PART A

TIME AND WORK

1. Time and Work formula and facts

1. If A can do a piece of work in n days, then A's 1 day's work =

1/n

2. If A's 1 day's work =1/n, then A can finish the work in n days

3. If A is thrice as good a workman as B, then:

Ratio of work done by A and B = 3 : 1.

Ratio of times taken by A and B to finish a work = 1 : 3

Note: If you feel third formula a bit confusion, then please check

following explanation.

If your speed is 3 times than your friend, then you will be able to

do three times work when your friend will able to do it 1 time.

So work done ratio of you and your friend is 3:1

Also you will take 1/3 time to finish the work than your friend,

so time ratio will be 1:3

Some Important tips on Time and Work

1. If A can do a piece of work in n days, then A’s 1 day work = 1/n

2. If A’s 1 day’s work = 1/n, then A can finish the work in n

days.

Example: If A can do a piece of work in 4 days,then A’s 1 day’s

work = 1/4. If A’s 1 day’s work = 1/5, then A can finish the

work in 5 days

3. If A is thrice as good workman as B,then: Ratio of work

done by A and B = 3:1. Ratio of time taken by A and B to finish

a work = 1:3

4. Definition of Variation: The change in two different

variables follow some definite rule. It said that the two variables

vary directly or inversely. Its notation is X/Y = k, where k is

called constant. This variation is called direct variation. XY = k.

This variation is called inverse variation.

5. Some Pairs of Variables: i. Number of workers and their wages. If the number of workers

increases, their total wages increase. If the number of days

reduced, there will be less work. If the number of days is

increased, there will be more work. Therefore, here we have

direct proportion or direct variation.

ii. Number workers and days required to do a certain work is an

example of inverse variation. If more men are employed, they

will require fewer days and if there are less number of workers,

more days are required.

iii. There is an inverse proportion between the daily hours of a

work and the days required. If the number of hours is increased,

less number of days are required and if the number of hours is

reduced, more days are required.

6. Some Important Tips: More Men - Less Days and Conversely More Day - Less Men.

More Men - More Work and Conversely More Work - More

Men.

More Days - More Work and Conversely More Work - More

Days.

Number of days required to complete the given work = Total

work/One day’s work.

Since the total work is assumed to be one(unit), the number of

days required to complete the given work would be the

reciprocal of one day’s work. Sometimes, the problems on time

and work can be solved using the proportional rule

((man*days*hours)/work) in another situation.

7. If men is fixed,work is proportional to time. If work is

fixed, then time is inversely proportional to men therefore, (M1*T1/W1) = (M2*T2/W2)

hours to do a job. How long should it take both A and B,

working together to do same job.

1. 49

2. 249

3. 349

4. 449

Answer: Option D

Explanation:

In this type of questions, first we need to calculate 1 hours work,

then their collective work as,

A's 1 hour work is 1/8

B's 1 hour work is 1/10

(A+B)'s 1 hour work = 1/8 + 1/10

= 9/40

So both will finish the work in 40/9 hours

=

449

days. If A alone can complete the same work in 12 days, in

how many days can B alone complete that work ?

1. 4 days

2. 5 days

3. 6 days

4. 7 days

Answer: Option C

Explanation:

(A+B)'s 1 day work = 1/4

A's 1 day work = 1/12

B's 1 day work =

(14−112)=3−112=16

So B alone can complete the work in 6 days

. A does a work in 10 days and B does the same work in

15 days. In how many days they together will do the same

work ?

1. 5 days

2. 6 days

3. 7 days

4. 8 days

Answer: Option B

Explanation:

Firstly we will find 1 day work of both A and B, then by adding

we can get collective days for them,

So,

A's 1 day work = 1/10

B's 1 day work = 1/15

(A+B)'s 1 day work =

(110+115)=(3+230)=16

So together they can complete work in 6 days.

in half the time taken by A. then working together, what

part of same work they can finish in a day

1. 1\5

2. 1\6

3. 1\7

4. 1\8

Answer: Option B

Explanation:

Please note in this question, we need to answer part of work for

a day rather than complete work. It was worth mentioning here

because many do mistake at this point in hurry to solve the

question

So lets solve now,

A's 1 day work = 1/18

B's 1 day work = 1/9 [because B take half time than A]

(A+B)'s one day work =

(118+19)=(1+218)=16

So in one day 1/6 work will be done.

has two punctures. The first puncture alone

would have made the tyre flat in 9 minutes and the second

alone would have done it in 6 minutes. If air leaks out at a

constant rate, how long does it take both the punctures

together to make it flat ?

1. 315min

2. 325min

3. 335min

4. 345min

Answer: Option C

Explanation:

Do not be confused, Take this question same as that of work

done question's. Like work done by 1st puncture in 1 minute and

by second in 1 minute.

Lets Solve it:

1 minute work done by both the punctures =

(19+16)=(518)

So both punctures will make the type flat in

(185)mins=335mins

finish a piece of work in 18 days. In how many days will B

alone finish the work.

1. 27 days

2. 54 days

3. 56 days

4. 68 days

Answer: Option B

Explanation:

As per question, A do twice the work as done by B.

So A:B = 2:1

Also (A+B) one day work = 1/18

To get days in which B will finish the work, lets calculate work

done by B in 1 day =

=(118∗13)=154

[Please note we multiplied by 1/3 as per B share and total of

ratio is 1/3]

So B will finish the work in 54 days

help of his son he can do it in 3 days. In what time can the

son do it alone ?

