# 2IIM Quant Notes - Time, Speed and Distance - Part (1/2)

• Two friends A and B simultaneously start running around a circular track . They run in the same direction. A travels at 6 m/s and B runs at b m/s. If they cross each other at exactly two points on the circular track and b is a natural number less than 30, how many values can b take?
3
4
7
5

Let track length be equal to T.
Time taken to meet for the first time = T/relative speed= T/(6−b) or T/(b−6)
Time taken for a lap for A = T/6
Time taken for a lap for B = T/b
So, time taken to meet for the first time at the starting point = LCM (T/6,T/b)=T/HCF(6,b)

Number of meeting points on the track = Time taken to meet at starting point/Time taken for first meeting = Relative speed / HCF (6,b).

So, in essence we have to find values for b such that (6−b)/HCF(6,b) = 2 or (b−6)/HCF(6,b) = 2

b = 2, 10, 18 satisfy this equation. So, there are three different values that b can take.

Consider a square ABCD. EFGH is another square obtained by joining the midpoints of the sides of the square ABCD where E, F , G amd H are the midpoints of AB, BC, CD and DA respectively. Lakshman and Kanika start from points B and D respectively at speeds ‘l’ kmph and ‘k’ kmph respectively and travel towards each other along the sides of the square ABCD. Jagadeesh starts from Point E and travels along the Square EFGH in the anti-clockwise direction at ‘j’ kmph. Lakshman and Kanika meet for the second time at H where Jagadeesh also meets them for the first time. If l : k : j is 1: 3 : 5√2, then the distance travelled by Jagadeesh is
a) 7.5 × √2 times the side of the square ABCD
b) 7.5 × √2 times the side of the square EFGH
c) 7.5 times the side of the square ABCD
d) 7.5 times the side of the square EFGH

Lakshman and Kanika meet for the second time at H where Jagadeesh also meets them for the first time.

Let us assume that the side of the big square ABCD to be ‘2a’; Half of this length, given by AE would be a, and distance EF would be √2a.

Lakshman and Kanika start from points B and D respectively at speeds ‘l’ kmph and ‘k’ kmph respectively and travel towards each other along the sides of the square ABCD. They are at a distance of 4a from each other. Since they are at diametrically opposite points, the relative distance would be 4a irrespective of the directions they choose to travel in. So, to meet for the first time, they would have travelled a distance of 4a together.

To meet for the second time, they would have travelled a further 8a together. Essentially, between them, they would have to cover the entire perimeter of the square to meet again.

So by the time they meet for the second time, they would have covered a distance of 12a together. Their speeds are in the ratio 1: 3. So, Lakshman would have travelled 3a and Kanika would have travelled 9a.

Or, Lakshman travels in the direction BADC, while Kanika would have travelled in the direction DABC. They meet for the first time at E and the second time at H.

In the same time, Jagadeesh travels along the square EFGH in the anti-clockwise direction at ‘j’ kmph and meets Lakshman and Kanika. While Jagadeesh meets the other two for the first time, we do not know how many laps he has completed by then.

The ratio of Lakshman’s speed to that of Jagadeesh is 1: 5√2 so, they would have travelled distances in the same ratio as well. So, if Lakshman has travelled a distance of 3a, Jagadeesh should have travelled a distance of 3a x 5√2 to reach H.

Three cars leave A for B in equal time intervals. They reach B simultaneously and then leave for Point C which is 240 km away from B. The first car arrives at C an hour after the second car. The third car, having reached C, immediately turns back and heads towards B. The first and the third car meet a point that is 80 km away from C. What is the difference between the speed of the first and the third car?
60 kmph
80 kmph
20 kmph
40 kmph

