Math, asked by Anonymous, 2 days ago

 \huge{\mathtt{\underline{\underline{Question}}}}
Twenty years ago, a father was six times as old as his daughter. After 20 years, he will be twice as old as his daughter. Determine their present ages.

 \red{\fbox{→ Spammers not allowed}}
 \green{\fbox{→ Please give explanation too}}

Answers

Answered by taneruragavendra
2

Answer:

Let the present age of father and his daughter are x and y respectively.

Given that ten years ago a father was six times as old as his daughter.

(x−10)=6(y−10)

⇒x−10=6y−60

⇒x−6y+50=0⟶(1)

Also given that after 10 years, he will be twice as old as his daughter.

(x+10)=2(y+10)

⇒x+10=2y+20

⇒x−2y−10=0⟶(2)

Subtracting eq

n

(1) from (2), we have

x−2y−10−(x−6y+50)=0

⇒x−2y−10−x+6y−50=0

⇒4y−60=0

⇒y=15

Substituing the value of y in eq

n

(2), we have

x−2×15−10=0

⇒x−30−10=0

⇒x=40

Hence, the present ages of father and his daughter is 40 and 15 respectively.

∴ sum of their present ages =x+y=40+15=55

Hence, the correct answer is 55.

Answered by itzmedipayan2
4

Answer:

Let the father's age be x

Let the daughter's age be y

Atq,

 \frac{x - 20}{y - 20}  =  \frac{6}{1}  \\  \\ x - 20 = 6y - 120 \\  \\ x - 6y = 100  \dashrightarrow (i) \\

Again

 \frac{x + 20}{y + 20}  =  \frac{2}{1}  = x + 20 = 2y + 40 \\  \\ x - 2y = 20 \dashrightarrow \: (ii)

From equation (i) and (ii)

x - 6y =  - 100 \\  \frac{x - 2y =  - 20}{ =  - 4y =  - 120}  \\  \\  - 4y = 120 \\  \\ y =  \frac{ { \cancel{120}} \:  \: ^{30} }{ \cancel4}

30 years.

putting the value of y in eq (ii)

x - 2 \times 30 = 20 \\  \\ x = 20 + 30 = 50

Hence

Age of daughter is 30 years

Age of father is 50 years

Hope it helps you from my side :)

Similar questions