Understanding the Time and Work Problem

This question is a classic example of a time and work problem. These problems often involve calculating the time taken by individuals or groups to complete a task, based on their work rates.

The core concept here is that the amount of work done per unit of time is called the work rate. If a person can complete a task in $T$ days, their work rate per day is $\frac{1}{T}$. The total work is usually considered '1 unit' (completing the task).

Setting Up the Equations for Time and Work

We are given the following information about completing a task:

• A can complete the task in the same time as B and C together. Let $T_A$, $T_B$, and $T_C$ be the time taken by A, B, and C alone, respectively. This means $T_A = T_{B+C}$. In terms of work rates, this translates to the work done by A in a day being equal to the combined work done by B and C in a day. If $W_A, W_B, W_C$ are the daily work rates of A, B, C respectively, then $W_A = W_B + W_C$.
• A and B together can complete the task in 10 days. Their combined work rate per day is $\frac{1}{10}$. So, $W_A + W_B = \frac{1}{10}$.
• C alone can complete the task in 60 days. C's work rate per day is $\frac{1}{60}$. So, $W_C = \frac{1}{60}$.

We need to find the time taken by B alone to complete the task, which is $T_B$. This means we need to find B's work rate, $W_B$, because $T_B = \frac{1}{W_B}$.

Solving for B's Work Rate using Equations

We have the following relationships based on the given information:

1. $W_A = W_B + W_C$
2. $W_A + W_B = \frac{1}{10}$
3. $W_C = \frac{1}{60}$

We can use these equations to find $W_B$. Let's substitute the value of $W_C$ from equation (3) into equation (1):

Substituting $W_C = \frac{1}{60}$ into the equation $W_A = W_B + W_C$, we get:

$\qquad W_A = W_B + \frac{1}{60}$

Now we have an expression for $W_A$ in terms of $W_B$. We can substitute this expression for $W_A$ into equation (2), which is $W_A + W_B = \frac{1}{10}$:

Substituting $W_A = W_B + \frac{1}{60}$ into $W_A + W_B = \frac{1}{10}$ gives:

$\qquad (W_B + \frac{1}{60}) + W_B = \frac{1}{10}$

Now, let's solve this equation for $W_B$. Combine the $W_B$ terms:

$\qquad 2W_B + \frac{1}{60} = \frac{1}{10}$

To isolate the term with $W_B$, subtract $\frac{1}{60}$ from both sides of the equation:

$\qquad 2W_B = \frac{1}{10} - \frac{1}{60}$

To perform the subtraction on the right side, we need a common denominator for 10 and 60. The least common multiple is 60. So, rewrite $\frac{1}{10}$ as $\frac{6}{60}$:

$\qquad 2W_B = \frac{6}{60} - \frac{1}{60}$

Now subtract the numerators:

$\qquad 2W_B = \frac{6-1}{60}$

$\qquad 2W_B = \frac{5}{60}$

Simplify the fraction $\frac{5}{60}$ by dividing both numerator and denominator by 5:

$\qquad 2W_B = \frac{1}{12}$

Finally, divide both sides by 2 to find $W_B$:

$\qquad W_B = \frac{1}{12} \div 2$

$\qquad W_B = \frac{1}{12} \times \frac{1}{2}$

$\qquad W_B = \frac{1}{24}$

Calculating the Time Taken by B Alone

We have found that B's work rate is $W_B = \frac{1}{24}$ task per day. The time taken by B alone to complete the entire task is the reciprocal of B's daily work rate.

$\qquad T_B = \frac{1}{W_B} = \frac{1}{\frac{1}{24}} = 24$ days.

Thus, B alone can complete the task in 24 days.

Revision Table: Key Time and Work Concepts

Additional Information: Time and Work Problem-Solving Tips

Solving time and work problems often involves setting up equations based on the work rates of individuals or groups. Here are some useful tips:

• Represent the total work as 1 unit.
• If someone takes $T$ days, their daily work rate is $\frac{1}{T}$.
• If multiple people work together, add their individual work rates to get the combined work rate.
• If someone stops working or joins later, calculate the work done in phases.
• Problems involving pipes filling or emptying tanks are similar; filling is positive work rate, emptying is negative.

Always check if the question asks for the time taken by an individual, a group, or the time to complete a specific fraction of the work.