# Age - MCQ 3

## 20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Age - MCQ 3

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Attempt Age - MCQ 3 | 20 questions in 15 minutes | Mock test for Quant preparation | Free important questions MCQ to study Quantitative Aptitude for Competitive Examinations for Quant Exam | Download free PDF with solutions
QUESTION: 1

### The average score of a cricket player after 48 innings is 49 and in the 49th innings the player scores 98 runs. In the 50th innings the minimum number of runs required to increase his average score by 2 than it was before the 50th innings?

Solution:

Average score of Dhoni after 48th Innings = 49
Average score of Dhoni after 49th Innings = (48*49 + 98)/49 = 50
Average score of Dhoni after 50th Innings ⇒ 50 + 2 = 52
Score in 50th Innings = (50 * 52) – (49 * 50) = 150

QUESTION: 2

### The average age of board of directors of a company having 15 directors was 48 years. When a director aged 56 resigned from the board of directors another director died on the same day. A new director joined board of directors aged 36. Next year the average age of all 14 directors was found to be 48 years. The age of late director at the time of his death was _______

Solution:

Before Death = 48*15 = 720
The age of late director at the time of his death
720 - (56+x) + 36 + 14 = 672
X = 42

QUESTION: 3

### The average expenditure of the hotel when there are 10 guests is Rs. 80 per guests and the average expenditure is Rs.40 when there are 30 Guests. If it is known that there are some fixed expenses irrespective of the number of guests then the average expenditure per guest when there are 50 guests in the hotel.?

Solution:

Let the fixed expenditure of the hotel be Rs.x and the variable expenditure is Rs.y , then
x + 10y = 800 ---------(1)
x + 30y = 1200 ----------(2)
20y = 400
y = Rs. 20 and x= 600
Hence the total expenditure when there are 50 guests = 600 + 50 x 20

= 1600
Average expenditure = 1600 /50

= Rs. 32

QUESTION: 4

The average marks of Sumit decreased by one, when he replaced the subject in which he has scored 40 marks by the other two subjects in which he has just scored 23 and 25 marks respectively. Later he has also included 57 marks of Computer Science, then the average marks increased by two. How many subjects were there initially?

Solution:

Total subjects = x ; Average Marks = y
(x + 1)(y – 1) = (xy – 40) + (23 + 25)
y – x = 9 –(1)
(x + 2)(y + 1) = (xy – 40) + (23 + 25)+ 57
xy + 2y + x + 2 = xy + 65
2y + x = 63 –(2)
From Equation (1) and (2) ⇒ x = 15; y = 24

QUESTION: 5

The average runs scored by a cricketer is 42 innings, is 40. The difference between his maximum and minimum scores in an inning is 100. If these two innings are not taken into consideration, then the average score of remaining 40 inning is 38. Calculate the maximum runs scored by him in an innings ?

Solution:

Let the minimum score = x
∴ Maximum score = x + 100
∴ x + (x + 100) = 40 x 42 – 40 x 38
⇒ 2x + 100 = 1680 – 1520 = 160
⇒ 2x = 160 – 100 = 60
⇒ x = 30
Maximum score = x + 100 = 30 + 100 = 130

QUESTION: 6

The average weight of 4 persons, Amit, Bala, Catherine and David is 65 kg. The 5th person Elina is included and the average weight decreses by 2 kg. Amit is replaced by Francis. The weight of Francis is 4 kg more than Elina. Average weight decreases because of the replacement of Amit and now the average weight is 64 kg. Find the weight of Amit.

Solution:

Sum of the weight of Amit, Bala, Catherine and David = 65 x 4 = 260 kg
Average weight of Amit, Bala, Catherine and David = 65 – 2 = 63kg
Sum of the weight of Amit, Bala, Catherine and David and Elina = 63 x 5 = 315 kg
Weight of Elina = 315 – 260 = 55 kg
Weight of Francis = 55 + 4 = 59 kg
Average weight of Francis, Bala, Catherine, David and Elina = 64 kg
Sum of the weight of Francis, Bala, Catherine, David and Elina = 64 x 5 = 320 Kg
Sum of the weights of Bala, Catherine, David = 320 – 55 – 59 = 206 kg
Weight of Amit = 260 – 206 = 54 kg

QUESTION: 7

The average weight of the students in four sections Red, Black, Green and Yellow is 60 kg. The average weight of the students of Red, Black, Green and Yellow individually are 45 kg, 50 kg, 72 kg and 80 kg, respectively. If the average weight of the students of section Red and Black together is 48 kg and that of Black and Green together is 60 kg. What is the ratio of the number of the students in sections Red and Yellow?

Solution:

Average weight of students of sections of Red and Black = 48 kg
⇒ (45a + 50b)/(a + b) = 48
⇒ 15a = 10b
Average weight of students of sections of Black and Green = 60 kg
⇒ 50b + 72c = 60(b + c)
⇒ 10b = 12c
Average weight of students of Red, Black, Green and Yellow = 60 kg
45a + 50b + 72c + 80d = 60(a + b + c + d)
⇒ 15a + 10b – 12c – 20d = 0
⇒ 15a = 20d
⇒ a : d = 4 : 3

QUESTION: 8

Out of the three annual examinations, each with a total of 500 marks, a student secured average marks of 45% and 55% in the first and second annual examinations. To have an overall average of 60%, how many marks does the student need to secure in the third annual examination ?

