Question

The product of the present ages of two brother is 160 and 4 years ago elder brother was twice as old as his younger brother. Find their present ages.​

Answer 1

Step-by-step explanation:

let the present age of younger brother =a year

and elder brother b year

4 years ago

age of elder brother=b-4

age of younger brother=a-4

A/q,2(a-4)=(b-4)

2a-b=4----(1)

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Answer 2

Given :

• The product of the present ages of two brother is 160.
• 4 years ago elder brother was twice as old as his younger brother.

To find :

• Their present ages.

Solution :

Consider,

• Present age of elder brother = x years.
• Present age of younger brother = y years.

According to the 1st condition:-

• The product of the present ages of two brother is 160.

$\implies\sf{xy=160................eq[1]}$

According to the 2nd condition :-

• 4 years ago elder brother was twice as old as his younger brother.

4 years ago ,

• Age of elder brother = (x-4) years
• Age of younger brother = (y-4) years

$\implies\sf{x-4=2(y-4)}$

$\implies\sf{x-4=2y-8}$

$\implies\sf{x=2y-4...................eq[2]}$

Now taking eq [1].

$\implies\sf{xy=160}$

$\implies\sf{(2y-4)y=160\:[Put\:x=2y-4\: from\:eq(1)]}$

$\implies\sf{2y^2-4y=160}$

$\implies\sf{2(y^2-2y)=160}$

$\implies\sf{y^2-2y=80}$

$\implies\sf{y^2-2y-80=0}$

$\implies\sf{y^2-(10-8)y-80=0}$

$\implies\sf{y^2-10y+8y-80=0}$

$\implies\sf{y(y-10)+8(y-10)=0}$

$\implies\sf{(y-10)(y+8)=0}$

Either,

$\implies\sf{y-10=0}$

$\implies\sf{y=10}$

Or,

$\implies\sf{y+8=0}$

$\implies\sf{y=-8}$

Age can't be negative .

• Present age of younger brother = 10 years.

Now put y = 10 in eq [2] for getting the value of x.

$\implies\sf{x=2y-4}$

$\implies\sf{x=2\times\:10-4}$

$\implies\sf{x=16}$

• Present age of elder brother = 16 years.

Therefore, the present age of elder brother is 16 years and the present age of younger brother is 10 years.