Average - Aptitude MCQ Questions and Solutions with Explanations

81

A student finds the average of 10, 2 digits numbers. If the digits of one of the numbers interchanged, the average increases by 3.6. The difference between the digits of the 2 digits number is :

A) 4 B) 3 C) 2 D) 5

Discuss

Answer & Solution

Answer: Option A

Solution:

According to the question,
Let as consider by mistake he writes 10th number with its digits interchanged
∴ $$\frac{{10x + y - \left( {10y + x} \right)}}{{10}} = 3.6$$
∴ In these remaining nine numbers are same and they cancel out
$$\eqalign{ & \Rightarrow \frac{{10x + y - \left( {10y + x} \right)}}{{10}} = 3.6 \cr & \Rightarrow 9x - 9y = 36 \cr & \Rightarrow x - y = 4 \cr} $$

82

The mean of 100 observations was calculated as 40. It was founder later on that one of the observation was misread as 83 instead of 53. The correct mean is :

A) 39 B) 39.7 C) 40.3 D) 42.7

Discuss

Answer & Solution

Answer: Option B

Solution:

Difference = 53 - 83 = - 30
Change in average
= $$\frac{- 30}{100}$$
= - 0.3
So, correct average
= 40 - 0.3
= 39.7

83

Sachin Tendulkar has a certain average for 11 innings. In the 12th innings he scores 120 runs and thereby increases his average by 5 runs. His new average is :

A) 60 B) 62 C) 65 D) 66

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the average of '11' innings is = x
According to the question,
$$\frac{{11x + 120}}{{12}} = x + 5$$
11x + 120 = 12x + 60
x = 60
∴ His new average
= 60 + 5
= 65

84

The average age of husband and his wife was 23 years at the time of their marriage. After five tears they have a one-year old child. The average age of the family of three, when the child was born, was-

A) 23 years B) 24 years C) 18 years D) 20 years

Discuss

Answer & Solution

Answer: Option C

Solution:

According to the question,
$$\frac{H + W}{2}$$ = 23 years
H + W = 46 years.....(i)
As given in the question after five years they have a one-year old child.
i.e., After 4 years of marriage child was born
∴ Average of (H + E + C)
= $$\frac{46 + 8}{3}$$
= $$\frac{54}{3}$$
= 18 years

85

The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. The father's age is :

A) 33 years B) 39 years C) 45 years D) 40 years

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the present age of Son = x years
And the father's age = 3x + 3
According to the question,
2 (x + 3) + 10 = 3x + 3 + 3
2x + 6 + 10 = 3x + 6
2x + 16 = 3x + 6
x = 10 years
Father's age
= 3 × 10 + 3 = 33 years

86

The average (arithmetic mean) of 3³⁰, 3⁶⁰ and 3⁹⁰ is :

A) $${3^{27}} + {3^{57}} + {3^{87}}$$ B) $${3^{29}} + {3^{59}} + {3^{89}}$$ C) $${3^{60}}$$ D) $${3^{177}}$$

Discuss

Answer & Solution

Answer: Option B

Solution:

According to the question,
$$\eqalign{ & \Rightarrow \frac{{{3^{30}} + {3^{60}} + {3^{90}}}}{3} \cr & \Rightarrow {3^{29}} + {3^{59}} + {3^{89}} \cr} $$

87

A set A consists of integers 27, 28, 30 and 33. If integer k is included in the set, the average of set A will increase by 30%. What is the value of integer k ?

A) 68 B) 79 C) 73.75 D) 75.25

Discuss

Answer & Solution

Answer: Option C

Solution:

Average of set A
= $$\frac{27 + 28 + 30 + 33}{4}$$
= $$\frac{118}{4}$$
= 29.5
After increase
= $$\frac{29.5 × 130}{100}$$
= 38.35
Now
$$\frac{{{\text{Sum of first four numbers + k}}}}{5}$$ $$ = {\text{New average}}$$
⇒ 118 + k = 38.35 × 5
⇒ 118 + k = 191.75
⇒ k = 191.75 - 118
⇒ k = 73.75

88

The average age of 40 students of class is 18 years. When 20 new students are admitted to the same class, the average age of the students of the class is increased by 6 months. The average age of newly admitted students is ?

A) 19 years B) 19 years 6 months C) 20 years D) 20 years 6 months

Discuss

Answer & Solution

Answer: Option B

Solution:

According to the question,
Average age of 40 students of class is = 18 years
Let the average age of 20 new students = x years
∴ $$\frac{40 × 18 + 20 × x}{60}$$ = $$\left( {18 + \frac{1}{2}} \right)$$ years
⇒ $$\frac{720 + 20x}{60}$$ = $$\frac{37}{2}$$
⇒ $$\frac{720 + 20x}{30}$$ = 37
⇒ 20x = 390
⇒ x = 19.5
∴ Average age of newly admitted students is
= 19.5 = 19 years 6 months

89

The mean weight of 34 students of a school is 42 kg. If the weight of the teacher be included, the mean rises by 400 grams. Find the weight of the teacher (in kg.)

A) 55 kg B) 57 kg C) 66 kg D) 56 kg

Discuss

Answer & Solution

Answer: Option D

Solution:

According to the question,
Total increase in weight including teacher
= 400 × 35
= 14000 gm
= 14 kg
If the teacher's weight has been '42' kg so there would have not been any change in average weight.
∴ Teacher's weight = 42 + 14 = 56 kg.

90

Five years ago, the average age of P and Q was 25. The average age of P, Q and R today is 25. Age of R after 5 years will be :

A) 15 years B) 20 years C) 40 years D) 35 years

Discuss

Answer & Solution

Answer: Option B

Solution:

According to the question,
$$\frac{P + Q}{2}$$ = 25
P + Q = 50.....(i) (5 years ago)
$$\frac{P + Q + R}{3}$$ = 25
P + Q + R = 75.....(ii) (present age)
∴ Present age of P + Q = 50 + 10 = 60 years
Present age of R = 75 - 60 = 15 years
∴ Age of R after 5 years
= 15 + 5 = 20 years