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Section I

Group Question

Answer the following following questions questio ns based on the the inf i nformation ormation given gi ven below. A and B are a re runnin running g along a circul circ ular ar course of radius radi us 7 km in opposite opposi te directions direc tions such that when they meet, they reverse their directions and when they meet, A will run at the speed of o f B and a nd vice vice versa. Initially Initially,, the the speed spe ed of A is thrice thrice the the speed spe ed of B. 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 that coincides coi ncides with wi th M0 along the course? 1)

M1

2)

M2

3)

M3

4)

M4

Solution: Let the circumference circumference of the circul ci rcular ar 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 travels c km.

Then Then they they interchange interchange speeds speed s and directions di rections and meet at M 2, M3 and and M4 as shown. ∴

The The meeting meeti ng point M4 and the the initial i nitial point poi nt will coincide. coincide .

Hence, Hence, option opti on 4. 2. 3 Marks

What is the shortest distance between M1 and M2 along the course? 1)

11 km

2) 3)

7 km

4)

14 km

Solution: The The shortest s hortest distance between M1 and M2 will be one fourth of the circumference of the the circle ci rcle


What is the shortest distance between M1 and M3 along the course? 1)

22 km

2) 3) 4)

14 km

Solution: The The shortest s hortest distance between M1 and M3 will be half of the circumference of the the circle ci rcle


What is the total distance travelled by A when they meet at M 3? 1)

77 km

2)

66 km

3)

99 km

4)

88 km

Solution: The The total distance di stance travelled travelled by A till ti ll the the third meeting meeti ng point is = 33 + 11 11 + 33 = 77 kms


A man travels travels three-fifths three-fifths of a distance dis tance AB at a speed of 3a, and the

rema n ng a a spee o . e goes rom o an ac a a spee o 5c , it would take him the same time as he would have taken to go from A to B previously. 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 i n the the following equation, where the the individual i ndividual times are added up to obtain the the total travel travel time:


x kmph. He then rests at B for x x A man travels from A to B at a speed of x hours. hours. He then travels travels from B to C at a speed sp eed of o f 2 x kmph and and rests for 2 x hours. hours. He moves further further to D at a speed equal to twice of that taken to travel between B and C. He thus thus reaches D i n 16 hours. hours. If If the the distances di stances A-B, A-B , B-C, B-C , C-D are all equal to 12 km, the the time for which he he rested at a t B could be _____. ___ __.

1)

3 hours

2)

6 hours

3)

2 hours

4)

4 hours

Solution: (i) From A to B, 12 kms

Waitin aitin time ime at at B = x

(ii) From B to C, 12 kms

Waiting time at C = 2 x (iii) (iii ) From C to D, 12 kms

Total travel and wait time = 16 hrs

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

Hence, Hence, option opti on 1. 7. In a watch, the minute hand crosses the hour hand for the third time exactly after every 3 hours, hours, 18 minu mi nutes, tes, 15 seconds s econds of watch time. What is the time gained g ained or lost by this watch in one day?

3 Marks

1)

14 min min 10 se sec, lost

2)

13 min min 50 se sec, lost

3)

13 min 20 sec, sec, gain gained ed

4)

14 min 40 sec, sec, gain gained ed

Solution:

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

It is given that in this particular watch, the minute hand crosses the hour hand for the the third time after 3 hrs. 18 minu mi nutes tes 15 seconds, i.e., i .e., after 198.25 minu mi nutes. tes.

So, the time lost in 24 hrs is

= 13.833 minu mi nutes tes or 13 minu mi nutes tes 50 seconds s econds Hence, option 2. Group Question

Answer the following following questions questio ns based on the the inf i nformation ormation given gi ven below. Persons Perso ns X, Y and Z wish wish to go from place A to place B, B , which which are separated se parated by a distance dis tance of 70 km. k m. All the the three persons start off together from A, with X and a nd Y going goi ng by Luna Luna at a speed of 20 kmph. X drops drop s Y somewhere somewhere along the way, and and returns returns to pick pic k up Z, who who has already started walking towards B , at a speed s peed of 5 kmph. Y, Y, after being dropped dropp ed by X, starts walking towards B at a speed spee d of 5 kmph. In this manner, all three of them reach B at the same time. 8.

How much much distance is covered by Z on foot?

