Problems on Ages | Answer with Explanations

Problems on Age

Problems on Age-Answer with  Explanations

Problems on age/Age problems for all competitive exams are covered here.

For all competitive exams Ages problem is an important chapter. From Age chapter at least 2-3 problems are appear in all competitive exams like SSC and bank exams (SBI and IBPS-clerk and PO). This section will teach you to understand how to solve Age problems.

Problems on Age– Study material for all competitive exams

Age is an important topic for all competitive exams. This Age chapter will teach you to understand common formulas to solve questions for age chapter. Some important and repeated questions are covered this chapter.

Important formulas for “Problem on Ages”:

1. If the current age is x, then n times the age is nx.

2. If the current age is x, then age n years later/hence = x + n.

3. If the current age is x, then age n years ago = x – n.

4. The ages in a ratio a : b will be ax and bx.

5. If the current age is x, then 1/n of the age is x /n.

Problems on Age Answer with Explanation

Before go through the problems of Ages. Please understand and learn basic formulas of Age chapter.

Complexity of the question is differ for each exams. While learning solve high difficulty questions to boost your knowledge and success your exams.

1) If 6 years are subtracted from the present age of Babita and the remainder is divided by 18, then the present age of her granddaughter Geeta is obtained. If Geeta is 2 years younger to Sita whose age is 5 years, then what is Babita’s present age?
A) 77
B) 65
C) 84
D) 43
E) 79

Explanation:

Geeta’s age = (5-2) = 3 years
Let age of Babita = x years
So (x-6)/18 = 3
Solve, x = 60

2) A’s age is twice C’ age. Ratio of age of B 2 years hence to age of C 2 years ago is 5 : 2. C is 14 years younger than D. Difference in ages of D and A is 4 years. Find the average of their ages.
A) 36
B) 25
C) 27
D) 13
E) 18

Explanation:

A = 2C
(B+2)/(C-2) = 5/2
C = D – 14
D – A = 4
Solve, A = 20, B = 18, C = 10, D = 24

3) When the couple was married the average of their ages was 25 years. When their first child was born, the average age of family became 18 years. When their second child was born, the average age of the family became 15 years. Find the average age of the couple now.
A) 31
B) 27
C) 28
D) 29
E) 30

Explanation:

Sum of ages of couple = 25*2 = 50
When 1st child born, total age of 3 = 18*3 = 54 years
At this time the child’s age was 0, so age of father and mother would have increased by same.
So increased by 2 years each.
So 50 +2 + 2 = 54
Now when 2nd child born, total age of 4 = 15*4 = 60
So this time second child’s age = 0 and age of father, mother and first child would have increased by same.
So increased by 2 each such that 54 + 2+2+2 = 60
So now this time (after 4 years from age 50),
total age of couple is 50+4+4 = 58
So average = 29 years

4) Ratio of age of A to B is 3 : 2 and that of A to C is 1 : 2. Difference in ages of B and C is 24 years. Find the average of their present ages.
A) 24
B) 22
C) 14
D) 26
E) 31

Explanation:

B/A = 2/3 and A/C = 1/2
So B : A : C = 2*1 : 3*1 : 3*2 = 2 : 3 : 6
So 6x – 2x = 24, 4x = 24, x = 6
So total of their present ages = (2+3+6)*6.
So average = (2+3+6)*6 / 3 = 22 years

5) Ratio of ages of A 5 years hence to B’s age 3 years ago is 5 : 3. Also ratio of ages of A 4 years ago to B’s age 2 years hence is 4 : 5. Find the age of the elder.
A) 15
B) 18
C) 21
D) 20
E) 24

Explanation:

(A+5)/(B-3) = 5/3
(A-4)/(B+2) = 4/5
Solve A = 20, B = 18

6) The ratio of ages of Sneha to Bhavna is 6 : 13. Also ratio between Kritika’s age two years after and Bhavna’s age 4 years after will be 2 : 3. If the average age of Sneha and Kritika is 15 years, what will be Kritika’s age three years hence?
A) 23
B) 19
C) 12
D) 15
E) 8

Explanation:

S/B = 6/13
(K+2)/(B+4) = 2/3
S + K = 2*15 = 30
Solve the equations,
K = 12, So K + 3 = 15

7) Difference between ages of Raman and Preet is 16 years. If Raman’s age ten years hence will be two times the age of Preet, find Raman’s age.
A) 26
B) 19
C) 42
D) 38
E) 46

Explanation:

R – P = 16
(R + 10) = 2P
Solve, R = 42

8) Three years ago, Pihu was thrice old as Ravi that time. How old is Pihu today if ratio of age of Pihu six years hence to Ravi’s age four years ago is 9 : 2?
A) 25
B) 30
C) 33
D) 22
E) 28

Explanation:

(P – 3) = 3* (R-3)
(P + 6)/(R – 4) = 9/2
Solve both equations, P = 30

9) The average age of Yogita, Kanika and Prachi is 14 years. The ratio of Yogita’s age one year ago to Kanika’s age one years hence to Prachi’s age three years hence is 5 : 6 : 4. Find ratio of Yogita’s age two years hence to Prachi’s age.
A) 8 : 5
B) 5 : 3
C) 1 : 4
D) 2 : 1
E) 4 : 7

Explanation:

Y + K + P = 42
(Y – 1) : (K + 1) : (P + 3) = 5 : 6 : 4
So (5x + 1) + (6x – 1) + (4x – 3) = 42
Solve, x = 3
So Yogita’s age 1 ago hence = 5x = 15,
so present age = 16
Prachi’s age 3 years hence = 4x = 12,
so present age = 9
So (Y+2)/P = 18/9 = 2/1

10) Sneha’s mother’s age is five years more than twice the age of Sneha. When Sneha was born, her brother Rahul was four years old and her father two years older than her mother. If the average age of her mother and father is 46 years. Find the ratio of age of Rahul to that of Sneha.
A) 3 : 7
B) 7 : 4
C) 6 : 5
D) 8 : 11
E) 3 : 10

Explanation:

Let age of Sneha = x,
So age of Mother = 2x+5,
Rahul = x+4,
Father = 2x+7
(2x+5 + 2x+7) = 2*46
So x = 20
So (x+4)/x = 24/20 = 6/5

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