AB/V1−AB/V2=AB/V2−AB/V3
240/V1−240/V2 = 1
v3 = 2v1

Let v1, v2 and v3 be the speeds of the cars.
Condition I states that the cars leave in equal intervals of time and arrive at the same time. Or, the difference in the time taken between cars 1 and 2 should be equal to the time taken between cars 2 and 3.
We get AB/V1−AB/V2=AB/V2−AB/V3
As the second car arrived at C an hour earlier than the first, we get a second equation 240/V1−240/V2 = 1
The third car covered 240 + 80 kms when the first one covered 240 – 80 kms. Therefore, 320/V3=160/V1
This gives us v3 = 2v1
From condition 1, we have AB/V1−AB/V2=AB/V2−AB/V3
Substituting v3 = 2v1, this gives us AB/V1−AB/V2=AB/V2−AB/2V1
or 3xAB/2xV1=2xAB/V2
or v2 = 4xv1/3
Solving 240/v1−240/v2 = 1, we get 60/v1 =1 or, v1 = 60 kmph
=> v2 = 80 kmph and v3 = 120 kmph

Three friends A, B and C decide to run around a circular track. They start at the same time and run in the same direction. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Find the ratio of the speeds of A, B and C?
5 : 4 : 2
4 : 3 : 2
5 : 4 : 3
3 : 2 : 1

Let track length be equal to T. When A completes a lap, let us assume B has run a distance of (t - d). At this time, C should have run a distance of (t - 2d).
After three laps C would have travelled a distance of 3 * (t - 2d) = 3t - 6d.
After 3 laps C is in the same position as B was at the end of one lap. So, the position after 3t - 6d should be the same as t - d. Or, C should be at a distance of d from the end of the lap. C will have completed less than 3 laps (as he is slower than A), so he could have traveled a distance of either t - d or 2t - d.
=> 3t - 6d = t - d
=> 2t = 5d
=> d = 0.4t
The distances covered by A, B and C when A completes a lap will be t, 0.6t and 0.2t respectively. Or, the ratio of their speeds is 5 : 3 : 1.
In the second scenario, 3t - 6d = 2t - d => t = 5d => d = 0.2t.
The distances covered by A, B and C when A completes a lap will be t, 0.8t and 0.6t respectively. Or, the ratio of their speeds is 5 : 4 : 3.
The ratio of the speeds of A, B and C is either 5 : 3 : 1 or 5 : 4 : 3.

Mr. X decides to travel from Delhi to Gurgaon at a uniform speed and decides to reach Gurgaon after T hr. After 30 km, there is some engine malfunction and the speed of the car becomes 4/5th of the original speed. So, he travels the rest of the distance at a constant speed 4/5th of the original speed and reaches Gurgaon 45 minutes late. Had the same thing happened after he travelled 48 km, he would have reached only 36 minutes late. What is the distance between Delhi and Gurgaon?
90 kmph
80 kmph
20 kmph
40 kmph

Let the distance from Delhi to Gurgaon be ‘d’ km. The first 30 km he travels at his usual speed. However, the remaining ‘d-30’ km he travels at a reduced speed.
To travel ‘d’ km he usually takes T hr. Therefore, to travel ‘d - 30’ km he should ideally take (d−30)xT/d hr. However, this is only if he travels at his usual speed. It is given that he travelled only at 4/5th of his usual speed. Because of this he would have taken 5/4th of the time to travel the remaining distance, i.e., he takes 1/4th of the time extra. This is given to be 45 minutes (or 34thhr)
(1/4)x(d−30)xT/d=3/4 ........ (1)
On the other hand, had the same thing happened after he travelled 48 km, he would have reached only 36 minutes or 3/4 hrs late. Hence,
(1/4)x(d−48)xT/d=3/5 ........ (2)
Dividing (1) by (2) and solving for d, we get d = 120 km.

