Question

The ratio of ages of fatther and son is 7:2 five years ago the product of their age was 150 what is the age of the father​

Answer 1

Answer:

Appropriate Question :-

• The ratio of ages of father and son is 7 : 2. Five years ago the product of their ages was 150. What is the age of the father.

Given :-

• The ratio of ages of father and son is 7 : 2.
• Five years ago the product of their ages was 150.

To Find :-

• What is the age of father.

Solution :-

Let,

$\mapsto$ The age of father be 7x years

$\mapsto$ The age of son will be 2x years

$\leadsto \sf\bold{Five\: years\: ago\: :-}$

$\mapsto$ The age of father is (7x - 5) years

$\mapsto$ The age of son is (2x - 5) years

$\clubsuit\: \: \sf\bold{\underline{According\: to\: the\: question\: :-}}$

$\implies \sf (7x - 5)(2x - 5) =\: 150$

$\implies \sf 14x^2 - 35x - 10x + 25 =\: 150$

$\implies \sf 14x^2 - 45x + 25 =\: 150$

$\implies \sf 14x^2 - 45x + 25 - 150 =\: 0$

$\implies \sf 14x^2 - 45x - 125 =\: 0$

$\implies \sf 14x^2 - (70 - 25)x - 125 =\: 0$

$\implies \sf 14x^2 - 70x + 25x - 125 =\: 0$

$\implies \sf 14x(x - 5) + 25(x - 5) =\: 0$

$\implies \sf (14x + 25)(x - 5) =\: 0$

$\implies \sf (14x + 25) =\: 0$

$\implies \sf 14x + 25 =\: 0$

$\implies \sf 14x =\: - 25$

$\implies \sf \bold{\purple{x =\: \dfrac{- 25}{14}}}\: \: \bigg\lgroup \sf\bold{Age\: can't\: be\: taken\: as\: negetive\: (- ve)}\bigg\rgroup$

Either,

$\implies \sf (x - 5) =\: 0$

$\implies \sf x - 5 =\: 0$

$\implies \sf\bold{\purple{x =\: 5}}$

We have to take the positive (+ ve) sign.

Hence, the required ages of father and son are :

$\bigstar\: \sf\bold{\green{Age\: of\: father\: :-}}$

$\longrightarrow \sf 7x\: years$

$\longrightarrow \sf 7(5)\: years$

$\longrightarrow \sf\bold{\red{35\: years}}$

$\bigstar\: \: \sf\bold{\green{Age\: of\: son\: :-}}$

$\longrightarrow \sf 2x\: years$

$\longrightarrow \sf 2(5)\: years$

$\longrightarrow \sf\bold{\red{10\: years}}$

$\therefore$ The age of father is 35 years.

Answer 2

$\huge\blue{ANSWER:-}</p><p>$

• Let the present age of Father = 7x.
• and present age of son = 2x.

$\huge\red{A.}\green{T.}\red{Q.}$

$\small\red{(7x-5)(2x-5)= 150}$

$\small\pink{=> 14x^2 - 45x + 25 = 150}$

$\small\green{=> 14x^2-45x-125=0}$

$\small\blue{ = > 14x^2-70x+25x-125=0}$

$\small\purple{=> 14x(x-5)+25(x-5)=0}$

$\small\orange{=> (x-5) (14x+25)}$

$\small\red{∴x=5}$

$So, present \: age \: of \: father = 7x .\\ \: \: \: \: \: \: \: \: \: \:$

$\longmapsto\tt\boxed\small\pink{35 years}$

$\huge\green{Alternate:-}$

Father Son

Present 7 : 2

↓ × 5 : ↓ × 5

Or 35 : 10

- 5 ↓ ↓ - 5

Before 30 : 5

$\huge\blue{A.T.Q,}$

$\small\red{(30×5) unit = 150 years}$

$\small\pink{∴35 \: units = 35 \: years}$

$\longmapsto\tt\boxed{∴ Present \: Age \: of \: Father = 35 years. }$