Question

If a person reduces his selling price by an article by Rs. 40, then the suggested profit of $33\dfrac{1}{3}\% $ converts into loss of $20\% $. Find the cost price?

Answer 1

Hint:
We can take the cost price as x and sell price at profit as y, then we can find the selling price at loss using the relation given in the question. Then we can form 2 equations in 2 variables, then we can solve the equations to get the value of the variables. Then the value of x will give the required cost price.

Complete step by step solution:
Let x be the cost price of the article.
Let the selling price when there is profit of $33\dfrac{1}{3}\% $ be y.
Then the selling price when there is loss of $20\% $ is $y - 40$ .
Consider the 1st case where there is profit of $33\dfrac{1}{3}\% $ .
As the profit is $33\dfrac{1}{3}\% $ , then the profit is given by,
 $P = 33\dfrac{1}{3}\% x$
We can convert mixed fraction to improper fraction.
 $ \Rightarrow P = \dfrac{{33 \times 3 + 1}}{3}\% x$
On simplification we get,
 $ \Rightarrow P = \dfrac{{100}}{3}\% x$
On converting percentage to fraction, we get,
 $ \Rightarrow P = \dfrac{{100}}{{3 \times 100}}x$
On simplification, we get,
 $ \Rightarrow P = \dfrac{1}{3}x$
We know that the selling price is given by,
 $S.P = C.P. + P$
On substituting the values, we get,
 $ \Rightarrow y = x + \dfrac{1}{3}x$
On taking the LCM, we get,
 $ \Rightarrow y = \dfrac{{3x}}{3} + \dfrac{x}{3}$
On simplification, we get,
 $ \Rightarrow y = \dfrac{4}{3}x$ .. (1)

Now we can consider the case where loss of $20\% $ .
As the loss is $20\% $ , the loss is given by,
 $L = 20\% x$
On converting the percentage to fractions, we get,
 $ \Rightarrow L = \dfrac{{20}}{{100}}x$
On simplification, we get,
 $ \Rightarrow L = \dfrac{2}{{10}}x$
We know that the selling price is given by,
 $S.{P_2} = C.P. - L$
On substituting the values, we get,
 $ \Rightarrow y - 40 = x - \dfrac{2}{{10}}x$
On taking the LCM, we get,
 $ \Rightarrow y - 40 = \dfrac{{10x - 2x}}{{10}}$
On simplification, we get,
 $ \Rightarrow y - 40 = \dfrac{{8x}}{{10}}$ … (2)
Now we can substitute (1) in (2),
 $ \Rightarrow \dfrac{4}{3}x - 40 = \dfrac{{8x}}{{10}}$
On rearranging, we get,
 $ \Rightarrow \dfrac{4}{3}x - \dfrac{8}{{10}}x = 40$
On dividing throughout with 4, we get,
 $ \Rightarrow \dfrac{1}{3}x - \dfrac{2}{{10}}x = 10$
Now we can find LCM of the LHS. So, we get,
 $ \Rightarrow \dfrac{{10x - 6x}}{{30}} = 10$
On simplification we get,
 $ \Rightarrow \dfrac{{4x}}{{30}} = 10$
On cross multiplication, we get,
 $ \Rightarrow 4x = 10 \times 30$
On dividing by 4 we get,
 $ \Rightarrow x = \dfrac{{300}}{4}$
On simplification we get,
 $ \Rightarrow x = 75$

So, the cost price of the article is Rs. 75.

Note:
We only need to solve for one variable as the cost price is only asked. We must take the same variable for the cost price in both the cases. Similarly, we must write the selling price in one variable using the relation given in the question. We must note that the selling price when there is a profit is given by the sum of the profit and cost price. The selling price in the case of loss is given by subtracting the loss from the cost price. We must not interchange the operations in both the cases.