Question

divide 240 into three parts so that one upon three of the first one upon 4 of the second and one upon 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 = ?

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

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

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

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

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

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\:\:$

$\red{\underline \bold{To \: Find:}}$

$\:\:$

• All the three parts

$\:\:$

$\green{\underline \bold{Given :}}$

$\:\:$

• Sum of the three part = 240

$\:\:$

• $\sf \dfrac { 1 } { 3 }$ (part 1) = $\sf \dfrac { 1 } { 4 } (part 2)$ = $\sf \dfrac { 1 } { 5 } (part 3)$

$\:\:$

$\large{\orange{\underline{\tt{Solution :-}}}}$

$\:\:$

• Let the 1st part be = x

• Let the 2nd part be = y

• Let the 3rd part be = z

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

$\purple\longrightarrow$ $\bf \dfrac { 1 } { 3 } (x) = \dfrac { 1 } { 4 } (y) = \dfrac { 1 } { 5 } (z)$

$\:\:$

$\underline{\bold{\texttt{We know that :}}}$

$\:\:$

$\dag \bf \ x + y + z = 240$ ------(1)

$\:\:$

Here we will try to write the x and y part in term of z

$\:\:$

$\sf \longmapsto \dfrac { 1 } { 3 } (x) = \dfrac { 1 } { 5 } (z)$ [ Given ]

$\:\:$

$\rm\longmapsto x = \dfrac { 3 } { 5 } (z) ------(2)$

$\:\:$

$\sf\longmapsto \dfrac { 1 } { 4 } (y) = \dfrac { 1 } { 5 } (z)$ [ Given ]

$\:\:$

$\rm \longmapsto y = \dfrac { 4 } { 5 } (z) ------(3)$

$\:\:$

$\underline{\bold{\texttt{Putting equation (2) \& (3) in (1)}}}$

$\:\:$

$\sf \dashrightarrow x + y + z = 240$

$\:\:$

$\bf \implies \dfrac { 3 } { 5 } (z) + \dfrac { 4 } { 5 } (z) + z = 240$

$\:\:$

$\underline{\bold{\texttt{Taking LCM as 5}}}$

$\:\:$

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

$\:\:$

$\sf \longmapsto \dfrac { 12z } { 5 } = 240$

$\:\:$

$\sf \longmapsto \dfrac { z } { 5 } = \dfrac { 240 } { 12 }$

$\:\:$

$\sf \longmapsto z = 20 \times 5$

$\:\:$

$\red{\bold{z \ = \ 100}}$

$\:\:$

$\underline{\bold{\texttt{Putting z = 100 in (2)}}}$

$\:\:$

$\sf \longmapsto x = \dfrac { 3 } { 5 } (z)$

$\:\:$

$\sf \longmapsto x = \dfrac { 3 } { 5 } (100)$

$\:\:$

$\sf \longmapsto x = 3 \times 20$

$\:\:$

$\red{\bold{x \ = \ 60}}$

$\:\:$

$\underline{\bold{\texttt{Putting z = 100 in (3)}}}$

$\:\:$

$\sf \longmapsto y = \dfrac { 4 } { 5 } (z)$

$\:\:$

$\sf \longmapsto y = \dfrac { 4 } { 5 } (100)$

$\:\:$

$\sf \longmapsto y = 4 \times 20$

$\:\:$

$\red{\bold{y \ = \ 80}}$

$\:\:$

So,

$\:\:$

• x = 60

$\:\:$

• y = 80

$\:\:$

• z = 100

$\:\:$

$\rule{200}5$