1. 712days

2. 612days

3. 512days

4. 412days

Answer: Option A

Explanation:

In this type of question, where we have one person work and

together work done. Then we can easily get the other person

work just by subtracting them. As,

Son's one day work =

(13−15)=(5−315)=215

So son will do whole work in 15/2 days

which is =

712days

days. With the help of C they did the job in 4 days. C alone

can do the same job in how many days ?

1. 612days

2. 712days

3. 835days

4. 935days

n

Answer: Option D

Explanation:

In this question we having, A's work, B's work and A+B+C

work. We need to calculate C's work.

We can do it by,

(A+B+C)'s work - (A's work + B's work).

Let's solve it now:

C's 1 day work =

14−(116+112)=(14−748)=548

So C can alone finish the job in 48/5 days,

Which is =

935days

takes 12 days, A,B and C takes 6 days. How much time A

and C will take

1. 24 days

2. 16 days

3. 12 days

4. 8 days

Answer: Option D

Explanation:

A+B 1 day work = 1/8

B+C 1 day work = 1/12

A+B+C 1 day work = 1/6

We can get A work by (A+B+C)-(B+C)

And C by (A+B+C)-(A+B)

So A 1 day work =

16−112=112

Similarly C 1 day work =

16−18=4−324=124

So A and C 1 day work =

112+124=324=18

So A and C can together do this work in 8 days

-fourth of the

time. If together they take 18 days to complete the work,

how much time shall B take to do it

1. 40 days

2. 35 days

3. 30 days

4. 25 days

Answer: Option C

Explanation:

Suppose B takes x dys to do the work.

As per question A will take

2∗34∗x=3x2days

(A+B)s 1 days work= 1/18

1/x + 2/3x = 1/18 or x = 30 days

less to do a piece of work than B takes. B alone can do the

whole work in

1. 15 days

2. 10 days

3. 9 days

4. 8 days

Answer: Option A

Explanation:

Ratio of times taken by A and B = 1:3

Means B will take 3 times which A will do in 1 time

If difference of time is 2 days, B takes 3 days

If difference of time is 10 days, B takes (3/2) * 10 =15 days

do it in 10 days. B works at it for 5 days and then leaves. A

alone can finish the remaining work in

1. 5 days

2. 6 days

3. 7.5 days

4. 8.5 days

Answer: Option C

Explanation:

B's 5 days work =

110∗5=12Remaining work =1−12=12A can finish work

=15∗12=7.5days

can do it in just 2 hours, while B and C together need 3

hours to finish the same work. In how many hours B can

complete the work ?

1. 10 hours

2. 12 hours

3. 16 hours

4. 18 hours

Answer: Option B

Explanation:

Work done by A in 1 hour = 1/4

Work done by B and C in 1 hour = 1/3

Work done by A and C in 1 hour = 1/2

Work done by A,B and C in 1 hour = (1/4)+(1/3) = 7/12

Work done by B in 1 hour = (7/12) (1/2) = 1/12

=> B alone can complete the work in 12 hour

joins and A and B together finish the remaining work in 3

days. How long does it need for B if he alone completes the

work?

1. 3512

2. 3612

3. 3712

4. 3812

Answer: Option C

Explanation:

Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because

remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 ---(2)

Work done by B in 1 day = 1/15 � 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 (1/2) days

s and B alone

in 8 days. A and B undertook to do it for Rs. 3200. With the

help of C, they completed the work in 3 days. How much is

to be paid to C

1. Rs. 300

2. Rs. 400

3. Rs. 500

4. Rs. 600

Explanation:

C's 1 day's work =

13−(16+18)=(13−724)=124A:B:C=16:18:124=4:3:1C′sShare=1

8∗3200=400

If you are confused how we multiplied 1/8, then please study

ratio and proportion chapter, for small information, it is the C

ratio divided by total ratio.

and 7 women finish it in 10 days. In how many days will 10

women working together finish it ?

1. 30 days

2. 40 days

3. 50 days

4. 60 days

Answer: Option B

Explanation:

Let 1 man's 1 day work = x

and 1 woman's 1 days work = y.

Then, 4x + 6y = 1/8

and 3x+7y = 1/10

solving, we get y = 1/400 [means work done by a woman in 1

day]

10 women 1 day work = 10/400 = 1/40

10 women will finish the work in 40 days

6 days or 3 men and 4 women in 10 days. It can be done by 9

men and 15 women in how many days ?

1. 3 days

2. 4 days

3. 5 days

4. 6 days

Answer: Option A

Explanation:

To calculate the answer we need to get 1 man per day work and

1 woman per day work.

Let 1 man 1 day work =x

and 1 woman 1 days work = y.

=> 6x+5y = 1/6

and 3x+4y = 1/10

On solving, we get x = 1/54 and y = 1/90

(9 men + 15 women)'s 1 days work =

(9/54) + (15/90) = 1/3

9 men and 15 women will finish the work in 3 days

do in 20 days. A started the work and was joined by B after

10 days. The number of days taken in completing the wotk

were ?

1. 1423kmph

2. 1523kmph

3. 1623kmph

4. 1723kmph

Answer: Option C

Explanation:

Work done by A in l0 days = (1/25) *10 = 2/5

Remaining work = 1 - (2/5) = 3/5

(A+B)s 1 days work = (1/25) + (1/20) = 9/100

9/100 work is done by them in 1 day.

hence 3/5 work will be done by them in (3/5)*(100/9)

= 20/3days.

Total time taken = (10 + 20/3) = 16*(2/3) days

children take 14 days to complete the work. How many days

will 5 women and 10 children take to complete the work?

