Average - Aptitude MCQ Questions and Solutions with Explanations

When the first child was born, the total age of all the family members
= (16 × 3) years
= 48 years
When the second child was born, the total age of all the family members
= (15 × 4) years
= 60 years
By the time the second child was born, each one of the 3 family members had grown by
$$\eqalign{ & = \left( {\frac{{60 - 48}}{3}} \right) \cr & = \frac{{12}}{3} \cr} $$
= 4 years
Hence, the age of eldest son when the second child was born = 4 years
When the third child was born, the total age of all the family members
= (16 × 5) years
= 80 years
By the time, the third child was born, each one of the four family members had grown by
= $$\left( {\frac{{80 - 60}}{4}} \right)$$
= 5 years
So, the age of the eldest son when the third child was born
= (4 + 5) years
= 9 years
When the fourth child was born, the total age of all the family members
= (15 × 6) years
= 90 years
By the time, the fourth child was born, each of the five family members had grown by
= $$\left( {\frac{{90 - 80}}{5}} \right)$$
= 2 years
So, the age of the eldest son when the fourth child was born
= (9 + 2) years
= 11 years
At present, the total age of all the 6 family members
= (16 × 6) years
= 96 years
By now, each one of the 6 members have grown by
= $$\left( {\frac{{96 - 90}}{6}} \right)$$  years
= 1 year
Hence, the present age of the eldest son
= (11 + 1) years
= 12 years

Total number of teachers = 10
Age of new teacher = 25 years
Age of the retire teacher
= (25 + 3 × 10) years
= 55 years

Let the total number of workers be x.
Then,
⇒ 8000x = (12000 × 7) + 6000(x - 7)
⇒ 2000x = 42000
⇒ x = 21

Sum of the present ages of husband, wife and child
= (23 × 2 + 5 × 2) + 1
= 57 years
∴ Required average
= $$\left( {\frac{{57}}{3}} \right)$$
= 19 years

Let the original average weight of the class be x kg and let there be n students.
Then, sum of weights of n students = (nx) kg
$$\eqalign{ & \therefore \frac{{nx + 50}}{{n + 1}} = x + 1 \cr & \Rightarrow nx + 50 = \left( {n + 1} \right)\left( {x + 1} \right) \cr & \Rightarrow nx + 50 = nx + x + n + 1 \cr & \Rightarrow x + n = 49 \cr & \Rightarrow 2x + 2n = 98.....(i) \cr & {\text{And,}} \cr & \Rightarrow \frac{{nx + 100}}{{n + 2}} = x + 1.5 \cr & \Rightarrow nx + 100 = \left( {n + 2} \right)\left( {x + 1.5} \right) \cr & \Rightarrow nx + 100 = nx + 1.5n + 2x + 3 \cr & \Rightarrow 2x + 1.5n = 97.....(ii) \cr} $$
Subtracting (ii) from (i), we get: 0.5n = 1 or n = 2
Putting n = 2 in (i), we get: x = 47

Sum of height of the 5 boys
= (25 × 1.4 - 20 × 1.55) m
= 4 m
∴ Required average
= $$\left( {\frac{4}{5}} \right)$$
= 0.8 m

Monthly income of the married daughter
= Rs. (15130 × 4 - 14660 × 3)
= Rs. (60520 - 43980)
= Rs. 16540

Mark obtained by the 11th candidate
= [(45 × 22) - (55 × 10 + 40 × 11)]
= (990 - 990)
= 0

Correct sum
= (40.2 × 10 - 18 + 33 - 13)
= 404
∴ Correct average
= $$\left( {\frac{{404}}{{10}}} \right)$$
= 40.4

Correct sum
= (160 × 35 + 104 - 144) cm
= 5560 cm
∴ Actual average height
= $$\left( {\frac{{5560}}{{35}}} \right)$$  cm
= 158.857 cm $$ \approx $$ 158.86 cm