# Aptitude::Problems on Ages

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Example 1 / 22

Let Jack's present age be x

Jack's age before 5 years = (x - 5)

Jack's age after 22 years = (x + 22)

We are given that, Jack's age after 22 years (x + 22) is 4 times his age 5 years back (x – 5)

Therefore,

(x + 22) = 4(x – 5)

Solving the equation, we get

x + 22 = 4x – 20

3x = 42

x = 14 years

Jack's age before 5 years = (x - 5)

Jack's age after 22 years = (x + 22)

We are given that, Jack's age after 22 years (x + 22) is 4 times his age 5 years back (x – 5)

Therefore,

(x + 22) = 4(x – 5)

Solving the equation, we get

x + 22 = 4x – 20

3x = 42

x = 14 years

**Ans.**
If the current age of a person be y, then

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A: B, then A: B will be Ay and By

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A: B, then A: B will be Ay and By

NA

NA

Example 2 / 22

Let age of Rohan be y

Rahul is 16 years elder than Rohan = (y + 16).

So Rahul's age 4 years ago = (y + 16 – 4)

Rohan's age before 4 years = (y – 4)

4 years ago, Rahul is 5 times as old as Rohan

Thus, we can have the equation as,

(y + 16 – 4) = 5 (y – 4)

(y + 12) = (5y – 20)

4y = 32

y = 8

Rohan's age = 8 years

Rahul's age = (y + 16) = (8+16) = 24 years

Rahul is 16 years elder than Rohan = (y + 16).

So Rahul's age 4 years ago = (y + 16 – 4)

Rohan's age before 4 years = (y – 4)

4 years ago, Rahul is 5 times as old as Rohan

Thus, we can have the equation as,

(y + 16 – 4) = 5 (y – 4)

(y + 12) = (5y – 20)

4y = 32

y = 8

Rohan's age = 8 years

Rahul's age = (y + 16) = (8+16) = 24 years

**Ans.**
If the current age of a person be y, then

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 3 / 22

We are given that age ratio of Paul: Peter = 5:6

Let,

Paul's age = 5x and Peter's age = 6x

1 year ago, their age was 5x and 6x.

Hence at present,

Paul's age = 5x +1

Peter's age = 6x +1

Now after 4 years,

Paul's age = (5x +1) + 4 = (5x + 5)

Peter's age = (6x +1) + 4 = (6x + 5)

Hence,

After 4 years, this ratio becomes 6 : 7. Therefore,

\(\frac{Paul's age}{6}\)= \(\frac{Peter's age}{7}\)

\(\frac{(5x+5}{6x+5}\)= \(\frac{6}{7}\)

7 (5x + 5) = 6 (6x + 5)

X = 5

Peter's present age = (6x + 1) = (6 x 5 + 1) = 31 years

Paul's present age = (5x + 1) = (5 x 5 + 1) = 26 years

Let,

Paul's age = 5x and Peter's age = 6x

1 year ago, their age was 5x and 6x.

Hence at present,

Paul's age = 5x +1

Peter's age = 6x +1

Now after 4 years,

Paul's age = (5x +1) + 4 = (5x + 5)

Peter's age = (6x +1) + 4 = (6x + 5)

Hence,

After 4 years, this ratio becomes 6 : 7. Therefore,

\(\frac{Paul's age}{6}\)= \(\frac{Peter's age}{7}\)

\(\frac{(5x+5}{6x+5}\)= \(\frac{6}{7}\)

7 (5x + 5) = 6 (6x + 5)

X = 5

Peter's present age = (6x + 1) = (6 x 5 + 1) = 31 years

Paul's present age = (5x + 1) = (5 x 5 + 1) = 26 years

**Ans.**
If the current age of a person be y, then

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 4 / 22

Age of y= \(\frac{ratio of y}{sum of ratios}\)* sum of ages

Age of y= \(\frac{q}{p+q}\)* A

Age of y= \(\frac{q}{p+q}\)* A

**Ans.**
If the current age of a person be y, then

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 5 / 22

Let 10 years ago, age of son be x and so mother's age= 6x.

