# Time Speed and Distance Questions for CAT

CAT 2020 MOCK Test Series at INR 1000/- only.
Question 16: Raghunandan completes 9/30th of his total journey by ship, 5/18th by a horse carriage and the remaining 10 km on foot. The length of the total journey is
[1] 450/19 km
[2] 24.62 km
[3] 27.2 km
[4] 92 km

Option: 1

Explanation:

Let the total length of journey, performed by Raghunandan be x km.

According to the given condition, $x - \frac{9}{{30}}x - \frac{5}{{18}}x = 10$

$\Rightarrow \frac{{90x - 27x - 25x}}{{90}} = 10 \Rightarrow 38x = 10 \times 90$

Or $x = \frac{{10 \times 90}}{{38}} = \frac{{450}}{{19}}km$

Question 17: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Find the time he would have taken walking at his usual speed.
[1] 25 min
[2] 20 min
[3] 30 min
[4] 24 min

Option: 2

Explanation:

Speed = 4/7th of the usual speed.

=>Time required = 7/4th of the usual time.

=> (7/4th of the usual time)- (usual time required) = 15/60 hours.

=> usual time = 1/3 hours = 20 min.

Question 18: If I take as much time in running 15 m as a car takes in covering 36 m, then find the distance I can cover in the time taken by the car to travel 1 km.
[1] 416.6 m
[2] 41.6 m
[3] 4.16 km
[4] 41.6 km

Option: 1

Explanation:

36m = 15 m.

$1000m \equiv \frac{{15}}{{36}} \times 1000 = 416.6m$

Question 19: Ramesh runs 22.4 kmph. How many metres does he run in five minutes?
[1] 1866m
[2] 18.66 km
[3] 2.866 km
[4] 1.432 km

Option: 1

Explanation:

Speed = 22.4 km/hr = 22.4 x 5/18 m/s = 6.22 m/s.

=>In 5 min (300 sec), distance covered = 6.22 x 300 = 1866 m.

Question 20: A handcart has to cover a distance of 120 km in 15 hours. If it covers half the distance in 4/7th of the time, then what speed should it maintain in order to cover the remaining journey in the scheduled time?
[1] 9.33 kmph
[2] 4.6 kmph
[3] 3.1 kmph
[4] None of these

Option: 1

Explanation:

Time left = 1 - 4/7 = 3/7th of the total scheduled time.

3/7 x 15 = 45/7 hrs; Distance left = 60 km.

Required speed = 60/(45/7) = 9.33 km/hr.