Solution:

Total marks for three examinations = 3x 500 = 1500
Total required marks in three examinations = 60% of 1500
= (3 x 500 x 60) / 100
= 900
Marks secured in first examination = 45 % of 500
= (500 x 45) / 100
= 225
Marks secured in third examination = 55 % of 500
= (500 x 55) / 100
= 275
Thus, the required marks in third examination
= 900 – (225 + 275)
= 900 – 500
= 400

QUESTION: 9

The average age of women and child workers in a factory was 20 year. The average age of all the 12 children was 8 year and average age of women workers was 26 year. If 9 women workers were married, then the number of unmarried women workers was.

Solution:

Women workers = x
26x + 12 x 8 = 20 (12+x)
⇒ 26x + 96 = 240 + 20x
⇒ 6x = 144
∴ x = 24
∴ Unmarried women workers = 24 – 9 = 15

QUESTION: 10

The average age of all the members of Mr. Ravi’s family is 25 year. The average age of males is 30 year while the average age of females is 20 year. If the number of females in the family is 10, then find out the number of males?

Solution:

Let the number of males = x
Total family members = 10 + x
25(10 + x) = 30x + 10*20
⇒ 250 + 25y = 30x + 200
⇒ 5y = 50
∴ y = 10

QUESTION: 11

James’ father was 30 years old when he was born. His mother’s age was 24 when his sister who is 5 years younger to him, was born. What is the difference between the age of James’ father and mother?

Solution:

James’age = F – 30
sister’s age = F – 35
M = 24 + Sister’s age
M = 24 + F – 35
F – M = 11

QUESTION: 12

The respective ratio between the present age of Monika and Deepak is 5:x. Monika is 9 years younger than Prem. Prem’s age after 9 years will be 33 years. The difference between Deepak’s and Monika’s age is same as the present age of Prem. What is the value of x?

Solution:

Prem’s age after 9 years = 33 years
Prem’s present age = 33 – 9 = 24 years
Monika’s present age = 24 – 9 = 15 years
Deepak’s present age = 15 + 24 = 39 years
Ratio between Monika and Deepak = 15 : 39 = 5 : 13
X = 13

QUESTION: 13

Three years ago, Poorvi was thrice as old as his sister Reena. After three years Poorvi will be twice as old as Reena. What is the present age of Reena?

Solution:

P – 3 = 3 *(R – 3) —(1)
P + 3 = 2 *(R + 3) —(2)
Solving eqn (1) and (2)  R = 9

QUESTION: 14

Fifteen years ago, Rita’s mother was thrice of Rita’s age and two years ago Rita’s Mother was twice of Rita’s age. What is the present age of Rita’s Mother ?

Solution:

Rita’s mother age = R1
Rita’s age = R2
R1 – 15 = 3 *(R2 – 15) —(1)
R1 – 2 = 2 *(R2 – 2) —(2) From eqn (1) and (2) R1 = 54

QUESTION: 15

Mr. Suresh has three daughters namely Ramya, Anita and Kiran. Ramya is the eldest daughter of Mr. Suresh while Kiran is the youngest one. The present ages of all three of them are square numbers. The sum of their ages after 5 years is 44. What is the age of Ramya after two years?

Solution:

Square numbers – a, b, c
(a + 5) + (b + 5)+ (c + 5) = 44
a + b + c = 44 – 15 = 29
Possible values of a, b, c = 4, 9, 16 [Out of 1, 4, 9, 16, 25] Ramya’s present age = 16; after two years = 18

QUESTION: 16

If 6 years are subtracted from the present age of Ravi and the remainder is divided by 12, then the present age of his grandson Pranav is obtained. If Pranav is 2 years younger to Mohit whose age is 8 years, then what is Ravi’s present age ?

Solution:
QUESTION: 17

One year back, Ria was six times as old as her daughter. Six years hence, Ria’s age will exceed her daughter’s age by 15 years. The ratio of the present ages of Ria and her daughter is?

Solution:

Ages of Ria and her daughter = 6x, x
[6x + 1 + 6] – [x + 1 + 6] = 15
5x = 15; x = 3
Ratio = 6x + 1 : x + 1 = 19 : 4

QUESTION: 18

Suresh age is 125% of what it was ten years ago, but 250/3% of what it will be after ten years. What is the present age of Suresh?

Solution:

Suresh’s age before 10 years = x
125x/100 = x + 10
125x = 100x + 1000 ⇒ x = 40
Present age = x + 10 = 50

QUESTION: 19

Ajay got married 6 years ago. His present age is 5/4 times his age at the time of his marriage. Ajay’s brother was 5 years younger to him at the time of his marriage. What is the present age of Ajay’s brother?

Solution:

Present age of Ajay = x ; Present age of Ajay’s sister = y
x = (x-6)(5/4)
x = 30
present age of Ajay’s brother = 30 – 5 = 25

QUESTION: 20

Rahul is as much younger than Shyam as he is older than Suresh. If the sum of the ages of Shyam and Suresh is 60 years, what is definitely the difference between Shyam and Rahul’s age?

Solution:

Shyam’s age – Rahul’s age = Rahul’s age – Suresh’s age
Suresh’s age + Shyam’s age = 2 Rahul’s age
Shyam’s age + Suresh’s age = 60
Rahul’s age = 30; We can not find the difference between Shyam and Rahul’s age

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