1. 6 days

2. 7 days

3. 8 days

4. 9 days

Answer: Option B

Explanation:

1 woman's 1 day's work = 1/70

1 Child's 1 day's work = 1/140

5 Women and 10 children 1 day work =

(570+10140)=17

So 5 women and 10 children will finish the work in 7 days.

can do in 12 days. Starting with A, they work on alternate

days. The total work will be completed in

1. 1314

2. 1312

3. 1334

4. 1344

Answer: Option C

Explanation:

A's 1 day work = 1/16

B's 1 day work = 1/12

As they are working on alternate day's

So their 2 days work = (1/16)+(1/12)

= 7/48

[here is a small technique, Total work done will be 1, right, then

multiply numerator till denominator, as 7*6 = 42, 7*7 = 49, as

7*7 is more than 48, so we will consider 7*6, means 6 pairs ]

Work done in 6 pairs = 6*(7/48) = 7/8

Remaining work = 1-7/8 = 1/8

On 13th day it will A turn,

then remaining work = (1/8)-(1/16) = 1/16

On 14th day it is B turn,

1/12 work done by B in 1 day

1/16 work will be done in (12*1/16) = 3/4 day

So total days =

1334

as much work as a man and a boy. Working capacity of man

and boy is in the ratio

1. 1:2

2. 1:3

3. 2:1

4. 2:3

Answer: Option C

Explanation:

Let 1 man 1 day work = x

1 boy 1 day work = y

then 5x + 2y = 4(x+y)

=> x = 2y

=> x/y = 2/1

=> x:y = 2:1

Part 2

1) If 5 women or 8 girls can do a work in 84 days. In how many

days can 10 women and 5 girls can do the same work?

ANSWER

Given that 5 women is equal to 8 girls to complete a work.

So, 10 women = 16 girls.

Therefore 10 women + 5 girls = 16 girls + 5 girls = 21 girls.

8 girls can do a work in 84 days then 21 girls can do a work in

(8*84/21) = 32 days.

Therefore 10 women and 5 girls can a work in 32 days

2) If 34 men completed 2/5th of a work in 8 days working 9

hours a day. How many more man should be engaged to finish

the rest of the work in 6 days working 9 hours a day?

ANSWER

From the above formula i.e (m1*t1/w1) = (m2*t2/w2)

so, (34*8*9/(2/5)) = (x*6*9/(3/5))

so, x = 136 men

number of men to be added to finish the work = 136-34 = 102

men

3) If 9 men working 6 hours a day can do a work in 88 days.

Then 6 men working 8 hours a day can do it in how many days?

ANSWER

From the above formula i.e (m1*t1/w1) = (m2*t2/w2)

so (9*6*88/1) = (6*8*d/1)

on solving, d = 99 days.

4) A is twice as good a workman as B and together they finish a

piece of work in 18 days. In how many days will A alone finish

the work?

ANSWER

If A takes x days to do a work then B takes 2x days to do the

same work.

--> 1/x+1/2x = 1/18

--> 3/2x = 1/18

--> x = 27 days.

Hence, A alone can finish the work in 27 days.

5) Worker A takes 8 hours to do a job. Worker B takes 10 hours

to do the same job. How long it take both A & B, working

together but independently, to do the same job?

ANSWER

: A's one hour work = 1/8.

B's one hour work = 1/10.

(A+B)'s one hour work = 1/8+1/10 = 9/40.

Both A & B can finish the work in 40/9 days

6) X can do ¼ of a work in 10 days, Y can do 40% of work in 40

days and Z can do 1/3 of work in 13 days. Who will complete

the work first?

ANSWER

Whole work will be done by X in 10*4 = 40 days.

Whole work will be done by Y in (40*100/40) = 100 days.

Whole work will be done by Z in (13*3) = 39 days

Therefore, Z will complete the work first.

7) A can do a piece of work n 7 days of 9 hours each and B

alone can do it in 6 days of 7 hours each. How long will they

take to do it working together 8 2/5 hours a day?

ANSWER

A can complete the work in (7*9) = 63 days

B can complete the work in (6*7) = 42 days

--> A’s one hour’s work = 1/63 and B’s one hour work = 1/42.

(A+B)’s one hour work = 1/63+1/42 = 5/126

Therefore, Both can finish the work in 126/5 hours.

Number of days of 8 2/5 hours each = (126*5/(5*42)) = 3 days

8) A can do a piece of work in 80 days. He works at it for 10

days & then B alone finishes the remaining work in 42 days. In

how much time will A and B, working together, finish the work?

ANSWER

Work done by A in 10 days=10/80=1/8

Remaining work=(1-(1/8))=7/8

Now, work will be done by B in 42 days.

Whole work will be done by B in (42*8/7)=48 days

Therefore, A's one day's work=1/80

B’s one day's work=1/48

(A+B)'s one day's work=1/80+1/48=8/240=1/30

Hence, both will finish the work in 30 days.

9) A and B are working on an assignment. A takes 6 hours to

type 32 pages on a computer, while B takes 5 hours to type 40

pages. How much time will they take, working together on two

different computers to type an assignment of 110 pages?

ANSWER

Number of pages typed by A in one hour=32/6=16/3

Number of pages typed by B in one hour=40/5=8

Number of pages typed by both in one hour=((16/3)+8)=40/3

Time taken by both to type 110 pages=110*3/40=8 hours.

10) A can finish a work in 18 days and B can do the same work

in half the time taken by A. Then, working together, what part of

the same work they can finish in a day?