At present,

Mother's age will be (6x + 10) and son's age will be (x + 10)

After 10 years,

Mother's age will be (6x + 10) +10 and son's age will be (x + 10) + 10

Mother's age is twice that of son

(6x + 10) +10 = 2 [(x + 10) + 10]

(6x + 20) = 2[x + 20]

Solving the equation, we get x = 5

Therefore, the present ratio.

(6x + 10) : (x + 10) = 70 : 15 = 14:3

At present,

Mother's age will be (6x + 10) and son's age will be (x + 10)

After 10 years,

Mother's age will be (6x + 10) +10 and son's age will be (x + 10) + 10

Mother's age is twice that of son

(6x + 10) +10 = 2 [(x + 10) + 10]

(6x + 20) = 2[x + 20]

Solving the equation, we get x = 5

Therefore, the present ratio.

(6x + 10) : (x + 10) = 70 : 15 = 14:3

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 6 / 22

Let us assume x years ago.

At present,

Raj's age= 40 years

Surya's age= 60 years

x years ago;

Raj's age = (40 – x)

Surya's age = (60 – x)

Also,

Ratio of their ages x years ago was 4 : 8

\(\frac{(40-x)}{(60-x)}\)= \(\frac{4}{8}\)

8(40 – x) = 4(60 – x)

320 – 6x = 240 – 4x

x = 40

Therefore, 40 years ago, the ratio of their ages was 4 : 8.

At present,

Raj's age= 40 years

Surya's age= 60 years

x years ago;

Raj's age = (40 – x)

Surya's age = (60 – x)

Also,

Ratio of their ages x years ago was 4 : 8

\(\frac{(40-x)}{(60-x)}\)= \(\frac{4}{8}\)

8(40 – x) = 4(60 – x)

320 – 6x = 240 – 4x

x = 40

Therefore, 40 years ago, the ratio of their ages was 4 : 8.

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 7 / 22

At present;

Ratio of their ages = 5 : 3.

Therefore, let it be 5x and 3x.

Rohan's age 4 years ago = 5x – 4

Rahul's age after 4 years = 3x + 4

Now;

Ratio of Rohan's age 4 years ago and Rahul's age after 4 years is 1 : 1

Therefore,

Solving, we get;;

\(\frac{(5x-4)}{(3x+4)}\)= \(\frac{1}{1}\)

So, x=4.

Now;

the ratio between Rohan's age 4 years hence and Rahul's age 4 years ago will be=

Rohan's age : (5x + 4)

Rahul's age: (3x – 4)

So, Rahul's age: Rohan's age

\(\frac{(5x+4)}{(3x-4)}\)

= \(\frac{24}{8}\)

= \(\frac{3}{1}\)= 3:1

Ratio of their ages = 5 : 3.

Therefore, let it be 5x and 3x.

Rohan's age 4 years ago = 5x – 4

Rahul's age after 4 years = 3x + 4

Now;

Ratio of Rohan's age 4 years ago and Rahul's age after 4 years is 1 : 1

Therefore,

Solving, we get;;

\(\frac{(5x-4)}{(3x+4)}\)= \(\frac{1}{1}\)

So, x=4.

Now;

the ratio between Rohan's age 4 years hence and Rahul's age 4 years ago will be=

Rohan's age : (5x + 4)

Rahul's age: (3x – 4)

So, Rahul's age: Rohan's age

\(\frac{(5x+4)}{(3x-4)}\)

= \(\frac{24}{8}\)

= \(\frac{3}{1}\)= 3:1

If sum of ages of x and y is A and ratio of their ages is p : q respectively, then u can determine age of y by using the formula shown below:

Age of y= \(\frac{ratio of y}{sum of ratios}\)* sum of ages

Age of y= \(\frac{q}{p+q}\)* A

Age of y= \(\frac{ratio of y}{sum of ratios}\)* sum of ages

Age of y= \(\frac{q}{p+q}\)* A

NA

NA

Example 8 / 22

We are given, 5 years ago sister's age was 5 times the age of her brother.

