Math, asked by Anonymous, 5 months ago

five year ago Sunny was three times as old as Mini. 10 year later Sunny would be twice as old as Mini. how old are they now.

Answers

Answered by evilqueen17
0

Step-by-step explanation:

ANSWER

A's age = a; B's age = b

∴2b=a...(1);a−5=3(b−5)

⇒a=3b−10...(2)

⇒2b=3b−10

⇒b=10 years

and a=2b=2×10=20 years

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Answered by ri4
0

Given:

Five years ago, sunny was three times as old as Mini.

10 years later,Sunny would be twice as old as Mini.

Find:

The present ages

Solution:

Let the present age of Sunny be 'x' years and that of Mini be 'y' years

Five years ago, sunny was three times as old as mini.

Sunny's age = (x – 5) years

Mini's age = (y – 5) years

=> (x - 5) = 3(y - 5)

=> x - 5 = 3y - 15

=> x = 3y - 15 + 5

=> x = 3y - 10 ........(i).

10 years later,Sunny would be twice as old as Mini.

Sunny's age = (x + 10) years

Mini's age = (y + 10) years

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

=> x + 10 = 2y + 20

=> x = 2y + 20 - 10

=> x = 2y + 10 ..........(ii).

Putting the value of 'x' from equation (ii) in equation (i).

=> (2y + 10) - 3y = -10

=> 2y + 10 - 3y = -10

=> 2y - 3y = -10 - 10

=> -y = - 20

=> y = 20

Putting the value of 'y' in equation (ii).

=> x = 2y + 10

=> x = 2 × 20 + 10

=> x = 40 + 10

=> x = 50

So,

Sunny's age = x = 50 years

Mini's age = y = 20 years

Hence, the Sunny's age = x = 50 years is present ages and Minu's age = y = 20 years is present ages.

I hope it will help you.

Regards.

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