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1.3 Marks

Four runners A, B, C and D are running around in a circle, 1215 meter incircumference, at 15 mpm, 12 mpm, 10 mpm and 6 mpm respectively. If they start atthe same time from the same point in the same direction, when will they be togetheragain?

1) 12 hours and 35 minutes





Solution:A, the quickest man gains one complete round on D, the slowest man, in

A gains one complete round on B, the next slowest man, in

A gains one complete round on C, in

∴ A, B, C, and D will be together LCM (135, 405 and 243) i.e. 1215 minutes or 20hours and 15 minutes.

Hence, option 4.

2.3 Marks

Four runners A, B, C and D are running around in a circle, 840 meter incircumference, at 4 mpm, 6 mpm, 8 mpm and 10 mpm respectively. If they start atthe same time from the same point in the same direction, when will they be togetheragain at the starting point for the first time?

1) 20 hours

2) 16 hours

3) 12 hours

4) 8 hours

5) 7 hours

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Section I

Solution:

A completes one round of the circle in (840/4) = 210 minutes

B completes one round of the circle in (840/6) = 140 minutes

C completes one round of the circle in (840/8) = 105 minutes

D completes one round of the circle in (840/10) = 84 minutes

They will meet at the starting point for the first time at LCM (210, 140, 105, 84) i.e.420 minutes or 7 hours.

Hence, option 5.

3.3 Marks

Two cars running on a circular track, 48 meter in circumference meet after every 12minutes when they are in the same direction. However when they are in oppositedirections, the faster car overtakes the slower one every 4 minutes. Find the speedsof the cars.

1) 12 mpm and 8 mpm

2) 8 mpm and 4 mpm

3) 4 mpm and 2 mpm

4) 16 mpm and 10 mpm

5) 20 mpm and 10 mpm

Solution:

Let the speed of the first car or the faster car be S1 while that of the second car be

S2.

When in the same direction, they meet after every 12 minutes,

∴ S1 – S2 = 48/12 = 4 mpm

When in opposite directions, the faster car overtakes the slower one every 4minutes,

∴ S1 + S2 = 48/4 = 12 mpm

Solving both the equations simultaneously,

S1 = 8 mpm and S2 = 4 mpm

Hence, option 2.

4.3 Marks

Two cars starting from the same point at the same time run in opposite directionsalong a circular track, 64 metre in circumference. Speed of car 1 doubles and car 2halves every time they meet. If 16 mpm and 32 mpm are the speeds of the cars, whatwill be the distance travelled by each when they meet for the fifth time?

1) 258.13 metre and 61.87 metre






Solution:

Let the speed of the car 1 be S1 and the car 2 be S2.

When they meet for the 1st time, S1 = 16 mpm and S2 = 32 mpm.

Ratio of their speeds = 1 : 2

∴ Distance covered by Car 2 = (64 × 2)/3 = 42.67 metre

When they meet for the 2nd time, S1 = 32 mpm and S2 = 16 mpm.


∴ Distance covered by Car 2 = (64/3) = 21.33 metre

When they meet for the 3rd time, S1 = 64 mpm and S2 = 8 mpm.



When they meet for the 4th time, S1 = 128 mpm and S2 = 4 mpm.



When they meet for the 5th time, S1 = 256 mpm and S2 = 2 mpm.


∴ Distance covered by Car 2= (64/129) = 0.496 ≈ 0.5 metre

∴ Total distance covered by the Car 2 = (42.67 + 21.33 + 7.11 + 1.94 + 0.5) = 73.55metre

∴ Total distance covered by Car 1 = Total distance covered by both the cars – Totaldistance covered by the Car 2

= (5 × 64) – 73.55 = 320 – 73.55 = 246.45 metre

Hence, option 4.

5.3 Marks

A policeman starts chasing a thief 35 minutes after the thief escapes. If the speed ofthe policeman is 18 mpm while that of the thief is 12 mpm, find the time in which thethief will be caught.

1) 105 minutes

1) 105 minutes

2) 90 minutes

3) 70 minutes

4) 65 minutes

5) 50 minutes

Solution:

Let the total time travelled by the policeman to catch the thief be x minutes.

∴ The total time travelled by the thief = (x + 35) minutes.

∴ 18x = 12(x + 35)

∴ x = 70 minutes

Hence, option 3.

