Question

A shopkeeper sells half of the grains plus \(3 \, \text{kg}\) of grains to Customer 1, and then sells another half of the remaining grains plus \(3 \, \text{kg}\) to Customer 2. When the 3rd customer arrives, there are no grains left. Find the total grains that were initially present.

• A. 10
• B. 36
• C. 42
• D. 18

Answer 1

The Correct Option is D

Let the initial amount of grains be \(x\) kg.

The shopkeeper first sells half of the grains plus 3 kg to Customer 1. This amount is:

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

The grains remaining after the first sale are:

\(x - \left(\frac{x}{2} + 3\right) = \frac{x}{2} - 3\)

Next, the shopkeeper sells half of these remaining grains plus 3 kg to Customer 2. This amount is:

\(\frac{1}{2}\left(\frac{x}{2} - 3\right) + 3\)

The grains remaining after the second sale are:

\(\frac{x}{2} - 3 - \left(\frac{1}{2}\left(\frac{x}{2} - 3\right) + 3\right) = \frac{x}{4} - \frac{3}{2}\)

The problem states that no grains are left after Customer 2's purchase:

\(\frac{x}{4} - \frac{3}{2} = 0\)

Solving this equation for \(x\):

1. \(\frac{x}{4} = \frac{3}{2}\)

2. Multiply both sides by 4:

\(x = 6\)

Upon re-checking the calculations, there was an error. Let's re evaluate the amounts sold and remaining:

Amount sold to Customer 1: \(\frac{x}{2} + 3\)

Remaining after Customer 1: \(\frac{x}{2} - 3\)

Amount sold to Customer 2: \(\frac{1}{2}\left(\frac{x}{2} - 3\right) + 3\)

Remaining after Customer 2: \(\left(\frac{x}{2} - 3\right) - \left(\frac{1}{2}\left(\frac{x}{2} - 3\right) + 3\right) = \frac{x}{4} - \frac{3}{2}\)

Setting the remaining amount to zero gives the correct equation:

\(\frac{x}{4} - \frac{3}{2} = 0\)

Solving this corrected equation:

\(\frac{x}{4} = \frac{3}{2}\)

Multiply both sides by 4:

\(x = 6\)

There seems to be a misunderstanding in the problem's interpretation. Let's reconsider the problem logic by working backward.

Let the amount of grains remaining after Customer 2 be 0.

Before Customer 2's purchase, the shopkeeper had sold half of the grains plus 3 kg. If \(R_1\) is the amount remaining after Customer 1, then Customer 2 bought \(\frac{R_1}{2} + 3\). The amount left after Customer 2 is \(R_1 - (\frac{R_1}{2} + 3) = \frac{R_1}{2} - 3\). So, \(\frac{R_1}{2} - 3 = 0\), which means \(\frac{R_1}{2} = 3\), and \(R_1 = 6\) kg.

Now, \(R_1\) is the amount remaining after Customer 1. Customer 1 bought half of the initial amount \(x\) plus 3 kg, so \(\frac{x}{2} + 3\). The remaining amount after Customer 1 is \(x - (\frac{x}{2} + 3) = \frac{x}{2} - 3\). We found that this remaining amount is 6 kg. So, \(\frac{x}{2} - 3 = 6\).

Solving for \(x\):

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

\(x = 18\)

The correct initial quantity of grains is therefore \(18\) kg.