Numerical reasoning II

Chain Rule
Direct Proportion :
A B
A B
Indirect Proportion:
A B
A B

Type - 1
Direct Proportion:
Price Quantity
100 15
150 x
Price increase no. of quantity will also increase
Direct proportion means Cross Multiply
100*x = 150*15 ;
X = 150*15/100 = 22.5

Type - 2
Indirect Proportion:
Men Days
15 50
x 10
To complete the work in less number of days,
they need to deploy more number of people.
15*50 = x*20;
X = 15*50/20 = 75

Tips
In given problem, if you have more than 2
variables, you need to,
1. Compare data you need to find with other
data which is given.
2. While comparing based on the proportion, do
either cross or direct multiply.

Problem - 1

If 6 men and 8 boys can do a piece of work in 10
days and 26 men and 48 boys can do the same
work in 2 days, what would be the time taken by
15 men and 20 boys to do the same type of work?

Solution
6 men + 8 Boys = 10 days, 26 men and 48 Boys = 2 days
60 men + 80 Boys = 52 Men + 96 Boys
8 M = 16 B
1 Man = 2 Boys
15 M + 20 B = 30 B + 20 B = 50 Boys
6 men + 8 boys = 12 Boys + 8 Boys = 20 Boys = 10 Days
Boy Days
20 10
50 x
50x = 20* 10
x = 20*10/50 = 4 days.

Problem - 2

20 men can complete one third of a work in 20
days. How many more men should be employed
to finish the rest of the work in 25 more days?

Solution
Men Work Days
20 1/3 20
x 2/3 25
x * 1/3 * 25 = 20*20*2/3
x = 20*20*2 /25 = 32
Already we have 20 men, so 12 more men
should be deployed

Problem - 3

95 men have provision for 200 days. After 5
days, 30 men die due to an epidemic. For how
many days will the remaining food last?

Solution
Remaining days = 200 – 5 = 95
Remaining men = 95 – 30 = 65
Men Days
95 195
65 x
65x = 195*95
x = 195*95/65 = 285 days

Problem - 4
A certain number of men completed a piece of
work in 60 days. If there were 8 more men, the
work could be finished in 10 days less. How
many men were originally there?

Solution
Days Men
60 x
50 x+8
60x = 50*x+8
60x/50 = x + 8
6x = 5x + 40
x = 40

Problem - 5
A contractor employs 60 people to complete a
task in 20 days. After 10 days, he finds that only
30% of the work is completed. In order to
achieve the target of completion of the task in 20
days, how many more people should he employ
now?

Solution
People Days Work
60 10 30
x 10 70
30*x*10 = 60*70*10
x = 60*70*10/30*10
x = 140
Already there are 60 men, so 80 more men are needed.

Problem - 6
If 6 engines consume 15 metric tons of coal,
when each is running 9 hours a day, how much
coal will be required for 8 engines each running
12 hours a day with the fact that 3 engines of the
former type consume as much as 4 engines of the
latter type?

Solution
Three engines of former type consume = 1 unit
One engine of former type consume = 1/3
Four engines of latter type consume = 1
1 engine of latter type consume = ¼
Engine Hours Consumption Coal
6 9 1/3 15
8 12 ¼ x
x *6*9*1/3 = 8*15*12*1/4
x = 20

Problem - 7
8 men working for 9 hours a day complete a
piece of work in 20 days. In how many days can
7 men working for 10 hours a day complete the
same piece of work?

Solution
Men Hours Days
8 9 20
7 10 x
7*x*10 = 8*20*9
x = 20 4/7 days

Problem - 8

35 cattle can graze a piece of land for 56 days.
How many cattle will graze a field three times as
large in 35 days?

Solution

Cattle Graze Days
35 1 56
x 3 35
x*35 = 35*3 *56
x = 56*35*3/35 = 168

Problem - 9
30 men can do a work in 50 days. After 20 days
10 men leave the work. The whole work will be
done in?

Solution
Men Days
30 30
20 x
X = 30*30/20 = 45 days

Problem - 10
The work done by a man, a woman and a boy are
in the ratio 3 : 2 : 1. There are in a factory 24
men, 20 women and 16 boys, whose weekly
wages amount to Rs. 224. What will be the
yearly wages of 27 men, 40 women, 15 boys?

Solution
1 man = 3 boys, 1 woman = 2 boys
24 M + 20 W + 16 B = 24*3 + 20*2 + 16*1 = 128 boys
27 M + 40 W + 15 B = 27*3 + 40*2 + 15*1 = 176 boys
Boys Duration Wages
128 1 224
176 52 x
x*128 = 52*224*176
x = 52*224*176/128 = Rs.16,016

Problem - 11
A contractor undertakes to dig a canal 12 km
long in 350 days and employs 45 men. He finds
after 200 days of working that only 4½ km of
canal has been dug. How many extra men must
be employed to finish the work on time?

Solution
Men Days KM
45 200 4½
x 150 7½
200*45*15/2 = 150*x*9/2
x= 100
Extra men = 100 – 45 = 55 men

Problem - 12
8 men can build a foundation 18 m long, 2m
broad and 12m high in 10 days, working 9 hours
a day. Find how many men will be able to build
a foundation 32m long, 3m broad and 9m high in
8 days, working 6 hours a day.

Solution
Men Metre Breadth Height Day Hour
8 18 2 12 10 9
x 32 3 9 8 6

x*18*2*12*8*6 = 32*8*3*9*10*9
x = 30

Type - 1
A can do work in 10 days
B can do work in 15 days
They can complete the work by working
together?
A’s 1 day work = 1/10
B’s 1 day work = 1/15
A + B = 1/10 + 1/15 = 25/150 = 1/6
6 days they will complete their work.

