac has £ 3 more than Ken, but then Ken wins on the horses and trebles his money, so that he now has £ 2 more than the original amount of money that the two boys had between them. How much money did Mac and Ken have between them before Ken's win ?

• (A) £ 9
• (B) £ 11
• (C) £ 13
• (D) £ 15

Show Answer

 Let money with Ken = x. Then, money with Mac = x + £ 3. Now, 3x = (x + x + £ 3) + £ 2 ⇔ x = £ 5. Therefore Total money with Mac and Ken = 2x + £ 3 = £ 13.

4.

An enterprising businessman earns an income of Re. 1 on the first day of his business. On every subsequent day, he earns an income which is just double of that made on the previous day. One the 10th day of business, his income is

• (A) Rs. 29
• (B) Rs. 210
• (C) Rs. 10
• (D) Rs. 102

Show Answer

 Income on the first day = Re. 1. Income on the 2nd day = Rs. (1x2) = Rs. 21. Income on the 3rd day = Rs. (21 x 2) = Rs. 22 and so on. Thus, Income on the rath day = Rs. 2n-1. Therefore Income on the 10th day = Rs. 29.

5.

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

Show Answer

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

6.

The total number of digits used in numbering the pages of a book having 366 pages is

• (A) 732
• (B) 990
• (C) 1098
• (D) 1305

Show Answer

 Total number of digits = (No. of digits in 1- digit page nos. + No. of digits in 2-digit page nos. + No. of digits in 3- digit page nos.) = (1 x 9 + 2 x 90 + 3 x 267) = (9 + 180 + 801) = 990.

7.

Ravi's brother is 3 years senior to him. His father was 28 years of age when his sister was born while his mother was 26 years of age when he was born. If his sister was 4 years of age when his brother was born, what were the ages of Ravi's father and mother respectively when his brother was born ?

• (A) 32 years, 23 years
• (B) 32 years, 29 years
• (C) 35 years, 29 years
• (D) 35 years, 33 years

Show Answer

 When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years. Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32. Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years. Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.

8.

After distributing the sweets equally among 25 children, 8 sweets remain. Had the number of children been 28, 22 sweets would have been left after equal distribution. What was the total number of sweets ?

• (A) 328
• (B) 348
• (C) 358
• (D) Data inadequate

Show Answer

 Let the total number of sweets be (25x + 8). Then, (25x + 8) - 22 is divisible by 28 ⇔ (25x - 14) is divisible by 28 ⇔ 28x - (3x + 14) is divisible by 28 ⇔ (3x + 14) is divisible by 28 ⇔ x = 14. Therefore Total number of sweets = (25 x 14 + 8) = 358.

9.

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.

10.

I have a few sweets to be distributed. If I keep 2, 3 or 4 in a pack, I am left with one sweet. If I keep 5 in a pack, I am left with none. What is the minimum number of sweets I have to pack and distribute ?

• (A) 25
• (B) 37
• (C) 54
• (D) 65

Show Answer

 Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25.