Question

divide 240 into three parts so that 1/3 of the first, 1/4 of the second and 1/5of the third part are equal solve the problem

Answer 1

Let the parts be a, b and c

a/3 = b/4 = c/5

=> a = 3b/4 -----(1)

=> c = 5b/4 -----(2)


Now,

a + b + c = 240

=> 3b/4 + b + 5b/4 = 240

=> ( 3b + 4b + 5b) /4 = 240

=> 12b / 4 = 240

=> 3b = 240

=> b = 80

a = 80 × 3 /4 = 60

c = 80 × 5 /4 = 100

First part = 60

Second part = 80

Third part = 100

Answer 2

Let the following three parts be x₁ , x₂ , and x₃

According to question ,

$\frac{x_1}{3} = \frac{x_2}{4} = \frac{x_3}{5} = k(say) \\ \\ x_1 = 3k \\ \\ x_2 = 4k \\ \\ x_3 = 5k \\ but \\ \\ x_1 + x_2 + x_3 = 240 \\ \\ 3k + 4k + 5k = 240 \\ \\ 12k = 240 \\ \\ k = 20 \\ \\ x_1 = 3*20 = 60 \\ \\ x_2 = 4*20 = 80 \\ \\ x_3 = 5*20 = 100$