Introduction / Context:
This is a straightforward time and work question in which one worker is twice as efficient as another. You are asked to determine the combined time taken when both work together, given the slower worker's time alone.
Given Data / Assumptions:
- B alone can finish the work in 18 days.
- A works twice as fast as B.
- Both A and B work together on the same piece of work.
- Each works at a constant daily work rate.
Concept / Approach:
Since A is twice as fast as B, A's daily work rate is double that of B. Convert B's 18-day completion time into a daily rate and multiply by 2 to find A's rate. Then add the two rates to get the joint rate and invert it to find the total time needed when both work together.
Step-by-Step Solution:
Let total work = 1 unit.
B's time alone = 18 days, so B's rate = 1 / 18 of the work per day.
A works twice as fast, so A's rate = 2 * (1 / 18) = 1 / 9 of the work per day.
Combined rate of A and B = 1 / 9 + 1 / 18.
Compute: 1/9 + 1/18 = (2 + 1) / 18 = 3 / 18 = 1 / 6.
Therefore, A and B together complete 1/6 of the work each day.
Time taken together to finish the entire work = 1 / (1 / 6) = 6 days.
Verification / Alternative check:
In 6 days, A does 6 * (1/9) = 6/9 = 2/3 of the work.
B does 6 * (1/18) = 6/18 = 1/3 of the work.
Total = 2/3 + 1/3 = 1 full work, so the answer is consistent.
Why Other Options Are Wrong:
4 days would require a combined rate of 1/4 which is larger than 1/6 and does not match A + B's computed rate.
8 or 10 days would imply too slow a combined rate, again contradicting the calculated sum of individual rates.
Common Pitfalls:
Some learners mistakenly assume A's time is half of B's and then average the two times, which is not correct for work rates.
Another error is adding days directly instead of adding rates.
Always convert days into daily work fractions and then combine the rates.
Final Answer:
A and B together can finish the work in 6 days.
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