Average - Aptitude MCQ Questions and Solutions with Explanations

71

The weight of a person was entered incorrectly as 83 kg instead of 63 kg. As a result, the average weight of a group of people increased by 500 gm. What is the total number of people in the group?

A) 44 B) 48 C) 40 D) 50

Discuss

Answer & Solution

Answer: Option C

Solution:

Let, number of peoples in group = x
Average = n
⇒ nx - 63 + 83 = n(x + 500 gm)
⇒ nx + 20 = nx + $$\frac{{\text{n}}}{2}$$ kg
⇒ n = 20 × 2
⇒ n = 40

OR
$$\eqalign{ & {\text{Average}} = \frac{{{\text{Sum}}}}{{{\text{Number}}}} \cr & 500\,{\text{gm}} = \frac{{20}}{n} \cr & \frac{1}{2} = \frac{{20}}{n} \cr & n = 40 \cr} $$

72

The average of 18 numbers is 37.5. If six numbers of average x are added to them, then the average of all the numbers increases by one. The value of x is:

A) 38.5 B) 42 C) 41.5 D) 40

Discuss

Answer & Solution

Answer: Option C

Solution:

After added 6 numbers, total numbers will be = 24
Now, the average of 24 numbers = 38.5
Increase '1' in every numbers
So, total increasation = 24
The average of 6 numbers (x) = 37.5 + $$\frac{{24}}{6}$$
x = 37.5 + 4
x = 41.5

73

The average of 44 consecutive odd numbers is 144. What is the largest number?

A) 189 B) 191 C) 187 D) 193

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the number is x, x + 2, x + 4, . . . . . . x + 86
Total value of 44 number is = 44x + 2 + 4 + . . . . . . 86
Value of 2 + 4 + 6 + . . . . . . 86
= 2(1 + 2 + 3 . . . . . . + 43)
= $$2\left( {\frac{{n \times \left( {n + 1} \right)}}{2}} \right)$$
= 43 × 44
= 1892
According to the question,
1892 + 44x = 144 × 44
44x = 144 × 44 - 1892
x = $$\frac{{44\left( {144 - 43} \right)}}{{44}}$$
x = 101
Largest number = 101 + 86 = 187

74

Average of all even numbers between 104 and 148 is.

A) 128 B) 130 C) 124 D) 126

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Average}} = \frac{{{\text{First number}} + {\text{Last number}}}}{2} \cr & {\text{So,}} \cr & {\text{First number}} = 104 \cr & {\text{Last number}} = 148 \cr & {\text{Average}} = \frac{{252}}{2} = 126 \cr} $$

Alternate:
Number of term between 104 and 148
146 = 106 + (n - 1)2
n = 21
Then, addition of all 21 term
$$\eqalign{ & = \frac{{21}}{2}\left[ {2 \times 106 + \left( {21 - 1} \right)2} \right] \cr & = \frac{{21}}{2}\left[ {212 + 40} \right] \cr & = \frac{{21}}{2} \times 252 \cr & {\text{Average}} = \frac{{21}}{2} \times \frac{{252}}{{21}} = 126 \cr} $$

75

In a one day match of 50 overs in an innings the team A had a run rate of 5.3 runs per over. Team B is playing and 5 overs are left and the required run rate to tie the match is 7.2 per over to match the score of team A. What is team B's score?

A) 265 B) 238 C) 254 D) 229

Discuss

Answer & Solution

Answer: Option D

Solution:

Run scored by team A $$ = \frac{{50 \times 53}}{{10}} = 265{\text{ runs}}$$
Run scored by team B in 5 over $$ = \frac{{72}}{{10}} \times 5 = 36{\text{ runs}}$$
Then, remaining run make by B = 265 - 36 = 229

76

The average of x numbers is y² and the average of y numbers is x². So the average of all the numbers taken together is:

A) $$\frac{{{x^3} + {y^3}}}{{x + y}}$$ B) $$xy$$ C) $$\frac{{{x^2} + {y^2}}}{{x + y}}$$ D) $$x{y^2} + y{x^2}$$

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{Average of }}x{\text{ number}} = {y^2} \cr & \therefore {\text{Sum of }}x{\text{ number}} = x{y^2} \cr & {\text{Average of }}y{\text{ number}} = {x^2} \cr & \therefore {\text{Sum of }}y{\text{ number}} = y{x^2} \cr & {\text{Average of all number}} = \frac{{x{y^2} + y{x^2}}}{{x + y}} \cr & = \frac{{xy\left( {x + y} \right)}}{{x + y}} \cr & = xy \cr} $$