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Three numbers are in the ratio

    \[\mathbf{1}/\mathbf{2}:\text{ }\mathbf{1}/\mathbf{3}:\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\]

. If the sum of their squares is

    \[244\]

, find the numbers.
Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
Three numbers are in the ratio

    \[\mathbf{1}/\mathbf{2}:\text{ }\mathbf{1}/\mathbf{3}:\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\]

. If the sum of their squares is

    \[244\]

, find the numbers.

It is given that

Ratio of three numbers

    \[=\text{ }1/2:\text{ }1/3:\text{ }1/4\]

    \[\begin{array}{*{35}{l}} =\text{ }\left( 6:\text{ }4:\text{ }3 \right)/\text{ }12 \\ =\text{ }6:\text{ }4:\text{ }3 \\ \end{array}\]

Consider first number =

    \[6x\]

Second number =

    \[4x\]

Third number =

    \[3x\]

So based on the condition

    \[\begin{array}{*{35}{l}} {{\left( 6x \right)}^{2}}~+\text{ }{{\left( 4x \right)}^{2}}~+\text{ }{{\left( 3x \right)}^{2}}~=\text{ }244 \\ 36{{x}^{2}}~+\text{ }16{{x}^{2}}~+\text{ }9{{x}^{2}}~=\text{ }244 \\ \end{array}\]

So we get

    \[\begin{array}{*{35}{l}} {{\left( 6x \right)}^{2}}~+\text{ }{{\left( 4x \right)}^{2}}~+\text{ }{{\left( 3x \right)}^{2}}~=\text{ }244 \\ 36{{x}^{2}}~+\text{ }16{{x}^{2}}~+\text{ }9{{x}^{2}}~=\text{ }244 \\ \end{array}\]

Here

First number

    \[=\text{ }6x\text{ }=\text{ }6\text{ }\times \text{ }2\text{ }=\text{ }12\]

Second number

    \[=\text{ }4x\text{ }=\text{ }4\text{ }\times \text{ }2\text{ }=\text{ }8\]

Third number

    \[=\text{ }3x\text{ }=\text{ }3\text{ }\times \text{ }2\text{ }=\text{ }6\]

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