Two workers a and b working together completed a job in 7 days. If a worked thrice as efficiently as he actually did and b worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days. A alone could complete the work in how many days?
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7
Two workers A & B working together completed a job in 5 days. If A worked twice as efficiently as he actually did & B worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days. When could A alone complete the work?
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45
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6 ANSWERS

Louis M. Rappeport, PhD. Unicorn Wrangler
Answered Jun 8, 2018
According to the first part of the question:
1/A + 1/B=1/5
Then the second part says:
2/A +1/3B=1/3
So:
1/A=1/5–1/B
2/A=2/5–2/B
2/5 -2/B + 1/3B=1/3
6B-30+5=5B
B=25
1/5–1/25=4/25
A=1 / 4/25=25/4 days working alone
Answer
45
Follow
Request
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6 ANSWERS

Louis M. Rappeport, PhD. Unicorn Wrangler
Answered Jun 8, 2018
According to the first part of the question:
1/A + 1/B=1/5
Then the second part says:
2/A +1/3B=1/3
So:
1/A=1/5–1/B
2/A=2/5–2/B
2/5 -2/B + 1/3B=1/3
6B-30+5=5B
B=25
1/5–1/25=4/25
A=1 / 4/25=25/4 days working alone
Answered by
0
Concept:
In mathematics, there are different operation like addition, subtraction, multiplication and division.
Given:
A and B can together complete a work in 7 days.
Find:
Time taken by A to complete the work alone.
Solution:
Let:
total work= W.
The speed of A be x work/day.
The speed of B be y work/day.
Two workers A and B working together completed a job in 7 days.
This means:
W=7x+7y ...(1)
Now, if A worked thrice as efficiently as he actually did and B worked 1/3 as efficiently as he actually did, then:
W= 9x+y
Multiplying 7 on both sides:
7W=63x+7y ...(2)
Solving (1) and (2) we get,
6W=56x
W/x=56/6=28/3=9 (1/3) days
Therefore, A alone can complete the work in 9 (1/3) days.
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