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Group Question

Answer the following questions based on the information given below.

A and B are running along a circular course of radius 7 km in opposite directionssuch that when they meet, they reverse their directions and when they meet, A willrun at the speed of B and vice versa. Initially, the speed of A is thrice the speed ofB. Assume that they start from M0 and they first meet at M1, then at M2, next at

M3, and finally at M4.

1.3 Marks

What is the point that coincides with M0 along the course?

1) M1

2) M2

3) M3

4) M4

Solution:

Let the circumference of the circular course be 4c km.

As A’s speed is thrice B’s, before they meet the first time at M1, A

travels 3c km and B travels c km.

Then they interchange speeds and directions and meet at M2, M3 and

M4 as shown.

∴ The meeting point M4 and the initial point will coincide.

Hence, option 4.

2.3 Marks

What is the shortest distance between M1 and M2 along the course?

1) 11 km

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Section I

2)

3) 7 km

4) 14 km

Solution:The shortest distance between M1 and M2 will be one fourth of the

circumference of the circle

Hence, option 1.

3.3 Marks

What is the shortest distance between M1 and M3 along the course?

1) 22 km

2)

3)

4) 14 km

Solution:The shortest distance between M1 and M3 will be half of the

circumference of the circle

Hence, option 1.

4.3 Marks

What is the total distance travelled by A when they meet at M3?

1) 77 km

2) 66 km

3) 99 km

4) 88 km

Solution:The total distance travelled by A till the third meeting point is = 33 + 11+ 33 = 77 kms

Hence, option 1.

5.3 Marks

A man travels three-fifths of a distance AB at a speed of 3a, and theremaining at a speed of 2b. If he goes from B to A and back at a speed of

remaining at a speed of 2b. If he goes from B to A and back at a speed of5c, it would take him the same time as he would have taken to go from A to Bpreviously. Then which of the following expressions must be true?

1)

2)

3)

4) None of these

Solution:

Let the total distance from A to be 1 unit.

Given data can be represented in the following equation, where the individualtimes are added up to obtain the total travel time:

Hence, option 3.

6.3 Marks

A man travels from A to B at a speed of x kmph. He then rests at B for xhours. He then travels from B to C at a speed of 2x kmph and rests for 2xhours. He moves further to D at a speed equal to twice of that taken to travelbetween B and C. He thus reaches D in 16 hours. If the distances A-B, B-C,C-D are all equal to 12 km, the time for which he rested at B could be _____.

1) 3 hours

2) 6 hours

3) 2 hours

4) 4 hours

Solution:

(i) From A to B, 12 kms

Waiting time at B = x

Waiting time at B = x

(ii) From B to C, 12 kms

Waiting time at C = 2x

(iii) From C to D, 12 kms

Total travel and wait time = 16 hrs

Solving the above equation for x, we get x = 3

Hence, option 1.

7.3 Marks

In a watch, the minute hand crosses the hour hand for the third time exactly afterevery 3 hours, 18 minutes, 15 seconds of watch time. What is the time gained orlost by this watch in one day?

1) 14 min 10 sec, lost

2) 13 min 50 sec, lost

3) 13 min 20 sec, gained

4) 14 min 40 sec, gained

Solution:

So, in a perfect watch, the minute hand crosses the hour hand for the third timeafter

It is given that in this particular watch, the minute hand crosses the hour hand for

the third time after 3 hrs. 18 minutes 15 seconds, i.e., after 198.25 minutes.

So, the time lost in 24 hrs is

= 13.833 minutes or 13 minutes 50 seconds

Hence, option 2.

Group Question


Persons X, Y and Z wish to go from place A to place B, which are separated by adistance of 70 km. All the three persons start off together from A, with X and Ygoing by Luna at a speed of 20 kmph. X drops Y somewhere along the way, andreturns to pick up Z, who has already started walking towards B, at a speed of 5kmph. Y, after being dropped by X, starts walking towards B at a speed of 5kmph. In this manner, all three of them reach B at the same time.

8.3 Marks

How much distance is covered by Z on foot?

1) 25 km

2) 10 km

3) 20 km

4) 15 km

Solution:

Let X and Y travel together on Luna for d km.