6.3 Marks

The police start chasing a thief 24 minutes after he escapes. Speed of the thief is 75kmph while that of the police is 95 kmph. A dog runs to and fro between the thief andthe police car with a speed of 120 kmph to show the direction. Find the distancetravelled by dog.

1) 180 km

2) 150 km

3) 240 km

4) 300 km

5) 320 km

Solution:

Let the total time travelled by the police to catch thief be x hours.

∴ Total time travelled by the thief = (x + 0.4) hours.

∴ 95x = 75(x + 0.4)

∴ x = 1.5 hours

The dog runs to and fro for 1.5 hours with a speed of 120 kmph.

∴ Total distance travelled by the dog = 120 × 1.5 = 180 km.

Hence, option 1.

7.3 Marks

To reach next floor of a building using an escalator moving upwards, a person takes56 seconds and climbs 42 steps of an escalator. When the same person increaseshis speed, he can climb 94 steps in 30 seconds to reach the next floor. How manysteps does the escalator have, between the two floors?

1) 75

2) 112

2) 112

3) 124

4) 144

5) 154

Solution:

Let escalator move x steps/sec.

So in 56 sec, the escalator covers 56x steps.

Hence the total number of steps on the escalator is 42 + 56x.

Similarly, in 30 sec the escalator covers 30x steps.

Hence total number of steps on the escalator = 94 + 30x

Solving both the equations

42 + 56x = 94 + 30x

∴ x = 2

∴ Total number of steps is 154.

Hence, option 5.

8.3 Marks A rabbit sees a greyhound 500 meters away from it, and scuds off in the opposite

direction at a speed of 75 kmph. 30 seconds later, the greyhound sees the rabbitand chases the rabbit at a speed of 120 kmph. After how much time does thegreyhound catch up with the rabbit?

1) 30 sec

2) 1 minutes

3) 90 sec

4) 3 minutes

5) 5 minutes

Solution:

∴ Distance between the rabbit and the greyhound after 30 sec = 500 + 625 = 1125meter

Hence, option 3.

9.3 Marks

A beats B in a 500 meter race by 25 meter. B beats C in a 400 meter race by 20meter. By how many meters will A beat C in an 800 meter race?

1) 40 meter

2) 45 meter

3) 55 meter

4) 66 meter

5) 78 meter

Solution:

Let speed of A, B and C be As, Bs and Cs respectively.

A beats B in a 500 m race by 25 m

B beats C in a 400 m race by 20 m

Multiplying both the equations,

∴ A beats C in an 800 meter race by = 800 – 722 = 78 meter.

Hence, option 5.

10.3 Marks

A gives B a start of 15 sec or 25 meters in an 800 m race, so that they can bothfinish the race simultaneously. How long does A take to complete the race?

1) 320 sec

2) 400 sec

3) 465 sec

4) 495 sec

5) 415 sec

Solution:Here B runs 25 meters in 15 sec.

Here B runs 25 meters in 15 sec.

∴ Time taken by A to complete the race = 480 – 15 = 465 sec

Hence, option 3.

11.3 Marks

Time in a clock is 2 hours and 25 minutes. Find the smaller angle between the twohands of the clock.

1) 90°

2) 92°

3) 96°

4) 77.5°

5) 106.5°

Solution:

The angle between 2 and 5 is 90°

∴ The angle between the minute hand and hour hand at 2:25 = 90 − 12.5 = 77.5°.

Hence, option 4.

12.3 Marks

The angle between the hours and minutes hands of a clock is 42°. What time couldthe clock be showing?

1) 10:03:10

2) 2:17:32

3) 4:29:27

4) 6:40:51

5) 5:32:54

Solution:

At 10 o’clock, the minute hand will gain 42° in





Hence, option 3.

13.3 Marks

At what time between 3 am and 4 am do the hours and minutes hands make anangle of 180 degrees?

1)

2)

3)

4)

5)

Solution:

This will happen when the minute hand gains (30 + 15) minutes = 45 minutes overthe hour hand.

The minute hand gains 45 minutes in

i.e. The hands will be 180l apart at

Hence, option 2.

14.3 Marks

At what time between 7 am and 8 am are the hours and minutes hands of a clocktogether?

1)

2)

3)

4)

5)

Solution:

At 7 am, the clock hands are 35 minutes apart. ∴ The minute hand must gain 35minutes before the hands can be together. The minute hand gains 35 minutes in

Hence, option 4.

