
Find the market price of the refrigerator, of its selling price along with goods and services tax is given. $S.P=Rs16,275$ and rate of goods and services tax is $5\%$.
Answer
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Hint: We have a relation between the Selling Price$\left( S.P \right)$ , cost price$\left( C.P \right)$ and tax percentage as $Tax\%=\dfrac{S.P-C.P}{C.P}\times 100$. In the above relation we will substitute the given values $S.P=Rs16,275$ and tax percentage$=5\%$ and simplify the above equation to get the value of Cost Price$\left( C.P \right)$.
Complete step by step answer:
Given that,
Selling Price of the refrigerator is $S.P=Rs16,275$.
Goods and service tax percentage is $Tax\%=5\%$.
Let the Cost Price is $C.P=x$.
Now we have relation between Selling Price$\left( S.P \right)$ , cost price$\left( C.P \right)$ and tax percentage as $Tax\%=\dfrac{S.P-C.P}{C.P}\times 100$. Substituting the values of $S.P$ and $Tax\%$ in the above equation, then we will have
$\begin{align}
& Tax\%=\dfrac{S.P-C.P}{C.P}\times 100 \\
& \Rightarrow 5=\dfrac{16,275-x}{x}\times 100 \\
\end{align}$
Dividing the above equation with $100$ on both sides, then we will have
$\begin{align}
& \dfrac{5}{100}=\dfrac{16,275-x}{x}\times \dfrac{100}{100} \\
& \Rightarrow \dfrac{1}{20}=\dfrac{16,275-x}{x} \\
\end{align}$
Multiplying $x$ on both sides, then we will have
$x\times \dfrac{1}{20}=\dfrac{16275-x}{x}\times x$
we know that $\dfrac{1}{x}\times x=1$, then we will get
$\Rightarrow \dfrac{x}{20}=16275-x$
Adding $x$ on both sides, then we will have
$\Rightarrow x+\dfrac{x}{20}=16275-x+x$
We know that $x-x=0$, then
$\Rightarrow x+\dfrac{x}{20}=16275$
Taking $x$ common on the left side of the equation, then we will have
$\begin{align}
& \Rightarrow x\left( 1+\dfrac{1}{20} \right)=16275 \\
& \Rightarrow x\left( 1+0.05 \right)=16275 \\
& \Rightarrow 1.05x=16275 \\
\end{align}$
Dividing the above equation with $1.05$ on both sides, then we will have
$\begin{align}
& x=\dfrac{16275}{1.05} \\
& \Rightarrow x=15,500 \\
\end{align}$
$\therefore $ Cost Price of the Refrigerator is Rs.$15,500$.
Note: In this problem we have given the values of Selling Price and the Tax percentage. In some cases, they give any two variables from the Selling Price, Cost Price and Tax percentage and ask to calculate the third one. In these cases, also we will use the above relation to solve the problem.
Complete step by step answer:
Given that,
Selling Price of the refrigerator is $S.P=Rs16,275$.
Goods and service tax percentage is $Tax\%=5\%$.
Let the Cost Price is $C.P=x$.
Now we have relation between Selling Price$\left( S.P \right)$ , cost price$\left( C.P \right)$ and tax percentage as $Tax\%=\dfrac{S.P-C.P}{C.P}\times 100$. Substituting the values of $S.P$ and $Tax\%$ in the above equation, then we will have
$\begin{align}
& Tax\%=\dfrac{S.P-C.P}{C.P}\times 100 \\
& \Rightarrow 5=\dfrac{16,275-x}{x}\times 100 \\
\end{align}$
Dividing the above equation with $100$ on both sides, then we will have
$\begin{align}
& \dfrac{5}{100}=\dfrac{16,275-x}{x}\times \dfrac{100}{100} \\
& \Rightarrow \dfrac{1}{20}=\dfrac{16,275-x}{x} \\
\end{align}$
Multiplying $x$ on both sides, then we will have
$x\times \dfrac{1}{20}=\dfrac{16275-x}{x}\times x$
we know that $\dfrac{1}{x}\times x=1$, then we will get
$\Rightarrow \dfrac{x}{20}=16275-x$
Adding $x$ on both sides, then we will have
$\Rightarrow x+\dfrac{x}{20}=16275-x+x$
We know that $x-x=0$, then
$\Rightarrow x+\dfrac{x}{20}=16275$
Taking $x$ common on the left side of the equation, then we will have
$\begin{align}
& \Rightarrow x\left( 1+\dfrac{1}{20} \right)=16275 \\
& \Rightarrow x\left( 1+0.05 \right)=16275 \\
& \Rightarrow 1.05x=16275 \\
\end{align}$
Dividing the above equation with $1.05$ on both sides, then we will have
$\begin{align}
& x=\dfrac{16275}{1.05} \\
& \Rightarrow x=15,500 \\
\end{align}$
$\therefore $ Cost Price of the Refrigerator is Rs.$15,500$.
Note: In this problem we have given the values of Selling Price and the Tax percentage. In some cases, they give any two variables from the Selling Price, Cost Price and Tax percentage and ask to calculate the third one. In these cases, also we will use the above relation to solve the problem.
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