Question

solve plzzzz kal exam hai ??? ​

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 = ?

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

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

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

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

$\bigstar \boxed{\sf{a+b+c = 240}}$....1)

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

According to First Condition :-

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

$\mapsto \boxed{\sf{a = \dfrac{3}{4}b}}$...2)

According to Second Condition :-

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

$\mapsto \boxed{\sf{b = \dfrac{4}{5}c}}$...3)

$\bigstar \boxed{\underline{\pink{\sf{Now,Put\:the\:Value\:of\:2\:and\:3\:Equation\: in\:1\:Equation}}}}$

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

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

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

Now, Put the Value of b :-

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

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

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

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

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

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

Therefore,

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

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

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

Now, Find Value of a :-

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

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

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

$\rule{200}2$

Answer 2

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

According to question ,

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