3 Marks

1)

25 km

2)

10 km

3)

20 km

4)

15 km

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

After this, X turn turns s back. ba ck. Let Le t Z travel a distance of say, sa y, m before meeting X . As Z ’s ’s speed is one-fourth X ’s, ’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, Hence, option op tion 3. 9.

What is the total distance di stance travelled travelled by b y X ?

3 Marks

1)

130 km

2)

140 km

3)

110 km

4)

90 km

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

Hence, Hence, option opti on 1. 10.

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

3 Marks

1)

6.0 hours

2)

6.5 hours

3)

7.0 hours

4)

14.0 hours

From equation (i) in the the answer to the first question of this set, the time taken to travel to B is


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

3 Marks

1)

2.0 hours

2)

3.0 hours

3)

2.5 hours

4)

1.5 hours

Solution: Y was dropped droppe d after 50/20 = 2.5 hrs.


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

3 Marks

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 hours

Hence, Hence, option opti on 4. Group Question

Answer the following following questions questio ns based on the the inf i nformation ormation given gi ven below.

A certain ce rtain race is i s made up of three stretches stretches A, B and C each eac h 2 km long long and to be covered by a certain ce rtain mode of transport. The The following table gives these modes of transport for the stretches, stretches, and the minimum and and maximum possible possi ble speeds spe eds (in kmph) over over these stretches. The The speed spe ed over a particular p articular stretch is assum ass umed ed to be be constant. The previous record for the race is ten minutes.

13. 3 Marks

Anshum Anshuman an travels travels at minimu mi nimum m speed spee d by car over A and a nd completes stretch B at the fastest speed. spee d. At what speed should should he cover stretch C in order to break brea k the previous previous record? 1)

Maxim aximu um spee speed d for for C

2)

Minim inimu um spee speed d for C

3)

This is not not poss possib iblle.

4)

None of these

Solution: Each stretch A, B and C is 2 kms long. Stretch A: Minimum speed of car = 40 km/hr km/hr

Stretch B: Maximum Maximum speed spee d of motorcycle motorcycle = 50 km/h k m/hr r

Stretch C:

To beat existing record, time ti me taken for stretch C shou s hould ld be = 10 – 3 – 2.4 = 4.6 mins. If stretch C (2 kms) has to be b e covered co vered in 4.6 mins, mi ns, then then the the required speed

But, the the maximum speed of the bicycl bic ycle e is i s 20 km/hr, km/hr, hence hence it i t is not possible to break the record. Hence, Hence, option opti on 3. 14. 3 Marks

Mr Hare completes the first stretch s tretch at the the minimum speed and a nd takes the the same time for stretch s tretch B. He takes 50% 5 0% more time ti me than the the previous record to complete c omplete the the race. What is Mr. Hare’s speed for the the stretch C? 1)

10.9 kmph

2)

13.3 kmph

3

17.1 km h

4)

None of these

Solution: Stretch A: Minimum speed of car = 40 km/hr. 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 time taken take n is 50% more than existi existing ng record = 10 × 1.5 = 15 1 5 mins Thus, Thus, time taken for stretch C should should be = 15 – 6 = 9 mins If stretch C (2 kms) has to be b e covered co vered in 9 mins, then then the the required req uired speed


Mr. Tortois Tortoise e completes the race at an average speed spee d of 20 kmph. k mph. His average speed spee d for the first two stretches stretches is i s 4 times ti mes that for for the last stretch. Find the the speed spe ed over stretch C. 1)

15 kmph

2)

12 kmph

3)

10 kmph

4)

This is not not poss possib iblle.

Solution: Overall Ov erall average speed spe ed = 20 km/h k m/hr. r.

Total distance = 6 km/hr. 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 = 4 x km/hr.

= 10 km/hr

∴ x


An express train travelling travelling at 80 kmph overtakes overtakes a goods goo ds train, trai n, twice twice as long and going at 40 kmph on a parallel track, track, in i n 54 seconds. How long long will the express express train take to cross c ross 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 km/hr

Total distance = L + 2L = 3L Total time taken by the the express train trai n to overtake overtake the goods good s train trai n

∴

L = 200 m

While crossing crossi ng a platform 400 m long: long: Total distance travelled travelled = 400 + 200 = 600 m

Hence, Hence, option opti on 3.

