# A man bought two old scooters for Rs. $18000$ . By selling one at a profit of $25\% $ and the other at a loss of $20\% $ , he neither gains nor loss . Find the cost price of each scooter

(a) $12000,6000$

(b) $8000,10000$

(c) $11000,7000$

(d) $13000,5000$

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**Hint:**In this type of question First suppose the cost price of one scooter is $x$ then the other ones price is $18000 - x$.For now consider these two values as Cost price and we know that the Selling price = Cost price + Profit amount and if it have loss then Selling price = Cost price - loss amount from these concept we will solve this question.

**Complete step-by-step answer:**

In question it is given that the man bought two scooter at a price of $18000$

So Let us suppose that the cost price of one scooter is = $x$

Than the Cost price of another is scooter is = $18000 - x$

Now he sells the one scooter for a profit of $25\% $ and we know that the Selling price = Cost price + Profit Amount .

Hence the profit amount is the $25\% $ of $x$ is $ = \dfrac{{25}}{{100}} \times x$ that is equal to $\dfrac{1}{4}x$ ;

Now Selling price = Cost price + Profit Amount .

Selling price =$x + \dfrac{1}{4}x$ = $\dfrac{5}{4}x$

Hence Selling price of scooter one is = $\dfrac{5}{4}x$

Now he sells the one scooter at a loss of $20\% $ and we know that the Selling Price = Cost price - Loss Amount .

Hence the loss amount is the $20\% $ of $18000 - x$ is $ = \dfrac{{20}}{{100}}(18000 - x)$ that is equal to $\dfrac{1}{5}(18000 - x)$ ;

Now Selling price = Cost price - Loss Amount

Selling Price $ = (18000 - x) - \dfrac{1}{5}(18000 - x)$

= $18000 - x - \dfrac{{18000}}{5} + \dfrac{x}{5}$

Multiple and divide by $5$ in the term $18000 - x$

$ = \dfrac{{5 \times 18000}}{5} - \dfrac{{5x}}{5} - \dfrac{{18000}}{5} + \dfrac{x}{5}$

Do not try to multiple $5 \times 18000$ otherwise calculation become tough to solve ;

Now after further solving we get

$ = \dfrac{{4 \times 18000}}{5} - \dfrac{{4x}}{5}$

now taking common $\dfrac{4}{5}$ we get

$ = \dfrac{4}{5}(18000 - x)$

Hence the selling price of scooter two is $ = \dfrac{4}{5}(18000 - x)$

Since man gets neither profit nor loss it is given in the question therefore

Selling price of scooter first + Selling price of scooter second is equal to the Cost price of scooter

$\dfrac{{5x}}{4} + \dfrac{4}{5}\left( {18000 - x} \right) = 18000$

Now opening the brackets

$\dfrac{{5x}}{4} + \dfrac{4}{5} \times 18000 - \dfrac{4}{5}x = 18000$

$\dfrac{{5x}}{4} + 14400 - \dfrac{4}{5}x = 18000$

$\dfrac{{5x}}{4} - \dfrac{4}{5}x = 18000 - 14400$

Now solving further and taking L.C.M we get ;

$\dfrac{{25x - 20x}}{{20}} = 3600$

$9x = 20 \times 3600$

$x = \dfrac{{72000}}{9}$ $ = 8000$

Hence the Cost Price of one scooter is Rs $8000$ and the other one is equal to $18000 - x$ that is Rs $10000$ .

**So, the correct answer is “Option B”.**

**Note:**Students should remember the formulas of profit and loss i.e If $S.P>C.P$ then it is Profit where $Profit=S.P-C.P$ and if ${SP}<{CP}$ then it is loss where $Loss=C.P-S.P$.In this problem they given that he gets neither profit nor loss Hence $S.P=C.P$.Therefore we can write ,Selling price of scooter first + Selling price of scooter second is equal to the Cost price of scooter.