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Problems On Ages Practice Quiz

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Problems On Ages Practice Quiz

shape Introduction

Problems on Ages are most frequently appearing questions in various competitive exams that include Quantitative Aptitude section. The article Problems On Ages Practice Quiz lists some of the most frequently asked model questions. Problems on Ages are often asked in the Quantitative Aptitude/ Numerical Ability sections of various competitive exams like SBI PO, SBI Clerk, IBPS PO, IBPS Clerk, SSC CGL, Railways RRB, NTPC, LIC AAO, UIIC AO, NICL AO, IBPS RRB, SSC Stenographer, CAT, XAT, CMAT, GMAT, SBI Associate PO, SBI Associate Clerk etc .


Algebra is a very powerful branch of Mathematics which can be used to solve the Problems on Ages. Algebra helps in transforming word problems into mathematical expressions in the form of equations using variables to denote unknown quantities or parameters and thus, providing numerous techniques to solve these mathematical equations and hence, determining the answer to the problem. Identifying key information, organizing information, using mathematical expressions to assume unknown values and thus solving mathematical expressions for the unknown values will help us identify solutions.


Some of the important concepts in Problems on Ages:

If the current age of a person be X, then


  • age after n years = X + n

  • age n years ago = X – n

  • n times the age = nX

  • If ages in the numerical are mentioned in ratio A : B, then A : B will be AX and BX


shape Quiz

Problems On Ages Practice Quiz – Exercise

1. The ratio of the ages of the father and the son at present is 7:1. After 4 years, the ratio will become 4 : 1. What is the sum of the present ages of the father and the son ?

    A. 29 years
    B. 35 years
    C. 32 years
    D. None of these


Answer – Option C

Explanation –

Father : Son = x : y = 7 : 1

After t = 4 year = α :β = 4 : 1

Son’s age = \(\frac{y * t(α – β) }{difference of cross product}\)

\(\frac{1 * 4(4 – 1)}{7 * 1 – 4 * 1}\)

\(\frac{1 * 4* 3}{3}\) = 4 years

and Father’s age = \(\frac{x * t(α – β) }{difference of cross product}\)

\(\frac{7 * 4(4 – 1)}{7 * 1 – 4 * 1}\)

\(\frac{7 * 4* 3}{3}\) = 28 years

Hence sum of the ages of father and son

= (28 + 4) = 32 years

2. If 10 years are subtracted from the present age of Ram and the remainder divided by 14, then you would get the present age of his grandson Shyam. If Shyam is 9 years younger to Sunder whose age is 14, then what is the present age of Ram?


    A. 80 years
    B. 70 years
    C. 60 years
    D. None of these


Answer – Option A

Explanation –

Let the Ram’s present age be x years, and Shyam’s present age be y years.

Now according to question

\(\frac{x – 10}{14}\) = y

Now, Shyam is 9 years younger to sunder whose age

i.e, Present age of Shyam (y) = 14 – 9 = 5 years.

Putting the value of y in equations (1) will get \(\frac{x – 10}{14}\) = 5

x = 80 years

3. A’s age is \(\frac{1}{6}\)1 of B’s age. B’s age will be twice of C’s age after 10 years. If C’s eighth birthday was celebrated two years ago, then the present age of A must be

    A. 5 years
    B. 10 years
    C. 15 years
    D. 20 years


Answer – Option A

Explanation –

A B C
10 60 20
5x 30x 10x
1x 6x 2x


Now C’s eight birth day is celebrated two years ago is present age of

C = 10 years = 2x

i.e, x = 5

i.e, Present age of A = 1 * x = 1 * 5 = 5 years

4. Suresh is half his father’s age. After 20 years, his father’s age will be one and a half times Suresh’s age. What is his father’s age now?

    A. 40 years
    B. 20 years
    C. 26 years
    D. 30 years


Answer – Option A

Explanation –

Let father’s present age be x and that of Suresh be y

i.e, y = \(\frac{1x}{2}\)

and x + 20 = \(\frac{3}{2}\)(y + 20)

From (i) and (ii), we have

x + 20 = \(\frac{3}{2}(\frac{x}{2} + 20)\)

4x + 80 = 3x + 120

x = 40.

5. If Dennis is\(\frac{1}{{3}^{rd}}\) the age of his father Keith now, and was \(\frac{1}{{4}^{th}}\) the age of his father 5 years ago, then how old will his father Keith be 5 years from now ?

    A. 20 years
    B. 45 years
    C. 40 years
    D. 50 years


Answer – Option D

Explanation –

Dennis = \(\frac{1}{3}\) Keith

Dennis – 5 = \(\frac{1}{4}\)(Keith – 5)

\(\frac{1}{3}\) Keith – 5 = \(\frac{1}{4}\)Keith – \(\frac{5}{4}\)

Keith = 45

Hence Keith is 50 year s old, 5 years from now.

6. The ages of the two persons differ by 20 years. If 5 years ago, the older one be 5 times as old as the younger one, then their present ages, in years, are

    A. 25, 5
    B. 30, 10
    C. 35, 15
    D. 50, 30


Answer – Option B

Explanation –

Let ages of two persons be x and y.

i.e, x – y = 20 …(i) [x > y]

x – 5 = 5(y – 5)

x – 5y = – 20 …(ii)

Solving equations (i) and (ii)

i.e, y = 10, x = 30

7. Rajan got married 8 years ago. His present age is \(\frac{6}{5}\) times his age at the time of his marriage. Rajan’s sister was 10 years younger to him at the time of his marriage. The age of Rajan’s sister is

    A. 32 years
    B. 36 years
    C. 38 years
    D. 40 years


Answer – Option C

Explanation –

Let the age at the time of marriage be x.

Then

x + 8 = \(\frac{6x}{5}\)

Solving, we get x = 40.

i.e, Age of his sister = 40 – 10 + 8 = 38

8. A father’s age is three times the sum of the ages of his two children, but 20 years hence his age will be equal to the sum of their ages. Then the father’s age is

    A. 30 years
    B. 40 years
    C. 35 years
    D. 45 years


Answer – Option A

Explanation –

Let the sum of ages of two children = x years,

Let the father’s age = y years

i.e, y = 3x …(i)

20 years hence, y + 20 = x + 20 + 20 …(ii)

Solving (i) and (ii), we get

y = 30 years.

9. Ratio of Ashok’s age to Pradeep’s age is 4 : 3. Ashok will be 26 years old after 6 years. How old is Pradeep now?

    A. 18 years
    B. 21 years
    C. 15 years
    D. 24 years


Answer – Option C

Explanation –

Ashok’s present age = 26 – 6 = 20 years

i.e, \(\frac{Ashok’s age}{Pradeep’s age}\) = \(\frac{4}{3}\)

Hence Pradeep’s age = 15 years

10. Sonu is 4 years younger than Manu while Dolly is four years younger than Sumit but\(\frac{1}{5}\)times as old as Sonu. If Sumit is eight years old, how many times as old is Manu as Dolly?

    A. 6
    B. \(\frac{1}{2}\)
    C. 3
    D. None of these


Answer – Option A

Explanation –

Age of Sonu + 4 years = Age of Monu

Age of Dolly+ 4 years = Age of Sumit

If Sumit is 8 year old, than Dolly is 4 year old and Sonu is 20 year old.

Monu = 24 year

then \(\frac{Monu}{Dolly}\) = \(\frac{24}{4}\) = 6