Question

Q10.The age of father is twice the sum of ages of his two children. Ten years hence, the age of father will be three-quarter of the sum of the ages of his children then. Find the present age of father.

Answer 1

Answer:

8 years

Step-by-step explanation:

Let the age of first son be x

Let the age of second son be y

Let the age of father be z

Now we are given that The age of father is twice the sum of ages of his two children.

$\Rightarrow z=2(x+y)$  --1

After 10 years

The age of first son =x+10

The age of second son = y+10

The age of father=z+10

Now we are given that Ten years hence, the age of father will be three-quarter of the sum of the ages of his children then

$\Rightarrow z+10=\frac{3}{4}(x+10+y+10)$ --2

Now substitute the value of z from 1 in 2

$\Rightarrow 2(x+y)+10=\frac{3}{4}(x+10+y+10)$

$\Rightarrow 2(x+y)+10=\frac{3}{4}(x+y+20)$

$\Rightarrow 8x+8y+40=3x+3y+60$

$\Rightarrow x+y=4$

Now substitute the value of x+y in 1

$\Rightarrow z=2(4)$

$\Rightarrow z=8$

Hence the age of father is 8 years.

Answer 2

Answer:

Let the age of first son be x

Let the age of second son be y

Let the age of father be z

Now we are given that The age of father is twice the sum of ages of his two children.

\Rightarrow z=2(x+y)⇒z=2(x+y) --1

After 10 years

The age of first son =x+10

The age of second son = y+10

The age of father=z+10

Now we are given that Ten years hence, the age of father will be three-quarter of the sum of the ages of his children then

\Rightarrow z+10=\frac{3}{4}(x+10+y+10)⇒z+10=

4

3

(x+10+y+10) --2

Now substitute the value of z from 1 in 2

\Rightarrow 2(x+y)+10=\frac{3}{4}(x+10+y+10)⇒2(x+y)+10=

4

3

(x+10+y+10)

\Rightarrow 2(x+y)+10=\frac{3}{4}(x+y+20)⇒2(x+y)+10=

4

3

(x+y+20)

\Rightarrow 8x+8y+40=3x+3y+60⇒8x+8y+40=3x+3y+60

\Rightarrow x+y=4⇒x+y=4

Now substitute the value of x+y in 1

\Rightarrow z=2(4)⇒z=2(4)

\Rightarrow z=8⇒z=8

Hence the age of father is 8 years.