Type - 2
A and B can complete the work in 10 days
B and C can complete the work in 15 days
C and A can complete the work in 20 days
How many days they will take to complete the work by working
all together.
A + B 1 day work = 1/10
B + C 1 day work = 1/15
C + A 1 day work = 1/20
A+B+B+A+C+A = 1/10 + 1/15 + 1/20
2(A+B+C) = 6/60 + 4/60 + 3/60 = 13/60
A+ B + C = 13*2/60 = 26/60 = 13/30
= 30 /13 = 2 4/13 days

Types - 3
A and B can complete the work in 10 days
B alone can complete the work in 30 days
How many days A will take to complete the
work?
A+B = 1/10
B = 1/30
A = A+B – B = 1/10 – 1/30 = 3/30 – 1/30 = 2/30
A = 1/15 = 15 days

Type - 4
A can complete the work in 30 days
B can complete the work in 50 days
C can complete the work in 40 days
A is working continuously from the starting, B
and C supporting A alternatively.
Day 1 A+B = 1/30 + 1/50 = 8/150
Day 2 A+ C = 1/30 + 1/40 = 7/120
2 days work = 8/150+ 7/120 = 67/600

Cont…

Type - 4
2* 8 days = 67*8/600 = 67/75
In 16 days they will complete 67/75 of the work
Remaining work = 1- 67/75 = 8/75
17th day A+ B = 8/150
Work left = 8/75 – 8/150 = 8 / 150 = 4/75
On 18th day A+C will work they will finish in
120/7 = 120 / 7 * 4/75 = 32 /35
Whole work will be done in 17 32/35 days

Type - 5
A is 60% more efficient than B.
B can complete the work in 20 days
A will take how many days to complete the work.
A : B
Eff 160 : 100
8 : 5
Days 5 : 8 (More efficient person will
take less Days)

Cont…

Types - 5
Use chain rule method
Effi Days
5 20
8 x
More efficient person takes less days
8x = 20*5
X = 20*5/8 = 12.5 days

Problem – 1

8 men and 12 children can do a work in 9 days.
A child takes double the time to do a work than
the man. In how many days 12 men can
complete double the work?

Solution
8 men and 12 children in 9 days
1 man = 2 children
6 men = 12 children
8 man + 6 man (12 Children) = 14 men
Men Days Work
14 9 1
12 men x 2
12x = 14*9*2
x = 14*9*2 /12 = 21 days

Problem - 2

A is three times more efficient than B, and is
therefore able to complete a work in 60 days
earlier than A. How long A and B together will
take to complete the work?

Solution
A is three times more efficient than B
A B
Effi 3 1
Days 1 3
If A takes 10 days
A B
10 30
20 60
30 90 (60 days difference)
So, A takes 30 days and B takes 90 days.
A and B together = 1/30 + 1/90 = 4/90 = 90/4 =21 ½ days

Problem – 3

A can do ½ of the work in 5 days, B can do 3/5
of the same work in 9 days and C can do 2/5 of
that work in 8 days. In how many days can three
of them together do the work?

Solution
The number of Days taken by A, B and C to complete
work
A = ½ work in 5 days = 1 work = 10 days
B = 3/5 of the work in 9 days = 1 work = 9*5/3
= 15 days
C = 2/3 of the work in 8 days = 1 work = 8*3/2
= 12 days
A + B + C = 1/10 + 1/15 + 1/12 = 6+4+5/60 = 15/60
=¼
A, B and C together take 4 days to complete the work

Problem - 4
A, B and C can do a piece of work in 6, 8 and 10
days respectively. They begin the work together.
A continues to work till it is completed. B leaves
off 1day before and C leaves off ½ day before
the work is completed. In what time is the work
completed?

Solution
A’s x days work = x/6
B’s (x-1) day work = x – 1/8
C’s (x-1/2) day work = 2x-1/2*10 = 2x-1/20
x/6 + x-1/8 + 2x -1/20 = 1
20x +15x – 15+12x – 6/120 = 1
20x + 15x – 15+12x – 6 = 120
47x = 120 + 21
x = 141/47 = 3 days

Problem - 5

A can do a piece of work in 12 days, while B
alone can do it in 15 days. With the help of C
they can finish it in 5 days. If all the three were
paid Rs. 960 for the whole work, what is the
share of A?

Solution
A + B = 1/12 + 1/15 = 9/60 = 3/ 20
A + B +C = 1/5
(A+B+C) – (A+B) = 1/3 – 3/20 = 4 -3/20 = 1/20
C will take 20 days to complete the work
A:B:C = 1/12 : 1/15 : 1/20
= 5/60 : 4/60 : 3/60 = 5 : 4 : 3
A’s Share = 960 * 5/12 = Rs.400

Problem - 6

A and B can do a piece of work in 40 days. After
working for 10 days, they are assisted by C and
work is finished in 20 days more. If C does as
much as work as what B does in 3 days, how
many days could A alone do the same work?

Solution
A and B’s one day work = 1/40
Day Work
1 1/40
10 x
x = 1/40*10 = ¼
Remaining work = 1- ¼ = ¾
A+B+C took 20 days to complete ¾ of the work.
Days Work
20 ¾
1 x
x = ¾ * 1/20 = 3/80

Solution
C’s one day work = 1 day (A+B+C) - 1 day (A+B)
= 3/80 – 1/40 = 1/80
C takes 80 days to complete the work
C B
Effi 1 3
Days 80 240
B takes 240 days to complete the work
A’s 1 day work = A + B – B = 1/40 – 1/240 = 6 – 1 /240
= 5/40 = 1/48
A will take 48 days to complete the work.