ANSWER

Given that B alone can complete the same work in days = half

the time taken by A

= 9days

A’s one day work = 1/18

B’s one day work = 1/9

(A+B)’s one day work = 1/18+1/9 = 1/6

Time and Distance

Important Formulae:

i) Speed=DistanceTime

ii) Time=Distancespeed

iii) Distance = speed*time

iv) 1kmhr=518ms

v) 1ms=185Kmhr

vi) If the ratio of the speed of A and B is a:b,then the ratio of the

time taken by them to cover the same distance is 1a:1b or b:a

vii) Suppose a man covers a distance at x kmph and an

equal distance at y kmph, then the AVERAGE SPEED during

the whole journey is (2xyx+y) kmph

Out of time, speed and distance we can compute any one of the

quantities when we happen to know the other two. For example,

suppose we drive for 2 hours at 30 miles per hour, for a total of

60 miles.

If we know the time and the speed, we can find the distance:

2 hour * 30 mileshour=60 miles

If we know the time and the distance, we can find the speed:

60 miles2 hours=30mileshour

Distance is directly proportional to Velocity when time is

constant:

Example: A car travels at 30km/hr for the first 2 hrs & then

40km/hr for the next 2hrs. Find the ratio of distance travelled

S1S2=V1V2=34

Example: Two cars leave simultaneously from points A & B

(100km apart) & they meet at a point 40 km from A. What is

VaVb?

Time is constant so V1V2=S1S2=4060=46

Example: A train meets with an accident and moves at (34)th its

original speed. Due to this, it is 20 min late. Find the original

time for the journey beyond the point of accident?

Method1: Think about 2 diff. situations, 1st with accident and

another w/o accident. As distance in both the cases is constant

So V1V2=T2T1

=>V1[34*V1]=T1+20T1

=> 43=T1+20T1 =>T1=60

Method 2: Velocity decreases by 25% (34 of original speed =>

decrement by 14) so time will increase by 33.3% (43 of original

time => increment by 13)

now, 33.3%=20 min =>100%=60 min

Relative Speed:

Case1: Two bodies are moving in opposite directions at speed

V1 & V2 respectively. The relative speed is defined

as Vr=V1+V2

Case2: Two bodies are moving in same directions at speed V1

& V2 respectively. The relative speed is defined as Vr=|V1–V2|

Train Problems:

The basic equation in train problem is the same

Speed=DistanceTime

The following things need to be kept in mind while solving the

train related problems.

When the train is crossing a moving object, the speed has to

be taken as the relative speed of the train with respect to the

object.

The distance to be covered when crossing an object,

whenever trains crosses an object will be equal to: Length

of the train + Length of the object

Boats & Streams:

Let U= Velocity of the boat in still water

V=Velocity of the stream.

While moving in upstream, distance covered, S=(U−V)T

In case of downstream, distance covered ,S=(U+V)T

Clock:

For clock problems consider the clock as a circular track of

60km.

Min. hand moves at the speed of 60km/hr (think min. hand as a

point on the track) and hour hand moves at 5km/hr and second

hand at the speed of 3600 km/hr.

Relative speed between HOUR hand and MINUTE hand = 55

Part 1

1. Walking at the rate of 4kmph a man cover certain

distance in 2hr 45 min. Running at a speed of 16.5 kmph

the man will cover the same distance in.

A. 12 min

B. 25 min

C. 40 min

D. 60 min

Answer – (C)

Solution:

Distance = Speed × time

Here time = 2hr 45 min =234 hr =114 hr

distance =4×114=11 km

New Speed =16.5 kmph

Therefore time =DS=1116.5= 40 min

2. Excluding stoppages, the speed of a bus is 54 kmph and

including stoppages, it is 45 kmph. For how many

minutes does the bus stop per hour?

A. 4

B. 6

C. 8

D. None of these

Answer – (D)

Solution:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km =(954)×60 min = 10 min.

3. 2 trains starting at the same time from 2 stations 200km

apart and going in opposite direction cross each other at

a distance of 110km from one of the stations. What is

the ratio of their speeds?

A. 11 : 9

B. 7 : 3

C. 18 : 4

D. None of these

Answer – (A)

Solution:

In same time, they cover 110km & 90 km

respectively.

For the same time speed and distance is inversely

proportional.

so ratio of their speed =110:90= 11: 9

A train covers a distance in 50 min, if it runs at a speed

of 48kmph on an average. The speed at which the train

must run to reduce the time of journey to 40min will be.

A. 45 kmph

B. 60 kmph

C. 75 kmph

D. None of these

Hide Ans

Discuss

Answer – (B)

Solution:

Time =5060 hr =56 hr

Speed = 48mph

distance =S×T=48×(56)=40 km

time =4060 hr =23 hr

New speed =40×(32) kmph = 60kmph

5. Sachin can cover a distance in 1hr 24min by covering

2/3 of the distance at 4 kmph

and the rest at 5kmph.the total distance is?

A. 5 km

B. 6 km

C. 7 km

D. 8 km

Answer – (B)

Solution:

Let total distance = D

Distance travelled at 4 kmph speed =(23)D

Distance travelled at 5 kmph speed =(1−23)D=(13)D

Total time =1 hr 24 min =(60+24) min =(8460) hr

=(2115) hr

We know, time=distancespeed

Total time =(2115)=(23)D4+(13)D5

2115=2D12+D15

2115=14D60

84=14D

D= 6km

6. Vikas can cover a distance in 1 hr 24 min by covering

23rd of the distance at 4 kmph and the rest at 5kmph.

The total distance is?