15.3 Marks

At what time between 5 am and 6 am do the hours and minutes hands of a clockmake an angle of 72 degrees?

1)

2)

3)

4)

5)

Solution:At 5 pm, the clock hands are 25 minutes apart. An angle of 72° will be formed whenthe minute hand gains (25 – 12) minutes = 13 minutes over the hour hand.

Hence, option 5.

16.3 Marks

A car after travelling 18 km from a point A developed some problem in the engineand speed became 4/5 of its original speed. As a result, the car reached point B 45minutes late. If the engine had developed the same problem after travelling 30 km

minutes late. If the engine had developed the same problem after travelling 30 kmfrom A, then it would have reached B only 36 minutes late. The original speed of thecar (in km per hour) and the distance between the points A and B (in km) is _____.

1) 25, 130

2) 30, 150

3) 20, 90

4) None of these

Solution:

Let the total distance between points A and B be d km and let the original speed ofthe car be x kmph.

Time taken to cover distance d with speed x = d/x

As per conditions in the question,

0.2d – 4.5 = 0.75x … (i)

0.2d – 7.5 = 0.6x … (ii)

Solving (i) and (ii) simultaneously,

x = 20 and d = 97.5

Hence, option 4.

17.3 Marks

Two trains are travelling in opposite direction at uniform speed 60 and 50 km perhour respectively. They take 5 seconds to cross each other. If the two trains hadtravelled in the same direction, then a passenger sitting in the faster moving trainwould have overtaken the other train in 18 seconds. What are the lengths of trains (inmetres)?

1) 112.78

2) 97.78, 55

3) 102.78, 50

4) 102.78, 55

Solution:

Let the lengths and speeds of the two trains be denoted by L1 and L2, s1 and s2

respectively.

Here, s1 = 60 km/hr and s2 = 50 km/hr.

(i) Consider the two trains moving in opposite directions:

Relative speed = s1 + s2 = 60 + 50 = 110 km/hr

(ii) Consider the two trains moving in the same direction:

Relative speed = s1 – s2 = 60 – 50 = 10 km/hr

The length to be covered by the 1st train if a passenger in it has to completely cross

the 2nd train = Length of the 2nd train = L2.

Thus, L1 = 152.78 – 50 = 102.78 m

Hence, option 3.

18.3 Marks

Every day Neera’s husband meets her at the city railway station at 6.00 p.m. anddrives her to their residence. One day she left early from the office and reached therailway station at 5.00 p.m. She started walking towards her home, met her husbandcoming from their residence on the way and they reached home 10 minutes earlierthan the usual time. For how long did she walk?

1) 1 hour

2) 50 minutes

3)

4) 55 minutes

Solution:

The time taken by the car in both the directions is the same. As Neera reaches 10minutes earlier than usual, the car saves 10 minutes of total travel, and 5 minutes oftravel in each direction.

∴ The car which usually meets Neera at 6:00 pm, meets her at 5:55 pm.

∴ Neera walks for 55 minutes.

Hence, option 4.

19.3 Marks

A man can walk up a moving ‘up’ escalator in 30 seconds. The same man can walkdown this moving ‘up’ escalator in 90 seconds. Assume that his walking speed is thesame both upwards and downwards. How much time will he take to walk up theescalator, when it is not moving?

1) 30 seconds

2) 45 seconds

3) 60 seconds

4) 90 seconds

Solution:

Let the number of steps on the escalator be n and let the speed of the escalator be ssteps per second

If the speed of the man on the escalator is x steps per second, then

Dividing (ii) by (i),

∴ x = 2s

Substituting in (i),

n = 90s = 45x

Time taken by the man to walk up the escalator, when it is not moving

Hence, option 2.

20.3 Marks

Mr. and Mrs. Shah travel from City A to City B and break the journey at City C inbetween. Somewhere between City A and City C, Mrs. Shah asks “How far have wetravelled?” Mr. Shah replies “Half as far as the distance from here to City C”.

travelled?” Mr. Shah replies “Half as far as the distance from here to City C”.Somewhere between City C and City B, exactly 200 km from the point where sheasked the first question, Mrs. Shah asks “How far do we have to go?” Mr. Shahreplies “Half as far as the distance from City C to here.” The distance between CitiesA and B is_______ (in km)

1) 200

2) 100

3) 400

4) 300

Solution:To find the distance between A and B consider the following diagram.

