Question

Rajveer started a business of selling second hand cars, which he had bought from the first owners of the respective cars. He bought two cars at ₹1,30,000 each. He sold one car at a profit of 10% and another car at a loss of 5%.

Based on the above, information, answer the following questions:

What was the selling price of the car Rajveer sold at a profit?

(a) ₹1,33,000 (b) ₹1,37,000

(c) ₹1,43,000 (d) ₹1,47,000

What was the selling price of the car Rajveer sold at a loss?

(a) ₹1,23,500 (b) ₹1,18,500

(c) ₹1,13,500 (d) ₹1,11,500

(iii) What is the profit in the whole transaction?

(a) ₹4500 (b) ₹6500

(c) ₹7500 (d) ₹9500

(iv) If Rajveer invest the profit made in the whole transaction at 10% compounded yearly,

then what will be the total sum after 3 years?

(a) ₹5989.5 (b) ₹8651.5

(c) ₹9982.5 (d) ₹12644.5​

Answer 1

$\large\underline{\sf{Solution-i}}$

Cost Price of a car = ₹ 130000

Profit % = 10 %

We know, Selling Price of a car is evaluated as

$\color{green}\boxed{ \rm{ \:Selling \: Price = \frac{(100 + Profit\%) \times Cost \: Price}{100} \: }}$

So, on substituting the values, we get

$\rm \: Selling \: Price = \dfrac{(100 + 10) \times 130000}{100}$

$\rm \: = \: 110 \times 1300$

$\rm \: = \: 143000$

Hence, Selling Price of a car = ₹ 143000

Option (c) is correct.

$\large\underline{\sf{Solution-ii}}$

Cost Price of a car = ₹ 130000

Loss % = 5 %

We know, Selling Price of a car is evaluated as

$\color{green}\boxed{ \rm{ \:Selling \: Price = \frac{(100 - Loss\%) \times Cost \: Price}{100} \: }}$

So, on substituting the values, we get

$\rm \: Selling \: Price = \dfrac{(100 - 5) \times 130000}{100}$

$\rm \: = \: 95 \times 1300$

$\rm \: = \: 123500$

Hence, Selling Price of a car = ₹ 123500

So, option (a) is correct.

$\large\underline{\sf{Solution-iii}}$

Case :- 1

Selling Price = ₹ 143000

Cost Price = ₹ 130000

Profit = 143000 - 130000 = ₹ 13000

Case :- 2

Cost Price = ₹ 130000

Selling Price = ₹ 123500

Loss = 130000 - 123500 = ₹ 6500

Now, there is a profit of ₹ 13000 in the first transaction.

And there is a loss of ₹ 6500 in the second transaction.

So, total profit in these two transactions = 13000 - 6500 = ₹ 6500.

So, option (b) is correct.

$\large\underline{\sf{Solution-iv}}$

Now, Sum invested, P = ₹ 6500

Rate of interest, r = 10 % per annum

Time, n = 3 years

We know,

Amount received on a certain sum of money of ₹ P invested at the rate of r % per annum compounded annually for n years is given by

$\color{green}\boxed{ \rm{ \:Amount \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: }}$

So, on substituting the values, we get

$\rm \: Amount \: = \: 6500 \: {\bigg[1 + \dfrac{10}{100} \bigg]}^{3}$

$\rm \: Amount \: = \: 6500 \: {\bigg[1 + \dfrac{1}{10} \bigg]}^{3}$

$\rm \: Amount \: = \: 6500 \: {\bigg[\dfrac{10 + 1}{10} \bigg]}^{3}$

$\rm \: Amount \: = \: 6500 \: {\bigg[\dfrac{11}{10} \bigg]}^{3}$

$\rm\implies \:Amount = 8651.5$

Hence, Amount received = ₹ 8651.5

So, option (b) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{Gain = \sf S.P. \: – \: C.P.} \\ \\ \bigstar \:\bf{Loss = \sf C.P. \: – \: S.P.} \\ \\ \bigstar \: \bf{Gain \: \% = \sf \Bigg( \dfrac{Gain}{C.P.} \times 100 \Bigg)\%} \\ \\ \bigstar \: \bf{Loss \: \% = \sf \Bigg( \dfrac{Loss}{C.P.} \times 100 \Bigg )\%} \\ \\ \\ \bigstar \: \bf{S.P. = \sf\dfrac{(100+Gain\%) or(100-Loss\%)}{100} \times C.P.} \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$