A. 4 km

B. 6 km

C. 8 km

D. 10 km

Answer – (B)

Solution:

Let total distance be S

total time = 1 hr 24 min = 84 min

=8460 hr =2115 hr

Let Vikas travels from A->T->S

A ------------------------ T ------------ S <---------- (23)S ---------><----(13)S---->

A to T :: speed = 4 kmph

Distance =(23)×S

T to S :: speed = 5 km

Distance =[1−(23)]S=(13)S

Total time:

2115hr=(23)S4+(13)S5

84=10S+4S {Multiply both sides by 15×4}

S=8414

= 6 km

7. A passenger train takes two hours less for a journey of

300km if its speed is increased by 5km/hr from its

normal speed. The normal speed is:

A. 35 km/hr

B. 50 km/hr

C. 25 km/hr

D. 30 km/hr

Answer – (C)

Solution:

Let the normal speed be 's' km/hr

Then new speed =(s+5) km/hr

300s−2=300(s+5)

On solving this equation we get:

s = 25 km/hr

8. A train covers a distance in 100 min, if it runs at a speed

of 48kmph on an average. The speed at which the train

must run to reduce the time of journey to 40min will be:

A. 30 kmph

B. 50 kmph

C. 80 kmph

D. 40 kmph

Answer – (A)

Solution:

Time =10060 hr = 53 hr

Speed =48 mph

Distance =S×T=48×53=80 km

Time =8060 hr =43 hr

New speed =40×34 kmph

= 30 kmph

9. A good train and a passenger train are running on

parallel tracks in the same direction. The driver of the

goods train observes that the passenger train coming

from behind overtakes and crosses his train completely

in 60 sec. Whereas a passenger on the passenger train

marks that he crosses the goods train in 40 sec. If the

speeds of the trains be in the ratio 1:2. Find the ratio of

their lengths.

A. 3 : 1

B. 2 : 1

C. 3 : 2

D. 4 : 3

Answer – (B)

Solution:

Let the speeds of the two trains be s and 2s m/s

respectively.

Also, suppose that the lengths of the two trains are P

and Q metres respectively.

Then, (P+Q)(2s−s)=60 -------- (i)

and P(2s−s)=40 -------- (ii)

On dividing these two equation we get:

(P+Q)P=6040

P:Q= 2 : 1

10. A race course is 400 m long. A and B run a race and A

wins by 5m. B and C run over the same course and B

win by 4m. C and D run over it and D wins by 16m. If

A and D run over it, then who would win and by how

much?

A. D by 7.2 m

B. A by 7.2 m

C. A by 8.4 m

D. D by 8.4 m

Answer – (A)

Solution:

If A covers 400m, B covers 395 m

If B covers 400m, C covers 396 m

If D covers 400m, C covers 384 m

Now if B covers 395 m, then C will cover

396400×395=391.05 m

If C covers 391.05 m, then D will cover

400384×391.05=407.24

If A and D run over 400 m, then D win by 7.2 m

(approx.)

11. The jogging track in a sports complex is 726 m in

circumference. Suresh and his wife start from the same

point and walk in opposite direction at 4.5 km/hr and

3.75 km/hr respectively. They will meet for the first

time in:

A. 5.5 minutes

B. 6 minutes

C. 4.9 minutes

D. 5.28 minutes

Answer – (D)

Solution:

Let both of them meet after T min

4500 m are covered by Suresh in 60 m.

In T min he will cover 4500T60

Likewise, In T min Suresh's wife will cover 3750T60

Given, 4500T60+3750T60=726

T = 5.28 minutes

12. A train starts from Delhi at 6:00 am and reaches Ambala

cantt. at 10am. The other train starts from Ambala cantt.

at 8am and reached Delhi at 11:30 am, If the distance

between Delhi and Ambala cantt is 200 km, then at what

time did the two trains meet each other?

A. 8:46 am

B. 8:30 am

C. 8:56 am

D. 8:50 am

Answer – (C)

Solution:

Average speed of train leaving Delhi =2004=50

km/hr

Average speed of train leaving Ambala

cantt.=200×27=4007 km/hr

By the time the other train starts from Ambala cantt,

the first train had travelled 100 km

Therefore, the trains meet after:

200(50+(4007))=1415 hr

=1415×60=56 minutes

Hence they meet at 8:56 am

13. Two stations A and B are 110 km apart on a straight

line. One train starts from A at 7 am and travel towards

B at 20 km/hr speed. Another train starts from B at 8 am

and travel towards A at 25 km/hr speed. At what time

will they meet?

A. 9 am

B. 10 am

C. 11 am

D. None of these

Answer – (B)

Solution:

A ----------------- C ---------------B

7am ------------------------------ 8am

AC =20 km, CB =90 km

Distance travelled in 1 hour =20 km

Remaining distance =110−20=90 km

Time taken= DistanceRelative speed

=90(20+25)=2 hours

So, time = 8 am + 2 = 10 am

14. A man can row 4.5 km/hr in still water and he finds that

it takes him twice as long to row up as to row down the

river. Find the rate of the stream.

A. 2 km/hr

B. 2.5 km/hr

C. 1.5 km/hr

D. 1.75 km/hr

Answer – (C)

Solution:

Let the speed of the current be x km/hr

Thus upward speed =(4.5+x) km/hr

and downward speed =(4.5−x) km/hr

Let distance travelled be y, then

y(4.5−x)=2y(4.5+x)

⇒ x= 1.5 km/hr

15. The circumference of the front wheel of a cart is 40 ft

long and that of the back wheel is 48 ft long. What is the

distance travelled by the cart, when the front wheel has

done five more revolutions than the rear wheel?

A. 950 ft

B. 1450 ft

C. 1200 ft

D. 800 ft

Answer – (C)

Solution:

Let the total distance travelled by the cart be x ft

Then, x40−x48=5

⇒ (6x−5x)240=5

⇒ x= 1200 ft

16. A train 120m in length passes a pole in 12sec and

another train of length 100m travelling in opposite

direction in 10sec. Find the speed of the second train in

km per hour.

