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Work

by heshamelaziry » Mon Nov 02, 2009 6:55 pm
Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

OA [spoiler]2XZ / (5X - 10Z)[/spoiler]
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Re: Work

by Resurgent » Tue Nov 03, 2009 1:17 am
A takes xhrs to complete the job
B takes 5yhrs to complete the same job (given, 1/5 of the work in y hrs)

Together, A & B can complete the job in 5xy/(x+5y) hrs, i.e. half of the job can be completed in 5xy/(2*(x+5y)) hrs. This is equal to z.

So, z=5xy/(2*(x+5y))

Bring y to one side of the equation & you will get 2xy/(5*(x-2z))

Hope this helps.
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Re: Work

by life is a test » Tue Nov 03, 2009 2:11 am
Resurgent wrote:A takes xhrs to complete the job
B takes 5yhrs to complete the same job (given, 1/5 of the work in y hrs)

Together, A & B can complete the job in 5xy/(x+5y) hrs, i.e. half of the job can be completed in 5xy/(2*(x+5y)) hrs. This is equal to z.

So, z=5xy/(2*(x+5y))

Bring y to one side of the equation & you will get 2xy/(5*(x-2z))

Hope this helps.
Resurgent could u pls explain how you rearrange the eqn z=5xy/(2*(x+5y)) for y?

thks.
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Re: Work

by Resurgent » Tue Nov 03, 2009 2:33 am
life is a test wrote:
Resurgent could u pls explain how you rearrange the eqn z=5xy/(2*(x+5y)) for y?

thks.
Bring all the y's to one end, take the common & equate.

z=5xy/(2*(x+5y))
==> 2xz+10zy = 5xy
==> 2xz = y(5x-10z)
==> y = 2xz/(5x-10z)
==> y = 2xz/(5*(x-2z))

Hope this helps.
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by sanjana » Tue Nov 03, 2009 3:06 am
Rate Time Work
Mach A 1/X X 1
Mach B 1/5Y Y 1/5

Combined rate : (1/x+1/5y)*z = 1/2 from the given statement.

Simplify and isolate y and you will get the OA.
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