Therefore,

(34 – x) – 5 = 5 (x – 5)

34 – x – 5 = 5x – 25

5x + x = 34 – 5 +25

6x = 54

x = 9

So, after 6 yrs

(x + 6)

= (9 + 6) = 15 years

Therefore,

(34 – x) – 5 = 5 (x – 5)

34 – x – 5 = 5x – 25

5x + x = 34 – 5 +25

6x = 54

x = 9

So, after 6 yrs

(x + 6)

= (9 + 6) = 15 years

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 9 / 22

Let daughter's age be x and father's age be 2x.

Father's age is 3 times more aged than his daughter, therefore father's present age = x + 2x = 3x

After 5 years, father's age is 3 times more than his daughter age.

(3x + 6) = 2(x + 6)

x = 6

After 6 years it is (3x + 5), then after further 6 years, father's age = (3x +12) and daughter's age = (x + 12)

\(\frac{(3x+12)}{(x+12)}\)= ?

Substitute the value of x, we get

\(\frac{30}{18}\)=1.6

After further 6 years, father will be 1.6 times of daughter's age.

Father's age is 3 times more aged than his daughter, therefore father's present age = x + 2x = 3x

After 5 years, father's age is 3 times more than his daughter age.

(3x + 6) = 2(x + 6)

x = 6

After 6 years it is (3x + 5), then after further 6 years, father's age = (3x +12) and daughter's age = (x + 12)

\(\frac{(3x+12)}{(x+12)}\)= ?

Substitute the value of x, we get

\(\frac{30}{18}\)=1.6

After further 6 years, father will be 1.6 times of daughter's age.

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 10 / 22

Let the son's present age be x years.

Then, (40 - x) = x

=> x= 20.

Son's age 5 years back = (20 - 5) = 15 years.

Then, (40 - x) = x

=> x= 20.

Son's age 5 years back = (20 - 5) = 15 years.

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 11 / 22

Let the present ages of son and father be x and (60 -x) years respectively.

Then, (60 - x) - 5= 4(x - 5)

55 - x = 4x - 20

5x = 75

Thus,

=> x = 15

Then, (60 - x) - 5= 4(x - 5)

55 - x = 4x - 20

5x = 75

Thus,

=> x = 15

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 12 / 22

let their present ages be 3X and 4X.

5 years ago, the ratio of their ages was 5 : 7,

Thus, (3x-5) : (4x-5) = 5 : 7.

Solving, we get X = 10.

Hence, their present ages are 3X = 30 and 4X = 40

5 years ago, the ratio of their ages was 5 : 7,

Thus, (3x-5) : (4x-5) = 5 : 7.

Solving, we get X = 10.

Hence, their present ages are 3X = 30 and 4X = 40

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 13 / 22

Let the present ages of Dhoni= 4x and Ganguly = 3x

So, 4x + 6 = 26

x = 5

Now,

Present age of Ganguly = 3x

So,

=> 3*5 = 15 years.

So, 4x + 6 = 26

x = 5

Now,

Present age of Ganguly = 3x

So,

=> 3*5 = 15 years.

- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 14 / 22

Let Riya's present age be x years.

Riya's age after 20 years = (x + 20) years.

Riya's age 4 years back = (x - 4) years

Then,

x + 20 = 5 (x - 4)

x + 20 = 5x - 20

=> x = 10

Riya's age after 20 years = (x + 20) years.

