Question

A bought a radio set and spent Rs.110 on its repairs. He then sold it to B at 20% profit, B sold it to C at a loss of 10% and C sold it for Rs.1188 at a profit of 10%. What is the amount for which A bought the radio set?
(a) Rs.850
(b) Rs.890
(c) Rs.1000
(d) Rs.950

Answer 1

Hint: In this question, let us assume the cost price of A as x. Now, the cost price of B will be the selling price of A which is at a 20% profit on x. Then again the selling price of B will be the cost price of C which is at 10% loss on cost price of B. Then again the 10% profit on this cost price of C is given as Rs.1188 which on further simplification gives the value of x. Now, on subtracting 110 from x gives the amount spent by A.

Complete step-by-step answer:
If profit is x% then
\[SP=\dfrac{100+x}{100}\times CP\]
 If loss is y% then
\[SP=\dfrac{100-x}{100}\times CP\]

Now, from the given condition in the question
Let us assume the cost price by A as Rs. x
Now, the selling price of A will be the cost price of B which is given by
As given A sells at a profit of 20% as we already know that formula for profit is given by
\[SP=\dfrac{100+x}{100}\times CP\]
Now, on substituting the respective values in the above formula we get,
\[\Rightarrow SP\text{ of A}=\dfrac{100+20}{100}\times x\]
Now, this can also be written as
\[\Rightarrow SP\text{ of A}=\dfrac{120}{100}\times x\]
Now, here again given that B sells at a loss of 10%
As we already know that formula for loss is given by
\[SP=\dfrac{100-x}{100}\times CP\]
Now, on substituting the respective values in the above formula we get,
\[\Rightarrow SP\text{ of B}=\dfrac{100-10}{100}\times \dfrac{120}{100}\times x\]
Now, on further simplification we get,
\[\Rightarrow SP\text{ of B}=\dfrac{90}{100}\times \dfrac{120}{100}\times x\]
Here, the selling price of B will be the cost price of C.
Now, given C sold it at a profit of 10% which is Rs.1188
SP of C =Rs.1188
Now, from the profit formula we get,
\[\Rightarrow SP\text{ of C}=\dfrac{100+10}{100}\times \dfrac{90}{100}\times \dfrac{120}{100}\times x\]
Now, on substituting the respective value we get,
\[\Rightarrow 1188=\dfrac{110}{100}\times \dfrac{90}{100}\times \dfrac{120}{100}\times x\]
Now, on cancelling out the common terms in the above equation we get,
\[\Rightarrow 108=\dfrac{1}{10}\times \dfrac{9}{5}\times \dfrac{3}{5}\times x\]
Now, on rearranging the terms we get,
\[\Rightarrow \dfrac{108\times 250}{27}=x\]
Now, on further simplification we get,
\[\therefore x=Rs.1000\]
Now, on subtracting the charges he spent on repair gives the price at which A bought.
\[\begin{align}
  & \Rightarrow Rs.1000-Rs.110 \\
 & \Rightarrow Rs.890 \\
\end{align}\]
Hence, the correct option is (b).

Note:Instead of finding the value of x we can also solve it by first finding the cost price of C which gives the cost price of C then again on further substitution in the respective formula accordingly gives the cost price of A.It is important to note that while finding the selling price we need to substitute the respective cost price which is the selling price of the before case. Because neglecting any of the cases or interchanging changes the result.