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
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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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.