Quiz 4

Let track length be equal to T.
Time taken to meet for the first time = TrelativespeedTrelativespeed = T6−bT6−b or Tb−6Tb−6
Time taken for a lap for A = T6T6
Time taken for a lap for B = TbTb
So, time taken to meet for the first time at the starting point = LCM (T6T6,TbTb) = THCF(6,b)THCF(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−bHCF(6,b)6−bHCF(6,b) = 2 or b−6HCF(6,b)b−6HCF(6,b) = 2

The question is If two people 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?
b = 2, 10, 18 satisfy this equation. So, there are three different values that b can take.

Hence, the answer is 3.

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.

Or, Jagadeesh travels 52a to reach H. Answer choice (a)

The question is If l : k : j is 1: 3 : 5√2 then the distance travelled by Jagadeesh is

Hence, the answer 7.5 × √2 times the side of the square ABCD

visual of where the three cars are and how they travel between the points.

Let V1, V2 and V3 be the speeds of the cars.

ABV1ABV1 - ABV2ABV2 = ABV2ABV2 - ABV3ABV3
240V1240V1 - 240V2240V2 = 1
V₃ = 2V₁

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 ABV1ABV1 - ABV2ABV2 = ABV2ABV2 - ABV3ABV3
As the second car arrived at C an hour earlier than the first, we get a second equation
240V1240V1 - 240V2240V2 = 1
The third car covered 240 + 80 kms when the first one covered 240 – 80 kms. Therefore, 320V3320V3 = 160V1160V1
This gives us V₃ = 2V₁
From condition 1, we have ABV1ABV1 - ABV2ABV2 = ABV2ABV2 - ABV3ABV3
Substituting V₃ = 2V₁, this gives us ABV1ABV1 - ABV2ABV2 = ABV2ABV2 - AB2V1AB2V1
or 3AB2V13AB2V1 = 2ABV22ABV2 or V₂ = 4V134V13
Solving 240V1240V1 - 240V2240V2 = 1 , we get 60V160V1 = 1 or V₁ = 60 kmph
=> V₂ = 80 kmph and V₃ = 120 kmph

The question is What is the difference between the speed of the first and the third car?

Hence, the answer is 60 kmph

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 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 question is \" Find the ratio of the speeds of A, B and C?\"
The ratio of the speeds of A, B and C is either 5 : 3 : 1 or 5 : 4 : 3.

Hence, the answer is 5 : 4 : 3

How time is related to speed?

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)Td(d−30)Td hr. However, this is only if he travels at his usual speed. It is given that he traveled only at 4545^th of his usual speed. Because of this he would have taken 5454^th of the time to travel the remaining distance, i.e., he takes 1414^th of the time extra. This is given to be 45 minutes (or 3434^th hr)
1414(d−30)Td(d−30)Td = 3434     ........(1)

The question is What is the distance between Delhi and Gurgaon? we get d = 120 km.

Hence, the answer is d = 120 km.

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 traveled by car A before meeting car B = a * t. Likewise distance traveled by car B before meeting car A = b * t.
Distance traveled by car A after meeting car B = a * 54. Distance traveled by car B after meeting car A = 24 * b.
Distance traveled by car A after crossing car B = distance traveled 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 * t² = 54 * 24 * ab
=> t² = 54 * 24
=> t = 36

The question is  How long did B take to cover the entire journey between City Q and City P?
So, both cars would have traveled 36 minutes prior to crossing each other. Or, B would have taken 36 + 24 = 60 minutes to travel the whole distance.

Hence, the answer is 60 mins.

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 503503 = 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

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

So, the question is, how many times will they meet before B finishes the race? 

Hence, they meet 3 times before B finishes the race.

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 = 99kmph.
= 99 * 518518m/s = 552552m/s = 27.5m/s

The relative distance = 220m. So, time they will take to cross each other = 518518 = 8 seconds

Now, car A has to overtake car B within 8 seconds. The relative distance = 50m
=> Relative speed should be at least 508508m/s = 6.25 m/s.
=>6.25 * 185185kmph = 22.5 kmph

The question is  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?
Car B travels at 45kmph, so car A should travel at at least 45 + 22.5 = 67.5kmph.

Hence, the answer is 67.5kmph.

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.

LaplengthRelativeSpeedLaplengthRelativeSpeed = 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.

The question is If B is quicker than A, find B’s speed.

Hence, the answer is 3 m/s

Boat P starts from city A and Q starts from city B.
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 3004030040 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 3003030030. 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 DistanceRelativeSpeedDistanceRelativeSpeed − 1001010010.
So, P and Q will meet after 20 hours at a point 200 kms from city B

The question is  When will the two boats meet for the first time? When and where will they meet for the second time?

Hence, the answer is 7.5 hours and 20 hours