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Question 1170885: Joe takes 30 minutes longer than Mary to do a task. If together they complete the same
task in 8 minutes, then how long does it take for Mary to finish the task by herself?
Found 2 solutions by josgarithmetic, ikleyn: Answer by josgarithmetic(39623) (Show Source): Answer by ikleyn(52835) (Show Source):
You can put this solution on YOUR website! .
Let m be the time for Mary to complete the job alone;
then the time for Joe is (x+30) minutes.
In each minute, Mary makes part of the job,
while Joe makes part of the job,
so together they make + part of the job per minute.
From the other side, the problem says that their combined rate of work is of the job per minute.
It gives us this equation
+ = .
It is the basic equation for the problem.
As soon as you get it (and as soon as you do understand on how to get it),
the setup is completed.
To solve the equation, multiply both sides by 8m*(m+30). You will get then
8(m+30) + 8m = m*(m+30)
8m + 240 + 8m = m^2 + 30m
m^2 + 14m - 240 = 0
(m+24)*(m-10) = 0. (after factoring left side)
So, you get two roots, m= -24 and m= 10, but only the positive root m= 10 minutes is meaningful.
ANSWER. Mary needs 10 minutes to complete the job working alone.
The problem is just solved.
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It is a standard and typical joint work problem.
There is a wide variety of similar solved joint-work problems with detailed explanations in this site. See the lesson
- Using quadratic equations to solve word problems on joint work
Read it and get be trained in solving joint-work problems.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this textbook under the topic
"Rate of work and joint work problems" of the section "Word problems".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
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