Math Problem Statement

Solve question 1 with xy/x*y method

A and B can complete a job together in 12.5 days, B and C together in 18.75 days, and C and A together in 15 days. If D is only 40% as efficient as C, how many days will A, B, C, and D working together be able to complete the same job?

Solution

To solve the problem using the work-rate method, we need to find out how many days A, B, C, and D working together can complete the job. The problem gives us the work rates of pairs of workers and individual efficiency ratios.

Let's define the rates of work for A, B, and C:

  • A and B together can complete the job in 12.5 days.
  • B and C together can complete the job in 18.75 days.
  • C and A together can complete the job in 15 days.

First, we convert these to rates of work:

  • Rate of (A + B) = 112.5=225\frac{1}{12.5} = \frac{2}{25} of the job per day.
  • Rate of (B + C) = 118.75=475\frac{1}{18.75} = \frac{4}{75} of the job per day.
  • Rate of (C + A) = 115=115\frac{1}{15} = \frac{1}{15} of the job per day.

Let's denote:

  • A's rate as aa,
  • B's rate as bb,
  • C's rate as cc.

From the given rates: a+b=225a + b = \frac{2}{25} b+c=475b + c = \frac{4}{75} c+a=115c + a = \frac{1}{15}

Next, we'll solve these equations to find the individual rates:

  1. a+b=225a + b = \frac{2}{25}
  2. b+c=475b + c = \frac{4}{75}
  3. c+a=115c + a = \frac{1}{15}

First, express each equation in terms of a common denominator if needed:

  1. a+b=225a + b = \frac{2}{25}
  2. b+c=475b + c = \frac{4}{75}
  3. c+a=115=575c + a = \frac{1}{15} = \frac{5}{75}

Now, let's subtract (2) from (3): (c+a)(b+c)=575475(c + a) - (b + c) = \frac{5}{75} - \frac{4}{75} ab=175a - b = \frac{1}{75}

Now, solve equations (1) and (4): a+b=225a + b = \frac{2}{25} ab=175a - b = \frac{1}{75}

Add these two equations: (a+b)+(ab)=225+175(a + b) + (a - b) = \frac{2}{25} + \frac{1}{75} 2a=675=2252a = \frac{6}{75} = \frac{2}{25} a=125a = \frac{1}{25}

Substitute a=125a = \frac{1}{25} back into (1): 125+b=225\frac{1}{25} + b = \frac{2}{25} b=125b = \frac{1}{25}

Now, substitute b=125b = \frac{1}{25} into (2): 125+c=475\frac{1}{25} + c = \frac{4}{75} c=475375c = \frac{4}{75} - \frac{3}{75} c=175c = \frac{1}{75}

Thus, we have: a=125,b=125,c=175a = \frac{1}{25}, \, b = \frac{1}{25}, \, c = \frac{1}{75}

Next, we need to consider the efficiency of D, who is 40% as efficient as C: d=0.4×c=0.4×175=0.475=2375d = 0.4 \times c = 0.4 \times \frac{1}{75} = \frac{0.4}{75} = \frac{2}{375}

To find the combined rate of A, B, C, and D: a+b+c+d=125+125+175+2375a + b + c + d = \frac{1}{25} + \frac{1}{25} + \frac{1}{75} + \frac{2}{375}

Convert all rates to a common denominator (375): 125=15375\frac{1}{25} = \frac{15}{375} 175=5375\frac{1}{75} = \frac{5}{375} 2375=2375\frac{2}{375} = \frac{2}{375}

a+b+c+d=15375+15375+5375+2375=37375a + b + c + d = \frac{15}{375} + \frac{15}{375} + \frac{5}{375} + \frac{2}{375} = \frac{37}{375}

Thus, the combined rate is 37375\frac{37}{375} of the job per day. To find the number of days to complete the job: Time=1Rate=3753710.135days\text{Time} = \frac{1}{\text{Rate}} = \frac{375}{37} \approx 10.135 \, \text{days}

The correct answer closest to this is 927=13\frac{9}{27} = \frac{1}{3} but this might be a mismatch due to a calculation error or misinterpretation of the answer format.

Would you like more details or have any questions?

Here are some related questions to explore further:

  1. How do you find the work rate when given individual times to complete a job?
  2. What is the general method for solving work problems with multiple workers?
  3. How does changing the efficiency of one worker affect the overall completion time?
  4. Can you derive the formula for combined work rates for 'n' workers?
  5. How would the solution change if the efficiency of D was different?

Tip: When solving work-rate problems, always ensure all rates are converted to a common unit before combining them.

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Math Problem Analysis

Mathematical Concepts

Algebra
Work-Rate Problems

Formulas

Work rate = 1 / Time
Combined work rate = Rate1 + Rate2 + Rate3 + ...

Theorems

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Suitable Grade Level

High School