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CAT 2024 Slot 2 — QA Question 18

Basics of TSD/ProportinalityEasy

A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is

Answer & solution

  • A

    7 pm

  • B

    9 pm

  • C

    10:30 pm

  • 7:30 pm

Solution

Medium

Let scheduled travel time be TT hours and distance DD. Day 1 gives one equation linking DD and TT; Day 2's "two-thirds of route in one-third of time, rest at 40, arriving on time" gives another. Solve, then add TT to the 9 am start.

1

Day 1 equation. At 60 km/h the trip took 3.53.5 h longer than scheduled.

D60=T+3.5⇒ D=60T+210\begin{aligned} &\frac{D}{60}=T+3.5\\ &\Rightarrow\ D=60T+210 \end{aligned}
2

Day 2 equation. First 23D\tfrac23 D takes 13T\tfrac13 T; remaining 13D\tfrac13 D at 40 km/h; total time =T=T (on time).

T3+13D40=T⇒ D120=2T3(subtract T3)⇒ D=80T\begin{aligned} &\frac{T}{3}+\frac{\tfrac13 D}{40}=T\\ &\Rightarrow\ \frac{D}{120}=\frac{2T}{3} \quad\text{(subtract }\tfrac{T}{3}\text{)}\\ &\Rightarrow\ D=80T \end{aligned}
3

Solve for TT. Equate the two expressions for DD (steps 1 and 2).

80T=60T+210⇒ 20T=210 ⇒ T=10.5 hours\begin{aligned} &80T=60T+210\\ &\Rightarrow\ 20T=210\ \Rightarrow\ T=10.5\text{ hours} \end{aligned}
4

Scheduled arrival. Start 9:00 am plus 10.510.5 h.

9:00 am+10 h 30 min=7:30 pm\begin{aligned} &9{:}00\text{ am}+10\text{ h }30\text{ min}=7{:}30\text{ pm} \end{aligned}
Scheduled arrival=7:30 pm\text{Scheduled arrival}=7{:}30\text{ pm}
CAT 2024 Slot 2 QA Q18: A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km pe — Solution | TheCATExam