Answer the following following questions questio ns based on the the inf i nformation ormation given gi ven below. Boston is 4 hours ahead of Frankfurt and two hours behind India. X leaves Frankfurt Frankfurt at 06.00 06 .00 p.m. on Friday Fri day and reaches Boston B oston the the next day. day. After waiting wai ting there for two hours, he leaves exactly at noon and reaches India at 01.00 a.m. On his return journey, he takes the same route as before, but halts at Boston for one hour less than his previous halt there. He then proceeds to Frankfurt. 17. 3 Marks

If his journey, journey, including including stoppage, stop page, was covered at an average speed spee d of 180 miles per hour, what was the distance between Frankfurt and India? 1)

3600 mi les

2)

4500 mi les

3)

5580 mi les

4)

Insuf sufficien icientt dat 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, Hence, option opti on 2. 18. 3 Marks

If X had started the return journey from India at 02:55 a.m. on the same day that he reached there, after how much time would he reach Frankfurt? 1)

24 hrs

2)

25 hrs

3)

26 hrs

4)

Insuf sufficien icientt dat data

Solution: Return journey journey is 1 hr less les s than the forward forwa rd journ jo urney, ey, thus thus return re turn journey journey is of 24 hrs.


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

3 Marks

1)

176 mph

2)

180 mph

3)

165 mph

Solution: Since Si nce the the distance di stance is not given in the the question, questio n, we we cann ca nnot ot find the average speed sp eed of the entire journey. journey.


Answer the following following questions questio ns based on the the inf i nformation ormation given gi ven below. A thief, after committing commi tting a burglary, burglary, started fleeing fleei ng at 12:00 noon, at the the speed spe ed of 60 kmph. He was then chased chased by a policeman police man X. X started s tarted the chase 15 minu mi nutes tes after the the thief had had started, at a speed of 65 kmph. 20.

At what time did di d X catch the the thief?

3 Marks

1)

3 : 30 p.m

2)

3 : 00 p.m.

3)

3 : 15 p.m.

4)

None of these

Solution:

Relative speed of o f the the policeman poli ceman X and the thief thief = 65 6 5 – 60 6 0 = 5 km/hr. km/hr.

Hence, Hence, police poli ce X shall catch the the thief at 3:15 p.m. Hence, Hence, option opti on 3. 21. 3 Marks

If another policeman had started the same chase along with X, but at a speed of 60 kmph. Then how far behind was he when X caught the thief? 1)

18.75 km

2)

15 km

3)

21 km

4)

37.5 km

Solution:

= Distance Di stance travelled travelled by X – Distance Di stance travelled travelled by the the other policeman. police man. = 3 × 65 – 3 × 60 = 15 km Hence, Hence, option opti on 2. 22. 3 Marks

The The distance di stance between A and B is 72 km. Two men started walking from A and B at the same time ti me towards each ea ch other. other. The The person pers on who who started from A travelled uniformly with an average speed of 4 kmph. The other man travelled with varying speeds as follows: In the first hour his speed was 2 kmph, in the second 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 idway bet between een A an and B

Solution: Let speeds of persons starting with A and B be denoted by v a and and v b respectively.

Here, v a = 4 km/hr and v b = 1.5 + 0.5t km/hr, where t is the number of hours of journey journey required for A and B to meet. mee t. Thus, Thus, the the relative speed spe ed of A and B = 5.5 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, Hence, option opti on 4. 23. 3 Marks

Three Three wheels can ca n complete 60, 36, 24 2 4 revolutions revolutions per p er minu mi nute te respectively resp ectively.. There is a red spot on each wheel that touches the ground at time zero. After how 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 would be the common multiple of the times taken by each of the 3 wheels


I started climbing up the hill at 6 a.m. and reached the temple at the top at 6 p.m. The next day I started coming down at 6 a.m. and reached the foothill at 6 p.m. I walked on the same road. The road is so short that only one person can walk on it. Although I varied my pace along the way, I never stopped on my way. On the basis of this, which of the following must be true? 1)

My average average speed speed downh downhill was was greater greater than than th that uph uphill ill..

2)

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

3)

There mu must be a point point wh which I reached reached at th the same time time on both the days.

4)

There cann cannot be a spot wh which I reached reached at the the same time time on both both the days.

Solution: Consider Option 1: ‘Average ‘A verage speed spee d down do wnhill hill is greater g reater than that that of uphill.’ uphill.’

In both cases, same distance d istance was covered in the same duration of time. Thus, average speed uphill and downhill is the same. Hence, the given statement 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 person is at the same spot on both the days at noon. Hence, the given statement is false.

