Quick Method to solve the Questions
SMART LEARNING WITH B M GAUR - 9462900411
If a person can do a piece of work in ‘m’ days, he can do (1/m) part of the work in 1 day.
Example 1: Ram will do a piece of work in 15 days, what part of work will he do in two days?
Solution : Here, Man × Days = Man days1 × 15 = 15, as given work will take 15 days
Therefore in one day, 1/15th of the work will be done
and in 2 days, 1×2/15=2/15of the work will be done.
If the number of persons engaged to do a piece of work be increased (or decreased) in a certain ratio the time required to do the same work will be decreased (or increased) in the same ratio.
If A is twice as good a workman as B, then A will take half the time taken by B to do a certain piece of work.Solution : Since Ram is twice as good, he will do the work in 30/2=15 days
If A and B can do a piece of work in X and Y days respectively while working alone, they will together take
xy/(x+y) days to complete it.
Proof: A’s 1 day of work = 1/x
B’s 1 day of work = 1/y
So (A + B)’s total 1 day work =(1/x+1/y) of the total work.
Let total work be W.
Now, (1/x+1/y) of W can be finished in 1 day
W (total work) can be finished in 1/(1/x+1/y = xy/(x+y) days.
Example 3: Shyam will do a piece of work in 30 days; Ram can do same work in 15 days, in how many days can both do the work?
Solution : As per the formula, required days
30×15
|
450
| ||
————
|
=
|
——
|
= 10 days
|
30-15
|
45
|
If A , B , C can do a piece of work in X , Y , Z days respectively while working alone , they will together take
xyz
| |
—————
|
Days to finish it
|
Xy+yz+zx
|
SMART LEARNING WITH B M GAUR - 9462900411
Example 4: Shyam will do a piece of work in 30 days; Ram can do same work in 15 days, Bhuvan can do the same work in 10 days, in how many days can all three do the work?
Solution : As per the formula, required days
30×15×10
|
4500
| ||
——————————
|
=
|
———
|
= 5 days
|
[30×15+15×10+30×10
|
900
|
If A can finish work in X days and B in Y days and A, B and C together in S day then :
C can finish work alone in sxy / xy-sy-sxB + C can finish in sx / x-y and
A + C can finish in sy / y-s
Example 5: A and B together can do a piece of work in 6 days and A alone can do it in 9 days. In how many days can B alone do it?
Solution : (A + B)’s 1 day’s work = 1/6thpart of the whole work.A’s 1 day’s work = 1/9th part of the whole work.
| 1 | 1 | 3-2 | 1 | ||||
| ∴ B’s 1 day’s work = | — | - | — | = | —— | —th | Tpart of the whole work. |
| 6 | 9 | 18 | 18 |
By Direct Formula : B alone can do the whole work in
| 6×9 | 54 | ||||
| —— | = | — | =18 | days | |
| 9-6 | 3 |
Example 6: A and B can do alone a job in 6 days and 12 days. They began the work together but 3 days before the completion of job, A leaves off. In how many days will the work be completed?
(a) 6 days (b) 4 days(c) 5 days (d) 7 days
Solution : (a) Let work will be completed in x days. Then,
work done by A in (x – 3) days + work done by B in x days = 1 i.e
| x-3 | x | 3x-6 | |||||
| —— | + | — | = 1 | → | —— | =1 | → x = 6 days |
| 6 | 12 | 12 |
By Direct Formula :
Required time =
| 12(6+3) | ||
| —— | = 6 | days |
| 12+6 |
Example 7: A is half good a workman as B and together they finish a job in 14 days. In how many days working alone will B finish the job.
(a) 20 days (b) 21 days(c) 22 days (d) None of these
Solution : (b) Let B can do the work in x days and A can do the work in 2x days, then
| 1 | 1 | 1 | ||
| — | + | — | = | — (given) |
| x | 2x | 14 |
| 3 | 1 | |||
| x = | — | × | 14 | = 21 days |
| 2 | 14 |
By Direct Formula :
Time taken by B =
| 1 | ||||
| 14( | — | + | 1) | = 21 days |
| 2 |
Example 8: 10 men can finish a piece of work in 10 days, where as it takes 12 women to finish it in 10 days. If 15 men and 6 women undertake to complete the work, how many days they will take to complete it?
(a) 7 days (b) 5 days(c) 4 days (d) 6 days
Solution : (b) It is clear that 10 men = 12 women or 5 men = 6 women
⇒ 15 men + 6 women = (18 + 6) i.e. 24 women
Now 12 women can complete the work in 10 days
∴24 women will do it in 5 days.
