A->15.
B->20.
LCM of 15,20 is 60.
A can do 4 parts and B can do 3 parts.
A and B together complete 7 parts in 1 day.
for 4 days 28 parts can complete.
60-28 = 32(leftover work).
32/60 = 8/15.

(12)

Bhasaid: 2 weeks ago

Thanks all for explaining the answer.

(8)

Anonymsaid: 3 weeks ago

A, completes the work in 15 days so, the work done by A in 1 day = 1/15.

Similarly, B completes the work in 20 days therefore work done by B in 1 day = 1/20.

A and b work together = 1/15+1/20 i.e. => 14/120 => 7/60.
Now, they both worked for 4 days which will be = 4*7/60 => 7/15.

Let's suppose the total work is x and they have already worked for 4 days.
Therefore remaining work is x -7/15 (for more clarity take x-7/15 = 0) => 8/15.



Which method is efficient to find the answer

Both methods seem to approach the problem of determining how much work A and B can complete together, but they are structured slightly differently. Let's break down both approaches to evaluate their efficiency:

### Method 1: 
This method approaches the problem by:
1. Finding the LCM of the total days it takes for A and B to complete the work.
2. Determining the parts each of them can do in a day (A does 4 parts, B does 3 parts).
3. Calculating the total parts done in 4 days and the leftover work (28 parts completed out of 60 total parts).
4. Then computing the remaining work as a fraction of the total work.

### Method 2:
This method uses the rate of work done by each person:
1. A completes the work in 15 days, so the work done by A in 1 day = \( \frac{1}{15} \).
2. B completes the work in 20 days, so the work done by B in 1 day = \( \frac{1}{20} \).
3. Adding their work rates together: \( \frac{1}{15} + \frac{1}{20} = \frac{7}{60} \) of the total work completed in one day.
4. For 4 days, they complete \( 4 \times \frac{7}{60} = \frac{7}{15} \) of the total work.
5. Then, the remaining work is \( 1 - \frac{7}{15} = \frac{8}{15} \).

### Comparison:

- **Clarity and Ease**: Method 2 directly uses work rates and gives a clearer picture of how much work is done daily by both A and B. It is easier to understand conceptually because it relies on the basic principle of work rates and fractions.
  
- **Steps Involved**: Method 2 has fewer steps and less manual computation than Method 1, which involves calculating parts, days, and the leftover work in terms of parts. In contrast, Method 1 may feel less straightforward because it converts the days into parts, which can add complexity.

- **Efficiency**: Method 2 is more efficient for this kind of problem because it directly calculates how much of the work is done by A and B together, avoids unnecessary intermediate steps, and leads quickly to the remaining fraction of work.

### Conclusion:
**Method 2 is more efficient** because it uses the fundamental work-rate formula and involves fewer steps. It directly solves for how much work A and B complete together and what remains, making it easier to follow and quicker to compute.

---

Do you want details on any part of this solution? Here are some related questions you might find interesting:

1. What is the concept of LCM in work problems, and how does it help?
2. How can you calculate the combined work rates of multiple workers?
3. How do you determine the remaining work when a portion is completed?
4. What is the significance of work rates like \( \frac{1}{15} \) or \( \frac{1}{20} \) in these problems?
5. Can this method be applied if more people are involved in the work?

**Tip**: When solving work problems, always focus on the rate of work (work done per day) for efficiency.

A->15. B->20. LCM of 15,20 is 60. A can do 4 parts and B can do 3 parts. A and B together complete 7 parts in 1 day. for 4 days 28 parts can complete. 60-28 = 32(leftover work). 32/60 = 8/15.

Learn how to solve a work and time problem where A completes work in 15 days and B in 20 days. They work together for 4 days, completing 7/15 of the total work. Understand LCM, combined work rates, and fractions in this detailed explanation. Suitable for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation