A can do a piece of work in 8 days which B can destroy in 3 days .A has worked for 6 days ,during the last 2 of which B has been destroying;how many days must A now work alone to complete the work?
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Answered by
51
A complete work in = 8 days
A complete part of work in 1 day = 1/8 part
A complete part of work in 6 day = 1/8×6
= 6/8 part
B destroy work in = 3 days
B destroy part of work in 1 day = 1/3 part
B destroy part of 1 work in 2 day = 1/3×2 = 2/3 part
After destroying remaining work of A =
6/8 - 2/3
(18-16)/24 = 2/24 = 1/12 part
Now A has to complete work ( 1 work)
So left over work = 1- (1/12) = 11/12 part
So he will take = Total work/ 1 day work
= 11/12 ÷ 1/8 days
= 11/12 × 8/1
= 88/12 = 22/3 days
A complete part of work in 1 day = 1/8 part
A complete part of work in 6 day = 1/8×6
= 6/8 part
B destroy work in = 3 days
B destroy part of work in 1 day = 1/3 part
B destroy part of 1 work in 2 day = 1/3×2 = 2/3 part
After destroying remaining work of A =
6/8 - 2/3
(18-16)/24 = 2/24 = 1/12 part
Now A has to complete work ( 1 work)
So left over work = 1- (1/12) = 11/12 part
So he will take = Total work/ 1 day work
= 11/12 ÷ 1/8 days
= 11/12 × 8/1
= 88/12 = 22/3 days
Answered by
4
Answer:total work done in 6days
A's 6day's work- B's 2day's work
=6*3- 2*8
=18-16
=2
Rest work=24-2=22
Time taken to complete rest work =work/efficiency
22/3
Step-by-step explanation:
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