Question

The product of the present ages of two sisters is 150. 5 years ago, the elder sister was twice as old as her younger sister. Find their present ages.

Answer 1

Answer: 10,15

Step-by-step explanation: Let the present ages be x&y,Let x>y

xy=150............(1)

2*(x-5)=y-5

Therefore 2x-y-5=0....(2)

Solving simultaneously

x=10

y=15

Answer 2

$\Large{\underline{\underline{\mathfrak{\red{\bf{Solution}}}}}}$

$\Large{\underline{\mathfrak{\orange{\bf{Given}}}}}$

• The product of the present ages of two sisters is 150
• 5 years ago, the elder sister was twice as old as her younger sister.

$\Large{\underline{\mathfrak{\orange{\bf{Find}}}}}$

• present age of two sister

$\Large{\underline{\underline{\mathfrak{\red{\bf{Explanation}}}}}}$

Let,

• Age of elder sister = x years
• Age of younger sister = y years

Case(1).

Now, A/C to question,

➩ x y = 150 -----------------(1)

Case(2).

Again,

➩ (x-5) = 2 × (y-5)

➩ x - 2y = - 10 + 5

➩ x - 2y = -5 -----------------(2)

By, equ(1)

➩ x = 150/y -----------------(3)

Keep value of x in equ(2)

➩ 150/y - 2y = -5

➩ 150 - 2y² = -5y

➩ -2y² + 5y + 150 = 0

Or,

➩ 2y² - 5y - 150 = 0

➩ 2y² - 20y + 15y - 150 = 0

➩ 2y(y - 10) + 15(y - 10) = 0

➩ (2y + 15)(y - 10) = 0

➩ (2y + 15) = 0 Or, (y - 10) = 0

➩ 2y = -15 Or, y = 10

➩ y = -15/2. Or, y = 10

Neglect , y = -15/2 .

because age be always positive .

So, take

• y = 10

keep value of y in equ(1)

➩ x * 10 = 150

➩ x = 150/10

➩ x = 15

$\Large{\underline{\underline{\mathfrak{\red{\bf{Hence}}}}}}$

• Age of elder sister (x) = 15 years
• Age of younger sister (y) = 10 years

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