After this, X turns back. Let Z travel a distance of say, m before

meeting X.

As Z’s speed is one-fourth X’s, distance travelled by X is four times

that travelled by Z.

∴ d + d – m = 4m

∴ 2d = 5m

Equating (i) and (ii),

d = 50 km

∴ m = 20 km

∴ Z walks for 20 kms.

Hence, option 3.

9.3 Marks

What is the total distance travelled by X?

1) 130 km

2) 140 km

3) 110 km

4) 90 km

Solution:

X travels 50 + 30 + 30 + 20 = 130 kms.

Hence, option 1.

10.3 Marks

How long does it take to go from A to B?

1) 6.0 hours

2) 6.5 hours

3) 7.0 hours

4) 14.0 hours

Solution:

Solution:From equation (i) in the answer to the first question of this set, the time

taken to travel to B is

Hence, option 2.

11.3 Marks

After how much time is Y dropped on the way by X?

1) 2.0 hours

2) 3.0 hours

3) 2.5 hours

4) 1.5 hours

Solution:

Y was dropped after 50/20 = 2.5 hrs.

Hence, option 3.

12.3 Marks

For how long does X travel alone over the entire journey?

1) 2.5 hours

2) 1.0 hour

3) 2.0 hours

4) 1.5 hours

Solution:

X travels alone for (d – m) = 30 km at speed of 20 kmph.

∴ Time taken = 30/20 = 1.5 hours

Hence, option 4.

Group Question


A certain race is made up of three stretches A, B and C each 2 km long and to becovered by a certain mode of transport. The following table gives these modes oftransport for the stretches, and the minimum and maximum possible speeds (inkmph) over these stretches. The speed over a particular stretch is assumed to beconstant. The previous record for the race is ten minutes.

13.3 Marks

Anshuman travels at minimum speed by car over A and completesstretch B at the fastest speed. At what speed should he cover stretch Cin order to break the previous record?

1) Maximum speed for C

2) Minimum speed for C

3) This is not possible.

4) None of these

Solution:

Each stretch A, B and C is 2 kms long.

Stretch A: Minimum speed of car = 40 km/hr

Stretch B: Maximum speed of motorcycle = 50 km/hr

Stretch C:

To beat existing record, time taken for stretch C should be = 10 – 3 –2.4 = 4.6 mins.

If stretch C (2 kms) has to be covered in 4.6 mins, then the requiredspeed

But, the maximum speed of the bicycle is 20 km/hr, hence it is notpossible to break the record.

Hence, option 3.

14.3 Marks

Mr Hare completes the first stretch at the minimum speed and takesthe same time for stretch B. He takes 50% more time than the previousrecord to complete the race. What is Mr. Hare’s speed for the stretchC?

1) 10.9 kmph

2) 13.3 kmph

3) 17.1 kmph

3) 17.1 kmph

4) None of these

Solution:

Stretch A: Minimum speed of car = 40 km/hr.

Stretch B: Same time as stretch A is taken to cover stretch B.

Time taken to cover stretch B = 3 mins

Stretch C:

Total time taken is 50% more than existing record = 10 × 1.5 = 15 mins

Thus, time taken for stretch C should be = 15 – 6 = 9 mins

If stretch C (2 kms) has to be covered in 9 mins, then the requiredspeed

Hence, option 2.

15.3 Marks

Mr. Tortoise completes the race at an average speed of 20 kmph. Hisaverage speed for the first two stretches is 4 times that for the laststretch. Find the speed over stretch C.

1) 15 kmph

2) 12 kmph

3) 10 kmph

4) This is not possible.

Solution:

Overall average speed = 20 km/hr.

Total distance = 6 km/hr.

Total time taken = (6/20) = 0.3 hrs

Let the average speed over C be x km/hr, then the average speed over

A and B = 4x km/hr.

∴ x = 10 km/hr

Hence, option 3.

16.3 Marks

An express train travelling at 80 kmph overtakes a goods train, twiceas long and going at 40 kmph on a parallel track, in 54 seconds. Howlong will the express train take to cross a platform 400 m long?

