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?

A. 12 days
B. 15 days
C. 16 days
D. 18 days

• Work done in 3 days=(3/20)+(1/30)+(1/60)=1/5
  Remaining work=4/5
  If 1/5 work is done in 3 days then,
  4/5 work is done in (3*5*4)/5=12days
  Total no. of days will become 12+3=15days for the complete work to be done.:)
• 12 years agoHelpfull: Yes(54) No(17)
• A's 2 day's work =2 * (1/20)= 1/10
  (A + B + C)'s 1 day's work =((1/20)+(1/30)+(1/60))= 1/10
  Work done in 3 days =(1/10)+(1/10)= 1/5
  Now, (1/5) work is done in 3 days.
  Whole work will be done in (3 x 5) = 15 days.
• 11 years agoHelpfull: Yes(45) No(9)
• The work done by A +B +C in 3 Days is 3/20+1/20+1/60=12/60 i.e 1/5
  hence 1/5 work done in 3 days
  so 1 work done in 3*5=15 days
  Option :B
• 11 years agoHelpfull: Yes(26) No(8)
• A can do 1/20 units of work per day.
  B can do 1/30 units of work per day.
  C can do 1/60 units of work per day.
  
  Here,B and C can able to assist A only on every 3rd day.
  So,we can express as:
  1/20+1/3(1/30+1/60)
  =1/20+1/3(3/60)
  =1/20+3/180
  =1/20+1/60
  =4/60
  =1/15
  
  So the work can be finished in 15 days if A is being assisted by B and C on every 3rd day.
• 8 years agoHelpfull: Yes(17) No(0)
• assume total work=lcm of(20,30,60=60) so total work=60 units
  a--20 60/20=3unit/day
  b--30 60/30=2 unit/day
  c--60 60/60=1 unit/day
  .............................
  a+b+c =6 units/day
  work done by a in 2 days=6 unit and next day assisted by b & c so total work on 3rd day=6 unit..thus in first 3 days work completed=6+6=12 units
  if 12 unit................3 days
  60 unit..............3*60/12=15days
• 9 years agoHelpfull: Yes(16) No(3)
• A completes 5% of total work in a day(100/20)...B->3.3% C->1.6%
  Work done by (B+C) in a day is 5%(approx)
  A does work alone for 2 days n completes 10% in 2 days.
  on 3rd day A is assisted by B and C so total work done on 3rd day is (5+5)=10%
  
  work done in 3 days=20%
  days required to complete 100% work=15 days.
• 9 years agoHelpfull: Yes(3) No(3)
• work done in 24/120= 1/5;
  then remaining work =4/5
  if 1/5 work is done in 3 days.
  4/5 work is done in (3*5*4)/5=12 days
  total no. of days will beccome 12+3= 15 days....
  so Ans is = 15 days..
• 7 years agoHelpfull: Yes(2) No(0)
• A can do the work in 20 days
  B can do the work in 30 days
  C can do the work in 60 days
  total work is 60(L.C.M of 20,30,60=60),
  so A`s 1 day work 3
  B`s 1 day work 2
  C`s 1 day work 1
  A`s 2 day work is 6
  on 3rd day B & C will work with A,then (A+B+C)`s total work in 3rd day is 6(3+2+1)
  then at total 3 day work is 12
  then 60 work can be done in 15 days..
• 7 years agoHelpfull: Yes(2) No(2)
• answer is 15days
• 9 years agoHelpfull: Yes(1) No(4)
• 15 days is ans
• 8 years agoHelpfull: Yes(1) No(2)
• LCM(20,30,60)=120
  A does 6 units/day
  B does 4 units/day
  C does 2 units/days
  
  so, 6+6+12,6+6+12,6+6+12,6+6+12,6+6+12
  24*5=120
  5*3=15 days
• 8 years agoHelpfull: Yes(1) No(4)
• 15 days is ans
• 8 years agoHelpfull: Yes(0) No(1)
• work done by A in one day =1/20
  work done by A in two days=2/20
  On third day A,B,C are working together, so
  work done by three of them in one day=(1/10)+(1/30)+(1/60)=(1/10)
  work completed in third day=(1/5)
  Remaining work to be completed=1-(1/5)=4/5
  Three days are required to complete 1/5th work, So numbr of days required to complete (4/5)th work is
  =4*3
  =12 days
• 8 years agoHelpfull: Yes(0) No(0)
• b.
  A's 1 day work = 1/20
  (A+B+C) 1 day work 2/20
  total work completed on 15th day
• 8 years agoHelpfull: Yes(0) No(0)
• AVG=SUM/8
  AVG+2=SUM-(W1+W2)+80/8
  W1+W2/2=32
• 7 years agoHelpfull: Yes(0) No(0)
• A's 2days work= 2*1/20= 1/10
  Third day i.e (A+B+C)'s 1days work= 1/20+1/30+1/60= 1/10
  Work done in 3 days= 1/10+1/10= 1/5
  Now, 1/5 work done in 3 days= 3*5= 15days
• 7 years agoHelpfull: Yes(0) No(0)
• total work= 60
  a=60/20=3
  b=60/30=2
  c=60/60=1
  3+3+3+2+1=12
  60/12*3=15
• 7 years agoHelpfull: Yes(0) No(0)
• LCM of 20,30,60 is 60, MEANS TOTAL WORK IS 60
  A does it work for - 60/20=3 units/day
  B does it work for -60/30=2 units /day
  c does it work for - 60/60=1 unit /day
  first day work done=3
  second day work done =3
  third day A+B+C=6
  FOR OVER ALL THREE DAY WORK DONE IS=12
  THEN, 3 DAYS*5 = WORK DONE FOR 3 DAYS IS 12*5
  15 DAYS = TOTAL WORK 60
  
  ANSWER IS 15 DAYS
• 7 years agoHelpfull: Yes(0) No(0)
• As the question says,
  A, B, C can do the work in 20, 30 and 60 days respectively,
  so taking a common multiple of these three numbers, say 60 (LCM can go fine with this too). And 60 is the total work to do in units.
  now we need to calculate efficiency, A can complete work in 20 days, so each day he needs to do (60/20)=3 unit work,
  similarly, B=2 unit/day
  C=1 unit/day
  now as question says,on third day both B and C joins A for his work;
  A+A+(A+B+C) this is for first three days
  on calculating we get,3+3+(3+2+1) ==12 units in three days.
  so multiplying 5 both sides,
  we get 60 units in 15 days,ANS
• 7 years agoHelpfull: Yes(0) No(0)
• Total units of work to be completed= LCM of 20,30,60 = 60 units
  So in 1 day A does=60/20= 3 units
  B= 2 units
  C= 1 unit
  & A is assisted by B and C on every third day
  So, 1st day 3 units (by A)
  2nd day 3 units (by A)
  3rd days 6 units (by A+B+C)
  In 3 days 12 units are completed, therefore in (60/12)*3 days = 15 days, the work can be completed.
• 6 years agoHelpfull: Yes(0) No(0)
• A=5%
  B=3.33%
  C=1.66%
  (Work done by them in a day)
  In 2 days A will do 10% work and on 3rd day A+B+C= 10%+10% = 20% work is done
  20% work = 3 day
  100% work = 100*3/20=15 days
  So, 15 day is the answer
• 6 years agoHelpfull: Yes(0) No(0)
• A=W/20 per day work
  B=W/30 C=W/60
  3x(w/20)+x(w/30+w/60)=w
  3x/20+3x/60=1
  12x/60=1
  x=5
  so 3(5)=15 days
  so Option B
• 6 years agoHelpfull: Yes(0) No(0)