A shopkeeper have 4 kg of pure sugar and 3 kg of impure sugar, if price of pure and impure sugar is Rs. 75 per kg and Rs. X per kg respectively. Shopkeeper sold mixture of all available sugar in Rs. 450 and made profit of 20%, then find the price of per kg of impure sugar?

Options

A. 10Rs
B. 15Rs
C. 20Rs
D. 25Rs

Answer

To solve the problem step-by-step, we will follow the information given in the question and use the formulas related to profit and loss.

### Step 1: Calculate the Total Cost Price of the Mixture
The shopkeeper has:
- 4 kg of pure sugar at Rs. 75 per kg
- 3 kg of impure sugar at Rs. X per kg

**Cost Price of Pure Sugar:**
\[ \text{Cost Price of Pure Sugar} = 4 \text{ kg} \times 75 \text{ Rs/kg} = 300 \text{ Rs} \]

**Cost Price of Impure Sugar:**
\[ \text{Cost Price of Impure Sugar} = 3 \text{ kg} \times X \text{ Rs/kg} = 3X \text{ Rs} \]

**Total Cost Price (CP) of the Mixture:**
\[ \text{Total CP} = \text{Cost Price of Pure Sugar} + \text{Cost Price of Impure Sugar} = 300 + 3X \text{ Rs} \]

### Step 2: Use the Selling Price and Profit to Find the Cost Price
The shopkeeper sold the mixture for Rs. 450 and made a profit of 20%.

Using the formula for selling price (SP):
\[ \text{SP} = \text{CP} \times \left(1 + \frac{\text{Profit \%}}{100}\right) \]

Substituting the values we have:
\[ 450 = (300 + 3X) \times \left(1 + \frac{20}{100}\right) \]
\[ 450 = (300 + 3X) \times 1.2 \]

### Step 3: Solve for the Total Cost Price
Now, we can simplify this equation:
\[ 450 = 1.2(300 + 3X) \]
Dividing both sides by 1.2:
\[ \frac{450}{1.2} = 300 + 3X \]
\[ 375 = 300 + 3X \]

### Step 4: Isolate X
Now, we isolate X:
\[ 375 - 300 = 3X \]
\[ 75 = 3X \]
\[ X = \frac{75}{3} = 25 \text{ Rs/kg} \]

### Conclusion
The price of per kg of impure sugar is Rs. 25.

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