CAT 2023 Question Paper | Quant Slot 3

CAT 2023 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 14 : 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

Video Explanation

Explanatory Answer

Since the mangoes make up 40% of his stock at the beginning of the day, and let the mangoes at the beginning of the day be 4x, So the bananas and apples will be equal to 6x. If the bananas equal y, then the apples would be 6x - y. Totally there are 10x number of fruits.
At the end of the day 50% of mangoes, 96 bananas and 40% of apples are sold and thereby he ends up selling 50% of the fruits.
\( 50 \% ( 4 x ) + 96 + 40 \% ( 6 x - y ) = 50 \% ( 10 x ) \)
\( 2 x + 96 + 2.4 x - 0.4 y = 5 x \)
\( 4.4 x + 96 - 0.4 y = 5 x \)
\( 96 - 0.4 y = 0.6 x \)
\( 480 - 2 y = 3 x \)
\( 3 x + 2 y = 480 \)
\( x = \frac { 480 } { 3 } - \frac { 2 } { 3 } y = 160 - \frac { 2 } { 3 } y \) (For x to be an integer, y should be a multiple of 3.)
Since y is the number of bananas, \( y \geq 96 \).
Also since the apples are (6x - y) in number...
(6x - y) = \( 6 \left( 160 - \frac { 2 } { 3 } y \right) - y \)= 960 - 5y
The number of apples must be non-zero.
960 - 5y = 0 when y = 192.
∴ 96 ≤ y < 192 (Also, y needs to be a multiple of 3.)
So, the maximum value that y can take is 189.
For the total number of fruits to be minimum, 10x should be minimum, which happens when x is minimum.
\( x = 160 - \frac { 2 } { 3 } y \); x is minimum when y is maximum.
So the minimum value of \( x = 160 - \frac { 2 } { 3 } \times 189 = 34 \)
So, the minimum number of total fruits = 10x = 340.

The question is " 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 "

The answer is '340'