Question

(ii) 2 years ago, my age was 4 1/2
times the age of my son then. 6 years ago, my age

was twice the square of the age of my son then. Find the present age of my son
The answer should be 10 years​

Answer 1

Let the present ages of

• Mine be x
• My son be y

2 years ago,

• My age was (x - 2)
• My son's age was (y - 2)

According to the question,

(x - 2) = 4.5(y - 2)

➜ x - 2 = 4.5y - 9

➟ x + 9 - 2 = 4.5y

➝ x + 7 = 4.5y

➝ x = 4.5y - 7          ----- [Equation 1]

6 years ago;

• My age was (x - 6)
• My son's age was (y - 6)

Now,

(x - 6) = 2 × (y - 6)²

➥ x - 6 = 2 × (y² + 36 - 12y)

➞ x - 6 = 2y² + 72 - 24y

Substitute the value of x from equation 1

4.5y - 7 - 6 = 2y² + 72 - 24y

⇒ 4.5y - 13 = 2y² + 72 - 24y

⇒ 4.5y = 2y² - 24y + 85y

➝ 9/2y = 2y² - 24y + 85y

➝ 9y = 2(2y² - 24y + 85)

➝ 9y = 4y² - 48y + 170

➝ 57y = 4y² + 170

➝ 4y² - 57y + 170 = 0

➝ 4y² - 40y - 17y + 170 = 0

➝ 4y(y - 10) - 17(y - 10) = 0

➝ (4y - 17) (y - 10)

So,

4y - 17 = 0

→ 4y = 17

→ y = 17/4

OR

(y - 10) = 0

→ y = 10

Age cannot be in fractions.

∴ Son's age = y = 10 years

Answer 2

Answer:

${\large{\sf{\pmb{\underline{Correct \; question...}}}}}$

★ Two year's ago, my age was ${4 \dfrac{1}{2}}$ times the age of my son then 6 years ago, my age was twice the square of my son then find the present age of my son.

${\large{\sf{\pmb{\underline{Given \; that...}}}}}$

★ Two year's ago, my age was ${4 \dfrac{1}{2}}$ times the age of my son

★ 6 years ago, my age was twice the square of my son.

${\large{\sf{\pmb{\underline{To \; Find...}}}}}$

★ The present age of my son.

${\large{\sf{\pmb{\underline{Solution...}}}}}$

★ The present age of my son = 10 year's

${\large{\sf{\pmb{\underline{Using \; concepts...}}}}}$

★ Quadratic equations.

★ Linear equations.

${\large{\sf{\pmb{\underline{Assumptions...}}}}}$

★ Let a be the age of son 2 year's ago.

${\large{\sf{\pmb{\underline{Full \; Solution...}}}}}$

~ As it's given that two year's ago, my age was ${4 \dfrac{1}{2}}$ times the age of my son. Henceforth, it means that

$:\implies \sf 4 \dfrac{1}{2} \times a \\ \\ :\implies \sf \dfrac{9}{2} \times a \\ \\ :\implies \sf \dfrac{9}{2}a \\ \\ :\implies \sf Henceforth, \: it \: comes \: as \: 4.5a$

~ Now as it's given that 6 years ago, my age was twice the square of my son. Henceforth, it means that

$:\implies \sf My \: age \: 6 \: years \: ago \: = 4.5a - 4 \\ \\ :\implies \sf My \: son \: age \: 6 \: years \: ago \: = a-4$

~ Now by the given question's information, we have to find that 6 years ago, my age was twice the square of my son. We have to use quadratic equations here, let's do it!

$:\implies \sf 4.5a-4 = 2(a-4)^{2} \\ \\ :\implies \sf 4.5a-4 = 2(a^{2} -8a +16) \\ \\ :\implies \sf 4.5a-4 = 2a^{2} - 16a + 32 \\ \\ :\implies \sf 2a^{2} - 16a - 4.5a + 32 + 4 = 0 \\ \\ :\implies \sf a = 8 \: and \: a = \dfrac{9}{4}$

~ Now let's solve again!

$:\implies \sf a = 8 \\ \\ :\implies \sf As \: a \: is \: the \: age \: of \: son \: 2\: year's\: ago \\ \\ :\implies \sf Present \: age \: of \: son \: = a+2 \\ \\ :\implies \sf Present \: age \: of \: son \: = 8+2 \\ \\ :\implies \sf Present \: age \: of \: son \: = 10$