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

To solve the problem, we begin by understanding that the man aims to make an overall profit of 15% on the entire 35 kg of sugar. Firstly, determine the cost price (CP) for the total sugar. Assume the CP per kg is $1, making the CP for 35 kg equal to $35.

The man intends to achieve a 20% initial profit, so the marked price (MP) per kg is:

MP = CP × (1 + Profit%) = 1 × (1 + 0.20) = $1.20

Now, proceed with the sales distribution:

• Sale 1: 5 kg sold at MP, yielding a revenue of 5 × $1.20 = $6.
• Sale 2: 15 kg sold at a 10% discount on MP:

Price per kg with discount = $1.20 × (1 - 0.10) = $1.08. Total revenue = 15 × $1.08 = $16.20.

• The total weight of sugar remaining is 35 kg - 5 kg - 15 kg - 3 kg (wasted) = 12 kg.

The man expects a total 15% profit from his initial investment of $35, so the required selling price (RSP) for all sugar is:

RSP = CP × (1 + Overall Profit%) = $35 × 1.15 = $40.25

Revenue from the first two sales = $6 + $16.20 = $22.20.

To attain an overall profit of 15%, the remaining revenue needed from the 12 kg is:

Required additional revenue = $40.25 - $22.20 = $18.05

Hence, the selling price per kg of the remaining sugar = $18.05 / 12 = $1.504

To find p% increase above the original MP ($1.20),

New price = MP × (1 + p/100)

So, $1.504 = $1.20 × (1 + p/100)

Solving for p:

1 + p/100 = 1.504 / 1.20 = 1.25333

p/100 = 1.25333 - 1 = 0.25333

p = 25.333%, rounded to the nearest integer: 25%

Thus, p is nearest to 25.

Answer 2

Let's assume the cost price (CP) of 1 kg of sugar is: \[ \text{CP}_{\text{per kg}} = \text{Rs } 100 \] Therefore, the total cost price for 35 kg is: \[ \text{Total CP} = 35 \times 100 = \text{Rs } 3500 \]

If the sugar is marked up by 20%, then: \[ \text{Marked Price (MP)} = 100 \times 1.2 = \text{Rs } 120 \text{ per kg} \]

A 15% overall profit is required. So the total selling price (SP) should be: \[ \text{Total SP} = 3500 \times 1.15 = \text{Rs } 4025 \]

Breakdown of Sales:

1. **First 5 kg sold at full marked-up price (no discount):** \[ \text{SP}_1 = 5 \times 120 = \text{Rs } 600 \]

2. **Next 15 kg sold at 10% discount:** \[ \text{SP}_2 = 15 \times 120 \times 0.9 = \text{Rs } 1620 \]

3. **3 kg of sugar is wasted** → only \( 35 - 5 - 15 - 3 = 12 \) kg left to be sold

Total revenue collected so far: \[ \text{Rs } 600 + \text{Rs } 1620 = \text{Rs } 2220 \]

Remaining revenue needed to meet total profit: \[ \text{Rs } 4025 - \text{Rs } 2220 = \text{Rs } 1805 \]

Determine Selling Price for Remaining 12 kg:

To get Rs 1805 from 12 kg: \[ \text{SP per kg} = \frac{1805}{12} \approx \text{Rs } 150.42 \] (Approximated to Rs 150 for simplicity.)

Required Further Markup:

Original marked price per kg = Rs 120. To sell at Rs 150: \[ \text{Required Markup \%} = \left( \frac{150 - 120}{120} \right) \times 100 = 25\% \]

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Final Answer:

\(\boxed{25\%}\) additional markup is required on the remaining 12 kg to ensure a total 15% profit.