Problem - 7
To complete a work A takes 50% more time than
B. If together they take 18 days to complete the
work, how much time should B take to do it?

Solution

A + B = 18 days
B takes = x days
A takes = x +x/2 days = 3x/2
1/x + 2/3x = 18 days
5/3x = 1/18
3x = 18*5
x = 18*5 /3 = 30 days

Problem - 8

A and B can complete a piece of work in 12 and
18 days respectively. If A begins to do the work
and they work alternatively one at a time for one
day each, then in how many days will the whole
work be completed?

Solution
1 day’s work of A = 1/12
1 day’s work of B = 1/18
2 days’ work of A and B = 1/12 + 1/18 = 5/36
7 times repeating 2 days work
7*2 = 5*7/36
Amount of work completed after 14 days = 35/36
Remaining work = 1 – 35/36 = 1/36
15th day A is working
= 1/36 *12 = 12/36 = 1/3 days
14 1/3 days to complete the work.

Problem - 9
12 men can complete a piece of work in 36 days.
18 women can complete the same piece of work
in 60 days. 8 men and 20 women work together
for 20 days. If only women were to complete the
remaining piece of work in 4 days, how many
women would be required?

Solution
12 men take 36 days and 18 women take 60 days to complete a work.
Ratio of efficiency = 12*36 = 18*60, 432 = 1080, 2 : 5
2 men = 5 women, 8 men = 20 women
Women Days
18 60
40 x
40*x = 18*60 = x = 18*60/4 = 27 days
They worked for only 20 days, 1 day’s work = 1/27, so 20 days’ work = 20/27
Remaining work = 7/27
Women Days Work
40 27 1
x 7 7/27
4x = 40*27*7/27 = 40*7/4 = 70 Women

Problem - 10
A, B and C are employed to do a piece of work
for Rs.529. A and B together are supposed to do
19/23 of the work and B and C together 8/23 of
the work. What is the share of A?

Solution
A + B = 19/23; remaining work by C = 1 – 19/23
= 4/23
B + C = 8/23, B = 8/23 – C = 8/23 – 4/23 = 4/23
A = A+B – B = 19/23 – B = 19/23 – 4/23
= 15/23
A : B : C = 15:4:4
A’s share = 529*15/23 = Rs. 345

Problem - 11

12 children take 16 days to complete a work
which can be completed by 8 adults in 12 days.
16 adults started working and after 3 days 10
adults left and 4 children joined them. How
many days will they take to complete the
remaining work?

Solution
12 children take 16 days, 8 Adults take 12 days to complete work
16 adults would complete the work in 6 days but they worked for 3 days, so
they completed ½ work. Remaining ½ work is yet to be completed.
12 children’s 1 day work = 12*16 = 192
8 adults’ 1 day work = 12 * 8 = 96
192 : 96 = 2 : 1
2 Children = 1 Adult
4 children are added instead of 10 adults
4 children = 2 Adults
2A +6A = 8 adults are going to complete ½ work
8 adults took 12 days to complete the whole work.
They will require 6 days to complete the remaining half work.

Problem - 12

Anu can complete a work in 10 days. Manu is
25% more efficient than Anu and Sonu is 60%
more efficient than Manu. Working together,
how long would they take to finish the job?

Solution
Anu takes 10 days to complete work;
Manu takes 10*100/160 = 8 days to
complete work;
Sonu takes 8*100/160 = 5 days to
complete work.
1/10 + 1/8 + 1/5 = 4 /40 + 5/40+8/40 =
17/40
Number of Days = 40/17 = 2 6/17

Pipes and Cistern
Pipes and Cistern and Time and Work problems
and method of solving are same.

Only difference, instead of Man and Days, in
pipes and cistern, they will give pipes / taps
and cistern / tank.

Pipes and Cisterns
• P1 and P2 both working simultaneously which fills in
x hrs and empties in y hrs resp ( y>x) then net part
filled is 1/x – 1/y

• P1 can fill a tank in X hours and P2 can empty the full
tank in y hours( where x>y), then on opening both
pipes, the net part empties in hour 1/y -1/x

Problem 1
Taps A and B fills a bucket in 12 and 15 minutes
respectively. If both are opened and A is closed
after 3 minutes, how much further time would it
take for B to fill the bucket?

Solution
1 minute = A + B = 1/12 + 1/15 = 9 /60
3 minutes = 3*9/60 = 9 /20
Remaining Bucket = 1 – 9/20 = 11/20
Pipe B time Bucket
15 1
x 11/20
x = 11/20 *15 = 33/4
Time required to fill remaining bucket by B
alone = 8 ¼ minutes.

Problem 2

Three pipes A, B and C working together can fill a
cistern in 6 hours. After working together for 2 hours C
is closed and A & B fill it in 7 hours more. How many
hours will C alone take to fill the cistern?

Solution
A + B = 7 hours
A + B + C = 6 hours
Hours Tank
6 1
2 x
6x = 2
x = 1/3 tank is full, remaining tank = 1 -1/3 = 2/3
2/3 *1/7 = 2/21
A+B+C – A+B = 1/6 – 2/21 = 21 – 12 /126 = 9/126 = 1/14
Time taken by C alone to fill the cistern = 14 Hours

Problem - 3
A swimming pool is filled with three pipes with
uniform flow. The first two pipes operating
simultaneously fill the pool in the same time
during which the pool is filled by the third pipe
alone. The second pipe fills the pool five hours
faster than first pipe and 4 hours slower than the
third pipe. What is the time required by each
pipe to fill the pool separately?

