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 13 : A boat takes 2 hours to travel downstream a river from port A to port B, and 3 hours to return to port A. Another boat takes a total of 6 hours to travel from port B to port A and return to port B. If the speeds of the boats and the river are constant, then the time, in hours, taken by the slower boat to travel from port A to port B is

Video Explanation

Explanatory Answer

Let B₁, B₂, S be the speeds of the faster boat in still water, speed of slower boat in still water and the speed of the river, and D be the distance between A and B.
The first boat takes (2 + 3 = 5) hours for the roundtrip while the second boat takes 6 hours for the roundtrip. This means that the first boat is the faster one.
Since the downstream and upstream times are in the ratio 2 : 3, the downstream and upstream speeds will be in the ratio 3 : 2
$\frac { B _ { 1 } + S } { B _ { 1 } - S } = \frac { 3 } { 2 }$
$2 \left( B _ { 1 } + S \right) = 3 \left( B _ { 1 } - S \right)$
$B _ { 1 } = 5 \times S$
$D = \left( B _ { 1 } + S \right) \times 2 = 6 S \times 2 = 12 S$
The second boat takes a total of 6 hours to complete the roundtrip.
$\frac { D } { B _ { 2 } + S } + \frac { D } { B _ { 2 } - S } = 6$
$\frac { 2 B _ { 2 } } { B _ { 2 } ^ { 2 } - S ^ { 2 } } = \frac { 6 } { D }$
$\frac { B _ { 2 } } { B _ { 2 } ^ { 2 } - S ^ { 2 } } = \frac { 1 } { 4 S }$
$B _ { 2 } ^ { 2 } - S ^ { 2 } = 4 B _ { 2 } S$
$\div b y S ^ { 2 }$
$\left( \frac { B _ { 2 } } { S } \right) ^ { 2 } - 1 = 4 \left( \frac { B _ { 2 } } { S } \right)$
let $\left( \frac { B _ { 2 } } { S } \right) = x$
$x ^ { 2 } - 4 x - 1 = 0$
$x = \frac { 4 \pm \sqrt { 16 + 4 } } { 2 } = 2 \pm \sqrt { 5 }$
$x$ is $+ v e$
$\therefore x = 2 + \sqrt { 5 }$
$B _ { 2 } = S ( 2 + \sqrt { 5 } )$
The time taken by the slower boat to travel from A to B (downstream) = $\frac { D } { B _ { 2 } + S } = \frac { 12 S } { ( 3 + \sqrt { 5 } ) S }$
$= \frac { 12 ( 3 - \sqrt { 5 } ) } { ( 3 + \sqrt { 5 } ) ( 3 - \sqrt { 5 } ) } = \frac { 12 } { 4 } ( 3 - \sqrt { 5 } ) = 3 ( 3 - \sqrt { 5 } )$

The question is " A boat takes 2 hours to travel downstream a river from port A to port B, and 3 hours to return to port A. Another boat takes a total of 6 hours to travel from port B to port A and return to port B. If the speeds of the boats and the river are constant, then the time, in hours, taken by the slower boat to travel from port A to port B is "

The answer is '$3(3-\sqrt{5})$'

Choice A is the correct answer.

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