Question

Divide 240 into three parts so that 1/3 of the first 1/4 of the second and 1/5 of the trd part are equal

Answer 1

Answer is
First Part = 60
Second part = 80
third part = 100

Let the 3 numbers be a, b, and c
According to problem statement
(1/3)a = (1/4)b = (1/5)c
∴ b = (4/3)a
and c = (5/3)a

Also, a + b +c  = 240
∴ a + (4/3)a + (5/3)a = 240
(3a+4a+5a)/3 = 240
12a = 720
a = 60
replacing a as 60, we get b = 80 and c = 100

Answer 2

$let \: the \: first \: part \: be \: x \\ second \: be \: y \\ and \: third \: be \: z. \\ atq \\ \frac{1}{3} x = \frac{1}{4} y \\ y = \frac{4}{3} x \\ also \: \frac{1}{3} x = \frac{1}{5} z \\ z = \frac{5}{3} x \\ now \: \\ x + y + z = 240 \\ x + \frac{4}{3} x + \frac{5}{3} x = 240 \\ \frac{3x + 4x + 5x}{3} = 240 \\ \frac{12x}{3} = 240 \\ 4x = 240 \\ x = 60 \\ so \: the \: first \: part \: is \: 60 \\ second \: part \: is \: 80 \\ and \: the \: third \: part \: is \: 100.$
hope this helps you.