The places where Mrs. Shah asks the questions the first and the second time are Dand E respectively.

Point D is half as far away from A as the distance from D to City C.

So, AD : DC = 1 : 2

Point E is as half as far away from B as the distance from City C to E.

∴ CE : EB = 2 : 1

If CD = x and CE = y

But,

DE = x + y = 200

∴ AB = 300 km

Hence, option 4.

21.3 Marks

A train without stopping travels at 60 km per hour and with stoppages at 40 km perhour. What is the time taken for stoppages on a route of 300 km?

1) 10 hours

2) 20 hours

3) 5 hours

4) 2.5 hours

Solution:

So the stoppages time is = 7.5 – 5 = 2.5 hrs

Hence, option 4.

22.3 Marks

A ship leaves on a long voyage. When it is 18 miles from the shore, a sea plane,whose speed is ten times that of the ship, is sent to deliver the mail. How far from theshore does the sea plane catch up with the ship?

1) 24 miles

2) 25 miles

3) 22 miles

4) 20 miles

Solution:

Let the speed of the ship be s mph and that of the plane be 10s mph.

So, the relative speed is 9s.

The distance after which the plane catches up with the ship

Hence, option 4.

Group Question

Answer the following questions based on the information given below.

Q started to move from point B towards point A. Exactly an hour later P started from A inthe opposite direction. P’s speed was twice that of Q. When P had covered one-sixth ofthe distance between the points A and B, Q had also covered the same distance.

23.3 Marks

The point where P and Q would meet is

1) Closer to A

2) Exactly between A and B

3) Closer to B

4) P and Q will not meet at all

Solution:

Let the distance between A and B be d.

If the speed of Q be s, then the speed of P is 2s

In the first hour, Q travels a distance of s.

By conditions,

This is closer to B than to A.

Hence, option 3.

24.3 Marks

How many hours would P take to reach B?

1) 2

2) 5

3) 6

4) 12

Solution:

To reach B, P would take

Hence, option 3.

25.3 Marks

How many more hours would Q (compared to P) take to complete his journey?

1) 4

2) 5

3) 6

3) 6

4) 7

Solution:

Q would take

This is 6 hours more than P.

Hence, option 3.

Group Question

Answer the following questions based on the information given below.

ABC forms an equilateral triangle in which B is 2 km from A. A person starts walking fromB in a direction parallel to AC and stops when he reaches a point D directly east of C. Hethen reverses direction and walks till he reaches a point E directly south of C.

26.3 Marks

Then D is:

1) 3 km east and 1 km north of A

2)

3)

4)

Solution:

The distance from B to D is equal to the distance from B to E, i.e. 2 km.

Hence, option 2.

27.3 Marks

The total distance walked by the person is:

1) 3 km

2) 4 km

3)

4) 6 km

Solution:The total distance walked by the person = BD + DB + BE = 2 + 2 + 2 = 6 km.

Hence, option 4.

28.3 Marks

Shyam went from Delhi to Shimla via Chandigarh by car.

The average speed from Delhi to Chandigarh was one and half as much as that fromChandigarh to Shimla. If the average speed for the entire journey was 49 kmph, whatwas the average speed from Chandigarh to Shimla?

1) 39.2 kmph

2) 63 kmph

3) 42 kmph

4) None of these

Solution:

Let Delhi, Chandigarh and Shimla and be denoted by D, C and S respectively.

Also, let the distance between D and C be 3x, then the distance between C and S = 4x.

Let the average speed between C and S be v,

Distance between D and S = 3x + 4x = 7x

Thus, the average speed between C and S = v = 42 kmph.

Hence, option 3.

29.3 Marks

It takes a pendulum clock 7 seconds to strike 4 o’clock. How much time will it take tostrike 11 o’clock?

1) 18 seconds

2) 20 seconds

3) 19.25 seconds

4) 23.33 seconds

Solution:

It takes 7 seconds to strike 4 o’clock,

At 4 o’ clock, there are 4 – 1 = 3 rounds between each strike. Thus, for each round, ittakes

To strike 11 o’ clock, there are 11 – 1 = 10 rounds, for which it takes (10 × 2.33) =23.33 s

Hence, option 4.

