Problems on Ages for IBPS Clerk Mains 2017

Dear Aspirants,

Time and Distance Questions for IBPS Clerk Mains 2017
Time and Distance Questions for IBPS Clerk Mains 2017

Quantitative Aptitude (Problems on Ages) is a very important section you must prepare if you are aiming for a job in Bank or Insurance sector. So, these 15 questions can help you practice three very important topics of Quant Section.

Q1. The average age of a group of 14 person is 27 years and 9 months. Two persons, each 42 years old, left the group. What will be the average age of the remaining persons in the group ?(a) 26.875 years
(b) 26.25 years
(c) 25.375 years
(d) 25 years
(e) None of these

Q2. The average age of some males and 15 females is 18 years. The sum of the ages of 15 females is 240 years and average age of males is 20 years. Find the number of males. 
(a) 8
(b) 7
(c) 10
(d) 15
(e) None of these

Q3. The difference between the present age of Arun and Deepak is 14 years. Seven years ago the ratio of their ages was 5 : 7 respectively. What is Deepak’s present age ?   
(a) 49 years
(b) 42 years
(c) 63 years
(d) 35 years
(e) 56 years

Q4. The sum of present ages of Ria and Abby is 48 years. Today Abby is 4 years older than Shweta. The respective ratio of the present ages of Ria and Shweta is 4 : 7. What was Abby’s age two years ago ? 
(a) 32 years
(b) 30 years
(c) 28 years
(d) 34 years
(e) None of these

Q5. The respective ratio of the present age of a mother and daughter is 7 : 1. Four years ago the respective ratio of their age was 19 : 1. What will be the mother’s age four years from now ?   
(a) 42 years
(b) 38 years
(c) 46 years
(d) 36 years
(e) None of these

S2. Ans.(d)
Sol.
Let there are n no. of males 
∴ (n + 15) × 18 = 240 + 20n
⇒ n = 15


S3. Ans.(e)
Sol.
Let Deepak’s present age = (7x + 7) years
Arun’s present age = (5x + 7) years
ATQ,
7x – 5x = 14
x = 7 
∴ Deepak’s present age = 49 + 7 = 56 years


S4. Ans.(b)
Sol.
Let present ages of Ria and Shweta be 4x and 7x respectively. 
∴ Abby’s present age = (7x + 4) years
ATQ,
4x +7x + 4 = 48 
⇒ x = 4 
∴ Abby’s age two years ago = 28 + 2 = 30 years



Q6. The present ages of Ranjana and Rakhi are in the ratio of 15 : 17 respectively. After 6 years, the respective ratio between the age of Ranjana and Rakhi will be 9 : 10. What will be the age of Ranjana after 6 years ?   
(a) Other than those given as options
(b) 40 years
(c) 34 years
(d) 30 years
(e) 36 years

Q7. The age of Sulekha and Arunima are in the ratio of 9 : 8 respectively. After 5 years the ratio of their age will be 10 : 9. What is the difference (in years) between their age  ?   
(a) 4 years
(b) 5 years
(c) 6 years
(d) 7 years
(e) 8 years

Q8. A boy was asked of his age by his friend. The boy said, ‘The number you get when you subtract 25 times my age from twice the square of my age will be thrice your age.’ If the friend’s age is 14, then the age of the boy is: 
(a) 28 years
(b) 21 years
(c) 14 years
(d) 25 years
(e) None of these

Q9. If the ages of P and R are added to twice the age of Q, the total becomes 59. If the ages of Q and R are added to thrice the age of P, the total becomes 68 and if the age of P is added to thrice the age of Q and thrice the age of R, the total becomes 108. What is the age of P? 
(a) 19 years
(b) 15 years
(c) 17 years
(d) 12 years
(e) None of these

Q10. If 6 years are subtracted from the present age of Shyam and the remainder is divided by 18, then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Mahesh whose age is 5 years, then what is the age of Shyam? 
(a) 48 years
(b) 60 years
(c) 84 years
(d) 96 years
(e) None of these 



S9. Ans.(d)
Sol.
P + R + 2Q = 59 ……….(i)
Q + R + 3P = 68 …………(ii)
P + 3Q + 3R = 108 ………….(iii)
From 3 × (ii) – (iii)
P = 12 years

S10. Ans.(b)
Sol.
Let A = Anup’s age
     M = Mahesh’s age
      S = Shyam’s age
(S-6)/18=A
Also, A = 3 years (∵ M = 5 years)
∴ S = 3 × 18 + 6 = 60 years

Q11.The ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born at the same time when person was died. After another three years, one more member died, again at 60, and a child was born at the same time when person was died. The current average age of this eight-member joint family is nearest to: 
(a) 22 years
(b) 25 years
(c) 20 years
(d) 23 years
(e) 24 years

Q12. Father is 5 years older than mother and mother’s age now is thrice the age of the daughter. The daughter is now 10 years old. What was father’s age when the daughter was born? 
(a) 20 years
(b) 15 years
(c) 25 years
(d) 30 years
(e) 36 years

Q13. The average age of the mother and her 6 children is 12 years which is reduced by 5 years if the age of the mother is excluded. How old is the mother?
(a) 42 years
(b) 40 years
(c) 48 years
(d) 50 years
(e) 55 years

Q14. Three times the present age of a father is equal to eight times the present age of his son. Eight years hence the father will be twice as old as his son at that time. What are their present ages? 
(a) 35, 15
(b) 32, 12
(c) 40, 15
(d) 27, 8
(e) None of these

Q15. The product of the present ages of Sarita and Gauri is 320. Eight years from now, Sarita’s age will be three times the age of Gauri. What was the age of Sarita when Gauri was born? 
(a) 40 years
(b) 32 years
(c) 48 years
(d) 36 years
(e) None of these

S11. Ans.(c)
Sol.
Total age of family = 231 years
Three years later total age of family = 231 + 8 × 3 - 60
= 195 years
After another three years total age of family = 195 – 60 + 8 × 3
= 159 years
∴ Average age ≈ 20 years

S12. Ans.(c)
Sol.
F = M + 5, M = 3D, and D = 10 years
∴ Father’s present age, F = 3 × 10 + 5 = 35 years
So, at time of birth of daughter, age of father = 25 years.

S13. Ans.(a)
Sol.
M + 6C = 12 × 7 = 84
And 6C = 7 × 6 = 42
∴ Mother’s age = 84 – 42 = 42 years

S14. Ans.(b)
Sol.
Let father’s age = F, Son’s age = y
3F = 8y
⇒ 3F-8y=0………(i)
⇒ (F+8)=2(y+8)^
 ⇒ F-2y=8 …………(ii)
From (i) – (ii) × 3
y= son’s age = 12 years
And F= father’s age = 32 years.

S15. Ans.(b)
Sol.
Sarita’s age × Gauri’s age = 320
S = 320/G
By question,
⇒ S + 8 = (G + 8)3
⇒ S – 3G = 16
⇒ 320/G-3G=16
⇒ 3G^2+16G-320=0
⇒ G(3G+40)-8(3G+40)=0
⇒ (G-8)(3G+40)=0
∴ Gauri’s age = 8
And Sarita’s age = 320/G=320/8=40
∴ Sarita was 32 years old when Gauri was born.