Question

8) A father is 30 years older than his son, and one year ago he was four times as old as his son.Find their present ages.?

Answer 1

AnsWer:

• Son's age = 11 years
• Father's age = 41 years

Step By step Explanation:

Given:

• Father is 30 years older than his son.
• One year ago he was four times old as his son.

To Find:

• Present ages of son and father.

Let son's age be 'x'.

And, father's age be 'x + 30'.

Now, according to question,

>> 1 year ago:

=> x + 30 - 1 = 4(x - 1)

=> x + 29 = 4x - 4

=> x + 33 = 4x

=> x + 33 - 4x = 0

=> -3x = -33

=> x = 33/3

=> x = 11

Hence,

• Son's age is = 11 years
• And father's age is = 30 + 11 = 41 years.

Answer 2

AnswEr:-

Present age of son = 11 years.

Present age of father = 41 years.

$\rule{200}{2}$

☯Given :-

• Father's present age = 30 + Son's present age.
• 1 year ago,father's age = 4 times son's age.

☯To find :-

• Present age of father = ?
• Present age of son = ?

☯Solution :-

Let the present age of son be S years & present age of father be F years.

First case:-

➩ F = S + 30 .................(Eq.1)

2nd case:-

➩ (F - 1) = 4 × (S - 1)

➩ F - 1 = 4S - 4

• Putting value from (Eq.1):-

➩ S + 30 - 1 = 4S - 4

➩ S + 29 = 4S - 4

➩ S - 4S = -4 - 29

➩ -3S = -33

➩ S = -33/-3

➩ S = 11 years.

$\therefore\underbrace{\textsf{Present \: age \: of \: son = {\textbf{11 \: years }}}}$

$\rule{100}{2}$

• Putting value of S in (Eq.1) :-

➩ F = 11 + 30

➩ F = 41 years.

$\therefore\underbrace{\textsf{Present \: age \: of \: father = {\textbf{41 \: years }}}}$