By Direct Formula :
| 10×12×10 | |||
| Required time = | —————— | = | 5 days |
| 10×6+12×15 |
M1 D1 W2 = M2 D2 W1
If T1 and T2 are the working hours for the two groups then
M1 D1 W2 T1 = M2 D2 W1 T2
Similarly, M1 D1 W2 T1 E1 = M2 D2 W1 T2 E2,
Where E1 and E2 are the efficiencies of the two groups.
If the number of men to do a job is changed in the ratio a : b, then the time required to do the work will be in the ratio b : a, assuming the amount of work done by each of them in the given time is the same, or they are identical.
A is K times as good a worker as B and takes X days less than B to finish the work. Then the amount of time required by A and B working together is K×X / K²-1 days.
If A is n times as efficient as B, i.e. A has n times as much capacity to do work as B, A will take 1/n of the time taken by B to do the same amount of work.
SMART LEARNING WITH B M GAUR - 9462900411
Solution : This example has an extra variable ‘time’ (hrs a day), so the ‘basic-formula’ can’t work in this case. An extended formula is being given :
M1 D1 T1 W2 = M2 D2 T2 W1
Here, 5 × 6 × 6 × 16 = 12 × D2 × 8 × 10
| 5×6×6×16 | |||
| ∴ D2 = | —————— | = | 3 days |
| 12×8×10 |
TIME AND WORK -DIFFERENT APPROACH
Problems on Time and Work
are a common feature in most of the standard BANKING exams. If you are well
versed with the basics and have practised these problems during your
preparation, they give you an easy opportunity to score and also save time.
Here, I will try and give you the basic fundas with the help of examples. Let
us start with a very basic problem:
NEXTGEN CAREER COACHING -for- SSC-BANK PO-CLERK -RAILWAY
bhagirathprayash@gmail.com - 9462900411 - www.facebook.com/bhagirathmal |
|---|
THE STANDARD APPROACH
Problem 1: A takes 5 days to complete a piece of work and B takes 15
days to complete a piece of work. In how many days can A and B complete the
work if they work together?
Standard Solution: Let us consider Work to be 1 unit. So if W = 1 Unit and A
takes 5 days to complete the work then in 1 day A completes 1/5th of the
work. Similarly B completes 1/15th of the work. If they work together, in one
day A and B can complete (1/5 + 1/15 = 4/15) of the work. So to complete 1 unit
of work they will take 15/4 days.
THE LCM APPROACH
New method: Let us assume W = 15 units, which is the LCM of 5 and 15.
Given that total time taken
for A to complete 15 units of work = 5 days
⇒ A’s 1 day work = 15/5 = 3 units
Given that total time taken
for B to complete 15 units of work = 15 days
⇒ B’s 1 day work = 15/15 = 1 unit
⇒ (A + B)’s 1 day work = 3 + 1 = 4 units⇒ 15 units of work can be done in 15/4 days.
Many solve Time and Work
problems by assuming work as 1 unit (first method) but I feel it is faster to
solve the problems by assuming work to be of multiple units (second method).
This would be more evident when we solve problems which are little more complex
than the above one.
Problem 2: X can do a work in 15 days. After working for 3 days he is
joined by Y. If they complete the remaining work in 3 more days, in how many
days can Y alone complete the work?
Solution: Assume W = 15 units.
(Note: You can assume work to be any number of units but it is
better to take the LCM of all the numbers involved in the problem so that you
can avoid fractions) X can do 15 units of work in 15 days
⇒ X can do 1 unit of work in 1 day
(Note: If I had assumed work
as 13 units for example then X’s 1 day work would be 13/15, which is a fraction
and hence I avoided it by taking work as 15 units which is easily divisible by
15 and 3)
Since X worked for 6 days,
total work done by X = 6 days × 1 unit/day = 6 units.
Units of work remaining = 15
– 6 = 9 units.
All the remaining units of
work have been completed by Y in 3 days
⇒ Y’s 1 day work = 9/3 = 3 units.
If Y can complete 3 units of
work per day then it would take 5 days to complete 15 units of work. So Y takes
5 days to complete the work.
Problem 3: A, B and C can do a piece of work in 15 days. After all the
three worked for 2 days, A left. B and C worked for 10 more days and B left. C
worked for another 40 days and completed the work. In how many days can A alone
complete the work if C can complete it in 75 days?
Solution: Assume the total work to be 600 units. (LCM of all the
numbers)Then C’s 1 day work = 8 units.⇒ (A + B + C)’s 1 day work = 40 units.