1) 36 sec

2) 45 sec

3) 27 sec

4) None of these

Solution:

Let the length of the express train be denoted by L.

Relative speed = 80 – 40 = 40 km/hr

Total distance = L + 2L = 3L

Total time taken by the express train to overtake the goods train

∴ L = 200 m

While crossing a platform 400 m long:

Total distance travelled = 400 + 200 = 600 m

Hence, option 3.Group Question

Group Question


Boston is 4 hours ahead of Frankfurt and two hours behind India. X leavesFrankfurt at 06.00 p.m. on Friday and reaches Boston the next day. After waitingthere for two hours, he leaves exactly at noon and reaches India at 01.00 a.m. Onhis return journey, he takes the same route as before, but halts at Boston for onehour less than his previous halt there. He then proceeds to Frankfurt.

17.3 Marks

If his journey, including stoppage, was covered at an average speed of180 miles per hour, what was the distance between Frankfurt andIndia?

1) 3600 miles

2) 4500 miles

3) 5580 miles

4) Insufficient data

Solution:

The average speed of the journey including stoppage = 180 km/hr

Total journey distance (from Frankfurt to India) = 180 × 25 = 4500 miles

Hence, option 2.

18.3 Marks

If X had started the return journey from India at 02:55 a.m. on the sameday that he reached there, after how much time would he reachFrankfurt?

1) 24 hrs

2) 25 hrs

3) 26 hrs


Solution:

Return journey is 1 hr less than the forward journey, thus return journeyis of 24 hrs.

Hence, option 1.

19.3 Marks

What was X’s average speed for the entire journey (to and fro)?

1) 176 mph

2) 180 mph

3) 165 mph



Solution:

Since the distance is not given in the question, we cannot find theaverage speed of the entire journey.

Hence, option 4.

Group Question


A thief, after committing a burglary, started fleeing at 12:00 noon, at the speed of60 kmph. He was then chased by a policeman X. X started the chase 15 minutesafter the thief had started, at a speed of 65 kmph.

20.3 Marks

At what time did X catch the thief?

1) 3 : 30 p.m

2) 3 : 00 p.m.

3) 3 : 15 p.m.

4) None of these

Solution:

Relative speed of the policeman X and the thief = 65 – 60 = 5 km/hr.

Hence, police X shall catch the thief at 3:15 p.m.

Hence, option 3.

21.3 Marks

If another policeman had started the same chase along with X, but at aspeed of 60 kmph. Then how far behind was he when X caught thethief?

1) 18.75 km

2) 15 km

3) 21 km

4) 37.5 km

Solution:The distance by which the other policeman would be behind

The distance by which the other policeman would be behind

= Distance travelled by X – Distance travelled by the other policeman.

= 3 × 65 – 3 × 60 = 15 km

Hence, option 2.

22.3 Marks

The distance between A and B is 72 km. Two men started walking from Aand B at the same time towards each other. The person who started from Atravelled uniformly with an average speed of 4 kmph. The other man travelledwith varying speeds as follows: In the first hour his speed was 2 kmph, in thesecond hour it was 2.5 kmph, in the third hour it was 3 kmph, and so on.When will they meet each other?

1) 7 hours

2) 10 hours

3) 35 km from A

4) Midway between A and B

Solution:

Let speeds of persons starting with A and B be denoted by va and vb

respectively.

Here, va = 4 km/hr and vb = 1.5 + 0.5t km/hr, where t is the number of hours

of journey required for A and B to meet.

Thus, the relative speed of A and B = 5.5 + 0.5t

Distance travelled in t hours = 72 kms

Solving for t, we get t = 9 hrs.

In 9 hrs, by the time they meet, person from A would have travelled (9 × 4) =36 kms, which is exactly mid-way between A and B.

Hence, option 4.

23.3 Marks

Three wheels can complete 60, 36, 24 revolutions per minute respectively.There is a red spot on each wheel that touches the ground at time zero. Afterhow much time, will all these spots simultaneously touch the ground again?

1)

2)

3) 5 seconds

4) 7.5 seconds

Solution:

Given: 3 wheels complete 60, 36 and 24 revolutions per minute.

