Average - Aptitude MCQ Questions and Solutions with Explanations

The total wages earned during the 15 days that the worker worked ,
= 15 × 90
= Rs. 1350
The total wages earned during the first 7 days
= 7 × 87
= Rs. 609
The total wages earned during the last 7 days
= 7 × 92
= Rs. 644

Total wages earned during the 15 days,
= wages during first 7 days + wage on 8^th day + wages during the last 7 days.

1350 = 609 + wage on 8^th day + 644
wage on 8^th day = 1350 - 609 - 644 = Rs. 97

Total temperature on Wednesday, Thursday and Friday was 25 × 3 = 75°
Total temperature on Thursday, Friday and Saturday was 24 × 3 = 72°
Hence, difference between the temperature on Wednesday and Saturday = 3°
If Saturday temperature =27°, then
Wednesday's temperature = 27 + 3 = 30°

Let the average weight of the 59 students be X Therefore, the total weight of the 59 of them will be 59X
The questions states that when the weight of this student who left is added, the total weight of the class = 59X + 45
When this student is also included, the average weight decreases by 0.2 kgs
59X + $$\frac{{45}}{{60}}$$ = X - 0.2
=> 59X + 45 = 60X - 12
=> 45 + 12 = 60X - 59X
=> X = 57

Let a and b be the two angles in the question, with a > b. We are given that the difference between the angles is 24°.
⇒ a – b = 24
Since the average of the two angles is 54°, we have $$\frac{{{\text{a}} + {\text{b}}}}{2}$$  = 54
Solving for b in the first equation yields b = a – 24, and substituting this into the second equation yields,
$$ {\frac{{\left\{ {{\text{a}} + \left( {{\text{a}} - 24} \right)} \right\}}}{2}} = 54$$
2a − 24 = 54 × 2
2a − 24 = 108
2a = 108 + 24
2a = 132
a = 66
Also,
b = a − 24 = 66 − 24 = 42
Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is
180°, a + b + c = 180°
Putting the previous results into the equation yields 66 + 42 + c = 180°
Solving for c yields c = 72°
Hence, the greatest of the three angles a, b and c is c, which equal.

At present the total age of the family = 5 × 20 =100
The total age of the family at the time of the birth of the youngest member,
= 100 - 10 - (10 × 4)
= 50
Therefore, average age of the family at the time of birth of the youngest member,
$$ = \frac{{50}}{4} = 12.5$$

The average of the elements in the original set S is:
$$\eqalign{ & \frac{{0 + 2 + 4 + 5 + 9}}{5} \cr & = \frac{{20}}{5} \cr & = 4 \cr} $$
If we remove an element that equals the average, then the average of the new set will remain unchanged. The new set after removing 4 is {0, 2, 5, 9}.
The average of the elements is,
$$\eqalign{ & \frac{{0 + 2 + 5 + 9}}{4} \cr & = \frac{{16}}{4} \cr & = 4 \cr} $$

Average cost of 5 apples and 4 mangoes = Rs. 36
Total cost = 36 × 9 = 324
Average cost of 7 apples and 8 mangoes = 48
Total cost = 48 × 15 = 720
Total cost of 12 apples and 12 mangoes = 324 + 720 = 1044
Therefore, cost of 24 apples and 24 mangoes = 1044 × 2 = 2088

In this question, sum of numbers is provided, net required sum (i.e. A + B+ C+ D + E) cannot be calculated by the given data.
Therefore the answer is Data inadequate.

Let 10 numbers be x1, x2, x3, . . . . . . . x10
According to question average of these 10 numbers is 10
$$ \Rightarrow \frac{{ {x1 + x2 + x3 + .... + x10} }}{{10}} = x$$
Now if each number is increased by 10%,
then new average, say y.
$$y = \frac{{ {1.1x1 + 1.1x2 + 1.1x3 + .... + 1.1x10} }}{{10}}$$
$${\kern 1pt} \Rightarrow y = 1.1 \times {\frac{{ {x1 + x2 + x3 + .... + x10} }}{{10}}} $$
$$\eqalign{ & \Rightarrow y = 1.1x \cr & \Rightarrow y\,{\text{is }}10\% \,{\text{increased}} \cr} $$

Total cost of 10 books = Rs. 120
Total cost of 8 books = Rs. 94
⇒ The cost of 2 books = Rs. 26
Let the price of each book be x and y.
⇒ x + y = 26 - - - - - -(1)
Given that the price of 1 book is 60% more than the other price
$$\eqalign{ & \left( {\frac{{160}}{{100}}} \right)y + y = 26 \cr & \Rightarrow y\left( {\frac{{160}}{{100}} + 1} \right) = 26 \cr & \Rightarrow y\left( {\frac{{160 + 100}}{{100}}} \right) = 26 \cr & \Rightarrow y = \frac{{\left( {26 \times 100} \right)}}{{260}} \cr & \Rightarrow y = 10 \cr & {\text{Substituting}}\,\,y = 10\,\,{\text{in }}{\kern 1pt} \left(1 \right){\text{we get}}, \cr & x + 10 = 26 \cr & x = 16 \cr} $$