Question

An iron box is sold at a gain of $ 16%. $ If it were sold for Rs. $ 20 $ more, there would have been a gain of $ 20%. $ Find the cost price of an iron box.
A. $ Rs.\text{ 420} $
B. $ Rs.\text{ 580} $
C. $ Rs.\text{ 480} $
D. $ Rs.\text{ 500} $

Answer 1

Hint: Write down the above information in mathematical notation. For this assume variables like x and y as selling price and cost price. Make two linear equation as per question using the relation
 $ gain\%=\left( \dfrac{selling\text{ price - cost price}}{\text{cost price}} \right)\left( 100 \right) $ .
 As we get the linear equation in two variables, we have to solve it in order to get the result.

Complete step-by-step answer:
The following information is from question
Gain % $ =16\% $
Also, if it were sold at Rs. 20 more there would be gain of $ 20% $
Let us assume that selling price of iron box $ =x $
As we know that
 $ gain\%=\left( \dfrac{selling\text{ price - cost price}}{\text{cost price}} \right)\left( 100 \right) $
As in first condition gain % = 16
So, we can use the above formula as
 $ 16=\left( \dfrac{x-\text{cost price}}{\text{cost price}} \right)\left( 100 \right) $
Let cost price is $ y $ , hence we can write the above as
 $ 16=\left( \dfrac{x-y}{y} \right)\left( 100 \right) $
On cross multiplication we can further write it as
 $ \begin{align}
  & 16y=100x-100y \\
 & \Rightarrow 116y=100x \\
 & \Rightarrow y=\dfrac{100x}{116}-------(a) \\
\end{align} $
Hence, we find here the relation between cost price and selling price
Now if selling price is 20 more then gain % is 20%
So, we can write
Selling price $ =x+20 $
Gain %= $ 20% $
So, substituting the value we can write it as
 $ 20=\left( \dfrac{\left( x+20 \right)-\text{cost price}}{\text{cost price}} \right)\left( 100 \right) $
As we already assume that cost price = y
Hence,
On cross multiplication we can write
 $ \begin{align}
  & 20(y)=\left( x+20-y \right)\left( 100 \right) \\
 & \Rightarrow 20y=100x+2000-100y \\
\end{align} $
On transposing $ -100y $ and $ 100x $ to left hand side we can write
 $ 120y-100x=2000------(b) $
Now we have two linear equation in one variable so in order to solve equation (a) and equation (b)
We can substitute the value of $ y=\dfrac{100x}{116} $ from equation (a) into equation (b) we can write
 $ \begin{align}
  & \left( 120 \right)\left( \dfrac{100x}{116} \right)-100x=2000 \\
 & \Rightarrow \left( 120 \right)\left( 100x \right)-\left( 100x \right)\left( 116 \right)=\left( 116 \right)\left( 2000 \right) \\
\end{align} $
We can take $ 100x $ common so we can write
 $ \begin{align}
  & \left( 100x \right)\left( 120-116 \right)=\left( 116 \right)\left( 2000 \right) \\
 & \Rightarrow \left( 100x \right)(4)=\left( 116 \right)\left( 2000 \right) \\
\end{align} $
On dividing both side by 400 hundred we can write
\[\begin{align}
  & \Rightarrow \dfrac{\left( 100x \right)(4)}{400}=\dfrac{\left( 116 \right)\left( 2000 \right)}{400} \\
 & \Rightarrow x=\left( 116 \right)(5) \\
 & \Rightarrow x=580 \\
\end{align}\]
Here we find the value of $ x $ , which we assumed as the selling price.
Hence the selling price of iron box is Rs. 580.
Now we substitute the value of $ x $ in equation (a) in order to find the value of $ y $ so we get the value of cost price
Hence, we can write
 $ \begin{align}
  & y=\left( \dfrac{100}{116} \right)\left( 580 \right) \\
 & \Rightarrow y=\left( 100 \right)(5) \\
 & \Rightarrow y=500 \\
\end{align} $
Hence, we get the value of $ y $ so cost price of iron box is Rs. 500
Hence, option D is correct.

Note: As we get linear equation in two variables in the form $ {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\text{ }and\text{ }{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0 $ . The equation represents two straight lines in the XY plane.
 it gives unique solution only if
 $ \dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}} $ , so that they cut each other at a point. Hence check the condition for the solution of the system of equations.
If
 $ \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}} $
Then lines are parallel, they do not cut each other so there is no unique solution.
We can solve the above equation by elimination method also.