Question

1. A woman is six times as old as her son. Two
years ago, the product of their ages was 84.
Find their present ages.
lor daughter​

Answer 1

$\huge{\sf{\underline{\red{Answer}}}}$

Present age of woman is 30 years and of her son is 5 years.

$\rule{200}2$

$\small{\sf{\underline{ASSUMPTION:}}}$

Let us assume that the present age of -

• Woman is W years
• Son is S years

$\small{\sf{\underline{SOLUTION:}}}$

A woman is six times as old as her son.

$\underline{\underline{\green{\sf{As\:per\:given\:condition}}}}$

Woman = 6 × Son

$\implies\:\sf{W\:=\:6\:\times\:S}$ -----------> [1]

Two years ago, the product of their ages was 84 years.

Two years ago -

• Age of woman = (W - 2) years
• Age of son = (S - 2) years

$\underline{\underline{\green{\sf{As\:per\:given\:condition}}}}$

Age of woman two years ago × Age of son two years ago = 84

$\implies\:\sf{(W-2)(S-2)\:=\:84}$

$\implies\:\sf{WS-2W-2S+4\:=\:84}$

Substitute value of W = 6S in above equation

$\implies\:\sf{S(6S)-2(6S)-2S+4\:=\:84}$

$\implies\:\sf{6S^2-12S-2S+4-84\:=\:0}$

$\implies\:\sf{6S^2-14S-80\:=\:0}$

Take 2 as common

$\implies\:\sf{2(3S^2-7S-40)\:=\:0}$

$\implies\:\sf{3S^2-7S-40\:=\:0}$

Now, solve it By Quadratic formula

$\implies\:\sf{D\:=\:\sqrt{(b)^2-4ac}}$

$\implies\:\sf{D\:=\:\sqrt{(-7)^2-4(3)(-40)}}$

$\implies\:\sf{D\:=\:\sqrt{529}\:=\:23}$

Now,

$\implies\:\sf{S\:=\:\dfrac{ - b \:\pm \:D }{2a}}$

$\implies\:\sf{S\:=\:\frac{7\:\pm \:23}{2(3)}}$

$\implies\:\sf{S\:=\:\frac{7+23}{6}}$

$\implies\:\sf{S\:=\:\frac{30}{6}\:=\:5}$

(Age can't be negative. So, negative one neglected.)

Substitute value of S = 5 in equation [1]

$\implies\:\sf{W\:=\:6(5)}$

$\implies\:\sf{W\:=\:30}$

Therefore,

Present age of of Woman is 30 years and Present age of her Son is 5 years.

Answer 2

||✪✪ GIVEN ✪✪||

• A woman is six times as old as her son.
• the product of their ages is 84.

|| ✰✰ ANSWER ✰✰ ||

Let us assume That , his son present age is = x years old .

So, women Present age = 6x years.

Now, 2 years ago, son was = (x -2) years old.

→ Women was = (6x - 2) years old.

A/q,

→ (x-2)(6x -2) = 84

→ 2(x-2)(3x-1) = 84

→ (x-2)(3x -1) = 42

→ 3x² - 7x + 2 = 42

→ 3x² - 7x - 40 = 0

→ 3x² - 15x + 8x - 40 = 0

→ 3x(x - 5) + 8(x - 5) = 0

→ (3x + 8) (x - 5) = 0

Putting both Equal to zero,

→ x = 5 or (-8/3)

(since -ve age not possible).

So, Present age of son = x = 5 years .

Present age of Mother = 6x = 5*6 = 30 years .