Question

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

Answer 1

$\huge{\underline{\pink{\tt{Given}}}}$

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

$\huge{\underline{\pink{\tt{To\:Find}}}}$

• The First Part = ?
• The Second Part = ?
• The Third Part = ?

$\huge{\underline{\pink{\tt{Solution}}}}$

$\longrightarrow \sf{Suppose\:the\:first\:part\:be\: \boxed{\sf{a}}}$

$\sf{And,Second\:part\:be\: \boxed{\sf{b}}\:,Third\:part\:be\: \boxed{\sf{c}}}$

$\boxed{\underline{\red{\sf{Now,According\:to\:the\:Question :}}}}$

$\bigstar \boxed{\sf{a+b+c = 240}}$

$\bigstar \boxed{\sf{\dfrac{1}{3}a = \dfrac{1}{4}b = \dfrac{1}{5}c}}$

According to First Condition :-

$\mapsto \sf{\dfrac{1}{3}a = \dfrac{1}{4}b}$

$\mapsto \boxed{\sf{a = \dfrac{3}{4}b}}$

According to Second Condition :-

$\mapsto \sf{\dfrac{1}{4}b = \dfrac{1}{5}c}$

$\mapsto \boxed{\sf{b = \dfrac{4}{5}c}}$

Now , Put the Value of 2 and 3 Equation in 1 Equation

$\longmapsto \sf{\dfrac{3}{4}b + \dfrac{4}{5}c + c = 240}$

$\longmapsto \sf{\dfrac{3}{4}b + \dfrac{4c + 5c}{5} = 240}$

$\longmapsto \sf{\dfrac{3}{4}b + \dfrac{9c}{5} = 240}$

Now, Put the Value of b :-

$\longmapsto \sf{\dfrac{3}{4} \times \dfrac{4}{5} + \dfrac{9c}{5} = 240}$

$\longmapsto \sf{\dfrac{3}{5}c + \dfrac{9}{5}c = 240}$

$\longmapsto \sf{\dfrac{3c + 9c}{5} = 240}$

$\longmapsto \sf{12c = 1200}$

$\longmapsto \sf{c = \dfrac{1200}{12}}$

$\longmapsto \boxed{\sf\orange{c = 100}}$

Therefore,

$\implies \sf{b = \dfrac{4}{5} c}$

$\implies \sf{b = \dfrac{4}{5} \times 100}$

$\implies \boxed{\sf\blue{b = 80}}$

Now, Find Value of a :-

$\implies \sf{a = \dfrac{3}{4}b}$

$\implies \sf{a = \dfrac{3}{4} \times 80}$

$\implies \boxed{\sf\green{a = 60}}$