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6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 5 men finish the same work?
- a)6
- b)8
- c)9
- d)15
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
6 children and 2 men complete a certain piece of work in 6 days. Each ...
6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
∴ 36+12X2 = 1 day
60 children can do the work in 1 day
Now, 5 men = 10 children
∴ 10 children can do the work in 6 days.
Most Upvoted Answer
6 children and 2 men complete a certain piece of work in 6 days. Each ...
Given: 6 children and 2 men complete a certain piece of work in 6 days.
Let's assume that a man can complete the work in 'm' days. Therefore, a child can complete the same work in '2m' days.
Total work done by 2 men in 1 day = 1/m
Total work done by 6 children in 1 day = 6/2m = 3/m
Total work done by 6 children and 2 men in 1 day = 1/m + 3/m = 4/m
As per the given question, 6 children and 2 men complete the work in 6 days. Therefore, total work done in 1 day = 1/6.
Equating the above two equations, we get:
4/m = 1/6
m = 24
Therefore, a man can complete the work in 24 days and a child can complete the work in 48 days.
Let's assume that 5 men can complete the work in 'd' days.
Total work done by 5 men in 1 day = 1/d
Total work done by 6 children in 1 day = 1/48
Total work done by 5 men and 6 children in 1 day = 1/d + 1/48
As per the given question, 6 children and 2 men complete the work in 6 days. Therefore, total work done in 1 day = 1/6.
Equating the above two equations, we get:
1/d + 1/48 = 1/6
1/d = 1/6 - 1/48
1/d = 7/48
d = 48/7 ≈ 6.86
Therefore, 5 men can complete the work in approximately 6.86 days, which can be rounded off to 6 days (Option A).
Let's assume that a man can complete the work in 'm' days. Therefore, a child can complete the same work in '2m' days.
Total work done by 2 men in 1 day = 1/m
Total work done by 6 children in 1 day = 6/2m = 3/m
Total work done by 6 children and 2 men in 1 day = 1/m + 3/m = 4/m
As per the given question, 6 children and 2 men complete the work in 6 days. Therefore, total work done in 1 day = 1/6.
Equating the above two equations, we get:
4/m = 1/6
m = 24
Therefore, a man can complete the work in 24 days and a child can complete the work in 48 days.
Let's assume that 5 men can complete the work in 'd' days.
Total work done by 5 men in 1 day = 1/d
Total work done by 6 children in 1 day = 1/48
Total work done by 5 men and 6 children in 1 day = 1/d + 1/48
As per the given question, 6 children and 2 men complete the work in 6 days. Therefore, total work done in 1 day = 1/6.
Equating the above two equations, we get:
1/d + 1/48 = 1/6
1/d = 1/6 - 1/48
1/d = 7/48
d = 48/7 ≈ 6.86
Therefore, 5 men can complete the work in approximately 6.86 days, which can be rounded off to 6 days (Option A).
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6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 5 men finish the same work?a)6b)8c)9d)15Correct answer is option 'A'. Can you explain this answer?
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