A retailer bought some apples rate of 7 apples for Rs 4 and sold them rate of 8 apples for Rs 5. If he gains Rs 30 on that day, the quantity of apples sold by him on that day is:

Options

A. 555
B. 560
C. 565
D. 570

Answer

To solve the problem step by step, we will follow the given information about the cost price (CP) and selling price (SP) of the apples, and then calculate the total quantity of apples sold based on the profit made.

### Step 1: Determine the Cost Price (CP) per apple
The retailer bought 7 apples for Rs. 4. Therefore, the cost price (CP) of one apple is:
\[ \text{CP} = \frac{4 \text{ Rs}}{7 \text{ apples}} = \frac{4}{7} \text{ Rs per apple} \]

### Step 2: Determine the Selling Price (SP) per apple
The retailer sold 8 apples for Rs. 5. Therefore, the selling price (SP) of one apple is:
\[ \text{SP} = \frac{5 \text{ Rs}}{8 \text{ apples}} = \frac{5}{8} \text{ Rs per apple} \]

### Step 3: Set up the equation for profit
The profit is calculated as:
\[ \text{Profit} = \text{SP} - \text{CP} \]
Given that the total profit made by the retailer is Rs. 30, we can express this as:
\[ \text{Profit} = \text{Total SP} - \text{Total CP} \]

Let \( X \) be the number of apples sold. Then:
\[ \text{Total SP} = X \times \frac{5}{8} \]
\[ \text{Total CP} = X \times \frac{4}{7} \]

Substituting these into the profit equation gives:
\[ \frac{5X}{8} - \frac{4X}{7} = 30 \]

### Step 4: Solve the equation
To solve the equation, we need a common denominator for the fractions. The least common multiple of 8 and 7 is 56. Therefore, we rewrite the equation:
\[ \frac{5X \times 7}{56} - \frac{4X \times 8}{56} = 30 \]
This simplifies to:
\[ \frac{35X - 32X}{56} = 30 \]
\[ \frac{3X}{56} = 30 \]

Now, multiply both sides by 56:
\[ 3X = 30 \times 56 \]
\[ 3X = 1680 \]

Now divide both sides by 3:
\[ X = \frac{1680}{3} = 560 \]

### Conclusion
The quantity of apples sold by the retailer on that day is:
\[ \text{Number of apples sold} = 560 \]