Problems on Average - Solved Examples

11. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is
A. $32\frac{2}{7}$ years	B. $31\frac{5}{7}$ years
C. $28\frac{1}{7}$ years	D. $30\frac{5}{7}$ years

answer with explanation

Answer: Option B
Explanation:
Total age of the grandparents = 67 × 2
Total age of the parents = 35 × 2
Total age of the grandchildren = 6 × 3

Average age of the family
$=\frac{(67\times 2)+(35\times 2)+(6\times 3)}{7}=\frac{134+70+18}{7}=\frac{222}{7}=31\frac{5}{7}$

12. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B?
A. $31$ kg	B. $28\frac{1}{2}$ kg
C. $32$ kg	D. $30\frac{1}{2}$ kg

answer with explanation

Answer: Option A
Explanation:
Let the weight of A, B and C are a,b and c respectively.

Average weight of A,B and C = 45
a + b + c = 45 × 3 = 135 --- equation(1)

Average weight of A and B = 40
a + b = 40 × 2 = 80 --- equation(2)

Average weight of B and C = 43
b + c = 43 × 2 = 86 --- equation(3)

equation(2) + equation(3) - equation(1)
=> a + b + b + c - (a + b + c) = 80 + 86 - 135
=> b = 80 + 86 -135 = 166 - 135 = 31

Weight of B = 31 Kg

13. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, what is the average marks of all the students?
A. 53.23	B. 54.68
C. 51.33	D. 50

answer with explanation

Answer: Option B
Explanation:
Average marks of batch1 = 50
Students in batch1 = 55
Total marks of batch1 = 55 × 50

Average marks of batch2 = 55
Students in batch2 = 60
Total marks of batch2 = 60 × 55

Average marks of batch3 = 60
Students in batch3 = 45
Total marks of batch3 = 45 × 60

Total students = 55 + 60 + 45 = 160

Average marks of all the students
$=\frac{(55\times 50)+(60\times 55)+(45\times 60)}{160}=\frac{275+330+270}{16}=\frac{875}{16}=54.68$

14. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband?
A. 40	B. 32
C. 28	D. 30

answer with explanation

Answer: Option A
Explanation:
Let the present age of the husband = h
Present age of the wife = w
Present age of the child = c

3 years ago, average age of husband, wife and their child = 27
=> Sum of age of husband, wife and their child before 3 years = 3 × 27 = 81
=> (h-3) + (w-3) + (c-3) = 81
=> h + w + c = 81 + 9 = 90 --- equation(1)

5 years ago, average age of wife and child = 20
=> Sum of age of wife and child before 5 years = 2 × 20 = 40
=> (w-5) + (c-5) = 40
=> w + c = 40 + 10 = 50 --- equation(2)

Substituting equation(2) in equation(1)
=> h + 50 = 90
=> h = 90 - 50 = 40

i.e., Present age of the husband = 40

15. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person?
A. 75 Kg	B. 50 Kg
C. 85 Kg	D. 80 Kg

answer with explanation

Answer: Option C
Explanation:
Total increase in weight = 8 × 2.5 = 20

If $x$ is the weight of the new person, total increase in weight = $x-65$
=> 20 = $x$ - 65
=> $x$ = 20 + 65 = 85

16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class?
A. 38.25	B. 37.25
C. 38.5	D. 37

answer with explanation

Answer: Option B
Explanation:
Total weight of students in division A = 36 × 40
Total weight of students in division B = 44 × 35

Total students = 36 + 44 = 80

Average weight of the whole class
$=\frac{(36\times 40)+(44\times 35)}{80}=\frac{(9\times 40)+(11\times 35)}{20}=\frac{(9\times 8)+(11\times 7)}{4}=\frac{72+77}{4}=\frac{149}{4}=37.25$

17. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning?
A. 39	B. 35
C. 42	D. 40.5

answer with explanation

Answer: Option A
Explanation:
Let the average after 17 innings = x
Total runs scored in 17 innings = 17x

Average after 16 innings = (x-3)
Total runs scored in 16 innings = 16(x-3)

Total runs scored in 16 innings + 87 = Total runs scored in 17 innings
=> 16(x-3) + 87 = 17x
=> 16x - 48 + 87 = 17x
=> x = 39

18. A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What is the value of x?
A. 12	B. 5
C. 7	D. 9

answer with explanation

Answer: Option C
Explanation:

$\frac{\text{3+11+7+9+15+13+8+19+17+21+14+}x}{12}$ $=12$

$\Rightarrow \frac{137+x}{12}=12\Rightarrow 137+x=144\Rightarrow x=144-137=7$

19. Arun obtained 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Chemistry, Biology and Physics. What is his average mark?
A. 53	B. 54
C. 72	D. 75

answer with explanation

Answer: Option D
Explanation:
Average mark = $\frac{76+65+82+67+85}{5}=\frac{375}{5}=75$

20. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of the train during the whole journey?
A. 69.0 km /hr	B. 69.2 km /hr
C. 67.2 km /hr	D. 67.0 km /hr

answer with explanation

Answer: Option C
Explanation:
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Solution 1
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If a car covers a certain distance at $x$ kmph and an equal distance at $y$ kmph. Then,
average speed of the whole journey = $\frac{2xy}{x+y}$ kmph.

By using the same formula, we can find out the average speed quickly.

Average speed
$=\frac{2\times 84\times 56}{84+56}=\frac{2\times 84\times 56}{140}=\frac{2\times 21\times 56}{35}=\frac{2\times 3\times 56}{5}=\frac{336}{5}=67.2$

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Solution 2
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Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics.

Let the distance between A and B = $x$

Train travels from A to B at 84 km per hour
Total time taken for traveling from A to B = $\frac{\text{distance}}{\text{speed}}=\frac{x}{84}$

Train travels from B to A at 56 km per hour
Total time taken for traveling from B to A = $\frac{\text{distance}}{\text{speed}}=\frac{x}{56}$

Total distance travelled = $x+x=2x$
Total time taken = $\frac{x}{84}+\frac{x}{56}$


Average speed $=\frac{\text{Total distance traveled}}{\text{Total time taken}}$

$=\frac{2x}{\frac{x}{84}+\frac{x}{56}}=\frac{2}{\frac{1}{84}+\frac{1}{56}}=\frac{2\times 84\times 56}{56+84}=\frac{2\times 84\times 56}{140}=\frac{2\times 21\times 56}{35}=\frac{2\times 3\times 56}{5}=\frac{336}{5}=67.2$