CAT — Time, Speed & Distance
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Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is
76800
112000
96000
86400
86400
Let the speeds (in m/min) be 960, 960+d, 960+2d, 960+3d and times (in minutes) be 30, 30+t, 30+2t, 30+3t.
Total time = 180 minutes: 30 + (30+t) + (30+2t) + (30+3t) = 180 → 120 + 6t = 180 → t = 10. So times are 30, 40, 50, 60 minutes.
Total distance = 224000 m: 28800 + 40(960+d) + 50(960+2d) + 60(960+3d) = 224000 → 172800 + 320d = 224000 → d = 160.
Fourth part: speed = 960 + 3×160 = 1440 m/min, time = 60 min. Distance = 1440 × 60 = 86400 meters.
− Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. She takes exactly 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, she would take to walk from A to B and run from B to C, is
Step 1: Let walking speed = v km/h. Running speed = 1.4v = 7v/5 km/h.
Let AB = x km, BC = y km.
Step 2: Given conditions
Walking from B to C: y/v = 3.5 → y = 7v/2
Running from B to A: x/(7v/5) = 3.5 → x = 49v/10
Step 3: Find the required time
Walking A to B: x/v = 49/10 hours
Running B to C: y/(7v/5) = (7v/2) × (5/7v) = 5/2 hours
Total time = 49/10 + 25/10 = 74/10 = 7.4 hours
Step 4: Convert to minutes
7.4 × 60 = 444 minutes
Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was
12
15
18
20
15
Step 1: Total journey time = 6 hours. Let initial speed = x km/h → Total distance = 6x km.
Step 2: Verify x = 15 km/h.
Total distance = 90 km.
After 20-min stop, travelling time = 17/3 hours.
Let t = time before stop.
15t + 18(17/3 − t) = 90 → 15t + 102 − 18t = 90 → t = 4 hours.
Rahul travels 60 km at 15 km/h, then 30 km at 18 km/h (1 hr 40 min).
Total = 4 hrs + 1h40m + 20min stop = 6 hours ✓
Step 3: Check second condition (30-min stop).
Remaining travelling time = 6 − 4 − 0.5 = 1.5 hours.
Remaining distance = 30 km.
Required speed = 30 ÷ 1.5 = 20 km/h.
Increase = 20 − 15 = 5 km/h ✓
Rita and Sneha can row a boat at 5 km/h and 6 km/h in still water, respectively. In a river flowing with a constant velocity, Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream. If Rita starts from a certain location in the river, and returns downstream to the same location, taking a total of 100 minutes, then the total distance, in km, Rita will cover is
Step 1: Find the speed of the river. Sneha's speed in still water = 6 km/h. Let the speed of the river be v km/h. Upstream speed = 6 − v. Downstream speed = 6 + v. Given that rowing 14 km upstream takes 48 minutes = 4/5 hour more than rowing 14 km downstream: 14/(6 − v) − 14/(6 + v) = 4/5.
Step 2: Solve for the speed of the river. Taking the LCM: 14[(6 + v) − (6 − v)]/[(6 − v)(6 + v)] = 4/5 → 28v/(36 − v²) = 4/5. Cross-multiplying: 140v = 144 − 4v² → v² + 35v − 36 = 0 → (v + 36)(v − 1) = 0. Since the speed of the river must be positive, v = 1 km/h.
Step 3: Find Rita's upstream and downstream speeds. Rita's speed in still water = 5 km/h. Upstream speed = 5 − 1 = 4 km/h. Downstream speed = 5 + 1 = 6 km/h.
Step 4: Let the one-way distance be d km. The total time for going upstream and returning downstream is 100 minutes = 5/3 hours. So, d/4 + d/6 = 5/3. Taking the LCM: 5d/12 = 5/3 → 5d = 20 → d = 4 km.
Step 5: Find the total distance covered. Rita rows 4 km upstream and 4 km downstream. Total distance = 4 + 4 = 8 km.
