Average - Aptitude MCQ Questions and Solutions with Explanations

Let the number of female students be x
Let weight of female students = 57x
Number of male students = 66
Total weights of male students = 65.9 × 66
Average weight of all the students = 60.3 kg
Total weights of all the students = 60.3 (66 + x)
According to given information,
Then,
⇒ 60.3 (66 + x) = 66 × 65.9 + 57x
⇒ 60.3 × 66 + 60.3x = 66 × 65.9 + 57x
⇒ 60.3x - 57x = 66 (65.9 - 60.3)
⇒ 3.3x = 66 × 5.6
∴ x = $$\frac{66 × 5.6}{3.3}$$
⇒ x = 2 × 56
⇒ x = 112

Average monthly income of P and Q = Rs. 6000
Average monthly income of Q and R = Rs. 5250
Average monthly income of P and R = Rs. 5500
Total income of P + Q
= 2 × 6000
= Rs. 12000.....(i)
Total income of Q + R
= 2 × 5250
= Rs. 10500.....(ii)
Total income of R + P
= 2 × 5500
= Rs. 11000.....(iii)
On adding equation (i) (ii) and (iii), we have
2 (P + Q + R) = 12000 + 10500 + 11000
⇒ P + Q + R = $$\frac{33500}{2}$$
⇒ P + Q + R = Rs. 16750.....(iv)
By equation (iv) - (ii)
P's monthly income
= Rs. (16750 - 10500)
= Rs. 6250

Average weight of 75 girls = 47 kgs
Total weight of 75 girls
= 47 × 75
= 3525 kgs
Actual weight of 75 girls = x
Correct weight of 75 girls
= 3525 - 45 + 25
= 3525 - 20
= 3505 kgs
∴ Required average weight
x =$$\frac{3505}{75}$$
x = 46.73 kgs

Average
$$ = \left( {\frac{{385 + 441 + 876 + 221 + 536 + 46 + 291 + 428}}{8}} \right){\text{kg}}$$
$$\eqalign{ & = \left( {\frac{{3224}}{8}} \right){\text{kg}} \cr & = 403{\text{ kg}} \cr} $$

25a + 25b = 115
⇒ 25 (a + b) = 115
⇒ a + b = $$\frac{115}{25}$$
⇒ a + b = $$\frac{23}{5}$$
∴ Average of a and b
= $$\frac{a + b}{2}$$
= $$\frac{23}{5}$$ × $$\frac{1}{2}$$
= $$\frac{23}{10}$$
= 2.3

Sum of numbers = 2XY
∴ Other number = 2XY - X

Weight of the teacher
= (35.4 × 25 - 35 × 24) kg
= 45 kg

Let the original strength of the class be x
Sum of ages of the whole class = 40x years
Sum of ages of 12 now students
= (12 × 32) years
= 384 years
∴ $$\frac{40x + 384}{x + 12}$$  = 36
⇒ 40x + 384 = 36x + 432
⇒ 4x = 48
⇒ x = 12
Hence, the original strength of the class = 12

Average
$$\eqalign{ & = \frac{{3\left( {1 + 2 + 3 + 4 + 5} \right)}}{5} \cr & = \frac{{45}}{5} \cr & = 9 \cr} $$

$$\eqalign{ & {{\text{X}}_1} + {{\text{X}}_2} + {{\text{X}}_3} \cr & = \left( {14 \times 3} \right) \cr & = 42 \cr} $$
According to the question,
$$\eqalign{ & 2\left( {{{\text{X}}_2} + {{\text{X}}_3}} \right) = 30 \cr & \Rightarrow {{\text{X}}_2} + {{\text{X}}_3} = 15 \cr & \therefore {{\text{X}}_1} = \left( {42 - 15} \right) = 27 \cr} $$