CAT 2025 Lesson : Profit & Loss - Fraudulent Trader

Ramlal purchases rice at Rs. $2.47$ per kg and marks up the price by $17 \%$ and sells it. He uses a defective weighing machine while selling the rice. It measures $900$ gm of rice as $1$ kg. What is Ramlal's profit percentage?

Solution

Note that we need to find the answer in percentage terms and $2.47$ is a difficult number. We need not use it as we are required to find the profit $\%$ and not the profit in Rupees.

∴ Let CP $=$ Rs. $100$ per $1$ kg

SP per stated unit $= 100 \times 1.17 =$ Rs. $\bm{117}$ per 1 kg

Due to the faulty weighing machine, trader actually sells $0.9$ kg instead of $1$ kg.

SP per actual unit $=$ Rs. $117$ per $0.9$ kg

$= \dfrac{\text{Rs.} 117}{0.9 \text{kg}} = \dfrac{\text{Rs.} 130}{1 \text{kg}} =$ Rs. $\bm{130}$ per $\bm{1}$ kg

We need to use the actual CP and SP to find profit $\%$.

Profit $\% = \dfrac{\text{SP} - \text{CP}}{\text{CP}} \times 100 \% = \dfrac{130 - 100}{100} \times 100 \% = 30 \%$

Answer: $30 \%$

Jairaman claims to purchase and sell Jowar at the same price per kg. However, he uses a defective pan balance, wherein, $1$ side weighs $200$ grams more than the other. He uses a $1$ kg weight for every purchase and sale, and he uses the balance to his advantage while buying and selling. What is his actual profit $\%$?

Solution

Let the CP and SP be Rs. $100$ per kg. It is given that one side of the pan balance weighs $0.2$ kg more than the other and he use the balance to his advantage.

While purchasing he would use the balance to his advantage by buying more, i.e., buy $1.2$ kgs instead of the stated $1$ kg.

Stated CP $=$ Rs. $100$ per $1$ kg
Actual CP $=$ Rs. $100$ per $1.2$ kg $= \dfrac{\text{Rs.}100}{1.2 \text{kg}}$

$= \bm{\text{Rs.} \dfrac{250}{3}}$ per $\bm{1}$ kg

While selling he would sell less, i.e., sell $0.8$ kgs instead of the stated $1$ kg.

Stated SP $=$ Rs. $100$ per $1$ kg
Actual SP $=$ Rs. $100$ per $0.8$ kg $= \dfrac{\text{Rs.} 100}{0.8 \text{kg}}$
$=$ Rs. $\bm{125}$ per 1 kg

Profit $\% = \dfrac{125 - \dfrac{250}{3}}{\dfrac{250}{3}} \times 100 \% = 50 \%$

Answer: $50 \%$