Thus, the time after which the red spots on all the three wheels meet wouldbe the common multiple of the times taken by each of the 3 wheels

Hence, option 3.

24.3 Marks

I started climbing up the hill at 6 a.m. and reached the temple at the top at 6p.m. The next day I started coming down at 6 a.m. and reached the foothill at6 p.m. I walked on the same road. The road is so short that only one personcan walk on it. Although I varied my pace along the way, I never stopped onmy way. On the basis of this, which of the following must be true?

1) My average speed downhill was greater than that uphill.

2) At noon, I was at the same spot on both the days.

3) There must be a point which I reached at the same time on boththe days.

4) There cannot be a spot which I reached at the same time on boththe days.

Solution:Consider Option 1: ‘Average speed downhill is greater than that of uphill.’

In both cases, same distance was covered in the same duration of time.Thus, average speed uphill and downhill is the same. Hence, the givenstatement is false.

Consider Option 2: ‘At noon, I was at the same spot on both the days.’

Since speeds could vary within the journey, it is not necessary that the personis at the same spot on both the days at noon. Hence, the given statement isfalse.

Consider Option 3: ‘There must be a point which I reached at the sametime on both the days.’

Assume that two people start walking at 6:00 a.m., one from the top of the hilland one from the bottom of the hill and reach their respective destinations at6:00 p.m.

It is obvious that they will meet at some point, which means that they willreach the same point at the same time. Extending the same logic, there hasto be a point at which both reached at the same time on both the days.

Thus option 3 is true and option 4 is false.

Hence, option 3.

25.3 Marks

Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6:30am and travels at 50 km per hour towards Baroda situated 100 kms away. At7:00 am Howrah - Ahmedabad Express leaves Baroda towards Ahmedabadand travels at 40 kms per hour. At 7:30 Mr. Shah, the traffic controller atBaroda realises that both the trains are running on the same tack. How muchtime does he have to avert a head-on collision between the two trains?

1) 15 minutes

2) 20 minutes

3) 25 minutes

4) 30 minutes

Solution:

Let Navjivan Express, Howrah Express, Ahmedabad and Baroda be

denoted by N.E, H.E., A and B respectively.

N.E. leaves A at 6.30 a.m. moving at 50 km/hr.

By 7 AM, it would have moved 50 × 1/2 = 25 kms from A towards B. So, it isat a distance of 100 – 25 = 75 kms from B.

H.E. leaves B towards A at 40 km/hr.

Now, relative speed = 40 + 50 = 90 km/hr.

Distance between the two trains at 7 a.m. = 75 kms.

Time taken for the two trains to meet = 75/90 = 5/6 hrs = 50 minutes

So, the time would be 50 minutes after 7 a.m. = 7.50 a.m.

Hence, the traffic controller has another (50 – 30) = 20 mins to stop thecollision.

Hence, option 2.

Group Question


A road network (shown in the figure below) connects A, B, C and D. All thesegments are straight lines. D is the midpoint on the road connecting A and C.Roads AB and BC are at right angles to each other with BC shorter than AB. Thesegment AB is 100 km long.

Ms. X and Mr. Y leave A at 8:00 a.m., take different routes to city C and reach atthe same time. X takes the highway from A to B to C and travels at an averagespeed of 61.875 km per hour. Y takes the direct route AC and travels at 45 kmper hour on segment AD. Y’s speed on segment DC is 55 km per hour.

26.3 Marks

What is the average speed of Y in km per hour?

1) 47.5

2) 49.5

3) 50

4) 52

Solution:

Hence, option 2.

27.3 Marks

The total distance travelled by y during the journey is approximately

1) 105 km

2) 150 km

3) 130 km

4) Indeterminate

Solution:

Solution:

AB2 + BC2 = AC2

Solving,

AC ≈ 105 km

Hence, option 1.

28.3 Marks

What is the length of the road segment BD?

1) 50 km

2) 52.5 km

3) 55 km

4) Indeterminate

Solution:

BD = 105/2 = 52.5 km.

Hence, option 2.

Group Question


Rajiv reaches city B from city A in 4 hours, driving at the speed of 35 km per hourfor the first 2 hours and at 45 km per hour for the next two hours.