Two friends A and B leave City P and City Q simultaneously and travel towards Q and P at constant speeds. They meet at a point in between the two cities and then proceed to their respective destinations in 54 minutes and 24 minutes respectively. How long did B take to cover the entire journey between City Q and City P?
60
36
24
48

Let us assume Car A travels at a speed of a and Car B travels at a speed of b.
Further, let us assume that they meet after t minutes.
Distance travelled by car A before meeting car B = a * t.
Likewise distance travelled by car B before meeting car A = b * t.
Distance travelled by car A after meeting car B = a * 54.
Distance travelled by car B after meeting car A = 24 * b.
Distance travelled by car A after crossing car B = distance travelled by car B before crossing car A (and vice versa).
=> at = 54b ---------- (1)
and bt = 24a -------- (2)
Multiplying equations 1 and 2
we have ab * t^2 = 54 * 24 * ab
=> t^2 = 54 * 24
=> t = 36
So, both cars would have travelled 36 minutes prior to crossing each other. Or, B would have taken 36 + 24 = 60 minutes to travel the whole distance. Choice (A).

A swimming pool is of length 50 m. A and B enter a 300 m race starting simultaneously at one end of the pool at speeds of 3 m/s and 5 m/s. How many times will they meet while travelling in opposite directions before B completes the race?
Twice
Thrice
Once
5 times

Let us tabulate this and then see if we can pick a pattern. Let us say the pool starts from P and ends at Q. B being the quicker one will reach Q first. B will reach Q after 10 seconds.

A will take 503 = 16.67 seconds to reach Q, so between 10s and 16.67 seconds they meet once.

First, let us have a table where we know times when A, B would be at the end points

AB
TimePositionTimePosition
(in sec)(in sec)
16.67Q10Q
33.33P20P
50Q30Q
66.67P40P
83.33Q50Q
100P60P

We see that the critical time slots are multiples of 10 seconds and multiples of 16.67 seconds.

So, let us break the time from 0 to 60 seconds into these slots and see in which direction A and B travel.

TimePosition APosition BDo they cross?
0 to 10P to QP to Q
10 to 16.67P to QQ to PYes
16.67 to 20Q to PQ to P
20 to 30Q to PP to QYes
30 to 33.33Q to PQ to P
33.33 to 40P to QQ to PYes
40 to 50P to QP to Q
50 to 60Q to PQ to P

So, they cross each other thrice totally. Between 10 and 16.67 seconds for the first time; between 20 and 30 seconds for the second time; and between 33.33 to 40 seconds for the third time.

Car A trails car B by 50 meters. Car B travels at 45km/hr. Car C travels from the opposite direction at 54km/hr. Car C is at a distance of 220 meters from Car B. If car A decides to overtake Car B before cars B and C cross each other, what is the minimum speed at which car A must travel?
36 km/hr
45 km/hr
67.5 km/hr
18 km/hr

To begin with, let us ignore car A. Car B and car C travel in opposite directions.
Their relative speed
= Sum of the two speeds
= 45 + 54 kmph.
= 99 kmph.
= 99 * 5/18 m/s
= 55/2 m/s = 27.5 m/s.

The relative distance = 220m. So, time they will take to cross each other = 220/27.5 = 8 seconds.
Now, car A has to overtake car B within 8 seconds. The relative distance = 50m
=> Relative speed should be at least 50/8 m/s = 6.25 m/s
=> 6.25 * 18/5 kmph = 22.5kmph
Car B travels at 45kmph, so car A should travel at at least 45 + 22.5 = 67.5kmph.

A and B stand at distinct points of a circular race track of length 120m. They run at speeds of a m/s and b m/s respectively. They meet for the first time 16 seconds after they start the race and for the second time 40 seconds from the time they start the race. Now, if B had started in the opposite direction to the one he had originally started, they would have meet for the first time after 40 seconds. If B is quicker than A, find B’s speed.
3 m/s
4 m/s
5 m/s
8 m/s

This is practically just a logical reasoning question.
They meet for the first time 16 seconds after they start the race and for the second time 40 seconds from the time they start the race.
Now, we do not know their relative positions when they start the race. But we know that the time between the first and second meeting is 24 seconds. This is the time when they cover a relative distance of one lap length.

Laplength/RelativeSpeed = 24; Or relative speed = 5 m/s.

