Question

A man buys 35 kg of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg at this price, and 15 kg at a 10% discount. Accidentally, 3 kg of sugar is wasted. He sells the remaining sugar by raising the marked price by p percent so as to make an overall profit of 15%. Then p is nearest to

• A. 31
• B. 22
• C. 35
• D. 25

Answer 1

The Correct Option is D

The correct answer is (D): \(25\)

Let the CP and MP of each kg of sugar be \(10x\) and \(12x\) respectively.

Total cost price = \(35×10x = 350x\)

Total selling price = \(35×10××1.15 = 402.5x\)

Selling price already realized = \(5×12x+15×12x×0.9+3×0+12×12x×\bigg(1+\frac{p}{100}\bigg)=402.5x\)

\(60+162+0+144\bigg(1+\frac{p}{100}\bigg)=402.5\)

\(p=25.34\%≈25\%\)

Answer 2

Let's assume the cost price of 1kg of sugar is Rs 100, making the total cost price of 35 kg Rs 3500. 
If the marked up price per kg is Rs 120, and the final profit is 15%, the final selling price of 35 kg would be \(\text{Rs } 4025 = 3500 \times 1.15\)

The first 5 kg are sold at a 20% marked up price, so SP₁ = \(5 \times 100 \times 1.2 = \text{Rs } 600\)
The next 15 kg are sold after applying a 10% discount, making SP₂ = \(15 \times 100 \times 1.2 \times 0.9 = \text{Rs } 1620\)

As 3 kg of sugar got wasted, 23 kg was sold for \(\text{Rs } (600 + 1620) = \text{Rs } 2220\)
The remaining 12 kg should be sold for \(\text{Rs } 4025 - 2220 = \text{Rs } 1805\)
Therefore, the SP of 1 kg would be \(\frac{1805}{12} \approx \text{Rs } 150\)

Hence, the seller should further mark up by \(\left( \frac{150 - 120}{120} \right) \times 100 = 25\%\)