Lesson Solving rate of work problem by reducing to a system of linear equations

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Solving rate of work problem by reducing to a system of linear equations


Problem 1

A piece of work can be done by  6  men and  5  women in  6  days or  3  men and  4  women in  10  days.
In how many days can it be done by  9  men and  15  women?

          Assume that all the men have the same qualification and, therefore, the same rate of work.  Similarly, all the women have the same rate of work,  perhaps,
          different from  (or distinct of)  that of the men.

Solution

Let  m  be the the man's rate of work and  w  be the woman's rate of work.

Then we have the linear system of two equations in two unknowns,  in accordance with the given data:

system%286%2A%286m+%2B+5w%29+=+1%2C%0D%0A10%2A%283m+%2B+4w%29+=+1%29,

or

system%2836m+%2B+30w+=+1%2C%0D%0A30m+%2B+40w+=+1%29.

To solve it,  subtract the second equation of the last system from the first equation.  You will get

6m = 10w,     or     w = 6%2F10m = 3%2F5m.

Next,  substitute it into the first equation of the system,  and you will get

w = 1%2F90,   and then   m = 1%2F54.

Very good.  Now,  calculate the value of the expression   9m + 15w,   which is the subject of the problem's question.  It is

9m + 15w = 9.1%2F54 + 15.1%2F90 = 9%2F54 + 15%2F90 = 1%2F6 + 1%2F6 = 1%2F3.

Hence,  9  men and  15  women will complete the work in  3  days.


Problem 2

8  women and  12  girls can paint a large mural in  10  hours.  6  women and  8  girls can paint the same mural in  14  hours.
How long it would take to paint the mural one woman?  How long it would take to paint the mural one girl?

Solution

Let  w = number of hours it would take to paint the mural one woman.
Let  g = number of hours it would take to paint the mural one girl.

Then one woman paints  1%2Fw  of the mural per hour.  8  women paint  8%2Fw  of the mural per one hour.
Next,  one girl paints  1%2Fg  of the mural per hour.  12  girls paint  12%2Fg  of the mural per one hour.

In this way you get the system of two equations

8%2Fw + 12%2Fg = 1%2F10,       (1)

6%2Fw + 8%2Fg = 1%2F14.         (2)

for two unknowns  w  and  g.  To solve it,  multiply equation  (1)  by  2  (both sides)  and equation  (2)  by  (-3),  then add.
Thus you exclude the variable  g and get a single equation for the variable  w  only:

16%2Fw+-+18%2Fw = 2%2F10+-+3%2F14,     or

-2%2Fw = 14%2F70+-+15%2F70 = -1%2F70.

It gives  w = 140.  Hence, it takes  140  hours for 1 woman to paint the mural.

Now please complete the solution yourself.


Problem 3

If  5  men and  3  boys can reap  23  acres in  4  days and  3  men and  2  boys can reap 7  acres in  2  days,
how many boys must assist  7  men in order that they may reap  45  acres in  6  days?

Solution

Let  m  be  (an unknown)  rate of work for one man, measured in  acres%2Fday,
and let  b  be  (an unknown)  rate of work for one boy,  measured in same units.

Then we have a system of two linear equations in two unknowns  m  and  b

4*(5m + 3b) = 23,     (1)
2*(3m + 2b) =   7.     (2)

Or

20m + 12b = 23,     (1')
  6m +   4b =  7.     (2')

To solve it,  multiply  (2')  by  3  and then distract it from  (1').  You will get

20m - 18m = 23 - 3*7,     or

2m = 2,   hence,  m = 1.

Substitute it into  (2'),  and you will get  4b = 7 - 6 = 1,  hence,  b = 1%2F4.
Thus you obtained that  1  man can reap  1  acr in  1  day,  while  1  boy can reap  1%2F4  of acr in  1  day.
Then  7  men in  6  days can reap  7*6*1 = 42  acres.

How many boys must assist them to reap remaining  45-42 = 3  acres acres in  6  days?

Their number is  3%2F%28%281%2F4%29%2A6%29 = 4%2F2 = 2.

Answer.  Two boys.


My other lessons on rate-of-work problems in this site are
    - Rate of work problems
    - Using Fractions to solve word problems on joint work
    - Solving more complicated word problems on joint work
    - Using quadratic equations to solve word problems on joint work
    - Solving joint work problems by reasoning
    - Selected joint-work word problems from the archive
    - Joint-work problems for 3 participants
    - HOW TO algebreze and solve these joint work problems ?
    - Had there were more workers, the job would be completed sooner
    - One unusual joint work problem
    - Special joint work problems that admit and require an alternative solution method
    - Snow removal problem
    - Entertainment problems on joint work
    - Joint work word problems for the day of April, 1
    - OVERVIEW of lessons on rate-of-work problems

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