A and B together can do a piece of work in 20 days, B and C together can do it in 15 days,
C and A together can do it in 12 days.
(i) How long will they take to finish the work, working all together?
(ii) How long would it take to do the same work?
The answer is:
i) 10 days
ii) A=30 days
B=60 days
C=20 days
Please give right answers with explanation and process
Correct answers will be marked as brainliest
Answers
Answer:
Always go with the LCM method, as answer can be calculated while the first reading of the question as the calculation involves integers instead of fractions and they are easily handled.
LCM(12,15,20)=60
Now this 60 units can be treated as the total work and we will now calculate the one day work
One day work of A and B = 60/12 = 5 units
One day work of B and C = 60/15 = 4 units
One day work of C and A = 60/20= 3 units
Now if we add all three we get:-
2(A+B+C)=12 units
So one day work of A, B and C put together is 12/2 = 6 units
To find one day work of A:-
One day work of A, B and C together minus one day work of B and C, that is 6–4=2 units
If the total work is of 60 units and the one day work of A is 2 unit, then alone A needs 60/2 = 30 days to finish the work.
Step-by-step explanation:
Time taken by
A
and
B
to complete a unit of work
=
12
days
Therefore, work done by
A
and
B
in
1
day
=
1
12
units of work
Similarly,
Work done by
B
and
C
is
1
day
=
1
15
units of work
Work done by
A
and
C
in
1
day
=
1
20
units of work
Therefore, total work done by
A
,
B
and
C
=
1
2
×
(
1
12
+
1
15
+
1
20
)
LCM of
12
,
15
and
20
is
60
=
1
2
×
(
1
×
5
+
1
×
4
+
1
×
3
60
)
=
1
2
×
(
12
60
)
=
1
10
units of work
Therefore, work done by
A
alone in
1
day
=
Total work done
−
Work done by
B
and
C
=
1
10
−
1
15
=
1
×
3
−
1
×
2
30
=
1
30
units of work
Hence, time taken by
A
to complete a unit of work
=
30
days