Quant · free practice

Time, Speed and Distance questions for IPMAT, with solutions

10 free Time, Speed and Distance practice questions for IPMAT Indore, Rohtak and JIPMAT, 3 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Time, Speed and Distance on Learn.

1. Question 1Type the answerWritten by IPMATwaySpeed, Time and Distance Basics

   Without stoppages a train runs at $60$ km/h; with stoppages its average speed over the same route is $40$ km/h. For how many minutes per hour does the train stop?

   Show the answer and solution

   Answer: 20

   Without stops the train covers $60$ km in an hour; with stops it covers only $40$ km per hour. So in each hour it loses $60-40=20$ km of running. At $60$ km/h, $20$ km takes $\frac{20}{60}$ h $=20$ minutes. So the train stops for $20$ minutes every hour (it runs $40$ minutes and covers $40$ km).

2. Question 2Type the answerWritten by IPMATwaySpeed, Time and Distance Basics

   During a journey Rishi drives for $20$ minutes at $30$ km/h, then $30$ minutes at $50$ km/h and finally $1$ hour at $60$ km/h. The total distance covered, in km, is

   Show the answer and solution

   Answer: 95

   Use distance $=$ speed $\times$ time for each leg, with time in hours. Leg 1: $30\times\frac{20}{60}=10$ km. Leg 2: $50\times\frac{30}{60}=25$ km. Leg 3: $60\times1=60$ km. Total distance $=10+25+60=95$ km.

3. Question 3Type the answerWritten by IPMATwayClocks

   A clock strikes the hour (once at $1$, twice at $2$, …, twelve times at $12$) and also strikes once at every half-hour. How many times does it strike in $12$ hours?

   Show the answer and solution

   Answer: 90

   Hour strokes in $12$ hours: $1+2+\dots+12=\frac{12\times13}{2}=78$. The clock also strikes once at each half-hour; there are $12$ half-hours ($12{:}30$, $1{:}30$, …, $11{:}30$) in $12$ hours. Total strokes $=78+12=90$.

4. Question 4Type the answerWritten by IPMATwayBoats and Streams

   A man rows $1.5$ km upstream in $15$ minutes and returns the same distance in $5$ minutes. The speed of the current, in km/h, is

   Show the answer and solution

   Answer: 6

   Convert each leg to km/h. Upstream: $1.5$ km in $15$ min $=\frac14$ h, so $1.5\div\frac14=6$ km/h. Downstream: $1.5$ km in $5$ min $=\frac1{12}$ h, so $1.5\times12=18$ km/h. The current is half the difference: $\frac{18-6}{2}=6$ km/h (and the boat's still-water speed is $12$ km/h).

5. Question 5Type the answerWritten by IPMATwayRaces and Circular Tracks

   A and B start together from the same point on a $400$ m circular track, running in the same direction at $10$ m/s and $6$ m/s. After how many seconds will they be together at the starting point for the first time?

   Show the answer and solution

   Answer: 200

   A is at the start at multiples of $40$ s, B at multiples of $\frac{200}{3}$ s. The first common time is $200$ s. (The first meeting anywhere is at $400/4=100$ s, but that is not at the start.)

6. Question 6Multiple choiceAsked in IPMAT Rohtak 2019Speed, Time and Distance Basics

   Train A takes $45$ minutes more than train B to travel $450$ km. Due to engine trouble, speed of train B falls by a quarter. So it takes $30$ minutes more than Train A to complete the same journey. Find the speed of Train A.

   1. A.$120$ km/hr
   2. B.$110$ km/hr
   3. C.$100$ km/hr
   4. D.$90$ km/hr
   Show the answer and solution

   Answer: C. $100$ km/hr

   Let A's time be $t$ h; B's time is $t-\frac34$. Reducing B's speed by a quarter multiplies its time by $\frac43$: $\frac43\left(t-\frac34\right)=t+\frac12$, so $\frac{4t}{3}-1=t+\frac12$, $\frac{t}{3}=\frac32$, $t=4.5$ h. Speed of A $=\frac{450}{4.5}=100$ km/h (B: $120$ km/h; at $90$ km/h B takes $5$ h $=4.5+0.5$).

