Understanding the Alternate Work Problem

This question asks us to determine the total time taken for two people, Pramod and Brajen, to paint a wall completely red when they work alternately for one hour each. We are given their individual times to paint the entire wall alone.

Problems involving alternate working require calculating the work done in a cycle (usually two turns, one for each person) and then figuring out how many cycles are needed to complete most of the work. The remaining work is then handled by the person whose turn it is.

Calculating Individual Painting Rates

First, let's find out how much of the wall each person can paint in one hour. This is their work rate.

• Pramod paints the wall in 12 hours. So, in 1 hour, Pramod paints $\(\frac{1}{12}\)$ of the wall.
• Brajen paints the wall in 16 hours. So, in 1 hour, Brajen paints $\(\frac{1}{16}\)$ of the wall.

Work Done in One Alternate Cycle

Pramod and Brajen work alternately for one hour each. A complete cycle consists of Pramod working for 1 hour and then Brajen working for 1 hour, totaling 2 hours.

In the first hour (Pramod's turn), work done is $\(\frac{1}{12}\)$.

In the second hour (Brajen's turn), work done is $\(\frac{1}{16}\)$.

Total work done in one 2-hour cycle = Work by Pramod + Work by Brajen

Total work in 2 hours = $\(\frac{1}{12} + \frac{1}{16}\)$

To add these fractions, we find a common denominator, which is the least common multiple (LCM) of 12 and 16. The LCM of 12 and 16 is 48.

$\(\frac{1}{12} = \frac{1 \times 4}{12 \times 4} = \frac{4}{48}\)$

$\(\frac{1}{16} = \frac{1 \times 3}{16 \times 3} = \frac{3}{48}\)$

Total work in 2 hours = $\(\frac{4}{48} + \frac{3}{48} = \frac{7}{48}\)$

So, in every 2-hour cycle of alternate work, $\(\frac{7}{48}\)$ of the wall is painted.

Calculating Number of Full Cycles

The total work to be done is painting the entire wall, which we represent as 1 unit of work.

In each 2-hour cycle, $\(\frac{7}{48}\)$ of the work is done.

We need to find out how many cycles can be completed before the work is finished or almost finished. Let $N$ be the number of cycles.

Total work after $N$ cycles = $N \times \(\frac{7}{48}\)$

We want to find the largest whole number $N$ such that $N \times \(\frac{7}{48}\)$ is less than or equal to 1. More specifically, we want to see how many full $\(\frac{7}{48}\)$ units fit into 1.

Divide total work (1) by work per cycle ($\(\frac{7}{48}\)$): $\(\frac{1}{7/48} = \frac{48}{7} \approx 6.857\)$

This means they can complete 6 full cycles before the work is finished. After 6 cycles (which take $6 \times 2 = 12$ hours), the work done is:

Work done after 6 cycles = $6 \times \(\frac{7}{48}\) = \(\frac{42}{48} = \frac{7}{8}\)$ of the wall.

Completing the Remaining Painting Work

After 12 hours (6 cycles), $\(\frac{7}{8}\)$ of the wall is painted. The remaining work is:

Remaining work = Total work - Work done = $\(1 - \frac{7}{8} = \frac{1}{8}\)$ of the wall.

After 6 complete cycles, it is the beginning of the 13th hour, and it is Pramod's turn to paint.

Pramod's rate is $\(\frac{1}{12}\)$ of the wall per hour.

In the 13th hour, Pramod works for 1 full hour and paints $\(\frac{1}{12}\)$ of the wall.

Work done after 13 hours (12 hours in cycles + 1 hour by Pramod) = Work done after 6 cycles + Work done by Pramod

Work done after 13 hours = $\(\frac{7}{8} + \frac{1}{12}\)$

Again, find a common denominator for 8 and 12, which is 24.

$\(\frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}\)$

$\(\frac{1}{12} = \frac{1 \times 2}{12 \times 2} = \frac{2}{24}\)$

Work done after 13 hours = $\(\frac{21}{24} + \frac{2}{24} = \frac{23}{24}\)$ of the wall.

The wall is not yet completely painted (it's $\(\frac{23}{24}\)$ done, less than 1). The remaining work is:

Remaining work = Total work - Work done = $\(1 - \frac{23}{24} = \frac{1}{24}\)$ of the wall.

It is now the beginning of the 14th hour, and it is Brajen's turn to paint the remaining $\(\frac{1}{24}\)$ of the wall.

Brajen's rate is $\(\frac{1}{16}\)$ of the wall per hour.

Time taken by Brajen to paint the remaining $\(\frac{1}{24}\)$ of the wall = $\(\frac{\text{Remaining Work}}{\text{Brajen's Rate}}\)$

Time taken by Brajen = $\(\frac{1/24}{1/16}\) = \(\frac{1}{24} \times \frac{16}{1}\) = \(\frac{16}{24}\)$ hours.

Simplify the fraction: $\(\frac{16}{24} = \frac{2 \times 8}{3 \times 8} = \frac{2}{3}\)$ hours.

Final Time Calculation

The total time taken is the sum of the time for the full cycles, Pramod's hour, and Brajen's final time.

Total time = 12 hours (for 6 cycles) + 1 hour (Pramod's turn) + $\(\frac{2}{3}\)$ hours (Brajen's turn)

Total time = 13 hours + $\(\frac{2}{3}\)$ hours.

Convert the fraction of an hour into minutes: $\(\frac{2}{3}\)$ hours = $\(\frac{2}{3} \times 60\)$ minutes = $\(2 \times 20\)$ minutes = 40 minutes.

So, the total time taken to paint the entire wall red is 13 hours and 40 minutes.

Revision Table: Key Steps

Additional Information on Alternate Work Problems

Alternate work problems are common in time and work topics. The key is to calculate the work done in one full cycle involving all individuals working alternately. Then, determine how many full cycles are needed without exceeding the total work (1 unit).

• Always calculate individual work rates first (work done per unit time, like per hour or per day).
• Identify the sequence of workers and the duration of each turn.
• Calculate the total work done in one full sequence or cycle.
• Divide the total work (1) by the work done in one cycle to find the number of cycles required. Use the floor value if it's not a whole number.
• Calculate the work done after completing the full cycles.
• Calculate the remaining work.
• Determine whose turn it is after the full cycles.
• Calculate the time taken by the next person(s) to complete the remaining work, considering they might not need their full turn duration.
• Sum up the time for full cycles and the time for the remaining work to get the total time.
• Pay attention to the units (hours, minutes, days) and convert if necessary.