Question

QUESTION:-

Ayush is four times as old as his brother Rohit at present After 10 years, he will be twice the age of his brother. Find their present ages. ​

Answer 1

Given:

• Ayush is four times as old as his brother Rohit at present
• After 10 years, he will be twice the age of his brother.

To Find:

• Find their present ages= ?

solution:

Let,

• the present age of Ayush be y
• the present age of Rohit be 4y

Then,

• Age of Ayush after 10 years be (y + 10)
• Age of Rohit after 10 years be (4y + 10)

$\\ \dag \: { \underline { \sf{ \: According \: \: to \: \: Question}}}$

$\sf\dashrightarrow 4y + 10 = 2(y + 10)$

$\sf\dashrightarrow 4y + 10 = 2y + 20$

$\sf\dashrightarrow 4y - 2y = 20 - 10$

$\sf\dashrightarrow 2y = 10$

$\sf\dashrightarrow y = \cancel\frac{10}{2}$

$\sf\dashrightarrow y = 5$

hence,

The present age of Ayush = y = 5 yrs

The present age of Rohit = 4 y = 4 × 5 = 20 yrs

Answer 2

$\rm \bigstar \: \underline{\underline{Solution}}$

Assume that :

Let present age of Rohit be x and present age of Ayush be y

According to question ,

$\rm \mapsto\: y = 4 \: x \: . \: .(i)$

After 10 years ,

Ayush age will be (y + 10) years

Rohit age will be (x+ 10) years

According to question,

$\rm \mapsto \: y + 10 = 2(x + 10). \: ..(ii)$

From equations (i) and (ii),

$\rm \longrightarrow \: 4x + 10 = 2x + 20$

$\rm \longrightarrow4x - 2x = 20 - 10$

$\rm \longrightarrow \: 2 \: x = 10$

$\rm \longrightarrow \: x = \dfrac{10}{2}$

$\bf \longrightarrow \: x = 5$

Putting the value of x in equation (i),

$\rm \mapsto \: y = 4 \times( 5) = 20$

Therefore :

• Ayush 's present age = 20 years

• Rohit's presnt age = 5 years