Two men and 7 boys can do a piece of work in 4 days. The same work
is done in three days by 4 men and 4 boys. How long would it take one man alone
to do it?
Answers
Step-by-step explanation:
Let the time taken by 1 man alone to do the work =x days and the time taken by 1 boy to do the work =y days.
Then, work done by 1 man in 1 day =
x
1
and work done by 1 boy in 1 day =
y
1
Therefore, work done by 2 men in 1 day =
x
2
and work done by 7 boys in 1 day =
y
7
According to the question,
Work done by 2 men and 7 boys in 1 day =
4
1
Thus,
x
2
+
y
7
=
4
1
....(i)
Similarly, work done by 4 men in 1 day =
x
4
and work done by 4 boys in 1 day =
y
4
Work done by 4 men and 4 boys in 1 day =
3
1
x
4
+
y
4
=
3
1
....(ii)
Putting
x
1
=pand
y
1
=q in equation (i) and (ii), we get,
2p+7q=
4
1
and4p+4q=
3
1
or,2p+7q=
4
1
=>8p+28q=1
=>p=
8
1−28q
....(iii)
Also,4p+4q=
3
1
=>12p+12q=1....(iv)
Substituting equation (iii) in equation (iv), we get,
12p+12q=1
=>12(
8
1−28q
)+12q=1
=>
2
3−84q
+12q=1
=>3−84q+24q=2
=>−60q=−1
=>q=
60
1
Substituting q=
60
1
in equation (iii), we get,
p=
8
1−28q
=>p=
8
1−28×
60
1
=>p=
8
1−
15
7
=>p=
15
8
×
8
1
=
15
1
p=
x
1
=
15
1
=>x=15
q=
y
1
=
60
1
=>y=60
Thus, time taken by 1 man alone to do the work =x=15 days and time taken by 1 boy alone to do the work =y=60 days
Work done by 1 man and 1 boy in 1 day
15
1
+
60
1
=
60
5
=
12
1
Time taken by 1 man and 1 boy to do the work =12 days.
A