Average - Aptitude MCQ Questions and Solutions with Explanations

Let the number be x.
Then, second number = 2x.
First number = 4x.
∴ $$\frac{1}{x}$$ + $$\frac{1}{2x}$$ + $$\frac{1}{4x}$$ = $$\frac{7}{72}$$ × 3
⇒ $$\frac{7}{4x}$$ = $$\frac{7}{24}$$
⇒ 4x = 24
⇒ x = 6
So, the number are 24, 12 and 6

According to given information
Average of odd numbers = Average of even numbers - 6
$$ \Rightarrow \frac{{{x_1} + {x_2} + {x_3} + {x_4}}}{4} = $$     $$\frac{{{y_1} + {y_2} + {y_3}}}{3} - 6$$
$$ \Rightarrow \frac{{{x_1} + {x_2} + {x_3} + {x_4}}}{4} = $$     $$\frac{{{y_1} + {y_2} + {y_3} - 18}}{3}$$
$$ \Rightarrow 3\left( {{x_1} + {x_2} + {x_3} + {x_4}} \right) = $$     $$4\left( {{y_1} + {y_2} + {y_3}} \right) - 72$$
Also,
$$ \Rightarrow {y_1} + {y_2} + {y_3} = $$     $${x_1} + {x_2} + {x_3} + {x_4} - 16$$
$$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} = $$     $${y_1} + {y_2} + {y_3} + 16$$   .....(i)
So we have,
$$ \Rightarrow 3\left( {{y_1} + {y_2} + {y_3} + 16} \right) = $$     $$4\left( {{y_1} + {y_2} + {y_3}} \right) - 72$$
$$ \Rightarrow 3{y_1} + 3{y_2} + 3{y_3} + 48 = $$     $$4{y_1} + 4{y_2} + 4{y_3} - 72$$
$$ \Rightarrow 4{y_1} + 4{y_2} + 4{y_3}$$   $$ - 3{y_1} - 3{y_2} - 3{y_3}$$     = 48 + 72
$$ \Rightarrow {y_1} + {y_2} + {y_3} = 120$$
$$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4}$$     = 120 + 16 = 136 [From (i)]
∴ Average of four odd numbers :
$$\eqalign{ & = \frac{{{x_1} + {x_2} + {x_3} + {x_4}}}{4} \cr & = \frac{{136}}{4} \cr & = 34 \cr} $$

Let the three positive numbers in increasing order be a, b and c and the average of these numbers be be A.
Then,
$$\frac{{a + b + c}}{3} = A.....(i)$$
Given,
$$\eqalign{ & c - \frac{A}{3} = 8 \cr & \Rightarrow c - \frac{{a + b + c}}{9} = 8.....(ii) \cr} $$
Also given,
$$\eqalign{ & \frac{{b + a}}{2} = 8 \cr & \Rightarrow a + b = 16.....(iii) \cr} $$
Putting the value of (a + b) in equation (ii), we get
$$\eqalign{ & \Rightarrow c - \left( {\frac{{16 + c}}{9}} \right) = 8 \cr & \Rightarrow 9c - 16 - c = 72 \cr & \Rightarrow 8c = 72 + 16 \cr & \Rightarrow 8c = 88 \cr & \Rightarrow c = 11 \cr} $$
∴ Highest number = 11

Total number of students in class = 60
Average score of passed student = 52
Average score of failed students = 16
By applying the rule of alligation,

∴ Number of students who failed in exam
= $$\frac{1}{4}$$ × 60
= 15

Let the number of boys and girls in the class are x and y.
According to given information,
⇒ 30x + 20y = 23.25 (x + y)
⇒ 30x + 20 y = 23.25x + 23.25y
⇒ 30x - 23.25x = 23.25y - 20y
⇒ 6.75x = 3.25y
⇒ $$\frac{x}{y}$$ = $$\frac{3.25}{6.75}$$
⇒ $$\frac{x}{y}$$ = $$\frac{13}{27}$$
Hence, possible number of boys and girls 13 and 27 respectively.

Let the number of students with work experience be x and those without work experience be y.
Then, 18000x + 12000y = 16000 (x + y)
⇒ 2000x = 4000y
⇒ $$\frac{x}{y}$$ = $$\frac{2}{1}$$
∴ Percentage of students with work experience
= $$\left( {\frac{2}{3} \times 100} \right)\% $$
= 66.67%
Percentage of students without work experience
= (100 - 66.67)%
= 33.33%

Arithmetic mean
$$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {\frac{{1132 + 1140 + 1144 + 1136 + 1148}}{5}} \right) \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{5700}}{5}} \right) \cr & = {\text{Rs}}{\text{. 1140}} \cr} $$

Total money paid for 115 books
= Rs. (1050 + 1020)
= Rs. 2070
∴ Average price paid per book
= Rs. ($$\frac{2070}{115}$$)
= Rs. 18

Age of the instructor
= (34 × 11 - 32 × 10) years
= (374 - 320) years
= 54 years

Total age decreased
= (24 × 1) months
= 24 months
= 2 years
∴ Age of the newcomer
= (18 - 2) years
= 16 years