7. Question 7Multiple choiceAsked in IPMAT Indore 2021Relative Speed and Trains

   A train left point A at $12$ noon. Two hours later, another train started from point A in the same direction. It overtook the first train at $8$ PM. It is known that the sum of the speeds of the two trains is $140$ km/hr. Then, at what time would the second train overtake the first train, if instead the second train had started from point A in the same direction $5$ hours after the first train?

   1. A.$3$ AM the next day
   2. B.$4$ AM the next day
   3. C.$8$ AM the next day
   4. D.$11$ PM the same day
   Show the answer and solution

   Answer: C. $8$ AM the next day

   First train runs $12$ noon–$8$ PM $=8$ h, second $2$ PM–$8$ PM $=6$ h, same distance, so speeds are $6:8=3:4$; with sum $140$: $60$ and $80$ km/h. If the second starts at $5$ PM the first is $5\times60=300$ km ahead; relative speed $20$ km/h; time $=15$ h; overtaking at $8$ AM the next day.

8. Question 8Multiple choiceWritten by IPMATwayBoats and Streams

   A boat takes $4$ hours longer to go $24$ km upstream than to go the same $24$ km downstream. The speed of the stream is $4$ km/h. How long will the boat take to go $40$ km in still water?

   1. A.$5$ hours
   2. B.$10$ hours
   3. C.$6$ hours
   4. D.$3\frac13$ hours
   Show the answer and solution

   Answer: A. $5$ hours

   Let the still-water speed be $v$ km/h and the stream $4$ km/h. Upstream time minus downstream time is $\frac{24}{v-4}-\frac{24}{v+4}=\frac{192}{v^2-16}$. Setting it equal to $4$ gives $v^2-16=48$, so $v=8$ km/h. Check: upstream $24/4=6$ h, downstream $24/12=2$ h, difference $4$ h. In still water $40$ km takes $40/8=5$ h. Dividing by the stream speed gives $10$ h, the upstream time is $6$ h, and using the downstream speed gives $3\frac13$ h.

9. Question 9Multiple choiceWritten by IPMATwayClocks

   Between $4$ o'clock and $5$ o'clock, the hands of a clock make an angle of $40^\circ$ at two moments. How many minutes after $4$ o'clock is the later of these two moments?

   1. A.$14\frac6{11}$
   2. B.$26\frac23$
   3. C.$29\frac1{11}$
   4. D.$7\frac3{11}$
   Show the answer and solution

   Answer: C. $29\frac1{11}$

   At $m$ minutes past $4$ the hour hand is at $120+0.5m$ degrees and the minute hand at $6m$, so the minute hand is behind by $120-5.5m$ degrees. The angle is $40^\circ$ when $120-5.5m=40$, i.e. $m=\frac{160}{11}=14\frac6{11}$, and again after the hands cross, when $120-5.5m=-40$, i.e. $m=\frac{320}{11}=29\frac1{11}$. Both lie before $5$ o'clock, and the later one is $29\frac1{11}$. Using $6^\circ$ per minute gives $26\frac23$, and dropping the initial $120^\circ$ gap gives $7\frac3{11}$.

10. Question 10Multiple choiceAsked in IPMAT Indore 2025Relative Speed and Trains

    Two swimmers, Ankit and Bipul, start swimming from the opposite ends of a swimming pool at the same time. Ankit can cover the length of the pool once in $10$ minutes. Bipul can cover the length of the pool once in $15$ minutes. They swim back and forth for $80$ minutes without stopping. The number of times they meet each other is

    1. A.$7$
    2. B.$6$
    3. C.$8$
    4. D.$5$
    Show the answer and solution

    Answer: A. $7$

    Ankit's round trip takes $20$ min, Bipul's $30$ min, so the pattern repeats every $60$ min. Tracking positions: they meet at $6$ min (head-on), $18$ min, $30$ min (both at Bipul's starting wall), $42$, $54$, $66$ and $78$ min. That is $7$ meetings before $80$ minutes.