Question

A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop sells half of the remaining quantity and an additional 3 kg of these grains to the third customer, there are no grains left. The initial quantity, in kg, of grains is

• A. 42
• B. 18
• C. 36
• D. 50

Answer 1

The Correct Option is A

Let \( x \) kg be the initial quantity of grains. We will work backward from the end, knowing that no grains remain after the third customer's purchase.

1. **Third Customer:** Before this sale, let the remaining quantity be \( y \) kg. This customer bought half of \( y \) plus 3 kg, leaving 0 kg. Therefore,

\(\frac{y}{2} + 3 = y \)

Solving for \( y \):

\(\frac{y}{2} = 3 \)

\(y = 6 \text{ kg}\)

So, 6 kg of grains remained before the third customer's purchase.

2. **Second Customer:** Let the quantity before this sale be \( z \) kg. This customer bought half of \( z \) plus 3 kg, leaving 6 kg. Therefore,

\(\frac{z}{2} + 3 = 6 \)

Solving for \( z \):

\(\frac{z}{2} = 3 \)

\(z = 12 \text{ kg}\)

So, 12 kg of grains remained before the second customer's purchase.

3. **First Customer:** Let the initial quantity be \( x \) kg. This customer bought half of \( x \) plus 3 kg, leaving 12 kg. Therefore,

\(\frac{x}{2} + 3 = 12 \)

Solving for \( x \):

\(\frac{x}{2} = 9 \)

\(x = 18 \text{ kg}\)

This calculation suggests the initial quantity was 18 kg. However, a re-evaluation indicates the correct initial quantity should be 42 kg. Let's retrace the steps carefully.

**Corrected Calculation:**

Let's work backward, assuming a final state of 0 kg after the last sale.

1. **Third Sale:** If \( Q_2 \) is the quantity before the third sale, and the customer buys \( \frac{Q_2}{2} + 3 \), then \( Q_2 - (\frac{Q_2}{2} + 3) = 0 \). This simplifies to \( \frac{Q_2}{2} = 3 \), so \( Q_2 = 6 \) kg. This represents the amount *before* the third sale.

2. **Second Sale:** If \( Q_1 \) is the quantity before the second sale, and the customer buys \( \frac{Q_1}{2} + 3 \), leaving 6 kg, then \( Q_1 - (\frac{Q_1}{2} + 3) = 6 \). This simplifies to \( \frac{Q_1}{2} = 9 \), so \( Q_1 = 18 \) kg. This is the amount *before* the second sale.

3. **First Sale:** If \( x \) is the initial quantity, and the customer buys \( \frac{x}{2} + 3 \), leaving 18 kg, then \( x - (\frac{x}{2} + 3) = 18 \). This simplifies to \( \frac{x}{2} = 21 \), so \( x = 42 \) kg.

Therefore, the initial quantity of grains was 42 kg.

To verify:

• Start with 42 kg.
• First customer buys \( \frac{42}{2} + 3 = 21 + 3 = 24 \) kg. Remaining: \( 42 - 24 = 18 \) kg.
• Second customer buys \( \frac{18}{2} + 3 = 9 + 3 = 12 \) kg. Remaining: \( 18 - 12 = 6 \) kg.
• Third customer buys \( \frac{6}{2} + 3 = 3 + 3 = 6 \) kg. Remaining: \( 6 - 6 = 0 \) kg.

The calculation is now consistent and verified.