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Question 975770: A and B can do a piece of work in 6 days,B and c together in 10 days,c and A together in 7 1/2 days.in how many days can c individually completes the work?
Found 4 solutions by lwsshak3, ikleyn, n2, greenestamps:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A and B can do a piece of work in 6 days,B and c together in 10 days,c and A together in 7 1/2 days.in how many days can c individually completes the work?
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let x=number of days A can individually complete the work
1/x=his work rate
let y=number of days B can individually complete the work
1/y=his work rate
let z=number of days C can individually complete the work
1/z=his work rate
..
1/x+1/y=1/6 (A & B)
1/z+1/y=1/10 (C & B)
1/x+1/z=1/7 (A & C)
..
1/x+1/y=1/6
1/z+1/y=1/10
subtract:
1/x-1/z=1/6-1/10=10/60-6/60=4/60=1/15
1/x-1/z=1/15
..
1/x-1/z=1/15
1/x+1/z=1/7
subtract:
-2/z=1/15-1/7=7/105-15/105-8/105
8z=2*105=210
z=210/8=26.25
How many days can C individually completes the work? 26.25
Answer by ikleyn(53879)  (Show Source):
You can put this solution on YOUR website! .
A and B can do a piece of work in 6 days, B and C together in 10 days, C and A together in 7 1/2 days.
in how many days can C individually completes the work?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The "solution" provided by @lwsshar3 is incorrect substantially and atrocious in form.
It is incorrect substantially because, in the course of solving the problem,
@lwsshar3 uses wrong number 7 instead of the given value of 7.5.
It is atrocious in form because nothing useful can be learned from his presentation.
Meanwhile, the problem itself is fantastically beautiful substantially
and admits an upper class solution, giving a food for a student's mind.
Let 'a' be the rate of work for A;
'b' be the rate of work for B;
'c' be the rate of work for C.
From the problem, we have this system of equations
a + b =  , (1)
b + c =  , (2)
a + c =  . (3)
Add equations (1), (2) and (3). You will get
2(a + b + c) = + + . (4)
Multiply both sides of (5) by 60. Notice that  = 8. You will get
120*(a + b + c) = 12 + 6 + 8 = 24,
which implies
a + b + c =  ,
or
a + b + c = . (5)
Now, to find 'c', subtract equation (1) from equation (5). You will get
c = -  =  =  .
Hence, the C's rate of work is 1/30 of the job per day,
which means that C can complete the entire job in 30 days working alone.
ANSWER. C can complete the entire job in 30 days working alone.
Solved correctly as it should be done and presented in a way as it is expected to be done
to teach you in a way how it should be done.
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Returning to the post by @lwsshar3, I only can say that it is a blatant gibberish,
inappropriate neither as a solution to the problem nor as a tool for teaching.
Answer by n2(89) (Show Source):
You can put this solution on YOUR website! .
A and B can do a piece of work in 6 days, B and C together in 10 days, C and A together in 7 1/2 days.
in how many days can C individually completes the work?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let 'a' be the rate of work for A;
'b' be the rate of work for B;
'c' be the rate of work for C.
From the problem, we have this system of equations
a + b = , (1)
b + c = , (2)
a + c = . (3)
Add equations (1), (2) and (3). You will get
2(a + b + c) = + + . (4)
Multiply both sides of (5) by 60. Notice that = 8. You will get
120*(a + b + c) = 12 + 6 + 8 = 24,
which implies
a + b + c = ,
or
a + b + c = . (5)
Now, to find 'c', subtract equation (1) from equation (5). You will get
c = - = = .
Hence, the C's rate of work is 1/30 of the job per day,
which means that C can complete the entire job in 30 days working alone.
ANSWER. C can complete the entire job in 30 days working alone.
Solved.
Answer by greenestamps(13355)  (Show Source):
You can put this solution on YOUR website!
Tutor @ikleyn has provided a solution using the standard formal method, using the rates of each worker.
Here is a solution by a different method that can usually be used to solve these "working together" problems.
Consider the least common multiple of the three given times needed for the three different pairs of workers to complete the job. That least common multiple is 30 days.
In 30 days...
(1) A and B together can do the job 30/6 = 5 times;
(2) B and C together can do the job 30/10 = 3 times; and
(3) A and C together can do the job 30/7.5 = 4 times
(4) From (2) and (3), A in 30 days can do the job 1 more time than B.
(5) Then, from (1) and (4), in 30 days A can do the job 3 times and B can do it 2 times.
Finally, from (5) and either (2) or (3), in 30 hours C can do the job 1 time.
ANSWER: 30 hours
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