Question

Five years hence, the age of Jacob will be three times that of his son. Five ago, Jacob age was seven times that of his son. Find their present age.

Answer 1

Step-by-step explanation:

Let 'x' be the present age of jacob and 'y' be the age of his son

As per question

x+5=3(y+5).....(1)

x−5=7(y−5).......(2)

Adding (1) and (2)

2x=3y+15+7y−35

2x=10y−20

x=5y−10...(3)

Putting (3) in (1)

5y−5=3y+15

2y=20

y=10

Putting in (3)

x=5×10−10

x=40

Therefore Present age of Jacob is 40 years and his son is 10 years.

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Answer 2

S O L U T I O N :

Let the x & y be the present age of Jacob & his son respectively.

A/q

$\underbrace{\bf{5\:years\:later\::}}$

The Jacob age will = (x + 5) years.

The son age will = (y + 5) years.

So,

$\longrightarrow\tt{(x+5) = 3(y+5) }$

$\longrightarrow\tt{x+5= 3y+15 }$

$\longrightarrow\tt{x-3y= 15-5 }$

$\longrightarrow\tt{x-3y= 10}$

$\longrightarrow\tt{x= 10+3y..............(1)}$

$\underbrace{\bf{5\:years\:ago\::}}$

The Jacob age was = (x - 5) years.

The son age was = (x - 5) years.

So,

$\longrightarrow\tt{(x-5) = 7(y-5)}$

$\longrightarrow\tt{x-5 = 7y-35}$

$\longrightarrow\tt{x-7y=-35+5}$

$\longrightarrow\tt{x-7y=-30}$

$\longrightarrow\tt{10+3y-7y=-30\:\:[from(1)]}$

$\longrightarrow\tt{10-4y=-30}$

$\longrightarrow\tt{-4y=-30 - 10}$

$\longrightarrow\tt{-4y=-40}$

$\longrightarrow\tt{y=\cancel{-40/-4}}$

$\longrightarrow\bf{y=10\:years}$

∴ Putting the value of y in equation (1),we get;

$\longrightarrow\tt{x = 10 + 3(10)}$

$\longrightarrow\tt{x = 10 + 30}$

$\longrightarrow\bf{x = 40\:years}$

Thus,

The present age of jacob will be 40 years & his son will be 10 years .