Question

Khushikumari sold a car at a profit of of 20%. If She sold the car for ₹52000 more , the profit would be have been 15%. Find the cost price of the car​ ? ​

Answer 1

Rs10,04,000

Since the difference between both profits is (20-15)=+5%

And the value of +5% is given as +52000

Since Cost price is 100% and

given that 5% is 52000

so 100%= 52000/5×100 (by cross multiplication)

=rs 10,04,000

Please mark it as brainliest

Answer 2

Appropriate Question :-

Khushi kumari sold a car at a profit of of 20%. If she sold the car for ₹ 52000 less, the profit would be have been 15%. Find the cost price of the car?

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

Given that,

Khushi kumari sold a car at a profit of of 20%.

Let assume that Cost Price of a car is ₹ x.

Profit % = 20 %

We know,

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

So, on substituting the values, we get

$\rm \: Selling \: Price \: = \: \dfrac{(100 + 20) \times x}{100}$

$\rm \: Selling \: Price \: = \: \dfrac{(120) \times x}{100}$

$\rm\implies \:\rm \: Selling \: Price \: = \: \dfrac{6x}{5}$

Now, further given that

If she sold the car for ₹52000 less, the profit would be have been 15%.

Now, we have

$\rm \: Cost \: Price \: = \: x$

$\rm \: Selling \: Price \: = \: \dfrac{6x}{5} - 52000$

$\rm \: Profit\% \: = \: 15 \: \%$

We know,

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

So, on substituting the values, we get

$\rm \: \dfrac{6x}{5} - 52000 = \dfrac{(100 + 15) \times x}{100}$

$\rm \: \dfrac{6x}{5} - 52000 = \dfrac{(115) \times x}{100}$

$\rm \: \dfrac{6x}{5} - 52000 = \dfrac{23x}{20}$

$\rm \: \dfrac{6x}{5} - \dfrac{23x}{20} = 52000$

$\rm \: \dfrac{24x - 23x}{20} = 52000$

$\rm \: \dfrac{x}{20} = 52000$

$\rm \: x = 1040000$

Hence, The cost price of car is ₹ 10, 40, 000.

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Additional Information :-

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