# RS Aggarwal Quantitative Aptitude PDF Free download: PROBLEMS ON AGES

** PROBLEMS ON AGES**

** **

**Ex. 1. Rajeev’s ****age after ****15 years ****will ****be 5 times his age 5 ****years ****back. ****What is the **

**present age of Rajeev ****?** (Hotel Management,2002)

Sol. Let Rajeev’s present age be x years. Then,

Rajeev’s age after 15 years = (x + 15) years.

Rajeev’s age 5 years back = (x – 5) years.

:. x + 15 = 5 (x – 5) óx + 15 = 5x – 25 ó 4x = 40 ó x = 10.

Hence, Rajeev’s present age = 10 years.

**Ex. 2. The ages of two persons differ by 16 years. If 6 years ago, the **

*elder***one be**

**3 times as old as the younger one, find their present ages.** (A.A.O. Exam,2003)

Sol. Let the age of the younger person be x years.

Then, age of the elder person = (x + 16) years.

:. 3 (x – 6) = (x + 16 – 6) ó 3x -18 = x + 10 ó 2x = 28 ó x = 14.

Hence, their present ages are 14 years and 30 years.

** **

**Ex. 3. The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita**

*is more than Ankit’s age by ***4 years, what is Nikita’s age? ** (S.B.I.P.O,1999)

Sol. Let Ankit’s age be x years. Then, Nikita’s age = 240/xyears.

\ 2 ´ (240 /x ) – x = 4 ó 480 – x^{2} = 4x ó x^{2} + 4x – 480 = 0

ó ( x+24)(x-20) = 0 ó x = 20.

Hence, Nikita’s age = __(22_0) __years = 12 years. 1

**Ex. 4. 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. Find the present age of the father. . (S.S.C, 2003)**

** **Sol. Let the son’s present age be x years. Then, father’s present age = *(3x *+ 3) years

\* (3x *+ 3 + 3) = 2 (x + 3) + 10 ó 3x + 6 = 2x + 16 ó x = 10.

Hence, father’s present age = *(3x *+ 3) = ((3 ´ 10) + 3) years = 33 years.

**Ex. 5. R ohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be**

*twice as old as his son. What are their present ages?*

Sol. Let son’s age 8 years ago be x years. Then, Rohit’s age 8 years ago = 4x years.

Son’s age after 8 years = (x + 8) + 8 = (x + 16) years.

Rohit’s age after 8 years = *(4x *+ 8) + 8 = *(4x+ *16) years.

\ 2 (x + 16) = 4x + 16 ó 2x = 16 ó x = 8.

Hence, son’s ‘present age = (x + 8) = 16 years.

Rohit’s present age = *(4x *+ 8) = 40 years.

** **

** **

** **

**Ex. 6. One ***year ago, the ratio ***of Gaurav’s and Sachin’s age was 6: 7 respectively.**

*Four years hence, this ratio would become ***7: 8. How old is Sa chin ?**

(NABARD, 2002)

** **

Sol:

. Let Gaurav’s and Sachin’s ages one year ago be 6x and 7x years respectively. Then, Gaurav’s age

4 years hence = *(6x *+ 1) + 4 = *(6x *+ 5) years.

Sachin’s age 4 years hence = *(7x *+ 1) + 4 = *(7x *+ 5) years.

__ __

__6x+5 __ = __7 __ ó 8(6x+5) = 7 *(7x *+ 5) ó *48x *+ 40 = *49x *+ 35 ó *x *= 5.

7x+5 8

Hence, Sachin’s present age = *(7x *+ 1) = 36 years.

,**7. Abhay’s age after six years will be three-seventh of his fathers age. Ten years **

**ago the**

*ratio***of**

*their ages was 1*: 5.*What is Abhay’s father’s age*at*present?*

Sol. Let the ages of Abhay and his father 10 years ago be x and 5x years respectively. Then,

Abhay’s age after 6 years = (x + 10) + 6 = (x + 16) years.

Father’s age after 6 years = *(5x *+ 10) + 6 = *(5x *+ 16) years.

*\x *+ 16) = __3__ *(5x *+ 16) ó 7 *(x *+ 16) = 3 *(5x *+ 16) ó *7x *+ 112 = *15x *+ 48

7

ó *8x *= 64 ó x = 8.

Hence, Abhay’s father’s present age = *(5x *+ 10) = 50 years.

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