A. 43.2 km/hr

B. 43 km/hr

C. 44 km/hr

D. 43.5 km/hr

Answer – (A)

Solution:

Let the speed of the train be x km/hr

Then, 120=x×518×12

⇒ x=36 km/hr

Let speed of the other train be y km/hr

Then, relative speed in opposite direction:

=(y+36)×(518)

So total distance:

(120+100)=(y+36)×(518)×10

y= 43.2 km/hr

17. A thief steals a car and drives it at 15 km/hr. The theft

has been discovered after one hour and the owner of the

car sets off in another car at 25 km/hr. When will the

owner overtake the thief from the starting point?

A. 1 hr

B. 1.5 hr

C. 2 hr

D. 2.5 hr

Answer – (B)

Solution:

Distance covered by the thief in one hour = 15 km

Now this distance is to be covered by the relative

speed of (25−15)=10km/hr

Hence, time required to cover this distance at a speed

of 10km/hr:

=1510

= 1.5 hr

18. A tower is 61.25m high. A rigid body is dropped from

its top and at the same instant another body is thrown

up-wards from the bottom of the tower with such a

velocity that they meet in the middle of the tower. The

velocity of projection of the second body is:

A. 24.5 m/s

B. 20 m/s

C. 25 m/s

D. 22 m/s

Answer – (A)

Solution:

Let the body moving down-wards take 't' sec to reach

half the height.

⇒ 2452=(12)×9.8×t2

⇒ t=52 sec

Again, assume that the second body is projected

minimum velocity 'v' up-wards

⇒ 2458=v×52−2458

⇒ v=492

= 24.5 m/sec

19. A car travels first half distance between two places with

a speed of 40 km/hr and rest of the half distance with a

speed of 60 km/hr. The average speed of the car is:

A. 48 km/hr

B. 37 km/hr

C. 44 km/hr

D. 45 km/hr

Answer – (A)

Solution:

Let the total distance covered be S km.

Total time taken=S(2×40)+S(2×60)=5S240 hr

Average speed =S×2405S

= 48 km/hr

20. A cyclist drove one kilometre, with the wind in his

back, in three minutes and drove the same way back,

against the wind in four minutes. If we assume that the

cyclist always puts constant force on the pedals, how

much time would it take him to drive one kilometre

without wind?

A. 73 min

B. 247 min

C. 177 min

D. 4312 min

Answer – (B)

Solution:

Let the speed of the cyclist be x km/h and wind be y

km/h

1(x+y)=360

⇒ x+y=20 -------- (i)

And, 1(x−y)=460

⇒ x−y=15 -------- (ii)

On solving both the equations we get:

x=352,y=52

Now, time taken by cyclist without wind:

=235×60=247 min

21. A ship 77 km from the shore, springs a leak which

admits to 9/2 tonnes of water in 11/2 minutes. 92 tonnes

of water would sink it. But the pumps can throw out 12

tonnes of water per hour. Find the average rate of sailing

so that the ship may just reach the shore as it begins to

sink.

A. 10.5 km/hr

B. 11 km/hr

C. 10 km/hr

D. 12.5 km/hr

Answer – (A)

Solution:

Given leak admits 94 tonnes of water in 112 min.

Leak admits one tank of water =112×49=229 min

Leak admits 922 tonne of water in one min

Now, pump throws 12 tonne of water in 60 min.

Pump throws 1 tonne of water in 5 min.

In 1 minutes it throws 15 tonne of water.

Water accumulated in the ship in 1 min:

=922−15=23110 tonnes or 92 tonnes

Water, sufficient to get the ship sunk can be

accumulated in:

=92(23110)=440 min =223 hours.

Rate of sailing in order that the ship may just reach

the shore:

=(7722)(223)

⇒ 10.5 km/hr

22. A train covered a certain distance at a uniform speed. If

the train had been 6 km/hr faster, it would have taken 4

hour less than the scheduled time. And, if the train were

slower by 6 km/hr, the train would have taken 6hr more

than the scheduled time. The length of the journey is:

A. 700 km

B. 740 km

C. 720 km

D. 760 km

Answer – (C)

Solution:

Let the length of the journey be d km and the speed

of train be S km/hr.

Then, d(S+6)=t−4 -------- (i)

and d(S−6)=t+6 --------- (ii)

Subtracting the 1 equation from another we get:

d(S−6)−d(S+6)=10 -------- (iii)

Now , t=dS

Substitute in equation (i) and solve for d and S

We get S=30

d= 720 km

23. Two identical trains A and B running in opposite

direction at same speed tale 2 min to cross each other

completely. The number of bogies of A are increased

from 12 to 16. How much more time would they now

require to cross each other?

A. 40 sec

B. 50 sec

C. 60 sec

D. 20 sec

Answer – (D)

Solution:

Total initial bogies is 12+12=24

Additional bogies =16−12=4

24 bogies take 2 minutes.

4 bogies will take:

=(2×60)(24)×4

= 20 sec.

24. A boatman rows to a place 45 km distant and back in 20

hour. He finds that he can row 12 km with the stream in

the same time as 4km against the stream. Find the speed

of the stream.

A. 3 km/hr

B. 2.5 km/hr

C. 4 km/hr

D. 3.5 km/hr

Answer – (A)

Solution:

Ratio of time taken for up and down =3:1

Out of 20 hr he took 15 hr for up and 5 for down.

Speed up =4515=3 and down =455=9

Hence speed of stream =(9−3)×(12)= 3 km/hr

25. A motorboat whose speed is 15 km/hr in still water goes

30km downstream and comes back in four and a half

hours. The speed of the stream is:

A. 4.5 km/hr

B. 6 km/hr

C. 7 km/hr

D. 5 km/hr

Answer – (D)

Solution:

Let the speed of the stream be 's' km/hr.

