SOLUTION: FOUR MEN AND 8 WOMEN CAN FINISH A PIECE OF WORK JOINTLY IN 5 DAYS WHILE 3 MEN AND 2 WOMEN CAN FINISH THE SAME WORK JOINTLY IN 10 DAYS. FIND THE TIME TAKEN BY ONE MAN ALONE AND TH

Algebra ->  Matrices-and-determiminant -> SOLUTION: FOUR MEN AND 8 WOMEN CAN FINISH A PIECE OF WORK JOINTLY IN 5 DAYS WHILE 3 MEN AND 2 WOMEN CAN FINISH THE SAME WORK JOINTLY IN 10 DAYS. FIND THE TIME TAKEN BY ONE MAN ALONE AND TH      Log On


   

Question 1188017: FOUR MEN AND 8 WOMEN CAN FINISH A PIECE OF WORK JOINTLY IN 5 DAYS WHILE 3 MEN AND 2
WOMEN CAN FINISH THE SAME WORK JOINTLY IN 10 DAYS. FIND THE TIME TAKEN BY ONE MAN
ALONE AND THAT OF ONE WOMAN ALONE TO FINISH THE SAME WORK.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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FOUR MEN AND 8 WOMEN CAN FINISH A PIECE OF WORK JOINTLY IN 5 DAYS WHILE 3 MEN AND 2
WOMEN CAN FINISH THE SAME WORK JOINTLY IN 10 DAYS. FIND THE TIME TAKEN BY ONE MAN
ALONE AND THAT OF ONE WOMAN ALONE TO FINISH THE SAME WORK.
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Let m be the rate of work of one man and let w be the rate of work of one woman.


Then we have this system of 2 equations in 2 unknowns


    4m + 8w = 1%2F5     (1)

    3m + 2w = 1%2F10    (2)


To run of the denominator, I will multiply eq(1) by 5 (both sides) and will multiply eq(2) by 20.
I will get then


    20m + 40w = 1           (3)

    60m + 40w = 2           (4)


Next, I will subtract eq(3) from eq(2).  I will get


    40m = 1,   hence  m = 1%2F40.


Now from eq((3),

    20%2A%281%2F40%29 + 40w = 1,


which gives then

    40w = 1 - 20%2F40 = 1%2F2,   w = 1%2F80.


The problem is just solved: a single man will do the job in 40 days, working alone;

                            a single woman will do it in 80 days.

Solved.