CSAT Solved Papers/ 2021/Q29

2021 CSAT — Q29

Quant Statement validity 2.5 marks Hard

A person $X$ from a place $A$ and another person $Y$ from a place $B$ set out at the same time to walk towards each other. The places are separated by a distance of $15$ km. $X$ walks with a uniform speed of $1\cdot 5$ km/hr and $Y$ walks with a uniform speed of $1$ km/hr in the first hour, with a uniform speed of $1\cdot 25$ km/hr in the second hour and with a uniform speed of $1\cdot 5$ km/hr in the third hour and so on.

Which of the following is/are correct?

They take $5$ hours to meet.
They meet midway between $A$ and $B$ .

A 1 only
B 2 only
C Both 1 and 2 Answer
D Neither 1 nor 2

Worked rationale

$X$ is constant at $1.5$ km/hr. $Y$ ’s hourly speeds form an AP: $1, 1.25, 1.5, 1.75, 2,\dots$ (step $0.25$ ). The combined closing distance in each hour is $1.5 + Y$ ‘s speed:

Hour	$X$	$Y$	closed this hour	cumulative
1	$1.5$	$1.00$	$2.50$	$2.50$
2	$1.5$	$1.25$	$2.75$	$5.25$
3	$1.5$	$1.50$	$3.00$	$8.25$
4	$1.5$	$1.75$	$3.25$	$11.50$
5	$1.5$	$2.00$	$3.50$	$15.00$

The cumulative closed distance reaches exactly $15$ km at the end of hour $5$ , so they meet after $5$ hours — Statement 1 is correct.

Distance covered by $X$ in $5$ hours $= 1.5\times 5 = 7.5$ km, exactly half of $15$ km. So they meet at the midpoint — Statement 2 is correct.

Answer: (c) Both 1 and 2.

Why the other options miss

A

answered the sub-step, not the question: confirms the $5$ -hour total but never checks $X$ ‘s own distance ( $7.5$ km $=$ half), so misses that the meeting is midway.
B

an arithmetic slip: gets the midpoint feel but mis-sums the closing distances and lands on a non- $5$ meeting time.
D

wrong formula: averages $Y$ ‘s speed wrongly or treats $Y$ as constant, breaking both claims.

Specialist insight

Variable-speed approach problems are tamed by cumulative closing distance per hour, not by chasing positions. The closing rate is itself an AP ( $2.5,2.75,3.0,3.25,3.5$ ), and its running total hits $15$ exactly at hour $5$ — no fractional-hour interpolation needed, which is what makes Statement 1 exact. Statement 2 is then a one-line check: $X$ alone covers $1.5\times 5 = 7.5 =$ half the gap. The trap is to verify only the time and forget the position split.

The trap, in one line

Closing distances sum to $15$ at hour $5$ , and $X$ covers $1.5\times5=7.5=$ half $\Rightarrow$ both hold $\Rightarrow$ (c).

← All 2021 CSAT questions