Answer
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Hint: We will assign variables to the initial sum of money that Kate and Nora have. We will form an equation according to the ratio given. Next, we will modify the amount of money of both Kate and Nora to get the second equation. We also have the ratio of the modified sum of money both of them have. Using this ratio, we will be able to find the initial sum of money possessed by both Kate and Nora.
Complete step by step answer:
Let the initial sum of money that Kate has be $x$ Rs. and let the initial sum of money Nora has be $y$ Rs. The ratio of the amount of money Kate has to that of Nora is $3:5$. Therefore, we have
\[\dfrac{x}{y}=\dfrac{3}{5}\] .
So, $x=\dfrac{3y}{5}$. Now, Nora gives Rs. 150 to Kate. So the modified amount of money that Nora has is $y-150$ Rs. and the modified amount of money Kate has is $x+150$ Rs. After Nora gives Rs. 150 to Kate, the ratio of the amount of money Kate has to that of Nora becomes $7:9$. Therefore, the new ratio gives us the following equation,
\[\dfrac{x+150}{y-150}=\dfrac{7}{9}\].
Simplifying this equation, we get
\[9(x+150)=7(y-150)\]
\[\therefore 9x+1350=7y-1050\]
Now we will rearrange the terms to get a linear equation in two variables as follows,
\[9x-7y=-2400\]
Substituting $x=\dfrac{3y}{5}$ in the above equation, we get
\[9\left( \dfrac{3y}{5} \right)-7y=-2400\]
We will simplify the above equation as follows,
\[\begin{align}
& \dfrac{27y}{5}-7y=-2400 \\
& \dfrac{27y-35y}{5}=-2400 \\
& -8y=-2400\times 5 \\
\end{align}\]
Solving for $y$, we get
\[\begin{align}
& y=\dfrac{-2400\times 5}{-8} \\
& y=300\times 5 \\
& y=1500 \\
\end{align}\]
We know that $x=\dfrac{3y}{5}$. Substituting $y=1500$ in this equation, we will get the value of $x$ as follows,
\[\begin{align}
& x=\dfrac{3\times 1500}{5} \\
& x=3\times 300 \\
& x=900 \\
\end{align}\]
Hence, the sum of money that Kate had initially was Rs. 900.
Note: The equations that we formed by using the ratios are the key step in solving this question. We should pay attention to the signs of the terms that are being shifted from one side of the equation to the other. Making the correct substitutions in correct equations is crucial.
Complete step by step answer:
Let the initial sum of money that Kate has be $x$ Rs. and let the initial sum of money Nora has be $y$ Rs. The ratio of the amount of money Kate has to that of Nora is $3:5$. Therefore, we have
\[\dfrac{x}{y}=\dfrac{3}{5}\] .
So, $x=\dfrac{3y}{5}$. Now, Nora gives Rs. 150 to Kate. So the modified amount of money that Nora has is $y-150$ Rs. and the modified amount of money Kate has is $x+150$ Rs. After Nora gives Rs. 150 to Kate, the ratio of the amount of money Kate has to that of Nora becomes $7:9$. Therefore, the new ratio gives us the following equation,
\[\dfrac{x+150}{y-150}=\dfrac{7}{9}\].
Simplifying this equation, we get
\[9(x+150)=7(y-150)\]
\[\therefore 9x+1350=7y-1050\]
Now we will rearrange the terms to get a linear equation in two variables as follows,
\[9x-7y=-2400\]
Substituting $x=\dfrac{3y}{5}$ in the above equation, we get
\[9\left( \dfrac{3y}{5} \right)-7y=-2400\]
We will simplify the above equation as follows,
\[\begin{align}
& \dfrac{27y}{5}-7y=-2400 \\
& \dfrac{27y-35y}{5}=-2400 \\
& -8y=-2400\times 5 \\
\end{align}\]
Solving for $y$, we get
\[\begin{align}
& y=\dfrac{-2400\times 5}{-8} \\
& y=300\times 5 \\
& y=1500 \\
\end{align}\]
We know that $x=\dfrac{3y}{5}$. Substituting $y=1500$ in this equation, we will get the value of $x$ as follows,
\[\begin{align}
& x=\dfrac{3\times 1500}{5} \\
& x=3\times 300 \\
& x=900 \\
\end{align}\]
Hence, the sum of money that Kate had initially was Rs. 900.
Note: The equations that we formed by using the ratios are the key step in solving this question. We should pay attention to the signs of the terms that are being shifted from one side of the equation to the other. Making the correct substitutions in correct equations is crucial.
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