A alone can do a piece of work in 10 days,B alone in 12 days,C alone in 15 days.If All start work at same day.A works for two days and C leaves before 3 days of remaining work.In how many days work will finish?

• B will work for all x days to complete the entire work
  2/10+x/12+(x-3)/15=1
  =>x=20/3
• 9 years agoHelpfull: Yes(62) No(9)
• LCM(10,12,15)=60
  Let total work required to be done=60
  A 1day work=6
  B 1 day work=5
  C 1 day work=4
  Let the whole work is completed in x days
  =>6(2)+4(x-3)+5x=60
  x=20/3
  
  
• 9 years agoHelpfull: Yes(27) No(5)
• A's work done =1/10
  B's work done =1/12
  C's work done =1/15 in 1 days
  
  A+B+C=1/10+1/12+1/15= 1/4
  A,B,C work for 2 days so 2*1/4=1/2
  
  C leaves before 3 days remaining work so 3*1/15=1/5
  
  So total work done in 5 days(3+2)=1/2+1/5=7/10
  
  remaining work = 1-7/10=3/10
  
  3/10 work done by B and C in x days
  3/10=x(1/12+1/15)
  
  x=2 days
  So A,B,C work for 2 days
  B,C WORKS for 2 days
  C works for 3 days
  total 2+3+2=7 Ans
  
  
• 9 years agoHelpfull: Yes(23) No(14)
• 1 day work of a,b&c=(1/10+1/12+1/15)= 1/4
  2 day work of a,b&c= 2*(1/4)= 1/2
  left work 1-1/2= 1/2
  3 day work of b= 3*1/12= 1/4
  b& c together do 1/4th work in 5/3 days
  total days =2+5/3+3= 20/3
• 9 years agoHelpfull: Yes(14) No(1)
• 7 days
  total work 60
  all work for 2 day ,work 30 done
  c work for 3 days alone ,work 12done
  
  remaining work to be done is 18,
  18/9=2 days
  so 2+2+3=7 is the ans
  
• 9 years agoHelpfull: Yes(7) No(4)
• take the lcm of all three numbers as number of units of work in this case 60
  so A can do 6 units of work in 1 day
  and B can do 5 units of work in 1 day
  and C can do 4 units of work in 1 day.
  Let work is completed in x days
  A worked for 2 days so he completes 12 units of work
  remaining is 45 units of work
  As per problem B worked for (x-3) days so B completes (x-3)5 units of work
  and C completes 4x units of work
  so solve (x-3)5+4x=48
  ans is 7 days
  
• 9 years agoHelpfull: Yes(3) No(0)
• 7 days , since a,b,c work together for 2 days , then work done =60 , reamining work 60 , c leaves before 3 days , hence work done by c=3*8=24 , remaining work is 36 which is completed in 2 days by b and c combined , hence total days =7
• 9 years agoHelpfull: Yes(2) No(3)
• A can do 1/10
  b 1/12
  c 1/15
  A,B,c to gether 1/10+1/12+1/15=1/4
  to gether 4
  
  A works 2 days nd c leaves bfore 3 days so 4+3=7
• 9 years agoHelpfull: Yes(2) No(4)
• let a work efficiency as 1/5
  b s work x/12
  c 's (X-3)/15
  by adding all, ans is : 20/3
• 9 years agoHelpfull: Yes(2) No(1)
• A does 10%
  B 8.33%
  C 6.66% work per day
  together in two days they do 50% work
  now 3 days before finish C leaves means B work alone for 3 days
  3*8.33=25% work
  left work is 25% that is being done by B and C
  in 25/(8.33+6.66) =5/3 days
  so total days=
  2+5/3+3
  20/3
  
• 9 years agoHelpfull: Yes(2) No(0)
• Answer=20/3
  
  In 1 day work of A=1/10
  In 1 day work of B=1/12
  In 1 day work of C=1/15
  
  Let x is working day of B and C
  So,
  2(A+B+C) + 3(B) + x(B+C)=1(=Total work)
  2(1/10+1/12+1/15) + x(1/12+1/15) + 3(1/12)=1
  x=5/3
  
  Total working day=2+3+5/3
  =20/3
• 9 years agoHelpfull: Yes(2) No(1)
• 2/10+(x-3)/12+x/15=1
  x=7
  
• 9 years agoHelpfull: Yes(1) No(5)
• A-10days,b-12days,c-15days
  let x be total no of days,then equationsare
  x/12+(x-3)/15+2/10=20/3=6.667days
• 9 years agoHelpfull: Yes(1) No(0)
• work completed by A,B,and C together is (1/10+1/12+1/15)=1/4
  As A works for two days so work done by A is ((1/10)*2)=1/5
  As C leaves before 3 days of remaining work ,so work will be done by B which is ((1/12)*3)=1/4
  Now works remain to do (1-(1/5+1/4))=11/20
  Now work done by B and C together in one day is (1/12+1/15)=9/60
  so 9/60 work is done in one days
  so 1 work will done in 60/9 days
  and 11/20 work will be done in 60/9*11/20
  so work will be finish in 11/3 +2+3=26/3 days
  
  
• 9 years agoHelpfull: Yes(1) No(0)
• let n is the total number of days, then work done by c is (n-3)/15.
  work done by A is 1/5. work done by B is n/12.
  (1/5)+(n/12)+((n-3)/15)=1
  on solving n=20/3=6.6666666
  i.e approximately 7 days of work.
• 9 years agoHelpfull: Yes(1) No(0)
• A=1/10 B=1/12 C=1/15
  1/10+1/12+1/15=1/4
  TWO DAYS WORK: 2(1/4)=1/2
  REMAINING WORK= 1-1/2=1/2
  LET X=REMAINING DAYS
  (X-3)(1/12)+(X/15)=1/2
  X=5................
• 6 years agoHelpfull: Yes(1) No(0)
• 29/60 days
• 9 years agoHelpfull: Yes(0) No(5)
• total no days=7
  
• 9 years agoHelpfull: Yes(0) No(4)
• ans is 6(2/3) days
• 9 years agoHelpfull: Yes(0) No(1)
• Ans:6.2days
• 9 years agoHelpfull: Yes(0) No(0)
• 34/5 days
  
• 9 years agoHelpfull: Yes(0) No(0)
• 
  A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :
• 9 years agoHelpfull: Yes(0) No(2)
• work is finished in 7 days
• 9 years agoHelpfull: Yes(0) No(7)
• 8.3 Days. As together a+b+c will complete it in 2d and b will complete it in 3 days so remaining work will be completed by (b+c) in 10/3 days. So total days required are 8.3 days.
• 9 years agoHelpfull: Yes(0) No(3)
• A 2 day work 2/10
  let work is completed x day
  2/10+x-3/12+x/15=1
  x=7
• 9 years agoHelpfull: Yes(0) No(1)
• B will work for all x days to complete the entire work
  2/10+x/12+(x-3)/15=1
  =>x=20/3
• 9 years agoHelpfull: Yes(0) No(0)
• 60/9 final answer
  
• 9 years agoHelpfull: Yes(0) No(0)
• 14 daysss
  
• 9 years agoHelpfull: Yes(0) No(1)
• The answer is 7 days
• 3 years agoHelpfull: Yes(0) No(0)