Math Problem Statement

A and B started working on a project and finished it in 5 days. If A had worked twice as efficiently and B worked at only half of his efficiency then the same job could have been competed in 4 days. A alone can could do it in how many days?

Solution

Let the work done by A in 1 day be aa and by B in 1 day be bb.

Scenario 1: They finish the project in 5 days

In 1 day, A and B together complete a+ba + b work. Since they finish the project in 5 days, the total work done is: 5(a+b)=1(1 unit of work represents the entire project)5(a + b) = 1 \quad \text{(1 unit of work represents the entire project)} This gives the equation: a+b=15(Equation 1)a + b = \frac{1}{5} \quad \text{(Equation 1)}

Scenario 2: A works twice as efficiently, and B works at half efficiency

In this case, A does 2a2a work per day, and B does b2\frac{b}{2} work per day. They complete the project in 4 days, so: 4(2a+b2)=14(2a + \frac{b}{2}) = 1 Simplifying: 2a+b2=14(Equation 2)2a + \frac{b}{2} = \frac{1}{4} \quad \text{(Equation 2)}

Solving the equations

From Equation 1: a+b=15a + b = \frac{1}{5} From Equation 2: 2a+b2=142a + \frac{b}{2} = \frac{1}{4} We now solve these equations simultaneously.

Step 1: Multiply Equation 1 by 2 to align terms with Equation 2:

2a+2b=25(Equation 3)2a + 2b = \frac{2}{5} \quad \text{(Equation 3)}

Step 2: Subtract Equation 2 from Equation 3:

(2a+2b)(2a+b2)=2514(2a + 2b) - (2a + \frac{b}{2}) = \frac{2}{5} - \frac{1}{4} Simplifying: 2bb2=820520=3202b - \frac{b}{2} = \frac{8}{20} - \frac{5}{20} = \frac{3}{20} Multiply through by 2 to eliminate the fraction: 4bb=6204b - b = \frac{6}{20} 3b=3203b = \frac{3}{20} b=120b = \frac{1}{20}

Step 3: Substitute b=120b = \frac{1}{20} into Equation 1:

a+120=15a + \frac{1}{20} = \frac{1}{5} a=15120=420120=320a = \frac{1}{5} - \frac{1}{20} = \frac{4}{20} - \frac{1}{20} = \frac{3}{20}

Final Step: Find the time A alone takes to complete the work

If A does 320\frac{3}{20} of the work per day, the time A alone would take to finish the project is: 1320=203=623 days\frac{1}{\frac{3}{20}} = \frac{20}{3} = 6 \frac{2}{3} \text{ days}

Thus, A alone can complete the project in 6236 \frac{2}{3} days.

Would you like more details, or have any questions?

Here are 5 related questions to expand on this:

  1. How many days would B alone take to complete the project?
  2. What if both A and B improved their efficiency by 50%? How long would it take them to finish the job?
  3. If A had worked thrice as efficiently, how would the total time to complete the project change?
  4. What is the work rate of A and B combined in the second scenario where efficiencies are altered?
  5. How would you model this situation if A and B’s efficiencies changed continuously over time?

Tip: To solve work problems, always relate efficiency to the total job, and try using simultaneous equations to break down complex scenarios.

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Math Problem Analysis

Mathematical Concepts

Work and Time
Efficiency
Simultaneous Equations

Formulas

Work done per day by A and B: a + b = Total work / Time
New efficiency: 2a + b/2 = Total work / New time
Time to complete work: Time = Total work / Rate

Theorems

Simultaneous Equations

Suitable Grade Level

Grades 9-12