The question says – Now, if B had started in the opposite direction to the one he had originally started, they would have met for the first time after 40 seconds.

Now, B would have crossed each other after 40 seconds if B had reversed direction. This is higher than the 24 seconds it takes them to cover the relative distance of a lap in the first instance.

Or, in the first instance they were travelling towards each other.
Or, a + b = 5 m/s.

They meet for the first time 16 seconds after they start the race. Means, they would have travelled 16 x 5 = 80 m to meet for the first time and this also implies they were originally 80 m apart.

When B started in the opposite direction, A is 120 - 80 = 40 m ahead of B and it is told that B catches up with A in 40 seconds.
Relative speed = 40m/40s = 1 m/s
So we have b - a = 1

Solve for a and b, a = 2 m/s and b = 3 m/s. Answer is 3 m/s

City A to City B is a downstream journey on a stream which flows at a speed of 5km/hr. Boats P and Q run a shuttle service between the two cities that are 300 kms apart. Boat P, which starts from City A has a still-water speed of 25km/hr, while boat Q, which starts from city B at the same time has a still-water speed of 15km/hr. When will the two boats meet for the first time? (this part is easy) When and where will they meet for the second time?
7.5 hours and 15 hours
7.5 hours and 18 hours
8 hours and 18 hours
7.5 hours and 20 hours

When boat P travels downstream, it will effectively have a speed to 30kmph. Likewise, Q will have an effective speed of 10kmph. The relative speed = 40kmph. So, the two boats will meet for the first time after 300/40 hours (Distance/relative speed) = 7.5 hours (Actually, for this part we do not need the speed of the stream)

The second part is more interesting, because the speed of the boats change when they change direction. Boat P is quicker, so it will reach the destination sooner. Boat P will reach City B in 10 hours 300/30. When boat P reaches city B, boat Q will be at a point 100kms from city B.

After 10 hours, both P and Q will be travelling upstream,
P's speed = 20 km/hr
Q's speed = 10 km/hr
Relative speed = 10km/hr
Q is ahead of P by 100 kms

P will catch up with Q after 10 more hours Distance/RelativeSpeed−100/10.

So, P and Q will meet after 20 hours at a point 200 kms from city B.

A car of length 4m wants to overtake a trailer truck of length 20m travelling at 36km/hr within 10 seconds. At what speed should the car travel?
12 m/s
14.8 m/s
12.4 m/s
7.6 m/s

Given :
car of length = 4 m
length of truck = 20 m
Speed of truck = 36 km/hr

Approach:
Total distance to be covered by the car 'd' = 20 + 4 = 24 m
Let speed of car be S1 m/s
Speed of truck S2 = 36 km/hr = 10 m/s
Both the car and the truck are travelling in the same direction and we know that to overtake the truck, the speed of the ca should be more than that of the truck.
Hence, relative speed = |S1 - S2|
Time to overtake = 10s

S = d/t
|S1 - 10| = 24/10
S1 - 10 = 2.4
S1 = 10 + 2.4 = 12.4 m/s

Train A travelling at 63 kmph takes 27 to sec to cross Train B when travelling in opposite direction whereas it takes 162 seconds to overtake it when travelling in the same direction. If the length of train B is 500 meters, find the length of Train A.
400 m
810 m
500 m
310 m

Let the length of Train A be x meters
Let speed of Train B be y kmph
Relative distance = Relative speed * time taken to cross/overtake

Crossing scenario:
Relative speed of 2 trains = 63 + y
Time taken to cross = 27 sec or 27/3600 hrs
Relative distance between 2 trains = Length of Train A + length of train B = (x + 0.5) km
Therefore, x + 0.5 = (63 + y) * 27 / 3600 ----- (1)

Overtaking scenario:
Relative speed of 2 trains = 63 – y
Time taken to overtake = 162 sec or 162/3600 hrs
Relative distance between 2 trains = x + 0.5
Therefore, x + 0.5 = (63 – y) * 162/3600 --- (2)

