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A is 20% more efficient than B, but he works for five hours a day, whereas B works for 6 hours a day. If both A and B work together, they complete the work in 6 days. How many days are required to complete the work, if only A works for 3 hours a day?
- a)8 days
- b)10 days
- c)12 days
- d)20 days
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A is 20% more efficient than B, but he works for five hours a day, whe...
If B can do 'x' work per hour, A can do 1.2x.
Since A works only for 5 hours and B works for 6 hours, both can do the same amount of work in a day.
They together take 6 days to complete the work.
So, A alone takes 12 days working 5 hours a day.
So, he takes
working 3 hours a day, i.e. 20 days.
Since A works only for 5 hours and B works for 6 hours, both can do the same amount of work in a day.
They together take 6 days to complete the work.
So, A alone takes 12 days working 5 hours a day.
So, he takes
working 3 hours a day, i.e. 20 days.Free Test
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Community Answer
A is 20% more efficient than B, but he works for five hours a day, whe...
To solve this problem, let's first calculate the efficiency of A and B.
Let's assume that B can complete the work in x days. Since A is 20% more efficient than B, A can complete the same work in 0.8x days.
Let's calculate the work completed by A and B in one day.
A completes 1/0.8x = 1.25/x of the work in one day.
B completes 1/x of the work in one day.
Combining their work, A and B complete 1.25/x + 1/x = 1.25 + 1/x of the work in one day.
Given that A works for 5 hours a day and B works for 6 hours a day, we can calculate their combined efficiency for one day.
A's efficiency = (work done by A) / (time taken by A)
A's efficiency = (1.25 + 1/x) / 5
B's efficiency = (work done by B) / (time taken by B)
B's efficiency = (1/x) / 6
Since they work together for 6 days to complete the work, the total work done in 6 days is 1.
Therefore, (work done by A in 6 days) + (work done by B in 6 days) = 1
Let's calculate the work done by A and B in 6 days.
Work done by A = (A's efficiency) * (time taken by A) * (number of days)
Work done by A = [(1.25 + 1/x) / 5] * 5 * 6
Work done by A = 1.25 + 1/x * 6
Work done by B = (B's efficiency) * (time taken by B) * (number of days)
Work done by B = [(1/x) / 6] * 6 * 6
Work done by B = 1/x * 6
Given that (work done by A) + (work done by B) = 1, we can substitute the values and solve for x.
(1.25 + 1/x * 6) + 1/x * 6 = 1
1.25 + 12/x = 1
12/x = 1 - 1.25
12/x = -0.25
x = -12 / -0.25
x = 48
Therefore, B can complete the work in 48 days.
Now, let's calculate the number of days required for A to complete the work if he works for 3 hours a day.
Work done by A = (A's efficiency) * (time taken by A) * (number of days)
Work done by A = [(1.25 + 1/48) / 3] * 3 * d
Work done by A = 1.25 + 1/48 * d
Given that (work done by A) + (work done by B) = 1, we can substitute the values and solve for d.
(1.25 + 1/48 * d) + 1/48 * 48 = 1
1.25 + d/48 = 1
d/48 = 1 - 1.25
d/48 = -0.25
Let's assume that B can complete the work in x days. Since A is 20% more efficient than B, A can complete the same work in 0.8x days.
Let's calculate the work completed by A and B in one day.
A completes 1/0.8x = 1.25/x of the work in one day.
B completes 1/x of the work in one day.
Combining their work, A and B complete 1.25/x + 1/x = 1.25 + 1/x of the work in one day.
Given that A works for 5 hours a day and B works for 6 hours a day, we can calculate their combined efficiency for one day.
A's efficiency = (work done by A) / (time taken by A)
A's efficiency = (1.25 + 1/x) / 5
B's efficiency = (work done by B) / (time taken by B)
B's efficiency = (1/x) / 6
Since they work together for 6 days to complete the work, the total work done in 6 days is 1.
Therefore, (work done by A in 6 days) + (work done by B in 6 days) = 1
Let's calculate the work done by A and B in 6 days.
Work done by A = (A's efficiency) * (time taken by A) * (number of days)
Work done by A = [(1.25 + 1/x) / 5] * 5 * 6
Work done by A = 1.25 + 1/x * 6
Work done by B = (B's efficiency) * (time taken by B) * (number of days)
Work done by B = [(1/x) / 6] * 6 * 6
Work done by B = 1/x * 6
Given that (work done by A) + (work done by B) = 1, we can substitute the values and solve for x.
(1.25 + 1/x * 6) + 1/x * 6 = 1
1.25 + 12/x = 1
12/x = 1 - 1.25
12/x = -0.25
x = -12 / -0.25
x = 48
Therefore, B can complete the work in 48 days.
Now, let's calculate the number of days required for A to complete the work if he works for 3 hours a day.
Work done by A = (A's efficiency) * (time taken by A) * (number of days)
Work done by A = [(1.25 + 1/48) / 3] * 3 * d
Work done by A = 1.25 + 1/48 * d
Given that (work done by A) + (work done by B) = 1, we can substitute the values and solve for d.
(1.25 + 1/48 * d) + 1/48 * 48 = 1
1.25 + d/48 = 1
d/48 = 1 - 1.25
d/48 = -0.25
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A is 20% more efficient than B, but he works for five hours a day, whereas B works for 6 hours a day. If both A and B work together, they complete the work in 6 days. How many days are required to complete the work, if only A works for 3 hours a day?a)8 daysb)10 daysc)12 daysd)20 daysCorrect answer is option 'D'. Can you explain this answer?
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