Then, upward speed =(15−s) km/hr

and downward speed =(15+s) km/hr

Therefore, 30(15+s)+30(15−s)=−4.5

On solving this equation we get, s = 5 km/hr

26. A hare, pursued by a grey-hound is 20 of her own leaps

ahead of him. While the hare takes 4 leaps, the grey-

hound takes 3 leaps. 3 leaps of grey-hound is equal to 2

leaps of hare. In how many leaps will the grey-hound

overtake the hare?

A. 180 leaps

B. 270 leaps

C. 360 leaps

D. 90 leaps

Answer – (A)

Solution:

When hare takes 4 leaps, the grey-hound takes 3

leaps

When grey hound takes 1 leap hare will take 43 leaps

3 leaps of grey-hound = 2 leaps of hare

1 leaps of grey-hound = 23 leaps of hare

43 leaps of hare = 43×23=89 leaps of grey-hound

Now the grey hound covers (1−89=19) leaps in his

every leap.

grey-hound covers 20 leaps in 20(19)

= 180 leaps

27. P and Q start running in opposite directions (towards

each other) on a circular track starting at diametrically

opposite points. They first meet after P has run for 75m

and then they next meet after Q has run 100 m after

their first meeting. Assume that both of them are

running a constant speed. The length of the track (in

metre) is:

A. 70

B. 175

C. 250

D. 350

Answer – (D)

Solution:

Both P and Q come equal distance as both have

equal constant speed.

Total distance covered by Q =75+100=175 m

Hence, total tracks length =175×2

= 350 m

28. A man sitting in train travelling at the rate of 50 km/hr

observes that it takes 9 sec for a goods train travelling in

the opposite direction to pass him. If the goods train is

187.5m long. Find its speed

A. 40 km/hr

B. 30 km/hr

C. 24 km/hr

D. 25 km/hr

Answer – (D)

Solution:

Let the speed of goods train be x km/hr.

Then, (50+x)×(518)=187.59

⇒ x= 25 km/hr

29. Two places A and B are separated by a distance of 200

m. Ajay and Jay have to start simultaneously from A, go

to B and return to A. In 10 s they meet at a place 10m

from B. If Ajay is faster than Jay, in how much time,

after they start, will Ajay return to A?

A. 19 sec

B. 20021 sec

C. 40021 sec

D. 19021 sec

Answer – (C)

Solution:

Since Ajay is faster than Jay and they start together,

to meet at 10m from B, Ajay would have covered a

distance from A to B and would meet Jay on his way

back to A.

Jay would be on his way from A to B.

So, Ajay covers 200+10=210 m in 10 sec

Hence, Ajay's speed = 21 m/sec

So he will take 19021 sec to cover the remaining 190

m.

The time required for Ajay to reach A will be:

=10+19021=40021 sec.

30. X and Y start walking towards each other at 10am at

speeds of 3km/hr and 4km/hr respectively. They were

initially 17.5 km apart. At what time do they meet?

A. 2:30 pm

B. 11 :30 pm

C. 1:30 pm

D. 12 :30 pm

Answer – (D)

Solution:

Let after T hours they meet

Then, 3T+4T=17.5

T=2.5

Time = 10:00 am + 2.5 hour = 12:30 pm

31. A car driver driving in fog, passes a pedestrian who was

walking at the rate of 2km/h in the same direction. The

pedestrian could see the car for 6 min and it was visible

to him up to a distance of 0.6 km. The speed of the car

would be :

A. 8 km/h

B. 800 m/h

C. 200 m/h

D. 15 km/h

Answer – (A)

Solution:

Traveller distance in 6 min =260×6=210 km.

Total distance in 6 min =210+610=810 km.

Speed =810×10

= 8 km/hr

32. Rahul can row a certain distance downstream in 6 hour

and return the same distance in 9 hour. If the speed of

Rahul in still water is 12 km/hr, Find the speed of the

stream.

A. 2 km/hr

B. 2.4 km/hr

C. 3 km/hr

D. 1.5 km/hr

Answer – (B)

Solution:

Let the speed of the stream be x km/hr

Thus, downward stream =(12+x)

and upward stream =(12−x)

suppose the distance travelled be y,Given:

y(12+x)=6 and y(12−x)=9

On solving these two equations we get:

x= 2.4 km/hr

33. In a race, the speeds of A and B are in the ratio 3:4. A

takes 30 minutes more than B to reach the destination.

The time taken by A to reach the destination is:

A. 1 hr

B. 1.5 hr

C. 2 hr

D. 2.5 hr

Answer – (C)

Solution:

Ratio of speeds = 3:4

Distance remaining constant, the ratio of time taken

= 4:3

A takes 0.5 hours more than B

Hence time taken by A=4×0.5= 2 hour

34. Jay started cycling along the boundaries of a square

field from corner point A. After half an hour he reached

the corner point C, diagonally opposite to A. If his

speed was 8km/hr, the area of the filed in square km is:

A. 64

B. 16

C. 9

D. 4

Answer – (D)

Solution:

Distance covered by Jay in 12 hr = 4 km

Therefore, side of the square =42=2 km

Hence, Area=2×2= 4 square km

35. Wheel of diameter 7cm and 14 cm start rolling

simultaneously from X and Y which are 1980cm apart

towards each other in opposite directions. Both of them

make same number of revolutions per second. If both of

them meet after 10s, the speed of the smaller wheel is:

A. 22 cm/s

B. 44 cm/s

C. 66 cm/s

D. 88 cm/s

Answer – (C)

Solution:

Circumference of the smaller wheel (X)

=2×π×3.5=22 m

Circumference of the bigger wheel (Y)=2×π×7=44 m

Let both the wheels make x revolutions in one

second

Distance covered by both the wheel in 1 second

=22x+44x=66x

Distance covered by both wheels in 10 second =660x

Given, 660x=1980

⇒ x=3

Speed of smaller wheel =22x=22×3

= 66 cm/s

36. A monkey tries to ascend a greased pole 14 m high. He

ascends 2 m in first two minutes and slips 1 m in

alternate minute. If he continues to ascend in this

fashion, how long does he take to reach the top?

