8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14
days. Find the time taken by one man alone and that by one boy alone to finish the work.
Answers
Answer:
Suppose that one man alone can finish the work in x days and one boy alone can finish it in y days. Then,
One man's one day's work =
x
1
One boy's one day's work =
y
1
∴ Eight men's one day's work =
x
8
12 boy's one day's work =
y
12
Since 8 men and 12 boys can finish the work in 10 days
10(
x
8
+
y
12
)=1⇒
x
80
+
y
120
=1 ..(i)
Again, 6 men and 8 boys can finish the work in 14 days.
∴14(
x
6
+
y
8
)=1⇒
x
84
+
y
112
=1 (ii)
Putting
x
1
=u and
y
1
=v in equations (i) and (ii), we get
80u+120v−1=0
84u+112v−1=0
By using cross-multiplication method, we have
−120+112
u
=
−80+84
−v
=
80×112−120×84
1
⇒
−8
u
=
−4
v
=
−1120
1
⇒u=
−1120
−8
=
140
1
and v=
−1120
−4
=
280
1
Now, u=
140
1
⇒
x
1
=
140
1
⇒x=140
and, v=
280
1
⇒
y
1
=
280
1
⇒y=280.
Thus, one man alone can finish the work in 140 days and one boy alone can finish the work in 280 days.
One man can finish the work in 140 days