A reduction of 20% in the price of sugar enables me to purchase 5 kg more for Rs. 600. Find the price of sugar per kg before reduction of price.

Options

A. Rs. 24
B. Rs. 30
C. Rs. 32
D. Rs. 36

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Define the Original Price
Let the original price of sugar per kg be \( X \) rupees.

**Hint:** Start by defining the variable for the unknown price.

### Step 2: Calculate the New Price After Reduction
Since there is a reduction of 20% in the price, the new price of sugar per kg will be:
\[ \text{New Price} = X - 0.2X = 0.8X \]

**Hint:** Remember that a 20% reduction means you are left with 80% of the original price.

### Step 3: Determine the Quantity of Sugar Purchased
With the original price, the amount of sugar that could be purchased for Rs. 600 is:
\[ \text{Quantity before reduction} = \frac{600}{X} \]

With the new price, the quantity of sugar that can be purchased for Rs. 600 is:
\[ \text{Quantity after reduction} = \frac{600}{0.8X} \]

**Hint:** Use the formula for quantity, which is total money divided by price per kg.

### Step 4: Set Up the Equation
According to the problem, the difference in quantity purchased before and after the price reduction is 5 kg:
\[ \frac{600}{0.8X} - \frac{600}{X} = 5 \]

**Hint:** This equation represents the increase in quantity due to the price reduction.

### Step 5: Simplify the Equation
To simplify the left side of the equation, find a common denominator:
\[ \frac{600X - 600(0.8X)}{0.8X \cdot X} = 5 \]
This simplifies to:
\[ \frac{600X - 480X}{0.8X^2} = 5 \]
\[ \frac{120X}{0.8X^2} = 5 \]

**Hint:** Combine the fractions carefully by subtracting the numerators.

### Step 6: Cross-Multiply to Solve for X
Cross-multiplying gives:
\[ 120X = 5 \cdot 0.8X^2 \]
\[ 120X = 4X^2 \]

**Hint:** When you cross-multiply, ensure both sides of the equation are balanced.

### Step 7: Rearrange the Equation
Rearranging the equation gives:
\[ 4X^2 - 120X = 0 \]
Factoring out \( X \):
\[ X(4X - 120) = 0 \]

**Hint:** Factor out common terms to simplify the equation.

### Step 8: Solve for X
Setting each factor to zero gives:
1. \( X = 0 \) (not a valid solution)
2. \( 4X - 120 = 0 \) leads to \( 4X = 120 \) or \( X = 30 \)

**Hint:** Only consider positive, meaningful solutions in context.

### Conclusion
The original price of sugar per kg before the reduction is:
\[ \boxed{30} \text{ rupees} \]