please help me to solve both the sums.....
Answers
Answer:
i ) 16 days
ii) A=30
B=60
C=20
Step-by-step explanation:
i) Time needed by A and B to finish the work = 20 days.
Time needed by B and C to finish the work = 15 days.
Time needed by C and A to finish the work = 12 days
Work done by A and B in one day = 1/20
Work done by B and C in one day = 1/15
Work done by C and A in one day = 1/12
2× work done by A, B and C in one day = 1/20 + 1/15 + 1/12
= 3+4+5/60
= 12/60
= 1/5
Work done by A, B and C in one day = 1/5
So, A,B and C working together will take 16 days to complete the work.
ii) (A + B) can complete the work in 20 days.
(B + C) can complete the work in 15 days.
(C+ A) can complete the work in 12 days.
(A + B) in one day = 1/20
(B + C) in one day = 1/15
(C+ A) in one day = 1/12
Adding, we get
2(A + B + C)'s 1 day's work = (1/20 + 1/15 + 1/12)
= 12/ 60
= 1/5
=>(A + B + C)'s 1 days work =(1/2 × 1/5)
= 1/10
=>A,B,C together can finish the work in 10 days
Now, A's 1 day's work
= {(A + B + C)'s 1 day's work} - {(B + C)'s 1 days work}
=(1/10 - 1/15)
= (3-2/30)
=1/30
Hence, A alone can finish the work in 20 days.
B's 1 day's work = {( A + B + C)'s 1 days work} - {(C + A)'s 1 day's work}
=1/10 - 1/12)
=(6-5/60)
=1/60
Hence, B alone can finish the work in 30 days.
C's 1 day's work = {(A + B + C)'s 1 day's work} - {(A+B)'s 1 days work}
=(1/10 - 1/20)
= (2-1/20)
=1/20
Hence, C alone can finish the work in 60 days.