CAT 2008 — QA Question 1

The initial quantity of rice is x kg.

The first customer buys half the total rice in the store, and another half kg.

∴ Rice purchased by the first customer = $\frac{x}{2}+\frac{1}{2}=\frac{x+1}{2}$
∴ Remaining rice = $x-\left(\frac{x+1}{2}\right)=\frac{x-1}{2}$

Now, the second customer buys half of this, and another half kg.

∴ Rice purchased by the second customer = $\frac{x-1}{4}+\frac{1}{2}=\frac{x+1}{4}$
∴ Remaining rice = $\frac{x-1}{2}-\frac{x+1}{4}=\frac{x-3}{4}$

Now, the third customer buys half the remaining rice, and another half kg.

∴ Rice purchased by the third customer = $\frac{x-3}{8}+\frac{1}{2}=\frac{x+1}{8}$

Since after this purchase, there is no rice left in the store, we conclude that:

$\frac{x-3}{4}-\frac{x+1}{8}=\frac{x-7}{8}=0$

∴ x = 7

Alternately,
If the shopkeeper had x kg rice before a customer comes, the amount of rice left after every time a customer buys half the amount of rice + half a kg more = $\frac{\mathrm{x}}{2}-\frac{1}{2}$

For 3^rd customer.
Now, $\frac{\mathrm{x}}{2}-\frac{1}{2}$ = 0 [No rice is left after third customer]
⇒ x = 1 kg.

Now, the shopkeeper had 1 kg rice when 3^rd customer came, hence after 2^nd customer left he had 1 kg of rice.

For 2^nd customer.
$\frac{\mathrm{x}}{2}-\frac{1}{2}$ = 1
⇒ x = 3 kg.

Now, the shopkeeper had 3 kg rice when 2^nd customer came, hence after 1^st customer left he had 3 kg of rice.

For 1^st customer.
$\frac{\mathrm{x}}{2}-\frac{1}{2}$ = 3
⇒ x = 7 kg.

Now, the shopkeeper had 7 kg rice when 1^st customer came.

Hence, option (b).