Time and Work (Part - 1) Video Lecture - CAT

hi welcome let's discuss about the
problems of time and work or you can say
pipe and cisterns
or you can say this kind of things using
the pipes and systems
and when discussing this topic or
discussion use three methods one like it
you can insane one day work or you can
say you need three method
Oh or I can say as a total
we'll discuss both the methods and we'll
try to solve the problem by using both
the methods what methods are essential
and according to the problem and the
conditions Barnum used the matter
whichever's whatever suits us at the
time for example let's take the example
they can do a piece of work in 10 days
you can do a piece of work in 10 days we
can do a piece of work in 15 days now
calculate calculate how long it will
take a plus B to do the work means a can
do the work Intendant speak and do the
work in 15 days and how long a plus B
can require to do them both knock
against a simple approach a can do the
work in 10 days it means in one day it
means in one day work or against a one
day worth of work done in a one day or
one day at work is 1 by 10 it means one
tenth of the work is done in one day
similarly for the be one day work is one
upon right now in one day it was run by
tenth of the work in one day do spun by
50 if I combine the a plus B write a
plus B work simultaneously then one day
of work would be one day work would be
summation of 1 by 10 plus 1 upon 15 now
here to take that here you can take the
calcium calcium is 30 but will not fit
but will just take the product here 10
into 15 I'll let you for a tell you to
have then this thing can entropy then
density the things are seen and I'll
tell you why then in this way instead of
lacking else 'i'm then perfectly the
content is 25 before my teeth or there
is one nice nowadays understand the
problem a + b combining a + b combining
in the working combined to the 1 by 6 of
the work in one day right so in one day
does 1 by 6 of the work therefore what
work will be done in 6 days therefore
the 8 + become an you if work together
the verb elimination 60s so now let me
tell you why I've said that do not take
the else you
we cross-multiply see here if i say t a
plus P time taken by a plus to be this
10 plus 15 P a plus B here is 6 days
it's a reciprocal of this quantity right
it's reciprocal then p FS B's 10 into 15
upon 10 plus 15 right now
an here's what T a TV into a pod PA so
this is the shortcut way to solve the
problems if you have the two persons
working together we can say the harmonic
mean or half of the harmony of T and T V
T or D FSDB is the Welkin together so
you can solve the problem by this method
but this is the second now second way
total work the method or what will work
method again same problem a can do the
piece of work in days
we can't do the piece of work in 15 days
right now now let assume the throttle
work now this is the work method let me
assume whatever now but as a mission the
rotten one we use usually take the LCM
of 10 in 15 or the rural work which
completely divided 10 as well as 15 and
gives the integer we do not divide any
fraction the plug-in complicated process
so generally we take the LCM or any
number which can be easily divisible 10
and 15 look at the integer so I assume
of 10 and 15 here is 30
I assume the total work is 30 units
fight so total work is 30 units the
computing that nine days so if one day
in one day it will finish 30 apart and
that is three units
similarly takes one day book it's thirty
upon 15 that is to Union it means be to
two units of work in a day a deuce that
is three units of work in effect now
combine a plus B T a plus B one day work
would be 3 plus 2 that is 5 unit so n
become manly work and work there five
units energy so number of days required
days require to finish how much work we
assume that 30 units right
we assumed 30 units yeah 30 units of
work to be done in 30 upon total work
done one day's five you that is 60s
answer this is a another way to solve
that problem right now
next it can be asked a can do the piece
of work against a 10 days be is unknown
quantity and a plus become man can do
the work and six days now how we solve
this method first thing first method
first method was Monday work to here we
can easily say the one day work of a is
1 upon 10 let's assume it for X days
here this is 1 plus X equal to 1 upon 6
right now 1 upon X is 1 upon 6 minus 1
upon and which is 1 upon 15 therefore X
is this is the first method right second
method was the formula you know that T a
plus B was de into PP upon PA plus PB
that is tan into X upon 10 plus X which
is equal to 6 on solving again we get X
equal to 15 now the third way total work
method a hence 10 days be workers are
non quantity and a plus B 60s now again
by the work that now LCM of n + 6 is 30
what about let's see with 30 units right
so total work is 30 units one day work
of a is
30 by 10 that is three units one day
worth of a plus B combined is a plus V
combined is 30 upon 6 that is five units
therefore one day work work of be alone
is five units three units is PI minus
three two years therefore number of days
required
thirty units of works one day work is
two units that is thirty upon two and
since fifty has the third right now
problem can be asked in another way also
discuss other way a can do the piece of
work in ten days we can do the work a
piece of work in twenty days right now a
and B work started together a and B has
started doing work together for the two
days
and then I left the job and we work
together for the two days and then I
left the job and B hello on finish the
work of calculate I'm required at the B
to finish the work I hope you understand
the problems let me again tell you you
can do a piece of work in ten days for
the piece of work in ten days he came
alone do the work in twenty days
now if ENB started the work and after
the two days he left the work that where
the time taken by B to finish the work
after that so we assume let me this X
stays this B X days no for the first
matter that is unitary method or one day
one in one day a can do the piece of
working we say what is 1 by 10 right B
can do the piece of work is one day for
a 120 right in this process they
together work for 2 days there for 2
days total work finish was a by a was to
a part-time hobby was to appoint 20
right and the left
he was alone working for the next s that
is X upon X upon 20 right which is equal
to total work that this one now this can
be calculated
easy right - upon 10 plus 2 upon 1 D
plus X 1 yd Esquire that is 4 plus 2 6
plus X is equal to 20 in an X equal to
therefore X require putting more days to
finish them right now other matter that
is a work done method can be started
together for the two days then he left
the job and be working for the XS we
have to go to the X now the vulgar
method lets has done work in 10 days we
alone can work in 10 days 20 days sorry
so I assume the work your total work yes
L seem up 10 and 20 that is 20 units
right in one day
a can work 20 a month and that is 2
units and in one day we can work twenty
of 120 that is one unit right so for the
two days the work done by envy together
is envy to get that work that is three
units for two - medicine for the two
days for the two days aim you're working
together the report they have raised
over 2
s 1 that is 3 but a one-day and for two
days which is 6 units right now total 1
this 20 units are left with the 620 -6
are left with the 14 units right 6 units
run/walk s 20 units that is body - six
days 20 minus 6 this 14 units now number
of days required than the 14 units work
probably be work one unit per day - what
in your super 14 days hence in both the
case answer is C so you apply the method
wherever you feel comfortable right in
the next lecture we'll discuss the more
complex problem of this time in work
please watch to another video also in
which you discuss about efficiency and
other parameters in volume solving the
time will work problem thank you