A and B together can do a piece of work in 15 days. If A’s one day work is 1½ times the one
day’s work of B, find in how many days can each do the work.
Let A’s one day work be x and B’s one day work be y.
Then according to the first condition given in the problem, we have
x = (3/2)y
2x = 3y
2x – 3y = 0 … (i)
Also given, in 15 days: A and B together can do a piece of work
So, according to this condition we have
x + y = 1/15
15 (x + y) = 1
15x + 15y = 1 … (ii)
Multiplying equation (i) by 5, we get
10x – 15y = 0
15x + 15y = 1
25x = 1
x = 1/25
On substituting the value of x in equation (i), we get
2(1/25) – 3y = 0
2/25 = 3y
y = 2/75
Therefore, Man A will do the work in 1/x days = 1/(1/25) = 25 days and Man B will do the work in (1/y)
days = 1/(2/75) = 75/2 = 37½ days.
it is 12 days. How long would it take for 1 man to do the work?
Solution:
Let’s assume that 1 man takes x days to do work and y days for women.
So, the amount of work done by 1 man in 1 day = 1/x
And the amount of work done by 1 woman in 1 day = 1/y
Now,
The amount of work done by 2 men in 1 day will be = 2/x
And the amount of work done by 5 women in 1 day = 5/y
Then according to the given conditions in the problem, we have
2/x + 5/y = ¼ … (i)
1/x + 1/y = 1/12 … (ii)
Multiplying equation (ii) by 5, we get
5/x + 5/y = 5/12
2/x + 5/y = ¼
(-)—(-)---(-)
3/x = 5/12 – ¼
3/x = (5 - 3)/12
3/x = 2/12 = 1/6
x = 18
Therefore, 1 man can do the work in 18 days.
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