work and time: A+B can finish, working on alt. days, in 17 d

[defeated_soldier]

A and B can finish a work, working on alternative days, in 17 days, where A works on the first day. Similarly they can finish the work, working on alternate days, in 53/3 days, where B works on the first day. C, working alone, can complete the work in 35 days. In how many days can the work be completed when A, B, and C work together?

A and B can finish a work , working on alternative days,in 17 days, where A works on the first day.

ok as A first starts, then B

So, in 17 days, A = 9 days and B = 8 days.

Similarly they can finish the work, working on alternate days, in 53/3 days, where B works on the first day

ok as B starts first , then B

in 53/3 days {B works 9 and A 16/3 days)

Ok...so there are just onformation about A, B's efficiency.

C working alone, can complete the work in 35 days. In how many days can the work be completed when A, B, and C work together

how do find now?

 

[stapel]

I'm not sure what you're doing on this exercise...? (Note: Complete sentences, with standard punctuation, can be very helpful in this regard.)

i) A and B complete the job in 17 days, with A working nine days and B working eight days. Do the standard "work problem" set-up:

. . .time to complete job (in days):
. . . . .A: a
. . . . .B: b
. . . . .A & B: [get back to this later]

. . .completed per units (days):
. . . . .A: 1/a
. . . . .B: 1/b
. . . . .A & B: 1/a + 1/b

In our case, we have A working one day more than B, so:

. . .(one day of A) + (8 days each for A and B)
. . . . .= 1(1/a) + 8(1/a + 1/b) = (one job done)

. . . . .1/a + 8/a + 8/b = 9/a + 8/b = 1

ii) Now we do the same thing, but with B working the extra day. Since 53/3 = 17²/₃, this means that 16²/₃ days were split between A and B, with A working 8²/₃ of them. Then:

. . .(one day of B) + (8²/₃ days for A and 8 days for B) = (one job completed)
. . . . .= 1(1/b) + (26/3)(1/a) + (8)(1/b)
. . . . .= 11/b + 26/(3a) = 1

This gives you two equations in two unknowns:

. . . . .9/a + 8/b = 1
. . . . .11/b + 26/(3a) = 1

Solve for the values of "a" and "b".

iii) If C can complete the job alone in 35 days, then clearly C's rate is 1/35 of the job per day. (That is, c = 1/35.)

Using the values you have found (above) for "a" and "b", complete the chart:

. . .completed per time unit (days):
. . . . .A: a = ??
. . . . .B: b = ??
. . . . .C: c = 1/35
. . . . .together: 1/a + 1/b + 1/35 = ??

Once you determine how much they finish per day, you can find how many days they take to complete the job.

Hope that helps!

Eliz.

 

[defeated_soldier]

Hi

I have been able to solve it now

i got A , B , C together can solve it in 7 days.

and individually A=35/3 days B=35 days,C=35 days .



Sometimes it becomes difficult task for me what Unknown variable needs to be assumed.......For example, in this case you instantly took A can complete work in a days ,B can complete work in b days individually and then it becomes easier to frame eqns.......and in fact this was the catch here..........and i was stuck near around that.


But how do i think what unknow variables i need to take .........How do i think ?

can i assume like this ,

WORK , TIME problem : always assume individual workers time to complete task...and then look for keys given in the question.

PROFT, SALES problem : always start with cost=C or cost=100 and then look for keys given in the question.

this way you dont have to cry for what unknown variable to should take .




Kindly , please validate my above statements and does that makes any sense ....if you add some comments /advice .....that would excellent.

thanks for the help.

 

[TchrWill]

Re: work and time: A+B can finish, working on alt. days, in

A and B can finish a work, working on alternative days, in 17 days, where A works on the first day. Similarly they can finish the work, working on alternate days, in 53/3 days, where B works on the first day. C, working alone, can complete the work in 35 days. In how many days can the work be completed when A, B, and C work together?

A and B can finish a work , working on alternative days,in 17 days, where A works on the first day.

ok as A first starts, then B

So, in 17 days, A = 9 days and B = 8 days.

Similarly they can finish the work, working on alternate days, in 53/3 days, where B works on the first day

ok as B starts first , then B

in 53/3 days {B works 9 and A 16/3 days)

Ok...so there are just onformation about A, B's efficiency.

C working alone, can complete the work in 35 days. In how many days can the work be completed when A, B, and C work together

how do find now?

Background
<< 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

Treating A and B as one person completing the job in 17 or 15/3 days together:

A, B and C will take (AB)C/((AB) + C) hours to complete the job, A working the first day and (BA)/(BA + C), B working the first day.

(AB)C/((AB) + C) = 17(35)/(17 + 35) = 11.44 days

(BA)C/((BA) + C) = (15/3)35/((15/3) + 35) = 7.95 days.

 

[sathikbasha]

alternate days

A and B can finish a work, working on alternative days, in 17 days, where A works on the first day. Similarly they can finish the work, working on alternate days, in 53/3 days, where B works on the first day. C, working alone, can complete the work in 35 days. In how many days can the work be completed when A, B, and C work together?

A and B can finish a work , working on alternative days,in 17 days, where A works on the first day.

ok as A first starts, then B

So, in 17 days, A = 9 days and B = 8 days.

Similarly they can finish the work, working on alternate days, in 53/3 days, where B works on the first day

ok as B starts first , then B

in 53/3 days {B works 9 and A 16/3 days)

Ok...so there are just onformation about A, B's efficiency.

C working alone, can complete the work in 35 days. In how many days can the work be completed when A, B, and C work together

how do find now?

here 8A+8B+17-8+8)A=1 =>9A+8B=1(if A start first ),8B+8A+(53/3-8-8)B=1(if B starts first)=>26/3A+8B=1, solve these two eq. A=1/35,B=26/(8*35)
and C=1/35(gn). so add these A,B,C we got 21/(35*4)=6and2/3 days if they worked together finished the job in 6+2/3 days