Question

two numbers are in ratio 8:3 and their sum is 132 find the number​

Answer 1

Answer:

The solution is as follows :

let the number be x

$8x + 3x = 132 \\ 11x = 132 \\ x = \frac{132}{11} \\ x = 12$

therefore,

8×12=96

3×12=36

Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• Two numbers are in ratio 8:3 and their sum is 132 find the number.

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

• Let the number be x
• Then, the first number be $\sf\green{ </strong><strong>8</strong><strong>x</strong><strong>}$
• The second number be $\sf\green{ </strong><strong>3</strong><strong>x</strong><strong>}$

$\begin{gathered}\begin{gathered}\underline{\boldsymbol{According\:\: to\: \:the\:\: Question :}} \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: 8x \: + \: 3x \: = 132 }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\sf{\: 8x \: + \: 3x \: = 132 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{\: 11x\: = 132 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{\: x \:= \cancel{ \frac{132}{11} } }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{x\: = \: 12 }}}$

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$\begin{gathered}\begin{gathered}\sf \to \: 8x \\ \sf \to \: 8 \times 12 \\ \sf \to \red{96 }\\ \\ \\ \sf \to \: 3x \\ \sf \to \: 3 \times 12 \\ \sf \to \orange{36} \\\end{gathered}\end{gathered}$

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$\large\underline{ \underline{ \sf \maltese{ \: Verification:- }}}$

$\quad {:} \longrightarrow \sf{\sf{\: 8 \times 12\: + \: 3 \times 12 \: = 132 }}$

$\quad {:} \longrightarrow \sf{\sf{\: 96\: + \: 36= 132 }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{132\: = \: 132 }}}$

━━━━━━━━━━━━━━━━━━━━━━━━━

$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ First \: number = \underline {\underline{ 96}}}\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ Second \: number = \underline {\underline{ 36}}}\\\end{gathered}\end{gathered}$

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