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CAT 2024 Slot 1 — QA Question 19

Profit & LossEasy

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

Answer & solution

  • A

    50

  • B

    18

  • C

    36

  • 42

Solution

Medium

"Half the remaining plus 33 kg" three times, ending at zero, is cleanest worked backwards: before each customer the stock was (after+3)×2(\text{after}+3)\times2. Start from 00 after the third customer.

1

Reverse the last sale. After customer 3 there are 00 kg. Selling "half +3+3" left 00, so before customer 3 the amount Q3Q_3 satisfies Q3−(Q32+3)=0Q_3-\left(\tfrac{Q_3}{2}+3\right)=0.

Q32−3=0⇒ Q3=6(stock before 3rd customer)\begin{aligned} &\tfrac{Q_3}{2}-3=0\\ &\Rightarrow\ Q_3=6 \quad\text{(stock before 3rd customer)} \end{aligned}
2

Reverse the second sale. Before customer 2, stock Q2Q_2 leaves Q3=6Q_3=6 behind.

Q2−(Q22+3)=6⇒ Q22−3=6(from step 1)⇒ Q2=18\begin{aligned} &Q_2-\left(\tfrac{Q_2}{2}+3\right)=6\\ &\Rightarrow\ \tfrac{Q_2}{2}-3=6 \quad\text{(from step 1)}\\ &\Rightarrow\ Q_2=18 \end{aligned}
3

Reverse the first sale. Initial stock Q1Q_1 leaves Q2=18Q_2=18 behind.

Q1−(Q12+3)=18⇒ Q12−3=18(from step 2)⇒ Q1=42\begin{aligned} &Q_1-\left(\tfrac{Q_1}{2}+3\right)=18\\ &\Rightarrow\ \tfrac{Q_1}{2}-3=18 \quad\text{(from step 2)}\\ &\Rightarrow\ Q_1=42 \end{aligned}
Initial quantity=42 kg\text{Initial quantity}=42\text{ kg}

Each reverse step is just (previous+3)×2(\text{previous}+3)\times2: 0→6→18→420\to6\to18\to42.

CAT 2024 Slot 1 QA Q19: A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg o — Solution | TheCATExam