From (1) and (2), solve for y.
(63 + y) * 27 = (63 – y) * 162
27y + 162 y = 63*162 – 63 *27
189y = 63 * 135 or y = 45 kmph

Substitute in (2) to get x.
x + 0.5 = (63 – 45) * 162/3600
Or x = 0.31 km or 310 meters

P cycles at a speed of 4 m/s for the first 8 seconds, 5 m/s for the next 8 seconds, 6 m/s for the next 8 and so on. Q cycles at a constant speed of 6.5 m/s throughout. If P and Q had to cycle for a 400 m race, how much lead in terms of distance, can P give Q and still finish at the same time as Q?
43.4 m
56.6 m
32.1 m
P cannot give a lead as Q is always ahead of P

Distance covered by P in 8 sec = 4 * 8 = 32 m
Distance covered by P in 16 sec = 32 + 5 *8 = 72 m
Distance covered by P in 24 sec = 72 + 6 * 8 = 120 m
Distance covered by P in 32 sec = 120 + 7 * 8 = 176 m
Distance covered by P in 40 sec = 176 + 8 * 8 = 240 m
Distance covered by P in 48 sec = 240 + 9 * 8 = 312 m
Total distance in 48 sec = 312m
To cover balance 86 m with speed = 10 m/sec, time taken = 86/10 = 8.6 sec
So P would finish 400 m in 48 + 8.6 = 56.6 seconds.

In 56.6 seconds, Q cycles 56.6 * 6.5 = 367.9 m or B should have a 32.1 m lead to result in a dead heat.

A bus starts from a bus stop P and goes to another bus stop Q. In between P and Q, there is a bridge AB of certain length. A man is standing at a point C on the bridge such that AC:CB = 1:3. When the bus starts at P and if the man starts running towards A, he will meet the bus at A. But if he runs towards B, the bus will overtake him at B. Which of the following is true?
Bus travels 3x times faster than the man
Bus travels 2x times faster than the man
The bus and the man travel at the same speed
4x the speed of the man is equal to 3x the speed of the bus

Let the speed of the bus be ‘b’ and man be ‘m’.
Let the distance between P and A be y.
AC:CB = 1:3 implies AC = x and CB = 3x

When bus goes from P to A and the man from C to A, time taken by both are equal and they meet at A.
y/b = x/m
or b/m = y/x --- (1)

When bus goes from P to B and the man from C to B, again time taken by both are equal and they meet at B.
(y+x+3x)/b = 3x/m
b/m = (4x+y)/3x ---(2)

Equating (1) and (2), we get y/x = (4x+y)/3x
3y = 4x + y or y = 2x

Therefore ratio of speeds b/m = 2x/x or 2 : 1

• Q 9 Circular Track of 120 m I Think the answer should be 3 m/s .

• @abhisharma Are you referrting to the question A and B stand at distinct points of a circular race track of length 120m. They run at speeds of a m/s and b m/s respectively... ?
Can you also share your approach please so that we can discuss based on that.

• @aneeeshp Suppose x is the gap between these two so x/a+b = 16 and when they meet for the second time 120/a+b = 24 . So a+b is 5 . In next case when they meet in 40 s .(L-x)/b-a = 40 so from 2 equations x is 80. So b-a come to be 1. and we have a+b = 5. So b=3 i.e. 3m/s

• Solution to the first question can be short and easy .
If the speeds are a and b ( a and b are the simplified ratios of the speed )
Then no of meetings if the person is moving in the same direction is given by a-b
And a-b should be equal be to 2 as they are meeting for 2 times
In this case a=6 we have to find the value of b .
So when b=2 then a:b = 3:1 therefore a-b = 2
Similarly for b=10 and 18

• @abhisharma This is correct and the answer should be 3 m/s only.
a + b = 5
b - a = 1
a = 2
b = 3

Even 2IIM video solution gives the same solution I feel the solution given here is not complete.

• @abhisharma Thanks :slight_smile:

Corrected the solution. Happy learning!

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