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

Assume that two people start s tart walking at 6:00 a.m., one from the the top of the hill and one from the the bottom bo ttom of the hill and and reach rea ch their their respective respec tive destinat desti nations ions at 6:00 p.m. It is obvious that they will meet at some point, which means that they will reach the the same point poi nt at the the same time. Extending the the same logic, there has to be a point poi nt at which which both reached reached at the same time ti me on both the the days. Thus Thus option 3 is i s true and option 4 is i s false. Hence, Hence, option opti on 3. 25. 3 Marks

Navjivan Navjivan Express Express from Ahmedabad to Chennai Chennai leaves Ahmedabad at 6:30 am and travels travels at 50 km per pe r hour hour towards Baroda Ba roda situ si tuated ated 100 kms away. At 7:00 am Howrah - Ahmedabad Express leaves Baroda towards Ahmedabad and travels travels at 40 4 0 kms per hour. At 7:30 Mr. Shah S hah,, the traffic controller at Baroda Barod a realises reali ses that that both the trains are running running on the same tack tack.. How much time does d oes he have have to avert a head-on collision collisi on between the the two trains? trai ns? 1)

15 mi nutes

2)

20 mi nutes

3)

25 mi nutes

4)

30 mi nutes

Solution: Let Navjivan Navjivan Express, Howrah Express, Ahm A hmedabad edabad and Baroda be

denoted by N.E, H.E., A and B respecti re spectivel vely. y. N.E. leaves A at 6.30 6 .30 a.m. moving at 50 km/hr. By 7 AM, it would have have moved moved 50 × 1/2 = 25 kms from A towards B. B . So, it i t is at a distance of 100 – 25 = 75 kms from B. H.E. leaves leaves B towards A at a t 40 km/hr. Now, relative relative speed = 40 + 50 = 90 9 0 km/hr. Distance Di stance between the the two trains at 7 a.m. = 75 kms. Time taken for the tw two o trains to meet mee t = 75/90 = 5/6 hrs = 50 minu mi nutes tes So, the time would be 50 minutes minutes after 7 a.m. = 7.50 a.m. Hence, Hence, the traffic control c ontroller ler has another another (50 – 30) 30 ) = 20 mins to stop the the collision.


Answer the following following questions questio ns based on the the inf i nformation ormation given gi ven below. A road roa d network (shown (shown in the the figure fig ure below) connects connects A, A , B, C and D. All A ll the the segments are straight strai ght lines. lines. D is i s the midpoint midpoi nt on the the road connecting connecting A and a nd C. Roads AB A B and a nd BC are at right ri ght angles angles to each ea ch other other with BC shorter than than AB. The segment AB is 100 km long. Ms. X and a nd Mr. Y leave A at 8:00 a.m., take different routes routes to city C and a nd reach at the the same time. X takes the highway from A to B to C and travels travels at an average speed of 61.875 km k m per hour. hour. Y takes the direct direc t route route AC and travels travels at 45 km per hour hour on segment AD. Y’s Y’s speed on segment DC is i s 55 km k m per hour. hour.

26.

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

3 Marks

1)

47.5

2)

49.5

3)

50

4)

52

Solution:


27.

The total distance travelled by y during the journey is approximately

3 Marks

1)

105 km

2)

150 km

3)

130 km

4)

Indeterm ermina inate

AB 2 + BC2 = AC2

Solving, AC ≈ 105 km Hence, Hence, option opti on 1. 28.

What is the length of the road segment BD?

3 Marks

1)

50 km

2)

52.5 km

3)

55 km

4)

Indeterm ermina inate

Solution: BD = 105/2 = 52.5 km.


Answer the following following questions questio ns based on the the inf i nformation ormation given gi ven below. Rajiv reaches city B from ci ty A in 4 hours, hours, drivin dri ving g at the speed of 35 km k m per hour for the first 2 hours and at 45 km per hour for the next two hours. Aditi Adi ti follows the the same route, but drives drives at three different speeds: 30, 30 , 40 and 50 km per hour, hour, covering an equal distance i n each speed segment. The The two cars ca rs are similar si milar with petrol consumption consumption characteristics characteristi cs (km per p er litre) shown in the the figure below.

29.