A+B=C
Solution
Let the first pipe be X
X+5=X=X–4
A + B = C ------ 1/(x + 5) + 1/x = 1/(x – 4)
1/(x+5) + 1/x = x/x(X+5) + (x+5)/x(x+5)
= 2x +5/x(x+5)
Time taken = x(x+5) /2x+5
x(x+5)/2x+5 = (x-4)
x2 – 8x – 20 = 0
(x+10) (x – 2)
x = 10, -2
x = 10
A = 10 +5 = 15 hours
B = 10 Hours
C = 10 – 4 = 6 hours

Problem - 4

A ship 55 km from the shore springs a leak
admits 2 tons of water in 6 minutes. The ship can
manage 80 tons of water from sinking. The
pumps can throw out 12 tons an hour. Find the
average rate of sailing such that the ship may
reach the shore just to avoid sinking.

Solution
Rate of admission of water = 2/6 = 1/3
Rate of pumping out = 12/60 = 1/5
Rate of accumulation of water = 1/3 – 1/5 = 2/15
Time to accumulate 80 tons of water = 80/2/15
= 80*15/2 = 600 min = 10 hours
Average rate of sailing = 55/10 = 5.5 km /hr

Problem - 5

Two pipes A and B can fill a tank in 12 and 16
minutes respectively. Both pipes are opened
together, but 4 minutes before the tank is full,
one pipe A is closed. How much time will they
take to fill the tank?

Solution
B’s 4 minutes’ work = 1/16*4 = ¼
Remaining tank = 1 – ¼ = ¾
A + B = 1/12 + 1/16 = 48/7 to fill 1 tank
Tank Time
1 48/7
¾ x
x = 48*4/7*3 = 64/7 = 9 1/7 minutes

Problem - 6
An electric pump can fill a tank in 3 hours,
because of leak in the tank it took 3½ hours to
fill the tank. If the tank is full, how much time
will the tank take to empty it?

Solution
1/3 – 1/x = x – 3 / 3x
Time Taken = 3x /x – 3
3x / x – 3 = 7/2
6x = 7x – 21
x = 21 hours

Problem - 7

Two pipes A and B can fill a tank in 24 minutes
and 32 min respectively. If both pipes are opened
simultaneously after how much time, B should
be closed so that the tank is full in 18 minutes?

Solution

Part filled by A + B in x min + part filled by A in 18-x
min = 1
x/24 + x/32 + (18-x)1/24 =1
7x/96 + (18-x)/24 = 1
7x – 4x – 72/96 = 1
7x – 4x – 72 = 96
3x = 96 -72
x = 8 min

Problem - 8
Two pips can fill a cistern in 14 hours and 16
hours respectively. The pipes are opened
simultaneously and it is found that due to
leakage in the bottom, 32 minutes extra are taken
for the cistern to be filled up. If the cistern is full,
in what time will the leak empty it?

Solution
A + B = 1/14 + 1/0 = 15/112 = 112 /15
= 7 hours 28 minutes
32 min extra due to leakage, so total time taken
to fill the tank
= 7.28 + 32 min = 8 hours
15/112 – 1/8 = 15/112 – 14/112 = 1/112
= 112 hours

Problem - 9
Two pipes A and B fill a cistern in 10 minutes
and 15 minutes respectively, and a tap C can
empty the full cistern in 20 minutes. All the three
were opened for 1 minute and then the emptying
tap was closed. How many minutes more would
it take for the cistern to be filled?

Solution
1 minute work of A+B –C = 1/10 + 1/15 – 1/20 = 7/60
Remaining tank = 1 – 7/60 = 53/60
A + B = 1/10 + 1/15 = 6 min
A+B take 6 minutes to fill the tank
Time Cistern
6 1
x 53/60
x = 53/60*6 = 5 3/10

Problem - 10
A pipe can fill a bath in 20 minutes and another
can fill it in 30 minutes. A person opens both the
pipes alternatively for 1 minute each. When the
bath was full, he finds that the waste pipe was
open. he then closes the waste pipe and in 3
more minutes the bath is full. In what time would
the waste pipe empty it?

Solution
1st min A = 1/20
2nd min B = 1/30
2 minutes A + B = 1/20 + 1/30 = 5/60 = 1/12
2*12 = 12/12 so 24 minutes 1 bath will be filled
2 minutes = 1/12
3rd minutes = 1/20
= 1/12 + 1/20 = 5+3/60 = 8/60 = 2/15
2/15 of bath in 24 minutes
Empty the full bath in = 24*15/2 = 180 = 3 hours

Problem - 11
A tap can fill a tank in 6 hours. After half the
tank is filled, three more similar taps are opened.
What is the total time taken to fill the tank
completely?

Solution
1 tank 6 hours: half tank = 3 hours
4 taps = 1/6 + 1/6 + 1/6 + 1/6 = 4/6 = 2/3
To fill one tank = 3/2 = 1 ½ hours
To fill half tank 45 minutes
Total time taken to fill the tank = 3 hours 45
minutes

Problem - 12
A tank 10m by 6m by 3m has an inlet pump
which pours water at the rate of ¼ cubic metres
in a minute. An emptying pipe can empty the
tank when it is full in 4 hours. If the tank is full
and both pipes are open, how long will it take to
empty it?

