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Rice Shop Customers

CAT 2008 Slot 1 · Quantitative Ability · Medium · Algebra

This is a medium Quantitative Ability question from the CAT 2008 Slot 1 paper. It tests Algebra. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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A shop stores xx kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of xx?

  1. A.

    2≤x≤62 \le x \le 6

  2. B.

    5≤x≤85 \le x \le 8

  3. C.

    9≤x≤129 \le x \le 12

  4. D.

    11≤x≤1411 \le x \le 14

  5. E.

    13≤x≤1813 \le x \le 18

Answer

B

Explanation

Amount bought by 1st customer =(x2+12)=x+12= \left(\frac{x}{2} + \frac{1}{2}\right) = \frac{x+1}{2}. Remaining =x−x+12=x−12= x - \frac{x+1}{2} = \frac{x-1}{2}.

Amount bought by 2nd customer =12(x−12)+12=x+14= \frac{1}{2}\left(\frac{x-1}{2}\right) + \frac{1}{2} = \frac{x+1}{4}. Remaining =x−12−x+14=x−34= \frac{x-1}{2} - \frac{x+1}{4} = \frac{x-3}{4}.

Amount bought by 3rd customer =12(x−34)+12=x+18= \frac{1}{2}\left(\frac{x-3}{4}\right) + \frac{1}{2} = \frac{x+1}{8}. Remaining =x−34−x+18=x−78= \frac{x-3}{4} - \frac{x+1}{8} = \frac{x-7}{8}.

Since no rice is left, x−78=0  ⟹  x=7\frac{x-7}{8} = 0 \implies x = 7.

Thus, 5≤x≤85 \le x \le 8.

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