Average - Aptitude MCQ Questions and Solutions with Explanations

Sum of odd numbers upto 100 = 1 + 3 + 5 +.....+ 99
= $$\frac{50}{2}$$ [2 + (50 - 1) × 2]
= 2500
[$$\because $$ Sum of n terms of an A.P. with first term a and common difference]
$$\left[ {d = \frac{n}{2}\{ 2n + \left( {n - 1} \right)d\} } \right]$$
∴ Required average
= $$\frac{2500}{50}$$
= 50

Number of men
= $$\frac{1}{5}$$ × 120
= 24
Number of women
= $$\frac{1}{4}$$ × 120
= 30
Number of children
= 120 - (24 + 30)
= 66
Average age of men = 60 years
Average age of children
= $$\frac{1}{4}$$ × 60
= 15 years
Average age of women
= $$\frac{5}{6}$$ × 60
= 50 years
∴ Avarage age of the group
$$\eqalign{ & = \left( {\frac{{60 \times 24 + 50 \times 30 + 15 \times 66}}{{120}}} \right){\text{ years}} \cr & = \left( {\frac{{3930}}{{120}}} \right){\text{ years}} \cr & = 32.75{\text{ years}} \cr} $$

Average of 10 numbers = 7
Sum of these 10 numbers = (10 × 7) = 70
$$\eqalign{ & \therefore {x_1} + {x_2} + ..... + {x_{10}} = 70 \cr & \Rightarrow 12{x_1} + 12{x_2} + ..... + 12{x_{10}} = 840 \cr & \Rightarrow \frac{{12{x_1} + 12{x_2} + ..... + 12{x_{10}}}}{{10}} = 84 \cr} $$
⇒ Average of new numbers is 84

M + T + W = (37 × 3)°C = 111°C.....(i)
T + W + Th = (34 × 3)°C = 102°C.....(ii)
Subtracting (ii) from (i), we get:
⇒ M - Th = 9°C
⇒ M - $$\frac{4}{5}$$M = 9
⇒ $$\frac{1}{5}$$ M = 9
⇒ M = 45
∴ Temperature on Thursday
= $$ \left( {\frac{{4}}{{5}}} \times 45 \right) $$ °C
= 36°C

Total weight of (A + B + C)
=( $$54\frac{1}{3}$$ × 3 ) kg
= 163 kg
Total weight of (B + D + E)
= (53 × 3) kg
= 159 kg
Adding both, we get:
= A + 2B + C + D + E
= (163 + 159) kg
= 322 kg
So, to find average weight of A, B, C, D and E, we ought to know B's weight, which is not given.
So, the data is inadequate.

Correct sum
= 36 × 100 + 90 - 40
= 3650
Correct average
= $$\frac{3650}{100}$$
= 36.5
Error = (36.5 - 36) = 0.5
∴ Error %
= ( $$\frac{0.5}{36.5}$$ × 100 )%
= $$\frac{100}{73}$$%
= 1.36%

Let the eight number be x
Then, sixth number = (x - 7)
Seventh number = (x - 7) + 4 = (x - 3)
So,
$$ \Leftrightarrow \left( {2 \times 15\frac{1}{2}} \right) + \left( {3 \times 21\frac{1}{2}} \right)$$     $$ + \left( {x - 7} \right)$$   $$ + \left( {x - 3} \right)$$   $$ + \,x = 8 \times 20$$
$$\eqalign{ & \Leftrightarrow 31 + 64 + \left( {3x - 10} \right) = 160 \cr & \Leftrightarrow 3x = 75 \cr & \Leftrightarrow x = 25 \cr} $$

Let the number of paper be x
Then, pupil's total score = 63x
∴ $$\frac{63x + 20 + 2}{x}$$   = 65
⇒ 2x = 22
⇒ x = 11

Let the average score of the remaining 6 batsmen be x runs.
Then, sum of their scroe = 6x
Captain's score = (x + 30)
∴ 6x + (x + 30) = 310
⇒ 7x = 280
⇒ x = 40
Hence, captain's score = x + 30 = 70

Total sale for 5 months
= Rs. (6435 + 6927 + 6855 + 7230 + 6562)
= Rs. 34009
∴ Required sale
= Rs. [(6500 × 6) - 34009]
= Rs. (39000 - 34009)
= Rs. 4991