2 men and 3 boys together can do a piece of work in 8 days. The same work is done in 6 days by 3 men and 2 boys together. How long would 1 boy alone or 1 man alone take to complete the work?
Answers
Answer:
Step-by-step explanation:
Let the number of days taken by a man and a boy be x and y respectively.
Therefore, work done by a man in 1 day = 1/x
Work done by a boy in 1 day = 1/y
According to the questions,
8[(2/x)(3/y)]=1
(2/x)(3/y)=1/8 (1)
6[(3/x)(2/y)] = 1
(3/x)(2/y)=1/6 (2)
Putting
1/x=p and 1/y=q
in these equations, we obtain
2p +3q = 1/8
16p +24q = 1 (3)
3p + 2q = 1/6
18p +12q = 1 (4)
p = (1-12q )/18 (5)
apply the (5) on (3)
16-192q + 432q = 18
q = 1/120 = 1/ y
y = 120
p = (1-12q)/18
= 1/20 = 1/x
x = 20
Answer:
Step-by-step explanation:
Let 1 boy alone can do a piece of work in x days and 1 man alone can do a piece of work in y days.
So, 1 day work of 1 boy = x /1
and 1 day work of 1 man = y /1
ATP
x /3 + y /2 = 1 /8 .........(i)
and
2 /x + y /3 = 1/6 .......(ii)
Multiplying equation (i) by 2 and equation (ii) by 3
6 /x + 4 /y = 1 / 4
6 /x + 9/y = 1 /2
------------------------------------------
− 5 /y =− 1 /4
⇒ y=20
substitute y=20 in equation (i)
3 /x + 3/20 = 1 /8
3/x = 1/8 − 1/10
3/x = 5−4 /40
3 /x = 1/40
⇒ x=120
So, one boy alone can do a piece of work in 120 days
and one man alone can do a piece of work in 20 days.