15 men take 21 days of 8 hrs. Each to do a piece of work. How many days of 6 hrs. Each would it take for 21 women if 3 women do as much work as 2 men?
Answers
Answer:
30 days of 6 hours is required by 21 women to do a piece of work.
Step-by-step explanation:
We have,
No. of men, m1 = 15
No. days taken by men, d1 = 21
No. of hours, h1 = 8
No. of women, m2 = 21
No. of hours, h2 = 6
To find: No. days taken by women, d2
3 women can do as much work as 2 men
∴ efficiency of women to do a piece of work, e2 = (1/3)
∴ efficiency of men to do a piece of work, e1 = (1/2)
We know,
Total work done = (no. of men or women) * (no. of days) * (no. of hours) * (efficiency)
Therefore,
m1 * d1 * h1 * e1 = m2 * d2 * h2 * e2
or, 15 * 21 * 8 * (1/2) = 21 * d2 * 6 * (1/3)
or, d2 = 30 days
Answer:30 days
Step-by-step explanation:
Total work in hours 21 *8 =168 by 15 men
As 3 Women = 2 Men,
so 21/3 =7 *2 =14 (Converted into Men )
Now 14 Men needs to do same work 168*15/14 = 12*15= 180 hours
Now each day working time is 6 hours
So, 180/6 =30 days is the answer.