Question

The Sum of the Present age of 5 brothers is 120 years. How many years ago was the sum of their ages was 80 years?

Answer 1

${\huge{\underbrace{\overbrace{\color{aqua}{Question}}}}}$

The sum of present age of 5 brothers is 120 years. How many years ago was the sum of their ages was 80 years?

$\huge{\mathcal\fcolorbox{black} {cyan} {Answer }}$

➪ Sum of present ages = a + b + c + d + e = 120

➪x years ago sum of ages = 80

⇒(a-x) + (b-x) + (c-x) + (d-x) + (e-x) = 80

⇒a + b + c + d + e - 5x = 80

⇒120 - 5x = 80 ( a + b + c + d + e = 120)

⇒120 - 80 = 5x

⇒40 = 5x

$⇒x \: = \cancel\frac{40}{5}$

⇒x = 8

Answer 2

Answer:

Sum of present ages = a + b + c + d + e = 120

➪x years ago sum of ages = 80

⇒(a-x) + (b-x) + (c-x) + (d-x) + (e-x) = 80

⇒a + b + c + d + e - 5x = 80

⇒120 - 5x = 80 ( a + b + c + d + e = 120)

⇒120 - 80 = 5x

⇒40 = 5x

⇒x \: = \cancel\frac{40}{5} ⇒x=

5

40

⇒x = 8