Solution
Capacity of the tank = 10*6*3 = 180 cu.m
The emptying pipe can remove 180 cu.m in 4 hours
In 1 minute it can remove 180/4*60 cu.m (i.e) 3/4 cu.m
The inner pipe pours ¼ cu.m in 1 minute
¾ - ¼ = 2/4 = ½ cu.m of water will be removed in 1 min.
Empty the tank = 10*6*3/ ½ = 360 minutes = 6 hours

Problems on Trains
Important Formula:
1. Distance = Speed * Time
2. Speed = Distance / Time
3. Time = Distance / Speed
Time taken to Cross Each Other:
1. Opposite Direction (Moving Object)
= S1 + S2
2. Same Direction (Moving Object)
= S1 – S2 Cont…

Problems on Trains
3. Opposite Direction (both object with Length)
= L1 + L2 / S1 + S2
4. Same Direction (both object with Length)
= L1 + L2 / S1- S2
5. Both object with Length / One non moving Object
= L1 + L2 / S1
6. One object with Length / both object are moving
(opp.dire) = L1 / S1 + S2
7. One object with Length / both object are moving
(same.dire) = L1 / S1- S2

Problems on Trains
• Speed from km/hr to m/sec - ( * 5/18).

• Speed from m/sec to km/h, - ( * 18/5).
• Average Speed:-
Average speed = Total distance traveled
Total time taken

Problem - 1

Two trains start from stations A and B and
travels towards each other at speeds of 50 km/hr
and 60 km/hr respectively. At the time of their
meeting the second train has traveled 120 km
more than the first train. What is the distance
between Stations A and B?

Solution
Speed difference in 1 hr = 10 km
120 km difference will come after 12 hrs of
traveling. So both the trains have traveled 12
hours each.
Distance covered in 12 hrs,
Train 1 = 12*50 = 600
Train 2 = 12*60 = 720
Distance between stations A and B is = 600 +
720 = 1320

Problem - 2

Trains are running with speeds of 30 km/hr and
58 km/hr in the same direction. A man in the
slower train passes by the faster train in 18
seconds. What is the length of the faster train in
meters?

Solution
Speed of the train (same direction)
= 58 – 30 = 28 Km/hr
= 28*5/18 = 70/9 m/sec
Distance (Length of the train) = S * time
= 70/9*18 = 140 metres

Problem - 3

A train passes two bridges of lengths 800m and
400m in 100 sec and 60 sec respectively. What is
the length of the train?

Solution
Let the length of the train be x.

Speed = Distance / Time
(x + 800)/100 = x+400/60
x = 200 m

Problem - 4
A train does a journey without stoppage in 8
hours. If it had traveled 5 km an hour faster, it
would have done the journey in 6 hours 40 min.
What is its slower speed?

Solution
Let the Speed be x
Speed Time
x 8
x+5 6 hr 40 min
Sp Time
x 480
x+5 400
480x = 400(x+5)
480x – 400x = 2000
80x = 2000
x = 25
Slower Speed = 25

Problem - 5
Without stoppage a train travels with an average
speed of 80 km/hr and with stoppages, it covers
the same distance with an average speed of 60
km/hr. How many minutes per hour the train
stops?

Solution
Stoppage time / hour =
= Difference in speed / Faster Speed
= 80 – 60/80 = 20/80
=1/4
= 15 minutes

Problem - 6
Two trains of lengths 190m and 210m
respectively start from a station in opposite
directions on parallel tracks. If their speeds are
40 km/hr and 32 km/hr respectively, in what
time will they cross each other?

Solution
Time = Distance / Speed
Opp. Direction = S1 + S2
= 190 +210/40+32
= 400/72*5/18
= 400/20
= 20 Sec

Problem - 7
A train 100 m long takes 6 sec to cross a man
walking at 5 km/hr in a direction opposite to that
of the train. Find the speed of the train.

Solution
Speed = Distance / Time
100/(x+5)*5/18= 6
= 30 (x+5) = 1800
X + 5 = 1800/30 = 60
x = 60 – 5 = 55 km/hr

Problem - 8
Two trains running in opposite directions cross a
man standing on the platform in 27 sec and 17
sec respectively and they cross each other in 23
seconds. What is the ratio of their speeds?

Solution
Sp. of Slower train Sp. of Faster train
17 27
23(cross Each other)
27-23 = 4 23-17 = 6
2 : 3
Ratio of Faster and Slower train = 3:2

Problem - 9
Two trains of equal lengths takes 10 seconds and
15 sec respectively to cross a telegraphic post. If
the length of each train be 120m, in what time
will they cross each other traveling in opposite
direction?

Solution
S1 = 120/10 = 12
S2 = 120/15 = 8
Opposite Direction = S1 + S2 = 12 +8 = 20

Time taken to cross each other = D/S
120 + 120/20 = 12 Sec

Problem - 10
A train consists of 12 bogies, each bogie is 15
metres long. The train crosses the telegraphic
post in 18 sec. Due to some problem, two bogies
were detached. In what time will the train cross
the Telegraph post now?

Solution
12 bogies each 15 m long
Total length = 12*15 = 180
12 bogies 180 m long
10 bogies 150 m long
Length Time
180 18
150 x
180x = 18*150
x = 18*150/180
x = 15 sec

Problem - 11
A train 300 m long overtook a man walking
along the line in the same direction at the rate 4
km/hr and passed him in 30 sec. The train
reached the station in 15 min after it had passed
the man. In what time will the man reach the
station?

Solution
L = 300 m
Speed of the man = 4 km/hr
Time took to pass = 30 sec
15 min to reach station after crossing a man
Relative Speed in same direction = x – 4*5/18
m/sec
= 300/(x-4)*5/18 = 30
= x – 4 = 180/5, x – 4 = 36, x = 40 km/hr

Solution
Speed of the Train = 40 km/hour
Distance Time
40 60
x 15
60x = 40*15; x = 10 km = Distance to reach the Station
Speed Time
40 km/hr 15
4 km/hr X
= 4x = 40*15
x = 40*15/4 = 150 min = 2 ½ hrs

Problem - 12
A train starts from station X at the rate of 80
km/hr and reaches station Y in 48 minutes. If the
speed is reduced by 8 km/hr, how much more
time will the train take to return from station Y
to station X?

