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

Answer: (D)

18

Answer 2

Let the total grains initially be $x$ kg.
Step 1: Sales to Customer 1
The shopkeeper sells half of the grains plus 3 kg to Customer 1. The remaining grains are:
$\text{Remaining grains} = x - \left(\frac{x}{2} + 3\right).$
Simplify:
$\text{Remaining grains} = \frac{x}{2} - 3.$
Step 2: Sales to Customer 2
The shopkeeper sells half of the remaining grains plus 3 kg to Customer 2. The remaining grains are:
$\text{Remaining grains} = \left(\frac{x}{2} - 3\right) - \left(\frac{\frac{x}{2} - 3}{2} + 3\right).$
Simplify step by step:
$\text{Remaining grains} = \frac{x}{2} - 3 - \frac{\frac{x}{2} - 3}{2} - 3.$
Combine terms:
$\text{Remaining grains} = \frac{x}{2} - 3 - \frac{x}{4} + \frac{3}{2} - 3.$
Simplify further:
$\text{Remaining grains} = \frac{2x}{4} - \frac{x}{4} - 6 + \frac{3}{2}.$
$\text{Remaining grains} = \frac{x}{4} - \frac{9}{2}.$
Step 3: Sales to Customer 3
When Customer 3 arrives, no grains are left:
$\frac{x}{4} - \frac{9}{2} = 0.$
Solve for $x$:
$\frac{x}{4} = \frac{9}{2}.$
Multiply through by 4:
$x = 18.$