A, B, C work together in the
first 2 days⇒ Work done in the first 2 days
= 40 × 2 = 80 units
C alone works during the last
40 days⇒ Work done in the last 40 days
= 40 × 8 = 320 units
Remaining work = 600 – (320 +
80) = 200 units
This work is done by B and C
in 10 days.
⇒ (B + C)’s 1 day work = 20 units
⇒ A’s 1 day work = (A + B + C)’s 1 day work – (B +
C)’s 1 day work = 40 units – 20 units = 20 units
⇒ A can do the work of 600 units in 30 days.
NEXTGEN CAREER COACHING -for- SSC-BANK PO-CLERK -RAILWAY
bhagirathprayash@gmail.com - 9462900411 - www.facebook.com/bhagirathmal |
|---|
Problem 4: Gerrard can dig a well in 5 hours. He invites Lampard and Rooney who can dig 3/4th as fast as he can to join him. He also invites Walcott and Fabregas who can dig only 1/5th as fast as he can (Inefficient gunners you see ) to join him. If the five person team digs the same well and they start together, how long will it take for them to finish the job?
Solution: Let the work be 100 units.
Gerrard’s 1 hour work = 100/5
= 20 units
Lampard and Rooney’s 1 day
work = 3/4 × 20 = 15 units.
Fabregas and Walcott’s 1 day
work = 1/5 × 20 = 4 units.
In one day all five of them
can do = 20 + 15 + 15 + 4 + 4 = 58 units of work. Hence they can complete the
work in 100/58 days.
THE MANDAYS APPROACH
I hope you got the knack of
it. Let us now see how to solve the second kind of problems in Time and Work –
the MANDAYS problems.
In these kinds of problems we
need to remember that the number of men multiplied by the number of days that
they take to complete the work will give the number of mandays required to
complete the work. The number of mandays required to complete a piece of work
will remain constant. We will try and understand this concept by applying it to
the next three problems.
A Very simple problem to
start with:
Problem 5: If 10 men take 15 days to complete a work. In how many days
will 25 men complete the work?
Solution: Given that 10 men take 15 days to complete the work. So the
number of mandays required to complete the work = 10 × 15 mandays. So
assume W = 150 mandays.
Now the work has to be done
by 25 men and since W = 150 mandays, the number of days to complete the work
would be 150/25 = 6 days.
Problem 6: A piece of work can be done by 8 boys in 4 days working 6
hours a day. How many boys are needed to complete another work which is three
times the first one in 24 days working 8 hours a day?
Solution: Assume the first piece of work to be
8 × 4 × 6 = 192 boy-day-hours.
The second piece of work = 3
(The first piece of work) = 3 × 192 = 576 boy-day-hours.
So W = 576 boy-day-hours.
If this work has to be completed in 24 days by
working 8 hours a day the number of boys required would be
576/(24 × 8) = 3 boys.
Problem 7: X can do a piece of work in 20 days working 7 hours a day.
The work is started by X and on the second day one man whose capacity to do the
work is twice that of X, joined. On the third day another man whose capacity is
thrice that of X, joined and the process continues till the work is completed.
In how many days will the work be completed, if everyone works for four hours a
day?
Solution: Since X takes 20 days working 7 hours a day to complete the
work, the number of day-hours required to complete this work would be 140
day-hours. Like in the two problems above, this is going to be constant
throughout. So, W = 140 day-hours.
Amount of work done in the
1st day by X = 1day × 4 hours = 4 day-hours2nd day, X does again 4
day-hours of work.
The second person is twice as
efficient as X so he will do 8 day-hours of work. Total work done on second day
= 8 + 4 = 12 day-hours.
Amount of work completed
after two days = 12 + 4 = 16 day-hours.3rd day, X does 4 day-hours of work.
Second Person does 8 day-hours of work.
Third person who is thrice as
efficient as X does 12 day-hours of work. Total work done on 3rd day = 4 + 8 +
12 = 24 day-hours
Amount of work completed
after 3 days = 16 + 24 = 40 day-hours
Similarly on 4th day the
amount of work done would be 4 + 8 + 12 + 16 = 40 day-hoursWork done on the 5th
day = 4 + 8 + 12 + 16 + 20 = 60 day-hours
Total work done after 5 days
= 4 + 12 + 24 + 40 + 60 = 140 day-hours = W.So it takes 5 days to complete the
work.
Remember that whenever there is money involved in a
problem, the money earned should be shared by people doing the work together in
the ratio of total work done by each of them.
Again I will explain this with the help of an example:
Problem 8: X can do a piece of work in 20 days and Y can do the same
work in 30 days. They finished the work with the help of Z in 8 days. If they
earned a total of Rs. 5550, then what is the share of Z?
