APTITUDE PREPARATION: Problems on Age


Introduction:

Concept of ratios and simple equations are used to solve problems related to age.
The ages are usually given as integers.
The difference in ages between individuals remains constant always.

Example 1:

The sum of ages of 5 children born at the intervals of 4 years each is 60 years. What is the age of the youngest child?

Solution:

Age of youngest child = X
X + (X+4) + (X+8) + (X+12) + (X+16) = 60
5X + 40 = 60
5X = 20
X = 4 years.

Example 2:

Five years ago Anand was 3 times as old as Barath. If Anand is 30 years older than Barath, find the present ages of both.

Solution:

Present age of Anand = A
Present age of Barath = B
Five years ago:
Age of Anand = (A-5)
Age of Barath = (B-5)
(A-5) = 3(B-5) or A = 3B – 10 … (i)
At Present:
A = B + 30 … (ii)
Using equation (i) in (ii);
3B – 10 = B + 30
2B = 40 or B = 20 years and A = 20 + 30 = 50 years.

Example 3:

One year ago, A was three times his sister’s age. Next year, he will be only twice her age. Find the age of A five years from now.

Solution:

Present age of A = A
His sister’s age: S
One year ago:
Age of A =(A-1)
His sister’s age = (S-1)
(A-1) = 3(S-1) or A = 3S-2 … (i)
After one year:
Age of A =(A+1)
His sister’s age = (S+1)
(A+1) = 2(S+1) or A = 2S + 1 … (ii)
Equating (i) in (ii);
A = 3S-2 = 2S + 1 or S = 3
A = (3×3) – 2 =7.
After 5 years from now:
Age of A = 7 + 5 = 12 years.


Example 4:

Present age of a Agalya is two-fifth of her Mother. She will be one-half of her Mother’s age after 8 years. Find the present age of mother.

Solution:

Present age of Agalya = A
Present age of mother = M
A = M(2/5) … (i)
After 8 years:
Agalya’s age = (A+8)
Mother’s age = (M+8)
(A+8) = (M+8)/2 … (ii)
Using equation (i) in (ii);
[M(2/5) + 8] = (M+8)/2
[2M+40]/5 = (M+8)/2
4M+80 = 5M+40
M = 40 years.

Example 5:

Total age of A, B and C is 93. The ratio of their ages was 2:3:4 ten years ago. Find the present age of C.

Solution:

Let the ages of A, B and C are A, B and C respectively 10 years ago. (this will be easier).
At Present:
Their ages: (A+10), (B+10) and (C+10) respectively
(A+10)+(B+10)+(C+10) = 93
A+B+C = 63 …(i)

Ten years ago:
A:B:C = 2:3:4
We have to find the age of C.
Convert others’ ages in terms of C’s age.
That is, if C = 4 then B = 3 and A =2.
A = (C/2) … (ii)
B = C(3/4) … (iii)
Using equations (ii) and (iii) in (i);
(C/2) + C(3/4) + C = 63
(2C+3C+4C)/4 = 63 … (LCM of 2 and 4 is 4)
9C = (4×63) = 252
C = 28, which is the age of C 10 years ago.
Present age of C is (28+10) = 38 years.

Example 6:

A father 35 years old has an 11 year old son. How many years ago was the father, 5 times as old as his son?

Solution:

Let the ages of A, B and C are A, B and C respectively 10 years ago. (this will be easier).
At Present:
Their ages: (A+10), (B+10) and (C+10) respectively
(A+10)+(B+10)+(C+10) = 93
A+B+C = 63 …(i)

Ten years ago:
A:B:C = 2:3:4
We have to find the age of C.
Convert others’ ages in terms of C’s age.
That is, if C = 4 then B = 3 and A =2.
A = (C/2) … (ii)
B = C(3/4) … (iii)
Using equations (ii) and (iii) in (i);
(C/2) + C(3/4) + C = 63
(2C+3C+4C)/4 = 63 … (LCM of 2 and 4 is 4)
9C = (4×63) = 252
C = 28, which is the age of C 10 years ago.
Present age of C is (28+10) = 38 years.Present age of father = 35
Present age of son = 11
X years ago:
Age of father = (35-X)
Age of son = (11-X)
(35-X) = 5(11-X)
5X –X = 55-35
4X = 20 and,
X = 5 years.


Example 7:

The difference between the ages of P and Q is 15. If P is four times older than Q, find the age of Q.

Solution:

P is 4 times older than Q.
Hence, P-Q = 15 … (i)
P = 4Q … (ii)
Using equation (ii) in (i);
4Q-Q = 15
Q = 5 years.

Example 8:

Today, B is 6 times older than A. After 10 years, B’s age will be twice that of C. C celebrated her eighth birthday two years ago. Find the present age of A.

Solution:

Ages of A, B and C are A, B and C respectively.
B = 6A … (i)
After 10 years:
(B+10) = 2(C+10)
B = 2C+10 … (ii)
Before two years:
C-2 = 8 and,
C = 10 … (iii)
Using (iii) in (ii);
B = (2×10)+10 = 30
From equation (i): B = 30 = 6A
Present age of A = 5 years.

Example 9:

Ajay’s mother is now four times his age. After 5 years, she will be three times his age. In how many years, will she be twice his age?

Solution:

At Present:
Age of Ajay = A
Age of mother = M
M=4A … (i)
After 5 years:
(M+5) = 3(A+5)
M = 3A+10 … (ii)
Using equation (ii) in (i);
4A = 3A + 10
A = 10 and M = 40.
After X years:
(M+X)=2(A+X)
(40+X) = 2(10+X)
X = 20 years.

Example 10:

In the year 1932, Raja was as old as the number formed by the last two digits of his year of birth. The same was applicable to his grandmother also. Find the year of birth of Raja and his grandmother.

Solution:

Raja’s year of birth should be in the 20th century, ie., after 1900 (otherwise, his age will be very high).
Similarly, the year of birth of his grandmother should be in the 19th century, ie., before 1900 (otherwise, her age will be very less).
In the year 1932:
Age of Raja = R
Age of grandmother = G
Considering the last two digits of 1932,
32-R = R and,
Age of Raja in 1932, R = 16.
His year of birth should be 1932-16 = 1916.
Now, add 100 to 32 = 132.
132-G = G and,
Age of grandmother in 1932, G = 66.
Her year of birth should be 1932-66 = 1866.

About Author

Dr. BASKAR .A,

a former Research Scientist Government of India, presently working as a Professor in the Dept of Mechanical Engineering, Authored books on Kinematics and Dynamics of Machinery, Also Published Numerous Research papers moreover a Geek in Mathematics

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