The amount of petrol consumed by Aditi for the journey is

3 Marks

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


Zoheb Zoheb would like to drive dri ve Aditi’s Aditi ’s car c ar over the the same route from A to B and minimize the petrol consumption for trip. The amount of petrol required by him is 1)

6.67 li tres

2) 3)

7 li tres 6.33 li tres

4)

6.0 litres

Solution:


31. 3 Marks

Arjun is traveling traveling from Andheri to Dadar D adar by b y car and Bharat is traveling traveling from Dadar Dad ar to Andheri by bike (on (o n the the same road). roa d). Speed Spee d of Arjun and Bharat is 60 kmph k mph and and 15 kmph. Arjun Arjun and and Bharat B harat meet at Bandra, B andra, somewhere somewhere between Andheri Andheri and a nd Dadar. After A fter reaching Dadar Arju A rjun n takes takes a rest of 1 hour and then return to Andheri. The total time taken by Arjun to travel from Andheri Andheri to Dadar Da dar and then from from Dadar Da dar to Andh A ndheri eri (including (including the halt) halt) is 10 10 minutes more than the time required by Bharat to travel from Bandra to Andheri. Andheri. What could be the distance between be tween Andheri Andheri and Dadar? Da dar? 1)

20 km

2)

10 km

3)

12.5 km

4)

8.33 km

5)

15 km

Solution:

Let the distance between be tween Andheri Andheri and Bandra be x km and the distance between Bandra and Dadar be y km.

∴

x + y ) + 60 = 4 x + 10 2( x

∴

2 x − 2y = 50 − y = 25,

∴ x

Now ratio of speed spe ed of Arju A rjun n and and Bharat is, is ,

Ratio of distance di stance travel travel by Arjun Arjun and and Bharat is same as ratio of speed s peed of o f Arjun and Bharat.

= 4y

∴ x

∴

4y ‒ y = 25

y = 8.33 km

, 32. 3 Marks

.

Moreshwar and Ganesh started travelling towards each other from their hometowns, hometowns, Hyderabad Hyderabad and Bangalore respecti res pectivel vely. y. They They met met at point p oint P in between for the first time. As soon as they met, they exchanged their cars (which could travel with their predefined speeds only) and turned back to travel travel towards their respective resp ective hometown cities. citi es. As soon as they reached their hometowns, they again started travelling back towards the other city and met at point poi nt Q for for the second time. Note that after meeting at poi nt P they they did not meet each other before they reached their their respective respecti ve hometown hometown cities. cities . What was the the ratio rati o of their their speeds speed s such that that the distance dis tance PQ was the the highest? 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 and v respectively respec tively.. Their Their relative speed was u + v and the the time ti me they took took for the first meeting meeti ng was d /( /(u + v ) where d is the the total di stance between the the two citie c ities. s. Therefore, Therefore, the distance dis tance of the point P from Hyderabad Hyderabad (H), where where they met for the the first fi rst time is

Similarly Si milarly,, after meeting meeti ng they they exchange exchange their cars and start travelling travelling back bac k to their own cities. Now the condition given is that they would not meet each other before reaching their their respecti res pective ve cities. cities . Let us say that that u > v . (We will get similar si milar results and the the same sa me ratio even if we consider c onsider v > u) This This means that Ganesh would not overtake Moreshwar before Moreshwar reaches Hyderabad.

This This distan dis tance ce is i s not more than the the distance di stance between the the point poi nt P to Bangalore and back to Hyderabad.

Solving this this we get, (u + v )( )(u − 2v ) < 0 Therefore u has to lie between be tween −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 fi nd the the distance di stance of the point poi nt Q from Hyderabad Hyderabad so that we can find the the distance di stance PQ. After leaving point P and meeting again at point Q they would have travelled a distance of 2d and their their relative re lative speed would be same as (u + v ). ).

This This is same as the the distance di stance between point P and Hyderabad and the distance dis tance between Hyderabad Hyderabad and point poi nt Q. Q.

Now, l (PQ) (PQ) = l (HP) (HP) − l (HQ) (HQ) Solving for l(PQ) l(PQ) from equations (i) and (ii) (ii ) we get,

Looking Looki ng at the the expression, expressi on, we can see that l (PQ) (PQ) is an increasing function function with the value value of o f u . Therefore given the range of u , we would see that the value value of o f l (PQ) (PQ) is maximum when u = 2v . Hence, Hence, option opti on 2.

TSD

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