Average - Aptitude MCQ Questions and Solutions with Explanations

According to the question,
Mean of 10 numbers is = 30
∴ Sum of 10 numbers is = 300
It was observed that numbers 15, 23 are wrongly taken as 51, 32
Difference
= (51 + 32) - (15 + 23)
= 83 - 38
= 45 (more)
∴ Actual sum of 10 numbers
= 300 - 45 = 255
∴ Actual average of 10 numbers
= $$\frac{255}{10}$$ = 25.5

According to the question,
Average age of a family of 10 members is = 20 years
Sum of the age of 10 members
= 20 × 10 = 200 years
If the age of youngest member is = 10 years
Sum of the age of 9 members at the time of birth of youngest member
= 200 - 10 × 10
= 200 - 100
= 100 years
∴ Average of 9 members is
= $$\frac{100}{9}$$ = $$11\frac{1}{9}$$ years

According to the question,
Consecutive even numbers
= a, b, c, d, e, f, g
Consecutive odd numbers
= j, k, l, m, n
Consecutive even numbers
2, 4, 6, 8, 10, 12, 14
$$\frac{2 + 4 + 6 + 8 + 10 + 12 + 14}{7}$$
= $$\frac{56}{7}$$
= 8 middle term
Consecutive odd numbers
1, 3, 5, 7, 9
$$\frac{1 + 3 + 5 + 7 + 9}{2}$$
= $$\frac{25}{5}$$
= 5 middle term
∴ Same as in above situation.
Average of even numbers = d
Average of odd numbers = 1
∴ Average of all numbers = $$\frac{1 + d}{2}$$

Let a, b, c, d are four number
∵ Average of first three number a, b, c = 16
Total of (a + b + c) = 16 × 3 = 48.....(i)
Again
∴ Average of last 3 numbers b, c, d = 15
⇒ Total of (b + c + d) = 15 × 3 = 45.....(ii)
⇒ Subtracting equation (i) - (ii), we get
⇒ (a + b + c) - (b + c + d) = 48 - 45
⇒ a - d = 3
⇒ a = - 20 = 3       [Given d = 20]
⇒ a = 23
⇒ Therefore, first number a = 23

M	:	S	:	P
2×3	:	1×3
		3	:	2
6	:	3	:	2	= 11

Average = $$\frac{11}{3}$$ unit
Average = $$\frac{11}{3}$$ unit = Rs. 110
1 unit = 10 × 3 = Rs. 30
∴ Mukesh has = 6 unit
= 6 × 30 = Rs. 180

According to the question,
⇒ Total temperature of first four days
Mon + Tue + Wed + Thu
= 25 × 4 = 100°C.....(i)
⇒ Total temperature of last four days
Thu + Fri + Sat + Sun
= 25.5 × 4 = 102°C .....(ii)
⇒ Total temperature of week
= 25.2 × 7 = 176.4°C .....(iii)
After adding equation (i) + (ii)
Mon + Tue + Wed + 2 × Thu + Fri + Sat + Sun
= 100° + 102° = 202°C.....(iv)
After subtracting equation (iv) - (iii)
= 202° - 176.4° = 25.6° C
⇒ Temperature of 4th day = 25.6° C

According to the question,
Average of $$\frac{{x + \frac{1}{x}}}{2} = M$$
Put x = 1
$$\eqalign{ & \therefore \,\,\frac{{1 + \frac{1}{1}}}{2} = M \cr & \Rightarrow M = 1 \cr & \therefore \,\,\frac{{{x^2} + \frac{1}{{{x^2}}}}}{2} \cr & = \frac{{{1^2} + \frac{1}{{{1^2}}}}}{2} \cr & = 1 \cr} $$
Now check from the option
Option: (C) 2M² - 1 (put M = 1)
= 2 × 1 - 1
= 1 (satisfied)

Alternate :
According to the question,
$$\eqalign{ & \Rightarrow \frac{{x + \frac{1}{x}}}{2} = M \cr & \text{Squaring the both sides} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {(2M)^2} - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 4{M^2} - 2 \cr} $$
Required average
$$\eqalign{ & = \frac{{{x^2} + \frac{1}{{{x^2}}}}}{2} \cr & = \frac{{4{M^2} - 2}}{2} \cr & = 2{M^2} - 1 \cr} $$

A + B + C = 84 × 3 = 252.....(i)
A + B + C + D = 80 × 4 = 320.....(ii)
On solving equation (i) and (ii)
D = 320 - 252
D = 68
E's weight = 68 + 3 = 71
B + C + D + E = 79 × 4 = 316
B + C + D + 71 = 316
B + C + D = 316 - 71 = 245
Now, from equation (ii)
(A + B + C + D) - (B + C + D)
A = 320 - 245
∴ A = 75 kg

Sum of two numbers = 8 ×2 = 16
Sum of other three numbers = 3 × 3 = 9
Total sum = 25
Average = $$\frac{25}{5}$$ = 5

Passengers	Weight	Total wight
16 ×	80	= 1280
20 ×	86	= 1720

Weight of 4 boys           = 440
Average weight of 4 boys
= $$\frac{440}{4}$$
= 110 kg