Exercise : 3


1. The average age of A and B is 20 years. If C were to replace A, the average would be 19 and if C were to replace B, the average would be 21. What are the age of A, B and C?
(a) 22, 18, 20
(b) 20, 20, 18
(c) 18, 22, 20
(d) 21, 20, 19
(e) None of the above
Ans.a

Given A + B = 40…(i)

C + B = 38 … (ii)

A + C = 42…(iii)

(i) + (ii) + (iii) ⇒ A + B + C = 60 …(iv)

from (i) and (iv), we get

C = 20 years

∴ B = 18 years and A = 22 years

2. 3 years ago the average age of a family of 5 members was 17 years. With the birth of a new baby, the average age of six members remains the same even today. Find the age of the new baby.
(a) 1 year
(b) 2 years
(c) 112 years
(d) cannot be determined
(e) None of the above
Ans.b

Sum of present ages of the six members

= (17 × 6) years = 102 years.

Sum of present ages of the 5 members (excluding baby)

= 5 × (17 + 3) years = 100 years.

∴ Age of the baby = 102 – 100 = 2 years

3. The average age of a group of person going for picnic is 16 years. Twenty new persons with an average age of 15 years join the group on the spot due to which their average becomes 15.5 years. Find the number of persons initially going for picnic.
(a) 20
(b) 18
(c) 22
(d) 19
(e) None of the above
Ans.a

Let the number of persons, initially going for Picnic = x

∴ Sum of their ages = 16x

Also, \(\frac{16x + 15 \times 20}{x + 20}\) = 15.5

⇒ 0.5x = 10 ⇒ x = 20 years.

4. A batsman in his 12th innings makes a score of 65 and thereby increases his average by 2 runs. What is his average after the 12th innings if he had never been ‘not out’?
(a) 42
(b) 43
(c) 44
(d) 45
(e) None of the above
Ans.b

Let ‘x’ be the average score after 12 th innings

⇒ 12x = 11 × (x – 2) + 65

∴ x = 43

5. A pupil’s marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of pupils in the class is:
(a) 10
(b) 20
(c) 40
(d) 73
(e) None of the above
Ans.c

Let there be x pupils in the class.

Total increase in marks = (x × 12) = x2.

x2 = (83 - 63) ⇒ x2 = 20 ⇒ x = 40.

6. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach a target of 282 runs ?
(a) 6.25
(b) 6.50
(c) 6.75
(d) 7.00
(e) None of the above
Ans.a

Total runs in the first 10 overs

= 10 × 3.2 = 32

Run rate required in the remaining 40 overs

= \(\frac{282 - 32}{40} = \frac{250}{40} = 6.25\) runs per over.

7. The average number of printing error per page in a book of 512 pages is 4. If the total number of printing error in the first 302 pages is 1,208, the average number of printing errors per page in the remaining pages is
(a) 0
(b) 4
(c) 840
(d) 90
(e) None of the above
Ans.b

Remaining pages = 512 – 302 = 210

Let average printing error in remaining pages = x

Then, \(\frac{1208 + 210 \times x}{512} = 4\)

⇒ 210x = 840 ⇒ x = 4

8. The average attendance in a school for the first 4 days of the week is 30 and for the first 5 days of the week is 32. The attendance on the fifth day is
(a) 32
(b) 40
(c) 38
(d) 36
(e) None of the above
Ans.b

Attendance on the fifth day = 32 × 5 – 30 × 4

= 160 – 120 = 40

9. The average expenditure of a labourer for 6 months was 85 and he fell into debt. In the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30. His monthly income is
(a) 70
(b) 72
(c) 75
(d) 78
(e) None of the above
Ans.d

Income of 6 months = (6 × 85) – debt

= 510 – debt

Income of the man for next 4 months

= 4 × 60 + debt + 30

= 270 + debt

∴ Income of 10 months = 780

Average monthly income = 780 ÷ 10 = 78

10. The average of a batsman for 40 innings is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, his average drops by 2 runs. Find his highest score.
(a) 172
(b) 173
(c) 174
(d) 175
(e) None of the above
Ans.c

Total runs = 40 × 50 = 2000

Let his highest score be = x

Then his lowest score = x – 172

Now \(\frac{200 - x - \left(x - 172 \right)}{38} = 48\)

⇒ 2x = 2172 – 1824

⇒ x = 174