TIME AND WORK , TIME SPEED AND DISTANCE FOR CAT , MAT , MBA , BANKING EXAM , RAILWAYS EXAM BY SOURAV SIRS CLASSES
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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
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