Riya's age 4 years back = (x - 4) years

Then,

x + 20 = 5 (x - 4)

x + 20 = 5x - 20

=> x = 10

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 15 / 22

Let A's age = X

Now,

C is 12 yrs older than A

So, C's age = x+12

B is 4 times as old as A

so B's age = 4x

The ages of A,B and C together are 57 years

x + (x+12) + 4x = 60

=> 6x = 48

x = 8

Thus,

A's age = 8

B's age = 4x= 32

C's age = x+12=20

Now,

C is 12 yrs older than A

So, C's age = x+12

B is 4 times as old as A

so B's age = 4x

The ages of A,B and C together are 57 years

x + (x+12) + 4x = 60

=> 6x = 48

x = 8

Thus,

A's age = 8

B's age = 4x= 32

C's age = x+12=20

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 16 / 22

Average age of 80 boys = 15 x 80 = 1200 yrs.-----(i)

average age of 15 boys = 15 *16 = 240 yrs-----(ii)

average age of 25 boys = 14 * 25 = 350 yrs.-----(iii)

Remaining boys= 80-(15+25)= 40 boys

Average age of remaining 40 boys =

\(\frac{1200 - (240+350)}{80 - (25+15)}\)

= \(\frac{610}{41}\)

= 15.25 years.

average age of 15 boys = 15 *16 = 240 yrs-----(ii)

average age of 25 boys = 14 * 25 = 350 yrs.-----(iii)

Remaining boys= 80-(15+25)= 40 boys

Average age of remaining 40 boys =

\(\frac{1200 - (240+350)}{80 - (25+15)}\)

= \(\frac{610}{41}\)

= 15.25 years.

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 17 / 22

Ram age when he was born = 0 years

=> His brother's age = 6 year

=> His father's age = brother age + 32years

= 6+32 = 38

=> His mother's age = father's age – 3

=38-3 = 35

So sister's age = 35-25 = 10 years.

=> His brother's age = 6 year

=> His father's age = brother age + 32years

= 6+32 = 38

=> His mother's age = father's age – 3

=38-3 = 35

So sister's age = 35-25 = 10 years.

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 18 / 22

Let Rishi's age now be A

Rahul's age= B

(A - 6) = X(B - 6)

But Rahul is 17 and therefore (17-6)=11 = X(B - 6)

\(\frac{11}{X}\) = B-6

\(\frac{11}{X}\) + 6 = B

B = \(\frac{11}{X}\) + 6

Rahul's age= B

(A - 6) = X(B - 6)

But Rahul is 17 and therefore (17-6)=11 = X(B - 6)

\(\frac{11}{X}\) = B-6

\(\frac{11}{X}\) + 6 = B

B = \(\frac{11}{X}\) + 6

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 19 / 22

Let Peter's age = x.

Then Paul's age = x + 2

After 6 years the total of their ages will be 7 times of their current age.

Not clear.

So the given data are inadequate.

Then Paul's age = x + 2

After 6 years the total of their ages will be 7 times of their current age.

Not clear.

So the given data are inadequate.

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 20 / 22

Let Ravi's age at the time of his marriage = X years

Then,

=> X + 10 = \(\frac{6}{5}\) X

X = 50.

Since got married 10 years ago,

So, his present age = 50+10= 60

His sister is 5 years younger => 60-5 = 55 years.

Then,

=> X + 10 = \(\frac{6}{5}\) X

X = 50.

Since got married 10 years ago,

So, his present age = 50+10= 60

His sister is 5 years younger => 60-5 = 55 years.

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 21 / 22

Let the Present age of Ram = x

Rahul's age = 5-2 = 3

So, as per the question:

\(\frac{x-6}{18}\) = 3

x-6 = 54

x = 60

Ram's present age = 60 years

Rahul's age = 5-2 = 3

So, as per the question:

\(\frac{x-6}{18}\) = 3

x-6 = 54

x = 60

Ram's present age = 60 years

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Example 22 / 22

As per given in question:

(A+B)=12+(B+C)-----(i)

So,

(A+B) - (B+C)= 12

A-C= 12

Thus;

C is younger than A by 12 years.

(A+B)=12+(B+C)-----(i)

So,

(A+B) - (B+C)= 12

A-C= 12

Thus;

C is younger than A by 12 years.

**Ans.**- age after n years = y+ n

- age n years ago = y – n

- n times the age = ny

- If ages in the numerical are mentioned in ratio A : B, then A : B will be Ay and By

NA

NA

Gurajala Mythili1 month ago

14