Some ques­tions do not tell you how long each per­son takes. Instead they com­pare the work­ers: "A is three times as good a work­man as B", "A is twice as good as C", "A can do half as much again as B". This les­son shows how to turn such a com­par­i­son into a ratio of times or of daily work, so that the ques­tion becomes an ordi­nary time and work prob­lem.

Effi­ciency and time

The effi­ciency of a worker means how much work they do in a fixed time, such as one day. A more effi­cient worker fin­ishes sooner, so effi­ciency and time are inversely pro­por­tional.

If A is kk times as good a work­man as B, then A's 1 day's work =k×= k \times B's 1 day's work, and A's time =1k×\displaystyle = \frac{1}{k} \times B's time.

For exam­ple, if A is three times as good as B and B takes 9090 days, A takes 3030 days. If the ratio of effi­cien­cies is 3:23 : 2, the ratio of times is 2:32 : 3.

Exam­ple 1

A is thrice as good a work­man as B and is there­fore able to fin­ish a piece of work in 6060 days less than B. Find the time in which they can do it work­ing together.

Sup­pose B fin­ishes the work in xx days. Then A fin­ishes it in (x−60)(x - 60) days. Being three times as good, A takes a third of B's time:

x−60=13x3x−180=x2x=180x=90\displaystyle \begin{aligned} x - 60 &= \frac{1}{3}x \\ 3x - 180 &= x \\ 2x &= 180 \\ x &= 90 \end{aligned}

So B takes 9090 days and A takes 90−60=3090 - 60 = 30 days. (Check: 3030 is a third of 9090.)

(A+B)’s 1 day’s work=130+190=3+190=490=245\displaystyle (A + B)\text{'s 1 day's work} = \frac{1}{30} + \frac{1}{90} = \frac{3 + 1}{90} = \frac{4}{90} = \frac{2}{45}

Together they fin­ish the work in 452=2212\displaystyle \frac{45}{2} = 22\frac{1}{2} days.

Exam­ple 2

A and B can do a piece of work in 1212 days, and B and C together can do it in 1515 days. If A is twice as good a work­man as C, find in what time B alone can do it.

(A+B)’s 1 day’s work=112,(B+C)’s 1 day’s work=115\displaystyle (A + B)\text{'s 1 day's work} = \frac{1}{12}, \qquad (B + C)\text{'s 1 day's work} = \frac{1}{15}

Since A is twice as good as C, A's 1 day's work equals C's 2 days' work. Replace A in the first equa­tion:

C’s 2 days’ work+B’s 1 day’s work=112\displaystyle \text{C's 2 days' work} + \text{B's 1 day's work} = \frac{1}{12}

Split C's two days as one day plus one day, and group one of them with B:

C’s 1 day’s work+(B+C)’s 1 day’s work=112\displaystyle \text{C's 1 day's work} + (B + C)\text{'s 1 day's work} = \frac{1}{12}

C’s 1 day’s work+115=112\displaystyle \text{C's 1 day's work} + \frac{1}{15} = \frac{1}{12}

C’s 1 day’s work=112−115=5−460=160\displaystyle \text{C's 1 day's work} = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}

Now B's share fol­lows:

B’s 1 day’s work=115−160=4−160=360=120\displaystyle \text{B's 1 day's work} = \frac{1}{15} - \frac{1}{60} = \frac{4 - 1}{60} = \frac{3}{60} = \frac{1}{20}

So B alone can fin­ish the work in 2020 days.

"Half as much again"

"Half as much again" means the amount itself plus half of it. If B does xx units of work in some time, then A, doing half as much again, does

x+x2=3x2\displaystyle x + \frac{x}{2} = \frac{3x}{2}

in the same time.

Exam­ple 3

A can do half as much again as B in the same time. B alone can do a piece of work in 1818 days. In how many days can they fin­ish it together?

