Motion and consecutive number....Help !!!

[brb1873]

I am stuck on this problem can someone help????

Three people together can do a certain job in three hours. If one of them can do the job in 6 hours and another can do it in 10 hrs, how long will the third person take to do the job alone?

Help Please !!!!

:roll:

 

[galactus]

\(\displaystyle \L\\\frac{1}{6}+\frac{1}{10}+\frac{1}{t}=\frac{1}{3}\)

solve for t.

 

[Deleted member 4993]

I'll put alittle explanation to Galactus's solution.

All 3 of those together can finish 1/3 of the work in 1 hour........(A)

given #1 takes 6 hours to complete the job.

Then #1 can finish 1/6 fraction of the job in 1 hour......(B)

given #2 takes 10 hours to complete the job.

Then #2 can finish 1/10 fraction of the job in 1 hour......(C)

Assume #3 takes 't' hours to complete the job.

Then #3 can finish 1/t fraction of the job in 1 hour......(D)

Using (B), (C) & (D) in (A),

1/6 + 1/10 + 1/t = 1/3

Now isolate '1/t'

1/t = 1/3 - 1/6 - 1/10

1/t = (10-5-3)/30 = 2/30 ... Now continue..

 

[Denis]

Another way; when they were working together:
#1 did 3/6 = 1/2 of job
#2 did 3/10 of job
So #3 did 1 - 1/2 - 3/10 = 1/5 of job

So #3 takes 3 / (1/5) = 15 hours

 

[TchrWill]

Three people together can do a certain job in three hours. If one of them can do the job in 6 hours and another can do it in 10 hrs, how long will the third person take to do the job alone?

The basics:


<< If it takes me 2 hours to paint a room and you 3 hours, ow long will it take to paint it together? >>

Method 1:

1--A can paint the house in 5 hours.
2--B can paint the house in 3 hours.
3--A's rate of painting is 1 house per A hours (5 hours) or 1/A (1/5) houses/hour.
4--B's rate of painting is 1 house per B hours (3 hours) or 1/B (1/3) houses/hour.
5--Their combined rate of painting is 1/A + 1/B (1/5 + 1/3) = (A+B)/AB (8/15) houses /hour.
6--Therefore, the time required for both of them to paint the 1 house is 1 house/(A+B)/AB houses/hour = AB/(A+B) = 5(3)/(5+3) = 15/8 hours = 1 hour-52.5 minutes.

Note - T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.

Three people version

It takes Alan and Carl 40 hours to paint a house, Bill and Carl 80 hours to paint the house, and Alan and Bill 60 hours to paint the house. How long, to the nearest minute, will it take each working alone to paint the house and how long will it take all three of them working together to paint the house?

1--The combined time of two efforts is derived from one half the harmonic mean of the two individual times or Tc = AB/(A + B), A and B being the individual times of each participant.
2--Therefore, we can write
AC/(A + C) = 40 or AC = 40A + 40C (a)
BC/(B + C) = 80 or BC = 80B + 80C (b)
AB/(A + B) = 60 or AB = 60A + 60B (c)
3--From (a) and (c), 40C/(C - 40) = 60B/(B - 60)
4--Cross multiplying, 40BC - 2400C = 60BC - 2400B or BC = 120(B - C)
5--Equating to (b) yields 120(B - C) = 80(B + C)
6--Expanding and simplifying, 40B = 200C or B = 5C
7--Substituting into (b), 5C^2 = 400C + 80C = 480C making 5C = 480 or C = 96.
8--Therefore, B = 480 and A = 68.571
9--The combined working time of three individual efforts is derived from Tc = ABC/(AB + AC + BC)
10--Therefore, the combined time for all three to paint the house is
Tc = 68.571(480)96/[(68.571x480) + (68.571x96) + 480x96)) = 36.923 hours = 36 hr - 55.377 min

Use Tc -- ABC/(AB + AC + BC) for your problem.

6(10)C /(6x10) + 6xC + 10xC) = 3

Solve for C