Question

A man buys a plot of land at $Rs.3,60,000$. He sells one third of the plot at a loss of $20\%$. Again, he sells one-third of the plot left at a profit of $25\%$. At what price should he sell the remaining plot in order to get a profit of $10\%$ on the whole?

Answer 1

Given:

Total cost of the Plot $=Rs.3,60,000$

$\Rightarrow$ Cost of $\frac{1}{3}\text{rd}$ of the plot $=\frac{1}{3}\times Rs3,60,000$

$=Rs1,20,000$

It is given that, $\frac{1}{3}\text{rd}$ sold out for a loss of $20\%$.

$\Rightarrow$ Selling price of $\frac{1}{3}\text{rd}$ of plot

$=Rs.(120000-(20\%\text{of}120000)$

$=Rs.(120000-(120000\times \frac{20}{100}\left)\right)$

$=Rs(120000-24000)$

$=Rs.96000$

$\Rightarrow$ Selling price of $\frac{1}{3}\text{rd}$ of the plot $=Rs.96000$....(1)

Since, $\frac{1}{3}\text{rd}$ of the plot is sold.

Remaining plot $=1-\frac{1}{3}=\frac{2}{3}$ of the plot

$\Rightarrow \frac{2}{3}$ of left plot area $=\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}$ part of initial plot area

$\Rightarrow$ Cost value of $\frac{4}{9}$ plot area $=Rs.360000\times \frac{4}{9}=Rs.1,60,000$

If $\frac{4}{9}$ part of plot is sold for $25\%$ profit, then

Selling price

$=Rs.(160000+25\%\text{of}160000)$

$=Rs.(160000+(160000\times \frac{25}{100}\left)\right)$

$=Rs.(160000+40000)$

$=Rs.200000$

$\Rightarrow$ Selling price of $\frac{4}{9}$ of the plot $=Rs.200000$.....(2)

Total selling price till now $=Rs.96000+Rs.200000=Rs.2,96,000$ ---(3)

Overall selling price for the profit of $10\%$

$=Rs.(360000+(360000\times \frac{10}{100}\left)\right)$ [Since, the CP of the plot is $Rs.360000$].

$=Rs.3,96,000$

i.e., We need to get the amount on whole transaction is $Rs.3,96,000$ for getting $10\%$ profit.

Till now we got the amount is $Rs.2,96,000$. [From (3)]

So, the selling price of remaining plot to get the $10\%$ profit is,

$Rs.3,96,000-Rs.2,96,000$

$=Rs.1,00,000$

Hence, the price of the remaining plot is $Rs.1,00,000$.