A, B and C can do a piece of work in 20, 30 and 60 days
respectively. In how many days can A do the work if he
is assisted by B and C on every third day?

• one day work done by A=1/20
  work done by A in 2 days=1/10
  work done by A,B,C on 3rd day=1/10+1/10=1/5
  1/5th of the work will be completed at the end of 3 days
  the remaining work =1-1/5=4/5
  1/5 of work is completed in 3 days
  total work wiil be done in 3*5=15 days
  
• 8 years agoHelpfull: Yes(26) No(4)
• lcm(20,30,60)=60.
  Let total work have to be done is 60 units.
  From question,A can do 3 unit of work in a day so that it can complete 60 unit of work in 20 days.
  similarly B can do 2 units a day and C 1 unit a day.
  Now the work done by A in 2 days are 3+3=6 units
  and with the help of B and C he can do 6 unit of work in a single day.
  So in first 3 days the 12 units of work can be done
  12*5=60 units
  therefore 5 times 0f 3 days will take to complete the 60 units work.
  that's 3*5=15 days
  hence 15 days is the correct answer.
  
• 8 years agoHelpfull: Yes(17) No(0)
• one day work of A=1/20
  one day work of B=1/30
  one day work of C=1/60
  one day work of A,B and C=1/20+1/30+1/60
  =1/10
  (2/20+1/10)x=1
  (2/10)x=1
  x=5
  therefore 3*5=15
• 8 years agoHelpfull: Yes(11) No(0)
• 3/20 + 1/30 + 1/60 = 1/5 (3 days work)
  1/5 work completed in 3 days
  so work completed in 3*5 = 15 days
• 8 years agoHelpfull: Yes(8) No(0)
• A's 2 day's work = 1 x 2 = 1 .
  20 10
  (A + B + C)'s 1 day's work = 1 + 1 + 1 = 6 = 1 .
  20 30 60 60 10
  Work done in 3 days = 1 + 1 = 1 .
  10 10 5
  Now, 1 work is done in 3 days.
  5
  Whole work will be done in (3 x 5) = 15 days.
• 8 years agoHelpfull: Yes(7) No(4)
• first 2 days A alone works..A's 1 day work is 1/20.
  work done by A in 2 days is 1/10.
  work done by A,B and C together is 1/10.
  So, total work done in 3 days (by A alone for 2 days and all together in 3rd day ) is 2/10.
  so in 3 days only 20% of work(i.e 2/10) is completed. So full work i.e 100% work(10/10=1) can be done in 5*3=15 days..
• 8 years agoHelpfull: Yes(6) No(0)
• a=1/20
  b=1/30
  c=1/60
  
  a's 2 days work=1/10
  
  (a+b+c')s 1 days work=1/20+1/30+1/60=1/10
  3 days work=1/10+1/10=1/5
  1/5 th work in 3days
  
  remaining 5*3=15 days
• 7 years agoHelpfull: Yes(2) No(0)
• A's 2 day's work = 1 x 2 = 1 .
  20 10
  
  (A + B + C)'s 1 day's work = 1 + 1 + 1 = 6 = 1 .
  20 30 60 60 10
  
  Work done in 3 days = 1 + 1 = 1 .
  10 10 5
  
  Now, 1 work is done in 3 days.
  5
  
  Whole work will be done in (3 x 5) = 15 days.
• 8 years agoHelpfull: Yes(1) No(0)
• A's one day work will be 5%
  As it is given that at every 3rd day B and C join A.
  hence A's two days work will be 10%.
  now (A+B+C)'s one day work will be 10%.
  now calculate the % when it will become 100% stop counting.
  ans. will be 15 days.
• 7 years agoHelpfull: Yes(1) No(0)
• they have to complete 60 units
  A can complete 3 units in a day
  so in 2 days a complete=6 units
  on third day he will be assisted by b and c
  so on third day=3+2+1=6
  in overall 3 days=12 units
  so in 15 days they complete 16 units
• 7 years agoHelpfull: Yes(1) No(0)
• in 15 days a will do his work completely..
  
• 8 years agoHelpfull: Yes(0) No(0)
• 15 days because on every 3 days total work is 120 units
• 8 years agoHelpfull: Yes(0) No(0)
• A =5% work in 1 day
  B=3.33% in 1 day
  C=1.66% in 1 day
  B+C= 5% in 1 day(approx)
  work done by A in 3 days = 15%
  Work done by C+B on 3rd day=5%
  Total % of work in 3 days =15+5=20%
  Hence, for 100% work total no. Of days required=3*5=15days
• 7 years agoHelpfull: Yes(0) No(0)
• given that a is assisted by b every 3rd day
  i.,3 days work=1/20+1/20+(1/20+1/30+1/40)=1/5
  by mul 5 to the above value+5/5=1
  then the whole work is completed in 5*3(for every 3 days1/5)=15days
• 6 years agoHelpfull: Yes(0) No(0)
• unit work of A, b and C= 3, 2, 1 respectively if we let total work to be 60
  Now,
  
  by the end of the third day, total work done is
  
  3 +3 +(3+2+1)=12
  So, 60/12=5 times is required to complete the work in chunks of 12 work
  Every chunk requires 3 days
  So, total days=5X3=15 Ans.
• 6 years agoHelpfull: Yes(0) No(0)
• Take lcm of 20 30 60
  A can do 3 unit per day
  B- 2 unit per day
  C- 1 unit per day..
  12 units are done in 3 days
  So 60×3%12
  And. Is 15
• 5 years agoHelpfull: Yes(0) No(0)
• one day work done by A=1/20
  work done by A in 2 days=1/10
  work done by A,B,C on 3rd day=1/10+1/10=1/5
  1/5th of the work will be completed at the end of 3 days
  the remaining work =1-1/5=4/5
  1/5 of work is completed in 3 days
  total work wiil be done in 3*5=15 days
• 5 years agoHelpfull: Yes(0) No(0)
• Work done by A = 1/20.
  Work done by B= 1/30.
  Work done by C= 1/60.
  Now, work done by A in 2 days is= (1/20) * 2 = 1/10.
  And work done by A,B,C in one day= 1/20 + 1/30 + 1/60 = (3+2+1)/60 = 6/60 = 1/10.
  
  And the work done in first 3 days = 1/10 + 1/10 = 1/5.
  
  So we can say that 1/5 part of work is done= in 3 days.
  And A complete work is done in=. 3/(1/5) = 3*5 = 15 days.
• 1 year agoHelpfull: Yes(0) No(0)