A can do in 6 day and B can do in 8 and C can do in 12 days but b left after working for 6 days for how many no of day
A and C shluld work ?

• a=1/6
  b=1/8
  c=1/12
  b left after 6 days so 6*1/8=3/4
  work remaining =1-3/4=1/4
  a+c=1/6+1/12=18/72=1/4 ( a &c take 4 days to complete the whole work ) 1 work=4 days
  no of days for 1/4th of work = 1 day
• 10 years agoHelpfull: Yes(38) No(3)
• B for one day work is 1/8
  he worked for 6 days, so 6*1/8=3/4 work will be completed.
  remaining work is 1-3/4=1/4
  A and C can complete the work in 4 days
  but there is only 1/4 work that is to be completed, so A and C can complete 1/4 work in 1 day..
  answer: 1 day
• 10 years agoHelpfull: Yes(32) No(1)
• try to find out their efficiencies and LCM first
  so we get
  LCM of 6,8,12=48 nothing but a total work to do
  efficiencies
  A=48/6=8
  B=48/8=6
  C=48/12=4
  therefore B finishes his work in 6 days 6*6=36
  remaining work=48-36=12
  that can be finished by A & C in
  12/(8+4)=12/12
  1 must be the answer
• 10 years agoHelpfull: Yes(24) No(13)
• let x be the no. of days..
  6/8+x/6+x/12=1
  solving we get x=1 ...so 1 day is the answer
• 10 years agoHelpfull: Yes(18) No(2)
• 6(a+b+c)+x(a+c)=1
  9/4+x/4=1
  x=5days,remaining days will be 5days
• 10 years agoHelpfull: Yes(6) No(1)
• this type of problems try like this. lets assume total work = total no of chocolates.
  a= 6 days
  b= 8 days
  c= 12 days
  find out LCM of a,b, and c = 48
  So total chocolates = 48 (assume total work = total chocolates).
  a can able to eat all chocolates in 6 days. So in one day he can able to eat = 8 choclates
  b can able to eat all chocolates in 8 days. So in one day he can able to eat = 6 choclates
  c can able to eat all chocolates in 12 days. So in one day he can able to eat = 4 choclates.
  
  
  Now look at the question b worked for 6 days, so he can eat 36 choclates in 6 days( 1 day = he can eat 6 chocs)
  remaining chocolates = 48-36 = 12.
  A and C can combinely eat 12 chocolates per day. So they will take 1 days to complete the remaining work
• 10 years agoHelpfull: Yes(6) No(3)
• a,b,c worked for 6days their work is 6*(1/6+1/8+1/12)= 9/4 remaning work is 1-9/4 = 5/4(work cannot be -ve) and now a and c has to complete the remaining work =efficiency /time ... so a+c eff= 1/6+1/12=1/4... time= 1/4*4/5=ie. .......5days............guys let me know if i am wrong ........
• 10 years agoHelpfull: Yes(4) No(3)
• A-1/6
  B-1/8
  C-1/12
  B(6days)--->>6*1/8--->>3/4
  Remaining work--->>1/4
  A+C(1day)--->>>1/6+1/12--->>1/4
  SO they complete it in 1 day
  
  
  
• 10 years agoHelpfull: Yes(3) No(0)
• No need to work because A lonely can do whole work in 6 days.
  They can do double work in 6 days.
  
  A do in 6 days, b Do in 8 days, c do in 12 days.
  B left after working 6 days,
  The work done by three of them is ( in 6 days)
  6(1/6+1/8+1/12) = 9/4
  
• 10 years agoHelpfull: Yes(2) No(9)
• Guys plz Try Understand the question...
  There is NO A+B+C work..
  Simply A do a WOrk in 6 days
  B do in 8 days
  c do in 12 days..
  do not combine three ...
  just consider B work only, If B continues to work it complete in 4 days,
  but due to some problem it only works for 6 days, remaing 1/4 work should b completed in 4 days...
  
  
• 10 years agoHelpfull: Yes(1) No(1)
• Take an LCM of (6,8,12)=24..
  So A can complete in 6days compare 6 wid 24==>6*4===>4.
  So B can complete in 8days compare 8 wid 24==>8*3===>3.
  So C can complete in 12days compare 12 wid 24==>12*2===>2.
  B left after 6days means they worked for 6days and then B left..
  So 6(A+B+C)===> 6(4+3+2)= =>6*9=54.
  A+B+C= 54-24===> i.e., 30.
  So after 6days B left.. A+C= 30/4+2
  A+C= 30/6====> A+C= 5days..
  
• 8 years agoHelpfull: Yes(1) No(3)
• FOR 6 days
  6(1/6+1/8+1/12)
  
  a&c can do it in
  
  ={1/6+1/12} - 6{1/6+1/8+1/12}
  = simplify this
• 10 years agoHelpfull: Yes(0) No(7)
• A and C working together only 1 day
• 10 years agoHelpfull: Yes(0) No(1)
• A+B+C = (1/6+1/8+1/12) = 9/24
  
  1 day work = 9/24*6 = 9/4
  
  The work will be done by three of them will be 9/4
  Remaining work done = 1-9/4 =5/4.
  multiply it with B days = 5/4*8 = 10 days.ans
  
• 10 years agoHelpfull: Yes(0) No(3)
• ans:1 day
  chacolate method:
  lcm of a,b,c=144
  so work = 144chacolates
  a can eat 144/6=24 chacolates per day
  b can eat 144/8=18 c/day
  c can eat 144/12=12 c/d
  b can do awork in 6 days=6*18=108
  a+c eat remaining chacolates=12+24=36chacolates/day
  so ans is 1day
  
  
• 9 years agoHelpfull: Yes(0) No(1)
• ans:1 day
  chacolate method:
  lcm of a,b,c=144
  so work = 144chacolates
  a can eat 144/6=24 chacolates per day
  b can eat 144/8=18 c/day
  c can eat 144/12=12 c/d
  b can do awork in 6 days=6*18=108
  a+c eat remaining chacolates=12+24=36chacolates/day
  so ans is 1day
  
  
• 9 years agoHelpfull: Yes(0) No(1)
• 3 women=2 man
  then 21 women =14 man
  then apply formula ( m 1 *d 1* h 1)/w 1=(m 2*d 2* h 2)/w 2
  15*8*21=14*6*d 2
  
  d 2= (15*8*21)/(14*6) =30
  
  
  
  
  
  
• 9 years agoHelpfull: Yes(0) No(3)