30.3 Marks

Along a road lie an odd number of stones placed at intervals of 10 m. These stoneshave to be assembled around the middle stone. A person can carry only one stone ata time. A man carried out the job starting with the stone in the middle, carryingstones in succession, thereby covering a distance of 4.8 km. Then the number ofstones is _____.

1) 35

2) 15

3) 29

4) 31

Solution:

Let the total number of stones be equal to (2x + 1), where there are x stones on each

side of the middle stone.

On each side, distance travelled = 20 × 1 + 20 × 2 +0 + 20 × x = 20(1 + 2 + … + x)

Thus, the total distance travelled = 2 × 10(x)(x+1) = 20(x)(x + 1)

= 4.8 kms = 4800 m

∴ x = 15

Thus, the total number of stones = 2x + 1 = 31

Hence, option 4.

31.3 Marks

A and B walk from X to Y, a distance of 27 km at 5 kmph and 7 kmph respectively. Breaches Y and immediately turns back meeting A at Z. What is the distance from X

to Z?

1) 25 km

2) 22.5 km

3) 24 km

4) 20 km

Solution:

The distance between X and Y is 27 kms.

Let the distance between Y and Z be d kms.

Since A travels faster than B, he reaches Y and then turns back to meet A at Z.

Here, A travels a distance of (27 + d) kms while B travels a distance of (27 – d) kms.

∴ d = 4.5 kms

Thus, the distance between X and Z = 27 – d = 27 – 4.5 = 22.5 kms.

Hence, option 2.

32.3 Marks

The winning relay team in a high school sports competition clocked 48 minutes for adistance of 13.2 km. Its runners A, B, C and D maintained speeds of 15 kmph, 16kmph, 17 kmph, and 18 kmph respectively. What is the ratio of the time taken by B tothat taken by D?

1) 5 : 16

2) 5 : 173) 9 : 8

4) 8 : 9

Solution:

Since, each runner is running the same distance; the ratio of the time taken for B andD will be the reverse of the ratio of their speeds (∵ Speed is inversely proportional totime, when the distance is constant).

The ratio of the time taken by B and D = 18 : 16 = 9 : 8

Hence, option 3.

33.3 Marks

In a race of 200 meters, A beats S and N by 20 metres and 40 metres respectively. If

S and N are running a race of 100 metres with exactly same speed as before, then,

by how many metres will S beat N?

1) 11.11 metres

2) 10 metres

3) 12 metres

4) 25 metres

Solution:

In a 200 m race, A beats S by 20 m and N by 40 m.

Thus, when A covers 200 m, S covers only 180 m and N covers only 160 m.

Hence, in a 100 m race, S beats N by (100 − 88.89) = 11.11 m

Hence, option 1.

34.3 Marks

Three bells chime at intervals of 18, 24 and 32 minutes respectively. At a certaintime they begin to chime together. What length of time will elapse before they chimetogether again?

1) 2 hours 24 minutes

2) 4 hours 48 minutes

3) 1 hour 36 minutes

4) 5 hours

Solution:

The three bells will ring together again at:

Least Common Multiple of (18, 24, 32).

18 = 2 × 3 × 3 = 21 × 32

24 = 2 × 2 × 2 × 3 = 23 × 31

32 = 2 × 2 × 2 × 2 × 2 = 25 × 30

Thus, LCM = 25 × 32 = 32 × 9 = 288 minutes

= 4 hrs, 48 minutes

Hence, option 2.

35.3 Marks

In a mile race, Akshay can be given a start of 128 metres by Bhairav. If Bhairav cangive Chinmay a start of 4 metres in a 100-metre dash, then who out of Akshay andChinmay will win a race of one and half miles, and what will the final lead given by thewinner to the loser be? (One mile is 1600 metres).

1)

2)

3)

4)

Solution:

Given: 1 mile = 1600 m.

(i) Akshay can be given a start of 128 m by Bhairav in a mile race:

When Bhairav covers 1600 m, then Akshay covers (1600 – 128) = 1472 m

(ii) Chinmay can be given a start of 4 m by Bhairav in a 100 m race:

When Bhairav covers 100 m, then Chinmay covers (100 – 4) = 96 m

Thus, when Bhairav covers 1600 m, then Chinmay covers 96 × 16 = 1536 m andAkshay covers 1472 m.

Hence, when Chinmay covers 2400 m (1.5 mile), then Akshay covers

Thus, Akshay can be given a start of

Hence, option 4.

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