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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.

Answer:

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.

- 2 men and 5 women can do a piece of work in 4 days, while one man and one woman can finish

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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