A, B and C working together completed a job in 10 days. However C only worked for the first three days when 37/100 of the job was done. Also, the work done by A in 5 days equals the work done by B in 4 days. How many days will the fastest worker take to complete the given work alone?
20 days 25 days 30 days 40 days...........

• A+B+C 3 day work = 37/100
  remaining work= 63/100
  A+B complete the work in 7 days= 9*7/100
  A+B 1 day work= 9/100
  
  Now A+B 1st 3 days work = 27 , remaining= 37-27=10
  so in 1st 3 days A work = 10 , A 1 day work = 10/3
  So, C alone needs 30 days to complete the work
  
  A 5 day work = B 4 day work (Clearly B is more efficient than A)
  A efficiency = 20%
  B efficiency = 25%
  
  Now A+B 1 day work = 9 {i.e A is doing 4/9 part and B is doing 5/9}
  If A work alone = 25 Days
  If B work alone = 20 Days
• 10 years agoHelpfull: Yes(16) No(5)
• work done by a,b,c in 3 days = 37/100
  work done by a+b in 7 days = 63/100
  so, 1/a+1/b=63/7*100 and 5*a=4*b
  By solving you get a=20 and b=25.
  By substitution of a,b in 1/+1/b+1/c=37/100*3 we get c=7/25
  so in 1 day a does 5%(1/20) work, b does 4%(1/25) work and c does 3.57% work
  so a is fastest and require 20 days to complete the work alone
• 10 years agoHelpfull: Yes(14) No(9)
• work done by a,b,c in three days =37/100
  remaining work done by a,b in 7 days =63/100
  1 day work done by a+b=9/100
  then 1 day work of c=1/100
  after calculation
  one day work of a=1/20
  one day work of b=1/25
  so A is the fastest worker done in 20 days...
  
• 10 years agoHelpfull: Yes(10) No(12)
• i think there is a mistake in the question
  as if (a+b+c) can complete the work in 10 days then
  3 days work (a+b+c)=3/10------(1)
  now its given that c's(alone) 3 days work is 37/100
  and 37/100 is greater than 3/10 !!!!!!!!! thats not possible
• 10 years agoHelpfull: Yes(8) No(5)
• Solution based on Efficiency
  Since A+B+C completed 37% of the work working together in 3 days
  so per day efficiency of A+B+C=37/3=12.33%
  and A & B Completed the remaining work i.e. 63% in 7 days
  so combined per day efficiency of A+B=9% ..............(1)
  Since work done by A in 5 days= work done by B in 4 days
  that means B is more efficient than A
  now look towards options
  lets say B Completes 100% work in 20 Days
  so per day Efficiency of B=5%
  from 1 we get
  per day efficiency of A=4%
  so days taken by A to complete 100% work=25 days
  and also the work done by A in 5 days(5*4=20)=work done by B in 4 days(4*5=20)
  hence B is the fastest worker and completes work in 20 days!!
• 9 years agoHelpfull: Yes(7) No(0)
• arun sharma l2 18th q
  3(a+b+c)=37%
  7(a+b)=63%
  a+b=9% work of one day after c leaving
  and also 5A=4B
  b=25%,a=20% devide this by 5 coz a takes 5
  hence one day of work a=4%,b=5%
  hence whole work can be done in 100%/5%=20 days
  
• 10 years agoHelpfull: Yes(4) No(1)
• IF A B,C COMPLETE=ING THE WORK IN 20,25,30 THEN A+B+C WON'T BE ABLE TO COMPLETE THE WORK IN 10 DAYS...
  
  WRONG QUESTION
  
• 10 years agoHelpfull: Yes(3) No(6)
• 1/a + 1/b + 1/c = 1/10 (Given)
  
  3(1/a + 1/b + 1/c) = 37/100 (given)
  and c lefts after this so,
  1/a + 1/b = 63/100
  5a = 4b (given)
  solving these two equstions
  a=20 and b= 25
  now solving first equation c=100
  so a is the fastest worker.
• 7 years agoHelpfull: Yes(1) No(0)
• Let work is 1000
  Work done by a+b in 7 days
  (A+B)*7=63
  One day work done by a+b=9
  Iab+c in three day. 3(a+b+c)=37
  But value of a+b will get c one day work. 10/3
  5A=4B = equal work &. A+b. =9 from above
  there for A one day work 4
  And b one day work. 5
  Total work done by. A=. 100/4=25 day
  B=100/5=20 day. And
  C=. 100/(10/3)=30 days
  Hence fastest is B
• 2 years agoHelpfull: Yes(1) No(0)
• It is right that answer can be 5E but here it can not be determined because how many times ratio becomes 16:9 is not given.
• 9 years agoHelpfull: Yes(0) No(1)