Understanding the Time and Work Problem

This question deals with the concept of time and work, specifically when the work rates of pairs of individuals are given, and we need to find the time taken by all three together.

Let's denote the amount of work A can do in one day as \(a\), the amount of work B can do in one day as \(b\), and the amount of work C can do in one day as \(c\). We can consider the total work to be completed as 1 unit.

Setting Up Equations from Given Information

We are given the time taken by pairs of individuals to complete the work:

• A and B together can do the work in 15 days. This means their combined work rate is \(\frac{1}{15}\) of the total work per day. So, we have the equation: \({a + b = \frac{1}{15}}\) (Equation 1)
• B and C together can do the work in 20 days. Their combined work rate is \(\frac{1}{20}\) of the total work per day. So, we have: \({b + c = \frac{1}{20}}\) (Equation 2)
• A and C together can do the work in 10 days. Their combined work rate is \(\frac{1}{10}\) of the total work per day. So, we have: \({a + c = \frac{1}{10}}\) (Equation 3)

Calculating the Combined Work Rate of A, B, and C

To find the combined work rate of A, B, and C together (\(a+b+c\)), we can add the three equations we formulated:

\({(a + b) + (b + c) + (a + c) = \frac{1}{15} + \frac{1}{20} + \frac{1}{10}}\)

This simplifies to:

\({2a + 2b + 2c = \frac{1}{15} + \frac{1}{20} + \frac{1}{10}}\)

We can factor out 2 on the left side:

\({2(a + b + c) = \frac{1}{15} + \frac{1}{20} + \frac{1}{10}}\)

Now, we need to add the fractions on the right side. The least common multiple (LCM) of 15, 20, and 10 is 60.

\({2(a + b + c) = \frac{1 \times 4}{15 \times 4} + \frac{1 \times 3}{20 \times 3} + \frac{1 \times 6}{10 \times 6}}\)

\({2(a + b + c) = \frac{4}{60} + \frac{3}{60} + \frac{6}{60}}\)

\({2(a + b + c) = \frac{4 + 3 + 6}{60}}\)

\({2(a + b + c) = \frac{13}{60}}\)

Now, divide by 2 to find the combined work rate of A, B, and C:

\({a + b + c = \frac{13}{60 \times 2}}\)

\({a + b + c = \frac{13}{120}}\)

This means that A, B, and C together complete \(\frac{13}{120}\) of the total work in one day.

Finding the Time Taken Together

If A, B, and C together complete \(\frac{13}{120}\) of the work in one day, the total time taken to complete the entire work (1 unit) is the reciprocal of their combined work rate.

Time taken by A, B, and C together = \(\frac{1}{\text{Combined work rate}} = \frac{1}{\frac{13}{120}}\)

Time taken = \(\frac{120}{13}\) days.

Calculating the Decimal Value

Let's calculate the decimal value of \(\frac{120}{13}\):

\({120 \div 13 \approx 9.2307...}\)

Rounding to two decimal places, the time taken is approximately 9.23 days.

Comparing with Options

Let's look at the given options:

1. 10.67 days
2. 10.91 days
3. 10.71 days
4. 9.23 days

Our calculated time of approximately 9.23 days matches option 4.

Revision Table: Key Time and Work Concepts

Additional Information: Solving System of Linear Equations

In this problem, we used a method of adding equations to find the combined work rate. Another way to approach problems like this, especially if individual rates (a, b, c) are needed, is to solve the system of linear equations:

1. \({a + b = \frac{1}{15}}\)
2. \({b + c = \frac{1}{20}}\)
3. \({a + c = \frac{1}{10}}\)

You could use substitution or elimination methods. For example, subtract Equation 2 from Equation 1:

\({(a+b) - (b+c) = \frac{1}{15} - \frac{1}{20}}\)

\({a - c = \frac{4-3}{60} = \frac{1}{60}}\) (Equation 4)

Now add Equation 3 and Equation 4:

\({(a+c) + (a-c) = \frac{1}{10} + \frac{1}{60}}\)

\({2a = \frac{6+1}{60} = \frac{7}{60}}\)

\({a = \frac{7}{120}}\)

Once you have 'a', you can substitute it back into Equation 1 and Equation 3 to find 'b' and 'c'. Then you can sum a+b+c. This method also works and might be useful if the question asked for individual times.