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(i) If \[\left( \mathbf{x}\text{ }\text{ }\mathbf{9} \right):\text{ }\left( \mathbf{3x}\text{ }+\text{ }\mathbf{6} \right)\] is the duplicate ratio of \[\mathbf{4}:\text{ }\mathbf{9}\], find the value of x. (ii) If \[\left( \mathbf{3x}\text{ }+\text{ }\mathbf{1} \right):\text{ }\left( \mathbf{5x}\text{ }+\text{ }\mathbf{3} \right)\] is the triplicate ratio of \[\mathbf{3}:\text{ }\mathbf{4}\], find the value of x.
(i) If \[\left( \mathbf{x}\text{ }\text{ }\mathbf{9} \right):\text{ }\left( \mathbf{3x}\text{ }+\text{ }\mathbf{6} \right)\] is the duplicate ratio of \[\mathbf{4}:\text{ }\mathbf{9}\], find the value of x. (ii) If \[\left( \mathbf{3x}\text{ }+\text{ }\mathbf{1} \right):\text{ }\left( \mathbf{5x}\text{ }+\text{ }\mathbf{3} \right)\] is the triplicate ratio of \[\mathbf{3}:\text{ }\mathbf{4}\], find the value of x.
Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
(i) If \[\left( \mathbf{x}\text{ }\text{ }\mathbf{9} \right):\text{ }\left( \mathbf{3x}\text{ }+\text{ }\mathbf{6} \right)\] is the duplicate ratio of \[\mathbf{4}:\text{ }\mathbf{9}\], find the value of x. (ii) If \[\left( \mathbf{3x}\text{ }+\text{ }\mathbf{1} \right):\text{ }\left( \mathbf{5x}\text{ }+\text{ }\mathbf{3} \right)\] is the triplicate ratio of \[\mathbf{3}:\text{ }\mathbf{4}\], find the value of x.

(i) \[\left( x\text{ }\text{ }9 \right)/\text{ }\left( 3x\text{ }+\text{ }6 \right)\text{ }=\text{ }{{\left( 4/9 \right)}^{2}}\]

So we get

\[\left( x\text{ }\text{ }9 \right)/\text{ }\left( 3x\text{ }+\text{ }6 \right)\text{ }=\text{ }16/81\]

By cross multiplication

\[\begin{array}{*{35}{l}}

81x\text{ }\text{ }729\text{ }=\text{ }48x\text{ }+\text{ }96  \\

81x\text{ }\text{ }48x\text{ }=\text{ }96\text{ }+\text{ }729  \\

\end{array}\]

So we get

\[\begin{array}{*{35}{l}}

33x\text{ }=\text{ }825  \\

x\text{ }=\text{ }825/33\text{ }=\text{ }25  \\

\end{array}\]

(ii) \[\left( 3x\text{ }+\text{ }1 \right)/\text{ }\left( 5x\text{ }+\text{ }3 \right)\text{ }=\text{ }{{3}^{3}}/\text{ }{{4}^{3}}\]

So we get

\[\left( 3x\text{ }+\text{ }1 \right)/\text{ }\left( 5x\text{ }+\text{ }3 \right)\text{ }=\text{ }27/64\]

By cross multiplication

\[\begin{array}{*{35}{l}}

64\text{ }\left( 3x\text{ }+\text{ }1 \right)\text{ }=\text{ }27\text{ }\left( 5x\text{ }+\text{ }3 \right)  \\

192x\text{ }+\text{ }64\text{ }=\text{ }135x\text{ }+\text{ }81  \\

192x\text{ }\text{ }135x\text{ }=\text{ }81\text{ }\text{ }64  \\

57x\text{ }=\text{ }17  \\

\end{array}\]

So we get

\[x\text{ }=\text{ }17/57\]

i

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