Average - Aptitude MCQ Questions and Solutions with Explanations

Income of A and B
= 2 × 14000
= Rs. 28000
Income of B and C
= 2 × 15600
= Rs. 31200
Income of A and C
= 2 × 14400
= Rs. 28800
Income of A, B and C
$$ = \frac{{\left( {28000 + 31200 + 28800} \right)}}{2}$$
= Rs. 44000
C's income
= 44000 - 28000
= Rs. 16000

According to the question,
Average of 25 observations = 13
Sum of 25 observations = 13 × 25 = 325
One observation entered wrongly 48 instead of 73.
∴ Difference = 73 - 48 = 25 (less)
∴ Actual sum of 25 observation = 325 + 25 = 350
Actual average = $$\frac{350}{25}$$ = 14

Average of twelve number = 15
Sum of twelve number = 15 × 12 = 180
Average of first two number = 14
Sum of first two number = 14 × 2 = 28
Sum of first two + Sum of rest = 180
Sum of rest = 180- 28 = 152
Average of rest = $$\frac{152}{10}$$ = $$15\frac{1}{5}$$

Let the no. of apples be 3x, 2x and 5x
∴ Average Cost
$$\eqalign{ & = \frac{{3x \times 15 + 2x \times 10 + 5x \times 5}}{{3x + 2x + 5x}} \cr & = \frac{{45x + 20x + 25x}}{{10x}} \cr & = \frac{{90x}}{{10x}} \cr & = {\text{Rs}}{\text{.}}\,{\text{9}} \cr} $$

Sum of ages of 6 sons of a family = 8 × 6 = 48
Sum of ages of 6 sons and their parents = 8 × 22 = 176
Parents's age = 176 - 48 = 128
Father's age - Mother's age = 8
x - y = 8
x + y = 128
x = 68
y = 60
∴ Mother's age = 60 years

Sum of age of 40 boys
= 16 × 40 = 640
New age of 40 boys
= 15.875 × 40 = 635
Difference = 640 - 635 = 5 years
17 - x = 5
x = 17 - 5 = 12 years

Alternate :
Average is decreased it means the boy who joined the class is younger than boy who leave the class.
Let the age of boy who join = x
17 - x = difference in average
$$\frac{17 - x}{40}$$ = 0.125
17 - x = 5
x = 12

According to the question,
$$\eqalign{ & \Rightarrow \frac{{{x_1} + {x_2} + {x_3} + {x_4} + .... + {x_{20}}}}{{20}} = y \cr & \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + .... + {x_{20}} = 20y \cr} $$

$$ = \frac{{{x_1} - 101 + {x_2} - 101 + {x_3} - 101 + {x_4} - 101 + .... + {x_{20}} - 101}}{{20}}$$
$$ = \frac{{\left( {{x_1} + {x_2} + {x_3} + {x_4} + .... + {x_{20}}} \right) - 20 \times 101}}{{20}}$$
$$\eqalign{ & = \frac{{20y - 20 \times 101}}{{20}} \cr & = y - 101 \cr} $$

According to the question,
Let M = 1
∴ 5 consecutive integers are = 1, 2, 3, 4, 5
∴ $$\frac{1 + 2 + 3 + 4 + 5}{5}$$   = n
n = $$\frac{15}{5}$$ = 3
∴ 6 consecutive integers starting with (m + 2) are = 3, 4, 5, 6, 7, 8
∴ $$\frac{3 + 4 + 5 + 6 + 7 + 8 }{6}$$    = $$\frac{33}{6}$$  = $$\frac{11}{2}$$
Now check from option to put n = 3
Option : (A) $$\frac{(2n + 5)}{2}$$
= $$\frac{2 × 3 + 5}{2}$$   = $$\frac{11}{2}$$ (satisfied)

According to the question,
Wrong marks = 42 + 74 = 116
Correct marks = 56 + 32 = 88
Difference = 116 - 88 = 28 marks
∵ This difference effect the 14 students = $$\frac{28}{14}$$ = 2
And, incorrect average = 71
∴ Correct average = 71 - 2 = 69

According to the question,
20 over match required run rate = 7.2
Total runs are = 7.2 × 20 = 144 runs
If the run rate is 6 at the end of the 15th over
∴ Required runs
= 144 - 90 = 54 runs
Required run rate
= $$\frac{54}{5}$$
= 10.8