In the same time, A does 3x2\displaystyle \frac{3x}{2} while B does xx, so the ratio of their work is

3x2:x=32:1=3:2\displaystyle \frac{3x}{2} : x = \frac{3}{2} : 1 = 3 : 2

For the same job, the ratio of their times is the reverse, 2:32 : 3. Write the times as 2k2k and 3k3k. B's time is 1818 days, so 3k=183k = 18, k=6k = 6, and A's time is 2k=122k = 12 days.

(A+B)’s 1 day’s work=112+118=3+236=536\displaystyle (A + B)\text{'s 1 day's work} = \frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36}

Together they fin­ish the work in 365=715\displaystyle \frac{36}{5} = 7\frac{1}{5} days.

Using one unknown rate

Another clean method is to give the slower worker a rate rr (work per day) and express every­one else in terms of rr.

Exam­ple 4

A is twice as good a work­man as B. Together they fin­ish a job in 1414 days. How long does each take alone?

Let B's 1 day's work be rr. Then A's is 2r2r, and together they do 3r3r per day.

3r=114  ⟹  r=142\displaystyle 3r = \frac{1}{14} \implies r = \frac{1}{42}

B alone takes 4242 days and A alone takes 2121 days.

Prac­tice

  1. A is twice as good a work­man as B. B takes 2424 days to do a job. How long does A take, and how long do they take together?
  2. A is thrice as good a work­man as B and takes 4040 days less than B to fin­ish a job. Find the time each takes, and the time they take together.
  3. The effi­cien­cies of A and B are in the ratio 4:34 : 3. A takes 66 days fewer than B to do a job. How long does each take?
  4. A can do half as much again as B in the same time. A alone takes 1010 days to do a job. How long does B take, and how long do they take together?
  5. P is one and a half times as effi­cient as Q. Together they fin­ish a job in 1010 days. How long does Q take alone?
  6. A and B together can do a job in 1515 days, and B and C together in 2020 days. A is twice as good a work­man as C. How long does B take alone?
  7. A is twice as good a work­man as B, and B is twice as good as C. Together the three fin­ish a job in 88 days. How long does C take alone?

Answers

Show answers
  1. A takes 1212 days. Together: 112+124=324=18\displaystyle \frac{1}{12} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}, so 88 days.
  2. If B takes xx days, x−40=x3\displaystyle x - 40 = \frac{x}{3}, so 2x=1202x = 120 and x=60x = 60. B 6060 days, A 2020 days. Together: 120+160=460=115\displaystyle \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}, so 1515 days.
  3. Times are in the ratio 3:43 : 4, say 3k3k and 4k4k. Then 4k−3k=64k - 3k = 6, so k=6k = 6: A 1818 days, B 2424 days.
  4. Work ratio A : B =3:2= 3 : 2, so time ratio 2:32 : 3. A's 1010 days is 2k2k, so k=5k = 5 and B takes 1515 days. Together: 110+115=16\displaystyle \frac{1}{10} + \frac{1}{15} = \frac{1}{6}, so 66 days.
  5. Let Q's rate be rr; P's is 32r\displaystyle \frac{3}{2}r. Then 52r=110\displaystyle \frac{5}{2}r = \frac{1}{10}, so r=125\displaystyle r = \frac{1}{25}. Q takes 2525 days.
  6. A's work == C's 2 days' work, so C's 1 day's work +120=115\displaystyle + \frac{1}{20} = \frac{1}{15}, giv­ing C's 1 day's work =160\displaystyle = \frac{1}{60}. B's 1 day's work =120−160=260=130\displaystyle = \frac{1}{20} - \frac{1}{60} = \frac{2}{60} = \frac{1}{30}, so 3030 days.
  7. Let C's rate be rr: B's is 2r2r, A's is 4r4r, total 7r=18\displaystyle 7r = \frac{1}{8}, so r=156\displaystyle r = \frac{1}{56}. C takes 5656 days.