Solving Time and Work Problems: Calculating Days to Complete Work

This problem involves calculating the total time taken to complete a piece of work when individuals work at different rates and one person leaves before the work is finished. We are given the time each person (P, Q, and R) takes to complete the work alone.

• P can do the work in 5 days.
• Q can do the work in 6 days.
• R can do the work in 4 days.

Calculating Individual Work Rates

The work rate of a person is the amount of work they can do in one day. It is the reciprocal of the time they take to complete the whole work alone.

• P's work rate = \(\frac{1}{5}\) of the work per day.
• Q's work rate = \(\frac{1}{6}\) of the work per day.
• R's work rate = \(\frac{1}{4}\) of the work per day.

Analyzing the Work Scenario

The problem states that all three (P, Q, and R) started the work together. However, Q left 1 day before the completion of the work. This means:

• P and R worked for the entire duration until the work was completed.
• Q worked for the entire duration except for the last day.

Let's assume the total number of days the work was completed in is \(T\) days.

• P worked for \(T\) days.
• Q worked for \(T-1\) days.
• R worked for \(T\) days.

Setting up the Equation

The total work done is the sum of the work done by P, Q, and R over the days they worked. The total work is considered as 1 unit.

Work done by P in \(T\) days = (P's rate) \(\times\) (Number of days P worked) = \(\frac{1}{5} \times T = \frac{T}{5}\)

Work done by Q in \(T-1\) days = (Q's rate) \(\times\) (Number of days Q worked) = \(\frac{1}{6} \times (T-1) = \frac{T-1}{6}\)

Work done by R in \(T\) days = (R's rate) \(\times\) (Number of days R worked) = \(\frac{1}{4} \times T = \frac{T}{4}\)

The total work done is the sum of these amounts, which equals 1 (the total work).

So, the equation is:

$$\frac{T}{5} + \frac{T-1}{6} + \frac{T}{4} = 1$$

Solving the Equation for T

To solve for \(T\), we need to find a common denominator for the fractions. The least common multiple (LCM) of 5, 6, and 4 is 60.

Multiply the entire equation by 60 to eliminate the denominators:

$$60 \times \left(\frac{T}{5}\right) + 60 \times \left(\frac{T-1}{6}\right) + 60 \times \left(\frac{T}{4}\right) = 60 \times 1$$

Simplify each term:

$$12T + 10(T-1) + 15T = 60$$

Distribute the 10 in the second term:

$$12T + 10T - 10 + 15T = 60$$

Combine the terms with \(T\):

$$(12T + 10T + 15T) - 10 = 60$$

$$37T - 10 = 60$$

Add 10 to both sides of the equation:

$$37T = 60 + 10$$

$$37T = 70$$

Divide by 37 to find \(T\):

$$T = \frac{70}{37}$$

So, the work was completed in \(\frac{70}{37}\) days.

Conclusion

The total number of days required to complete the work is \(\frac{70}{37}\) days. This fits one of the given options.

Revision Table: Time and Work Concepts

Additional Information on Time and Work Problems

Time and work problems often involve calculating the efficiency of individuals or groups and how long it takes them to complete a task, sometimes under varying conditions (like people joining or leaving). Key to solving these is understanding the inverse relationship between time and work rate: the faster someone works, the less time they take.

• Efficiency: Often synonymous with work rate. Higher efficiency means more work done per unit time.
• Work Done: Calculated by multiplying the work rate by the time spent working. If work rate is \(R\) and time is \(T\), work done is \(R \times T\).
• LCM Method: For problems involving fractions of work, assuming the total work to be the LCM of the individual times can simplify calculations, turning fractions into integers. In this problem, we could have assumed total work is 60 units (LCM of 5, 6, 4). P does 12 units/day, Q does 10 units/day, R does 15 units/day.
  • Let total days be T.
  • Work by P in T days = \(12T\)
  • Work by Q in T-1 days = \(10(T-1)\)
  • Work by R in T days = \(15T\)
  • Total work = \(12T + 10(T-1) + 15T = 60\)
  • \(12T + 10T - 10 + 15T = 60\)
  • \(37T - 10 = 60\)
  • \(37T = 70\)
  • \(T = \frac{70}{37}\)
• Leaving/Joining: When individuals leave or join, the calculation is split into phases, or you can account for the total work done by each person over the entire duration they were involved. The latter method was used in the primary solution above.