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31 May 2017, 00:54
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Hi, some where I found wonderful explanation on multiple rate problem, but they lack any further explanation. It would be grateful if the quant wizards can explain them with examples.
If A can finish a work in X time and B can finish the
same work in Y time then both of them together can
finish that work in (X*Y)/ (X+Y) time.
If A can finish a work in X time and A & B together can
finish the same work in S time then B can finish that
work in (XS)/(X-S) time.
If A can finish a work in X time and B in Y time and C in
Z time then all of them working together will finish the
work in (XYZ)/ (XY +YZ +XZ) time
If you found this post great, kindly appreciate me with an kudos
If A can finish a work in X time and B can finish the
same work in Y time then both of them together can
finish that work in (X*Y)/ (X+Y) time.
If A can finish a work in X time and A & B together can
finish the same work in S time then B can finish that
work in (XS)/(X-S) time.
If A can finish a work in X time and B in Y time and C in
Z time then all of them working together will finish the
work in (XYZ)/ (XY +YZ +XZ) time
If you found this post great, kindly appreciate me with an kudos
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
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31 May 2017, 07:54
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Bookmarks
Since the work done is same throughout the question so for simplicity work done can be taken as 1
Rate*Time = Work
Let Rate of A be A ,Rate of B be B and Rate of C be C
If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.
A can finish a work in X time
=> A * X = 1 (Work =1)
=> A = 1/X
B can finish the same work in Y
=> B * Y = 1 (Work =1)
=> B = 1/Y
When they both work together then their combined rate = Sum of their rates = A + B
(A+B)*T = 1 (T is the time they take to complete the work)
(1/X + 1/Y)*T =1
=> (X+Y)*T/XY = 1
=> T= XY/(X+Y)
If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.
A can finish a work in X time
=> A * X = 1 (Work =1)
=> A = 1/X
A & B together can finish the same work in S time
=> (A+B)*S = 1
=> A+B = 1/S
=> B = 1/S - A = 1/S - 1/X = (X-S)/XS
Let time taken by B to finish the work = T
B * T = 1
=> T = 1/B = 1/ (X-s)/XS = XS/(X-S)
If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time
A can finish a work in X time
=> A = 1/X (Same as above)
B in Y time and C in Z time
=> B = 1/Y and C = 1/Z
If they are working together then
(A+B+C)*T = 1
=> (1/X + 1/Y + 1/Z)*T=1
=> (YZ + XZ + XY)*T/XYZ = 1
=> T = XYZ/(YZ + XZ + XY)
Hope it helps!
Rate*Time = Work
Let Rate of A be A ,Rate of B be B and Rate of C be C
If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.
A can finish a work in X time
=> A * X = 1 (Work =1)
=> A = 1/X
B can finish the same work in Y
=> B * Y = 1 (Work =1)
=> B = 1/Y
When they both work together then their combined rate = Sum of their rates = A + B
(A+B)*T = 1 (T is the time they take to complete the work)
(1/X + 1/Y)*T =1
=> (X+Y)*T/XY = 1
=> T= XY/(X+Y)
If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.
A can finish a work in X time
=> A * X = 1 (Work =1)
=> A = 1/X
A & B together can finish the same work in S time
=> (A+B)*S = 1
=> A+B = 1/S
=> B = 1/S - A = 1/S - 1/X = (X-S)/XS
Let time taken by B to finish the work = T
B * T = 1
=> T = 1/B = 1/ (X-s)/XS = XS/(X-S)
If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time
A can finish a work in X time
=> A = 1/X (Same as above)
B in Y time and C in Z time
=> B = 1/Y and C = 1/Z
If they are working together then
(A+B+C)*T = 1
=> (1/X + 1/Y + 1/Z)*T=1
=> (YZ + XZ + XY)*T/XYZ = 1
=> T = XYZ/(YZ + XZ + XY)
Hope it helps!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