A. 26 min

B. 24 min

C. 22 min

D. 25 min

Answer – (D)

Solution:

In every two minutes he is able to ascend 1m. In this

fashion he ascends up to 12m because when he

reaches at the top he does not slip down. Thus, up to

12m he takes 12×2 = 24 min. and for the last 2 m he

takes 1m.

Therefore, total time taken by him is 24+1= 25 min

to reach the top.

37. A train leaves station X at 5am and reaches station Y at

9am. Another train leaves station Y at 7am and reaches

station X at 10:30 am. At what time do the two trains

cross each other?

A. 7:36 am

B. 7:56 am

C. 8:36 am

D. 8:56 am

Answer – (B)

Solution:

Let the distance between X and Y is d km

Then, speed of A is d4 km/hr and that of B is 2d7

km/hr.

X ------------------- Y (XY = d km)

Relative speed =(d4+2d7)=15d28 km/hr

Now distance between these trains at 7 am

=d−d2=d2 km

Hence, time =(d2)(15d28)=1415×60=56 min

Hence both of them meet at 7:56 am

38. A skating champion moves along the circumference of a

circle of radius 28m in 44 sec. How many seconds will

it take her to move along the perimeter of a hexagon of

side 48 m?

A. 90

B. 84

C. 68

D. 72

Answer – (D)

Solution:

Circumference of the circle =2πr=2×227×28=176 m

Given, side of the hexagon =48 m

So, perimeter of the hexagon =48×6=288 m

Skating champion moves 176 m in 44 sec

Skating champion moves hexagon (288m) in:

=44176×288

= 72 sec

39. A hare sees a dog 200m away from her and scuds off in

opposite direction at a speed of 24 km/hr. Two minutes

later the dog perceives her and gives chase at a speed of

32km/hr. How soon will the dog overtake the hare, and

what distance from the spot from where the hare took

flight?

A. 8 min, 2 km

B. 7.5 min, 2 km

C. 7.5 min, 3km

D. 7.5 min, 2.5km

Answer – (C)

Solution:

Distance covered by hare in 2 min =2460×2=800 m

Now to overtake the hare dog has to cover a distance

of (800+200)=1000 m with the relative speed of

(32−24)=8 km/hr

Time =18 hr =608=7.5 min

Now distance travelled by hare in 18 hr:

=18×24= 3 km

40. A boat covers a distance of 30km downstream in 2 hour

while it takes 6 hour to cover the same distance

upstream. What is the speed of the boat in km/hr

A. 5

B. 7.5

C. 10

D. 12

Answer – (C)

Solution:

Let b and s be the speed of boat and stream

respectively.

As per the given conditions:

30(b+s)=2 -------- (i)

And, 30(b−s)=6 -------- (ii)

From eq (i) and (ii):

s = 5 and b = 10 km/hr

41. A and B start running simultaneously. A runs from point

P to point Q and B from point Q to point P. A's speed is

6/5 of B's speed. After crossing B, if A takes 5/2 hr to

reach Q, how much time does B take to reach P after

crossing A?

A. 3 hr 6 min

B. 3 hr 16 min

C. 3 hr 26 min

D. 3 hr 36 min

Answer – (D)

Solution:

A->......................................<-B

VAVB=(tBtA)

⇒ (65)2=tBtA

⇒ tB=3625×52

=3.6 hour

= 3 hr 36 min

42. Walking at 3/4 of his usual place, a man reaches his

office 20 minute late. Find his usual time?

A. 2 hr

B. 1 hr

C. 3 hr

D. 1.5 hr

Answer – (B)

Solution:

Let the original speed be S and time be T

If new speed=S×34, then new time would be T×43.

(D = ST = Constant).

Given, 3T4−T=20T3

⇒ T=60 minutes

= 1 hour

43. A jet plane is rising vertically with a velocity of 10m/s.

It has reached a certain height when the pilot drops a

coin, which takes 4sec to hit the ground. Assuming that

there is no resistance to the motion of the coin, the

height of the place and the velocity of the coin on

impact with the ground are:

A. 38.4 m, 28.7 m/s

B. 38.4 m, 29.2 m/s

C. 26.5 m, 13.5 m/s

D. 26.5 m, 28.7 m/s

Answer – (B)

Solution:

The coin will move up with the initial velocity of 10

m/s till it comes to rest. Time taken is given by:

0=10−9.8t

⇒ t=109.8 s

Time taken to reach the ground from the highest

point:

=4−109.8=29.29.8 sec

Velocity of coin on impact =0+9.8×(29.29.8)=29.2

m/s

If 'h' is the height from which the coin dropped.

Then, 29.22−102=2×9.8×h

⇒ h= 38.4 m

44. A train running at the speed of 20m/s crosses a pole in

24 sec less than the time it requires to cross a platform

thrice its length at the same speed. What is the length of

the train?

A. 150 m

B. 200 m

C. 180 m

D. 160 m

Answer – (D)

Solution:

Let the length of the train be x m.

So, the length of the platform = 3x m.

Time taken in crossing the platform =4x20 sec

Time taken in crossing the pole =x20 sec

⇒ x20+24=4x20

⇒ x= 160 m