Question

two tables were bought at the same cost. One was sold at a profit of 5% and other was sold at a loss of 7%. If th actual difference of the selling price were Rs. 288 ,what is th cost price of each table?​

Answer 1

Step-by-step explanation:

Given:-

• The cost of two tables is same.

• One was sold at a profit of 5% and other was sold at a loss of 7%.

• The actual difference of the Selling price is Rs.288

To Find:-

• The cost price of each table.

Solution:-

Let the cost price of each table be Rs. x

For 1st table:-

$\sf Profit = 5\% \: and \: CP = x$

$\sf \implies SP = \dfrac{100 + Profit\%}{100} \times CP$

$\sf \implies SP = \dfrac{100 + 5}{100} \times x$

$\sf \implies SP = \dfrac{105x}{100}$

$\sf \dashrightarrow\underline{\: \: \underline{ \: \pink{SP\: of \: 1st \: table = \dfrac{105x}{100}}\:}\: \:}$

For 2nd table:-

$\sf Loss= 7\% \: and \: CP = x$

$\sf \implies SP = \dfrac{100 - Loss\%}{100} \times CP$

$\sf \implies SP = \dfrac{100 - 7}{100} \times x$

$\sf \implies SP = \dfrac{93x}{100}$

$\sf \dashrightarrow\underline{\: \: \underline{ \: \orange{SP\: of \: 2nd\: table = \dfrac{93x}{100}}\:}\: \:}$

Given, Difference in Selling price = Rs.288

SP of 1st table - SP of 2nd table = Rs.288

$\sf \implies \dfrac{105x}{100} -\dfrac{93x}{100} = 288$

$\sf \implies \dfrac{105x-93x}{100} = 288$

$\sf \implies \dfrac{12x}{100} = 288$

$\sf \implies x = 288\times \dfrac{100}{12}$

$\sf \implies x = 24 \times 100$

$\sf \implies x = 2400$

Cost price of each table = x = 2400

$\underline{\boxed{\purple{\rm \therefore Cost \: price \: of \: each \: table = Rs. 2400}}}$