Math, asked by harshitaxx, 9 months ago

7 years ago Varun's age was five times the square of Swati's age. 3 years hence, Swati's age will be two-fifth of Varun's age. Find their present ages.

Answers

Answered by Anonymous
9

➠ AnSwer :-

Let Swati's age 7 years ago be x years.

Then, Varun's age 7 years ago = 5x² years.

Swati's present age = (x + 7) years.

Varun's present age = (5x² + 7) years.

Swati's age 3 years hence = (x + 7 + 3) years =

(x + 10) years.

Varun's age 3 years hence =

➝(5x² + 7 + 3) years

➝(5x²+ 10) years.

➝(x + 10) = 2/5(5x² + 10)

➝5x + 50 = 10x² + 20 = 10x² – 5x - 30 = 0

➝2x² - x - 6= 0 = 2x² - 4x + 3x - 6 = 0

➝2x (x - 2) + 3(x - 2) = 0 = (x - 2)(2x + 3) = 0

➝x - 2 = 0 or 2x +3= 0 x = 2 or x = -3/2

➝x=2 [: age cannot be negative ]

Swati's present age = (2 + 7) years = 9 years.

Varun's present age = (5 x 22 + 7) years = 27 years.

Answered by Anonymous
2

Answer:

Let the present age of Varun and Swati are x and y years respectively.

Given – 7 years ago

x – 7 = 5(y-7)^2 ----------------1

Also given – 3 years hence

y + 3 = 2/5 (x + 3)

5y + 15 = 2x + 6

2x = 5y + 9

x = (5y + 9)/2 ------------------2

Substitute the value of x in equation 1

(5y + 9)/2 – 7 = 5 (y-7)^2

5y + 9 – 14 = 10 (y^2 – 14y + 49)

5y – 5 = 10y^2 – 140y + 490

10y^2 – 145y + 495 = 0

Dividing the equation by 5

2y^2 – 29y + 99 = 0

Solving above quadratic equation to find y

y = (-b + sqrt (b^2 – 4ac))/4ac or y = (-b - sqrt (b^2 – 4ac))/4ac

get y = 9 and y = 11/2

consider age as a whole number

Therefore, y = 9 years

Substitute the value of y in equation 2

Therefore, x = (5 * 9 + 9)/ 2 = 27 years

Answer – The present age of Varun is 27 years and present age of Swati is 9 years.

Step-by-step explanation:

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