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CAT 2023 Slot 3 — QA Question 14

PercentageEasy

A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is

Answer & solution

Answer: 340

Solution

Easy

Let the start stock be 5x5x so mangoes (40%40\%) are a whole 2x2x. Express bananas and apples with a parameter, write "sold =50%=50\% of stock", and reduce to one linear relation. Then push the divisibility/positivity constraints to make the total as small as possible.

Let total =5x=5x, so mangoes =2x=2x. Let apples =5a=5a (so 40%40\% of apples =2a=2a is whole); then bananas =5x−5a=5x-5a. Each fruit type needs at least one item.

1

Count fruits sold: half the mangoes, 96 bananas, 40%40\% of apples:

sold=12(2x)+96+40100(5a)=x+96+2a\text{sold}=\frac{1}{2}(2x)+96+\frac{40}{100}(5a)=x+96+2a
2

He sold half the total, so set sold =50%=50\% of 5x5x:

x+96+2a=2.5x⇒ 1.5x=96+2a ⇒ x=64+4a3\begin{aligned} &x+96+2a=2.5x\\ &\Rightarrow\ 1.5x=96+2a\ \Rightarrow\ x=64+\frac{4a}{3} \end{aligned}
3

Express the total and minimise. Multiply by 5:

5x=320+20a35x=320+\frac{20a}{3}

For 5x5x integer, aa must be a multiple of 33; the smallest valid aa is a=3a=3 (gives positive bananas), making 20a3=20\dfrac{20a}{3}=20.

4

Smallest total:

5x=320+20=3405x=320+20=340

Check: mangoes =2x=136=2x=136, apples =15=15, bananas =340−136−15=189>0 ✓=340-136-15=189>0\ \checkmark.

Smallest starting stock=340\text{Smallest starting stock}=\mathbf{340}
CAT 2023 Slot 3 QA Q14: A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginni — Solution | TheCATExam