Solution: Let work W =
120 units. (LCM of 20, 30 and 8)X’s 1 day work = 6 unitsY’s 1 day work = 4
units(X + Y + Z)’s 1 day work = 15 units.
So Z’s 1 day work = 15 – (6 +
4) = 5 unitsIn 8 days Z would have completed 5 units/day × 8 days =
40 units of workSince Z does 40/120 = 1/3rd of the work, he will receive 1/3rd
of the money, which is 1/3 x 5550 = Rs. 1850.
Problem 9: Sohan can
work for three hours non-stop but then needs to rest for half an hour. His wife
can work for two hours but rests for 15 min after that, while his son can work
for 1 hour before resting for half an hour. If a work takes 50 man-hours to get
completed, then approximately how long will it take for the three to complete
the same? Assume all of them all equally skilled in their work.
(a)
15
(b)
17
(c) 20 (d) 24
Solution: W = 50 man-hours
Since all of them are equally
skilled; in 1 hour they can do 3 man-hours of work if no one is resting.
It
will take them 50/3 = 16.6 hours to complete the work if they work
continuously.
But, since they take breaks
the actual amount of time would > 17 hours.
Option (a) and (b) are ruled
out.
Now let us calculate the
amount of work done in 20 hours.
Sohan does 3 man-hours in
every 3.5 hours (because he takes rest for half an hour on the 4th hour)
In 20 hours
(3.5 × 5 + 2.5) Sohan completes => 3 × 5 + 2.5 = 17.5
man-hours —- (1)
His wife completes 2
man-hours every 2.25 hours (because she rests on the 3rd hour)
In 20 hours
(2.25 × 8 + 2) she completes => 2 × 8 + 2 = 18
man-hours. —- (2)
Child completes 1 man-hours
every 1.5 hour.
In 20 hours (1.5 × 13
+ 0.5) he completes 1 × 13 + 0.5 = 13.5 man-hours of work. —— (3)
Adding 1, 2 & 3
In approximately 20 hours 49
man-hours will be completed; so the work can be completed in 20th hour.
PIPE AND APPROACH
Problem 10: There are three hoses, A, B and C, attached to a reservoir. A and B can fill the reservoir alone in 20 and 30 mins, respectively whereas C can empty the reservoir alone in 45 mins. The three hoses are kept opened alone for one minute each in the the order A, B and C. The same order is followed subsequently. In how many minutes will the reservoir be full?
Solution: These kinds of problems can be solved in the same way as we solve problems where one or more men are involved. A, B and C are equivalent to three people trying to complete a piece of work.SMART LEARNING WITH B M GAUR - 9462900411
The amount of work to be done would be the capacity of the reservoir. Lets assume capacity of the reservoir = W = 180 (LCM of 20, 30, 45) litres.
A can fill the reservoir in 20 mins Þ In 1 min A can fill 180/20 = 9 L. B can fill 180/60 = 6 L in a minute.
In one minute C can empty 180/45 = 4 L from the reservoir.
1st Minute => A is opened => fills 9 L
2nd Minute => B is opened =>fills another 6 L
3rd Minute => C is opened => empties 4 L
Hence every 3 minutes => (9 + 6 – 4 =) 11 litres are filled into the reservoir.
So in 45 minutes (11 × 15 =) 165 litres are filled.
In the 46th minute A is opened and it fills 9 litres. In the 47th minute B is opened and it fills 6 litres.
Hence the reservoir will be full in 47 minutes.
Problem 11: There is an empty reservoir whose capacity is 30 litres. There is an inlet pipe which fills at 5 L/min and there is an outlet pipe which empties at 4 L/min. Both the pipes function alternately for 1 minute. Assuming that the inlet pipe is the first one to function, how much time will it take for the reservoir to be filled up to its capacity?
Solution: The work to be done = Capacity of reservoir = W = 30 litres
1st Minute => inlet pipe opened => 5l filled
2nd minute => inlet pipe closed; outlet pipe opened => 4l emptied
In 2 minutes (5 litres -4 litres =) 1l is filled into the reservoir.
It takes 2 minutes to fill 1l => it takes 50 minutes to fill 25 litres into the tank.
In the 51st minute inlet pipe is opened and the tank is filled.
SMART LEARNING WITH B M GAUR - 9462900411
NEXTGEN CAREER COACHING -for- SSC-BANK PO-CLERK -RAILWAYSILENT-PURE-SURE-ONE TO ONE - LEARNING PROCESS AT ZERO COSTbhagirathprayash@gmail.com - 9462900411 - www.facebook.com/bhagirathmal |
|---|
No comments:
Post a Comment