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Hint: At first divide the numerator and denominator by y, then substitute $\dfrac{x}{y}=t$, hence, find the value of t. Then again put the value of t in the latter ratio to find the answer.
Complete step-by-step answer:
In the question, we are given a ratio of 2 quantities which are in terms of x, y which is $\left( 3x+4y \right):\left( x+2y \right)=9:4$. Now, we have to find the ratio of $\left( 3x+5y \right):\left( 3x-y \right)$
Let’s consider the given ratio and write it as,
$\dfrac{3x+4y}{x+2y}=\dfrac{9}{4}$
Now, we will divide by ‘y’ from both numerator and denominator we get,
$\dfrac{3\left( \dfrac{x}{y} \right)+4}{\left( \dfrac{x}{y} \right)+2}=\dfrac{9}{4}$
Let’s take $\left( \dfrac{x}{y} \right)$ as t, so we write equation as
$\dfrac{3t+4}{t+2}=\dfrac{9}{4}$
Now, we will do cross multiplication. So, we get,
$4\left( 3t+4 \right)=9\left( t+2 \right)$
On further simplification, we get
$12t+16=9t+18$
Now, taking all the ‘t’ term to left hand side and constant to right hand side, we get
$3t=2$
Hence, $t=\dfrac{2}{3}$
We assumed $\left( \dfrac{x}{y} \right)=t$
So, $\dfrac{x}{y}=\dfrac{2}{3}$
We are asked to find the ratio of $\left( 3x+5y \right)$ and $\left( 3x-y \right)$.
So, we have to find $\left( \dfrac{3x+5y}{3x-y} \right)$ in other words.
We can write $\left( \dfrac{3x+5y}{3x-y} \right)$ as $\left( \dfrac{3\left( \dfrac{x}{y} \right)+5}{3\left( \dfrac{x}{y} \right)-1} \right)$
We know that $\dfrac{x}{y}=\dfrac{2}{3}$ so, we will
Substitute it in $\left( \dfrac{3\left( \dfrac{x}{y} \right)+5}{3\left( \dfrac{x}{y} \right)-1} \right)$
So, we get
$\dfrac{3\left( \dfrac{2}{3} \right)+5}{3\left( \dfrac{2}{3} \right)-1}=\dfrac{2+5}{2-1}=\dfrac{7}{1}$
Hence, on calculation we get,
$\dfrac{3x+5y}{3x-y}=\dfrac{7}{1}$
The correct option is ‘c’.
Note: Students can also take $x=yt$ and substitute it in the former ratio to find the value of t. Then again substitute $x=yt$ and hence put the value of ‘t’ to get the final ratio.
Complete step-by-step answer:
In the question, we are given a ratio of 2 quantities which are in terms of x, y which is $\left( 3x+4y \right):\left( x+2y \right)=9:4$. Now, we have to find the ratio of $\left( 3x+5y \right):\left( 3x-y \right)$
Let’s consider the given ratio and write it as,
$\dfrac{3x+4y}{x+2y}=\dfrac{9}{4}$
Now, we will divide by ‘y’ from both numerator and denominator we get,
$\dfrac{3\left( \dfrac{x}{y} \right)+4}{\left( \dfrac{x}{y} \right)+2}=\dfrac{9}{4}$
Let’s take $\left( \dfrac{x}{y} \right)$ as t, so we write equation as
$\dfrac{3t+4}{t+2}=\dfrac{9}{4}$
Now, we will do cross multiplication. So, we get,
$4\left( 3t+4 \right)=9\left( t+2 \right)$
On further simplification, we get
$12t+16=9t+18$
Now, taking all the ‘t’ term to left hand side and constant to right hand side, we get
$3t=2$
Hence, $t=\dfrac{2}{3}$
We assumed $\left( \dfrac{x}{y} \right)=t$
So, $\dfrac{x}{y}=\dfrac{2}{3}$
We are asked to find the ratio of $\left( 3x+5y \right)$ and $\left( 3x-y \right)$.
So, we have to find $\left( \dfrac{3x+5y}{3x-y} \right)$ in other words.
We can write $\left( \dfrac{3x+5y}{3x-y} \right)$ as $\left( \dfrac{3\left( \dfrac{x}{y} \right)+5}{3\left( \dfrac{x}{y} \right)-1} \right)$
We know that $\dfrac{x}{y}=\dfrac{2}{3}$ so, we will
Substitute it in $\left( \dfrac{3\left( \dfrac{x}{y} \right)+5}{3\left( \dfrac{x}{y} \right)-1} \right)$
So, we get
$\dfrac{3\left( \dfrac{2}{3} \right)+5}{3\left( \dfrac{2}{3} \right)-1}=\dfrac{2+5}{2-1}=\dfrac{7}{1}$
Hence, on calculation we get,
$\dfrac{3x+5y}{3x-y}=\dfrac{7}{1}$
The correct option is ‘c’.
Note: Students can also take $x=yt$ and substitute it in the former ratio to find the value of t. Then again substitute $x=yt$ and hence put the value of ‘t’ to get the final ratio.
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