Solution
Speed Time
80 48
72 x
72x = 80*48 = x = 80*48/72 = 53 1/3
48 – 53 1/3 = 5 1/3 = 5 min 20 sec

Time and Distance

Speed:- Distance covered per unit time is called
speed.

Speed = distance/time (or)

Distance = speed*time (or)

Time = distance/speed

Problem - 1
In 1760m race, A can beat B by 44m while in a
1320m race, B can beat C by 30m. By what
distance will A beat C in a 880m race?

Solution
In 1760 m race, A = 1760m B = 1760 – 44 = 1716m
In 1320 m race, B = 1320, C = 1290m
B C
1716 x
1320 1290
C = 1716*1290/1320 = 1677m
A = 1760, B = 1716, C = 1677
In 1760 m race, A beats C by 83 m
In 880 m race, A will beat C by 41 ½ m

Problem - 2
Two trains traveling in opposite direction start
from a station at the same time with the speeds
of 40 km/hr and 120 km/hr respectively. What is
the distance between them after 15 minutes?

Solution
In 1 hour Both trains covers A + B = 40 + 120 =
160 km
Distance covered in 15 minutes = 160*15/60 =
16*15/6 = 40 km

Problem - 3
A hare preceded by a greyhound is 87.5m before
him. While the hare takes 4 leaps, the greyhound
takes 3 leaps. If the greyhound and the hare go
2¾ m and 1¾ m respectively in one leap, in how
many leaps will the greyhound overtake the
hare?

Solution
Dis. Covered by Hare = 4* 1¾ = 4*7/4 = 7
Dis. Covered by Greyhound = 3*11/4 = 33 / 4 = 8.25
Dis. gained by greyhound in 3 leaps
= 8.25 – 7 = 1.25 m
Dis. Leaps
1.25 3
87.5 x
x = 87.5 * 3 / 1.25 = 210 leaps

Problem - 4
A monkey is climbing up a greased pole ascends
5m and slips down 2m in alternate minute. If the
pole is 35 m high then when will the monkey
touch the top?

Solution
1st min = 5m up
2nd min = 2m down
= 2 minutes = 3m up
2*11 = 3*11 m
22 min = 33 m
Remaining = 35 -33 = 2m
Meter Min
5m 60 sec
2m x
x = 2*60/5 = 24 sec
Total time taken = 11 minutes 24 seconds

Problem - 5
Walking 6/7 of his usual rate a man is 25
minutes late. Find his usual time?

Solution
Usual time = Nr * Late
= 6*25 = 150 Min = 2 hr 30 Min

Problem - 6
A hare runs at 5 m/sec and a tortoise runs at 0.5
m/sec. They start from the same point in the
same direction. How far ahead will the hare be
after 30 min?

Solution
30*60 = 1800 Sec
Per sec difference = 5 – 0.5 = 4.5 m / sec
30 min = 4.5*30*60 / 1000 = 8.1 km

Problem - 7
Two cars A and B are running towards each
other from two different places 88 km apart. If
the ratio of the speeds of the cars A and B is 5 : 6
and the speed of the car B is 90 km / hr. At what
time will the two meet each other?

Solution
Ratio A : B=5 : 6
A : B = x : 90
x*6/11 = 90 x = 165
(Total Speed ) 165 – (B’s Speed) 90 = 75
Speed of car A = 75
Time to meet each other = 88/75+90 = 88/165hrs
88/165*60 = 32 min

Problem - 8
A train travelling at the rate of 60 km/hr, while
inside a tunnel meets another train of half its
length traveling at 90 km/hr and passes
completely in 4 ½ sec. Find the length of the
tunnel if the first train passes completely through
it in 4 minutes 37 ½ sec?

Solution
Relative Speed = 60 + 90 = 150
Dis. = 150*5/18*9/2 = 375/2 m
Length of the 1st train is = x
Length of the 2nd train is = x/2
X+ x/2 = 375/2
3x = 375
X = 375/3 = 125
Time taken by the first train to cross the tunnel,

Cont….

Solution
Speed = 60*5/18 = 50/3 m /sec
Time = 4 min 75/2 sec
= 4*60 +75/2 = 555/2
Distance traveled by 1st train is,
Sec Metre
1 50/3
555/2 x
50/3*555/2 = 4625
Length of the Tunnel = Total Dis. – Length of the Train
= 4625 – 125 = 4500m or 4.5 km

Problem - 9
A man standing on a 170 m long platform
watches that a train takes 7 ½ sec to pass him
and 21 seconds to cross the platform. Find the
Length of the train and its Speed?

Solution
Sp. to cross a man = x/15/2 = 2x/15
Sp. To cross a platform = 170+x/21
170 +x/21 = 2x/15
2550 + 15x = 42x
x = 850/9 m
Speed = 2* 850/15*9 = 12 16/27 m/sec

Problem - 10
A and B starts at the same time with speeds of 40
km/hr and 50 km/hr respectively. If in covering
the journey, A takes 15 minutes longer than B.
Find the Total distance of the Journey?

Solution
A (1 hour) covers = 40 km
B (I hour ) covers = 50 km
A took 15 minutes longer than B
Distance cover in 15 minutes = 40*1/4 = 10 km
Total Distance = 50 km

Problem - 11
A train 110 m in length passes a man at the rate
of 6 km/hr against it in 6 sec. How much time it
will take to pass another man walking at the
same speed in same direction?

