if P and Q together can complete a job 3 days while P alone can do the same in 12 days, then how many days would required by Q to do the job alone?
Answers
Answer 15.39
Step-by-step explanation:
Let P+Q, together, take x days to do the work.
P alone will do the same work in (x+12) days while Q does it in (x+3) days.
P does [1/(x+12)]th of the work in 1 day.
Q does [1/(x+3)]th of the work in 1 day.
So P and Q together do [1/(x+12)]+[1/(x+3)] or [x+3+x+12]/[(x+12)(x+3)] =(2x+15)/[(x+12)(x+3)]th part of the work in 1 day.
So P and Q will take [(x+12)(x+3)]/(2x+15) days which is the same as x.
Or, [(x+12)(x+3)]/(2x+15)=x, or
x^2+15x+36 = 2x^2+30, or
x^2–15x-6=0
x = [15+(225+24)^0.5]/2
= [15+15.78]/2
= 15.39
Answer:
Work done by P per day =
12
W
Let Work done by Q per day =
d
q
W
where d
q
is Number of days taken by Q to complete Work Alone
P and Q can together do the Work per day =
8
W
Now, using the Given Information
⇒
12
W
+
d
q
W
=
8
W
⇒
12
1
+
d
q
1
=
8
1
Solving Further we get
⇒d
q
=24
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