Ravi is driving at a speed of 40 km/h on a road. Vijay is 54 meters behind Ravi and driving in the same direction as Ravi. Ashok is driving along the same road from the opposite direction at a speed of 50 km/h and is 225 meters away from Ravi. The speed, in km/h, at which Vijay should drive so that all the three cross each other at the same time, is
58.8
67.2
61.6
64.4
61.6
Step 1: Find the time taken by Ravi to meet Ashok
Relative speed = 40 + 50 = 90 km/h = 90 × 5/18 = 25 m/s
Initial distance = 225 m
Time = 225/25 = 9 seconds
Step 2: Find the distance Vijay must travel
In 9 seconds, Ravi travels: 40 km/h = 100/9 m/s → distance = (100/9) × 9 = 100 m
Meeting point is 54 + 100 = 154 m ahead of Vijay.
Step 3: Find Vijay's speed
Speed = 154/9 m/s
Convert to km/h: (154/9) × 18/5 = 61.6 km/h
A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is
720
800
780
640
720
Given:
A train travels a certain distance at a uniform speed.
● If the speed were 6 km/h more, the journey would take 4 hours less.
● If the speed were 6 km/h less, the journey would take 6 hours more.
Find the distance travelled.
Step 1: Let the original speed be x km/h.
Let the original time taken be t hours.
Then,
Distance = xt
Step 2: Form the first equation
If the speed becomes
x + 6,
the time becomes
t − 4.
Hence,
xt = (x + 6)(t − 4)
Expanding,
xt = xt − 4x + 6t − 24
4x − 6t = −24
2x − 3t = −12
Step 3: Form the second equation
If the speed becomes
x − 6,
the time becomes
t + 6.
Hence,
xt = (x − 6)(t + 6)
Expanding,
xt = xt + 6x − 6t − 36
6x − 6t = 36
x − t = 6
Step 4: Solve the equations
From
x − t = 6,
x = t + 6
Substitute into
2x − 3t = −12,
2(t + 6) − 3t = −12
−t + 12 = −12
t = 24
Therefore,
x = 30
Step 5: Find the distance
Distance
= x × t
= 30 × 24
= 720 km
Answer:
A. 720
Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is
Let Surbhi's speed be v km/h.
Arvind's speed = 54 km/h
After meeting,
● Arvind takes 6 hours to reach
B.
● Surbhi takes 24 hours to reach
A.
Hence, Distance from meeting point to B = 54 × 6 = 324 km
Distance from meeting point to A = 24v
For two people starting at the same time and meeting, (speed ratio) = (distance covered before meeting ratio)
Therefore, 54 : v = 24v : 324
Cross multiply, 54 × 324 = 24v² → v² = (54 × 324)/24 = 729 → v = 27 km/h
Distance from meeting point to A = 24 × 27 = 648 km
Distance from meeting point to B = 324 km
Therefore, total distance = 648 + 324 = 972 km
Answer: 972
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
7 : 30 pm
7 : 00 pm
9 : 00 pm
10 : 30 pm
7 : 30 pm
Given:
The bus starts every day at 9:00 am.
One day, it travels at 60 km/h and reaches 3.5 hours late.
The next day,
● It covers two-thirds of the route in one-third of the scheduled travel time.
● It covers the remaining one-third of the route at 40 km/h.
● It reaches exactly on time.
Find the scheduled arrival time.
Step 1: Let the total distance be D km.
At 60 km/h, the bus takes
D/60 hours.
If the scheduled travel time is T hours, then
D/60 = T + 3.5
Step 2: Form the second equation
The first two-thirds of the journey is completed in
T/3 hours.
Hence, the speed during this part is
(2D/3)/(T/3)
= 2D/T
The remaining one-third of the journey is covered at 40 km/h.
Time taken for this part
= (D/3)/40
= D/120
Since the total journey is completed in the scheduled time,
T/3 + D/120 = T
D/120 = 2T/3
D = 80T
Step 3: Substitute into the first equation
D/60 = T + 3.5
80T/60 = T + 3.5
4T/3 = T + 3.5
T/3 = 3.5
T = 10.5 hours
Step 4: Find the scheduled arrival time
The bus starts at 9:00 am.
Adding 10.5 hours,
Scheduled arrival time
= 7:30 pm
Answer:
A. 7:30 pm
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