Solution
Opp. Dir = x + 6 km /hr
x+6 *5/18*6 = 110 m
x = 60
Time taken to pass 2nd man
= 110/60-6 = 7 1/3 sec

Problem - 12
A train running at the speed of 20 m/sec crosses
a pole in 24 seconds less than the time it required
to cross a platform thrice its length at the same
speed. What is the length of the train?

Solution
Distance = Speed * Time
Length of the train be x
Time = x/20 sec
To cross the platform 3x+x/20 = 4x/20
To cross the Pole = x/20
Diff. in time to cross the platform and pole,
4x/20 – x/20 = 24
3x/20 = 24
3x = 20*24; x = 160 m

Boats and Streams
• Up stream – against the stream
• Down stream – along the stream
• u = speed of the boat in still water
• v = speed of stream
• Down stream speed (a)= u+v km / hr
• Up stream speed (b)=u-v km / hr
• u = ½(a+b) km/hr
• V = ½(a-b) km / hr

Problem - 1
The speed of a boat in still water in 8 km/hr. If it
can travel 20 km downstream at the same time as
it can travel 12 km upstream, what is the rate of
stream (in km/hr)?

Solution
Rate of stream = x km/hr
Sp. Of boat in still water = 8 km/hr
Upstream speed = 8 – x km/hr
Downstream Speed = 8+x km/hr
Given, 20/8+x = 12/8-x
= 160 – 20x = 96+12x
32x = 64x
X = 64/32 = 2 km/hr

Problem - 2
A boat’s crew rowed down a stream from A to B

and up again in 7 ½ hours. If the stream flows at

3km/hr and speed of boat in still water is 5 km/hr.

find the distance from A to B ?

Solution
Solution:
Down Stream = Sp. of the boat + Sp. of the stream
= 5 +3 =8
Up Stream = Sp. of the boat – Sp. of the stream
= 5-3 = 2
Let distance be x
Distance/Speed = Time
X/8 + x/2 = 7 ½
X/8 +4x/8 = 15/2
5x / 8 = 15/2
5x = 15/2 * 8
x =12

Problem - 3
A boat goes 40 km upstream in 8 hours and 36
km downstream in 6 hours. Find the speed of the
boat in still water in km/hr?

Solution
Solution:
Speed of the boat in upstream = 40/8 = 5 km/hr
Speed of the boat in downstream = 36/6 = 6
km/hr
Speed of the boat in still water = 5+6/2 = 5.5
km/hr

Problem - 4

A man rows to place 48km distant and back in
14 hours. He finds that he can row 4 km with the
stream in the same time as 3 km against the
stream. Find the rate of the stream?

Solution
Solution:
Down stream 4km in x hours. Then,
Speed of Downstream = 4/x km/hr,
Speed of Upstream = 3/x km/hr
48/ (4/x) + 48/(3/x) = 14 , x = 1/2
Speed of Down stream = 8,
Speed of upstream = 6
Rate of the stream = ½ (8-6) km/hr = 1 km/hr

Problem - 5

In a stream running at 2 km/h a motor boat goes
10 km upstream and back again to the starting
point in 55 minutes. Find the speed of the motor
boat in still water.

Solution
Solution:
Let the speed be x km/h
10/(x+2) + 10/(x-2) = 55/60
Solving the above equation,
x = 22 km/h

Problem - 6
A swimmer can swim a certain distance in the
direction of current in 5 hours and return the
same distance in 7 hours. If the stream flows at
the rate of 1 km/h, find the speed of the swimmer
in still water.

Solution
Solution:
Let the speed of the swimmer in still water be x
Distance = speed x time
Distance covered at Downstream = (x+1) * 5
Distance covered at Upstream = (x-1) * 7
5(x+1) = 7(x-1)
Solving the equation, x = 6
Speed of the swimmer = 6 km/h

Problem - 7
A person can row a boat D km upstream and the
same distance downstream in 5 hours 15
minutes. Also, he can row the boat 2D km
upstream in 7 hours. How long will it take to row
the same distance 2D km in downstream?

Solution
Speed of boat and stream be x and y km/hr
respectively.
Rate of Downstream = (x+y) km/hr
Rate of Upstream = (x – y) km/hr
Given,
D / x+y + D / x-y = 21/4 hrs ---------1
And, 2D/x-y = 7; D / x-y = 7/2------2
Subtracting equation 2 from 1
D/x+y = 21/4 – 7/2; D/x+y = 7/4
2D/x+y = 7/2 hrs

Problem - 8
A boat takes 19 hours for traveling downstream
from point A to point B and coming back to a
point C, midway between A and B. If the
velocity of the stream is 4 km/hr and the speed of
the boat in still water is 14 km/hr. What is the
distance between A and B?

Solution
Speed of Downstream = 14 + 4 = 18 km/hr
Speed of Upstream = 14 – 4 = 10 km/hr
x/18 + x/2/10 = 19
x/18 + x/20 = 19
10x+9x/180 = 19
19*180/19 = 180 km

Problem - 9
Speed of a boat in standing water is 9 km/hr and
the speed of the stream is 1.5 km/hr. A man row
to a distance of 105 km and comes back to the
starting point. Find the total time taken by him?

Solution
Sp. of upstream = 7.5 km/hr
Sp. Of Downstream = 10.5 km/hr
Total time taken = 105/7.5 + 105/10.5 = 24 hrs

Problem - 10
A man can row 40 km upstream and 55 km
downstream in 13 hours. Also, he can row 30 km
upstream and 44 km downstream in 10 hours.
Find the speed of the man in still water and the
speed of the current?