Aditi follows the same route, but drives at three different speeds: 30, 40 and 50km per hour, covering an equal distance in each speed segment. The two carsare similar with petrol consumption characteristics (km per litre) shown in thefigure below.

29.3 Marks

The amount of petrol consumed by Aditi for the journey is

1) 8.3 litres

2) 8.6 litres

3) 8.9 litres

4) 9.2 litres

Solution:Total distance = 35 × 2 + 45 × 2 = 70 + 90 = 160 km.

Distance travelled by Aditi at each different speed = 160/3

Hence, option 3.

30.3 Marks

Zoheb would like to drive Aditi’s car over the same route from A to Band minimize the petrol consumption for trip. The amount of petrolrequired by him is

1) 6.67 litres

2) 7 litres

3) 6.33 litres

4) 6.0 litres

Solution:

Hence, option 1.

31.3 Marks

Arjun is traveling from Andheri to Dadar by car and Bharat is traveling fromDadar to Andheri by bike (on the same road). Speed of Arjun and Bharat is60 kmph and 15 kmph. Arjun and Bharat meet at Bandra, somewherebetween Andheri and Dadar. After reaching Dadar Arjun takes a rest of 1hour and then return to Andheri. The total time taken by Arjun to travel fromAndheri to Dadar and then from Dadar to Andheri (including the halt) is 10minutes more than the time required by Bharat to travel from Bandra toAndheri. What could be the distance between Andheri and Dadar?

1) 20 km

2) 10 km

3) 12.5 km

4) 8.33 km

5) 15 km

Solution:

Let the distance between Andheri and Bandra be x km and the distance

between Bandra and Dadar be y km.

∴ 2(x + y) + 60 = 4x + 10

∴ 2x − 2y = 50

∴ x − y = 25,

Now ratio of speed of Arjun and Bharat is,

Ratio of distance travel by Arjun and Bharat is same as ratio of speed ofArjun and Bharat.

∴ x = 4y

∴ 4y ‒ y = 25

y = 8.33 km

Hence, option 4.

Hence, option 4.

32.3 Marks

Moreshwar and Ganesh started travelling towards each other from theirhometowns, Hyderabad and Bangalore respectively. They met at point P inbetween for the first time. As soon as they met, they exchanged their cars(which could travel with their predefined speeds only) and turned back totravel towards their respective hometown cities. As soon as they reachedtheir hometowns, they again started travelling back towards the other city andmet at point Q for the second time. Note that after meeting at point P they didnot meet each other before they reached their respective hometown cities.What was the ratio of their speeds such that the distance PQ was thehighest?

1) 2 : 5

2) 2 : 1

3) 2 : 3

4) 5 : 6

5) 1 : 3

Solution:

Let the speed of Moreshwar and Ganesh be u and v respectively. Theirrelative speed was u + v and the time they took for the first meeting was d/(u

+ v) where d is the total distance between the two cities. Therefore, thedistance of the point P from Hyderabad (H), where they met for the first timeis

Similarly, after meeting they exchange their cars and start travelling back totheir own cities. Now the condition given is that they would not meet eachother before reaching their respective cities. Let us say that u > v. (We willget similar results and the same ratio even if we consider v > u) This meansthat Ganesh would not overtake Moreshwar before Moreshwar reachesHyderabad.

This distance is not more than the distance between the point P to Bangaloreand back to Hyderabad.

Solving this we get,

(u + v)(u − 2v) < 0

Therefore u has to lie between −v and 2v. But since we have assumed u > v,

the allowed interval for u is v to 2v, both inclusive.

Now we need to find the distance of the point Q from Hyderabad so that wecan find the distance PQ.

After leaving point P and meeting again at point Q they would have travelled

a distance of 2d and their relative speed would be same as (u + v).

This is same as the distance between point P and Hyderabad and thedistance between Hyderabad and point Q.

Now, l(PQ) = l(HP) − l(HQ)

Solving for l(PQ) from equations (i) and (ii) we get,

Looking at the expression, we can see that l(PQ) is an increasing functionwith the value of u. Therefore given the range of u, we would see that thevalue of l(PQ) is maximum when u = 2v.

Hence, option 2.

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