Question 66

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

Solution

Let us say the quantity of grains is X
For the first customer he sells $$\frac{X}{2}+3$$
Remaining is $$\frac{X}{2}-3$$

Second customer he sells: $$\frac{X}{4}-\frac{3}{2}+3=\frac{X}{4}+\frac{3}{2}$$
Remaining will be $$\frac{X}{4}-\frac{9}{2}$$

Third customer he sells: $$\frac{X}{8}-\frac{9}{4}+3$$ = $$\frac{X}{8}+\frac{3}{4}$$
Remaining will be $$\frac{X}{8}-\frac{21}{4}$$
Now, this is said to be 0,
$$\frac{X}{8}-\frac{21}{4}=0$$

X=42

Video Solution

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