Solution
Let the rate of upstream = x km/hr and
rate of downstream = y km/hr
40/x + 55/y = 13-------1
30/x + 44/y = 10-------2
1*3 = 120/x + 165/y = 39
2*4 = 120/x + 176/y = 40
Subtracting 1 and 2, we get y = 11
40/x + 55/y = 13
40/x + 55/11 = 13; x = 5 km/hr
Rate in still water = ½(11+5) = 8 km/hr
Rate of Current = ½(11 – 5) = 3 km/hr

Problem - 11
A man can row 7 ½ km/hr in still water. If in a
river running at 1.5 km an hour, it takes him 50
minutes to row to a place and back, how far off
is the place?

Solution
Sp. of Downstream = 7.5 + 1.5 = 9 km/hr
Sp. of Upstream = 7.5 – 1.5 = 6 km/hr
Let required distance = x km
x/9 + x/6 = 50/60 ; 5x/18 = 5/6;
x = 5/6*18/5 = 3 km

Problem - 12
The speed of a boat in still water is 15 km/hr and
the rate of current is 3 km/hr. Find the distance
traveled downstream in 15 min?

Solution
Distance = Speed * Time
= 15 + 13 *15/60
= 28*1/4 = 7 km

Simple Interest and Compound
Interest

Simple / Compound Interest

Simple Interest = PNR / 100

Amount A = P + PNR / 100

When Interest is Compound annually:

Amount = P (1 + R / 100)n

C.I = A-P

Simple / Compound Interest

• Half-yearly C.I.:
Amount = P (1+(R/2)/100)2n

• Quarterly C.I. :

Amount = P (1+(R/4)/100)4n

Simple /Compound Interest
Difference between C.I and S.I for 2 years
= P*(R/100)2.

Difference between C.I and S.I for 3 years
= P{(R/100)3+ 3(R/100)2 }

Problem - 1

A sum of money amounts to Rs. 6690 after 3
years and to Rs. 10035 after 6 years on C.I. Find
the sum.

Solution
Solution:
P(1 + R/100)3 = 6690 -----------1
P(1 + R/100)6 = 10035 -----------2
Dividing (1 + R/100)3 = 10035/6690 = 3/2
Substitute in equation 1 then P(3/2)=6690
P = 6690 * 2/3 = 4460
The sum is Rs.4460

Problem - 2

What will be the difference between S.I and C.I

on a sum of Rs. 4500 put for 2 years at 5% per

annum?

Solution

C.I – S.I = P (R/100)2

Difference = Rs. 11.25

Problem - 3

The simple interest on a certain sum is 16 / 25

of the sum. Find the rate percent and time if

both are numerically equal.

Solution

R = N and I = 16/25 P
R = I * 100 / P*N
R*N = 64 where R = N
R=8
The rate of interest is 8%

Problem - 4

The difference between the compound and
simple interest on a certain sum for 2 years at the
rate of 8% per annum is Rs.80,What is the sum?

Solution
C.I – S.I = P(R/100)2
80 = P*(8*8/100*100)
The sum is Rs.12,500

Problem - 5

If a sum of money compounded annually
amounts of thrice itself in 3 years, in how many
years will it become 9 times itself?

Solution

A = 3P and find n when A = 9P

The number of years required = 6 years

Problem - 6

What will be the C.I on Rs. 15625 for 2½ years
at 4% per annum?

Solution

A = P(1+R/100)n
A = 15625 ((1+4/100)2 (1+4*1/2/100))
= 17238
CI = A –P = 17238 - 15625
Compound interest = Rs. 1613

Problem - 7

What is the S.I. on Rs. 3000 at 18% per

annum for the period from 4th Feb 1995 to

18th April 1995

Solution

Time = 24+ 31+17 = 73 days = 73/365 = 1/5
P = 3000; R = 18%;
= PNR/100 = 3000*1*18/100*5
The simple interest is Rs. 108

Problem - 8

Raja borrowed a certain money at a certain

rate of S.I. After 5 years, he had to pay back

twice the amount that he had borrowed. What

was the rate of interest?

Solution

A = 2P
A = P + PNR/100
2P = P(1+NR/100)
2 = (1+5*R/100)
1 = R/20
The rate of interest is 20%

Problem - 9

In simple interest what sum amounts to Rs. 1120
in 4 years and Rs. 1200 in 5 years?

CTS Question

Solution

Interest for 1 year = 1200 – 1120 = 80
Interest for 4 year = 80*4 = 320
A = 1120
P = A – P = 1120 – 320
The Principal is Rs. 800

Problem - 10

A simple interest amount for Rs. 5000 for 6
months is Rs. 200. What is the annual rate of
interest?

CTS Question

Solution

P = 5000; N = 6/12 = ½
I = 200
R = I *100 / P*N
=200*100*2/5000*1
= 40/5 = 8%
The annual rate of interest is 8%

Problem - 11

A man earns Rs. 450 as an interest in 2 years on
a certain sum invested with a company at the rate
of 12% per annum. Find the sum invested.

Solution

P = I*100/R*N
= 450*100/12*2
Principal = Rs. 1875

Problem - 12

If Rs. 85 amounts to Rs. 95 in 3 years, what
Rs. 102 will amount in 5 years at the same rate
percent?

Solution

Let P = Rs. 85; A = Rs. 95; I = 10/3 in 1 year
Rate = I*100/P*N = 4% ( app)
Amount = P+PNR/100 = 102+20 =122
Hence the amount in 5 years = Rs. 122

Numerical reasoning II

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