1.A,B,C together can do a piece of work in 10 days.All the three started workingat it together and after 4 days,A left.Then,B and C together completed the work in 10 more days.In how many days can complete a work alone??
Options:
15 ,16,50,25 days

2.A train covers a distance between station A and station B in 45 min.If the speed of the train is reduced by 5 km/hr,then the same distance is covered in 48 min.what is the distance between the stations A and B?
Option: 64,55,80,60 km

• Ans1 (A+B+C) do 1 work in 10 days.
  So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5
  Remaining work=1-2/5=3/5
  (B+C) take 10 more days to complete 3/5 work.
  So( B+C)'s 1 day work=3/50
  Now A'S 1 day work=(A+B+C)'s 1 day work-(B+C)'s 1 day work=1/10-3/50=1/25
  A does 1/25 work in in 1 day
  Therefore 1 work in 25 days.(Ans)
• 9 years agoHelpfull: Yes(35) No(2)
• 2) d=s*t
  45mins=3/4 hr and 48mins=4/5 hr
  s*(3/4)=(s-5)*(4/5)
  3s/4=(4s-20)/5
  15s=16s-80
  s=80 km.
• 9 years agoHelpfull: Yes(27) No(16)
• S*T1=(S-5)8T2
  S*45=(S-5)48 SO S=80 PUT DIS IN FST EQN N DIVIDE BY 60
  SO THE ANS IS 60..
• 9 years agoHelpfull: Yes(15) No(3)
• 2. 60 k.m will be the ans .make a eqn with previous speed*time takin in hour = (speed-5)*time taking in hour .,.the eqn was posssible as the ditance it is going to n fro is same
• 9 years agoHelpfull: Yes(14) No(6)
• s1/s2=t2/t1;
  s1/s1-5=48/45
  from this equn s1=80kmph
  s=d/t
  d=s*t
  =80*45/60=60km
  
  
• 9 years agoHelpfull: Yes(9) No(1)
• 1) 25 days
• 9 years agoHelpfull: Yes(4) No(1)
• d=s*t
  d= s1*(45/60) that is s1*(3/4)
  d= s2*(48/60) that is s2*(4/5)
  s1*(3/4)=s2*(4/5)
  s1*(3/4)=(s1-5)*(4/5) that is speed reduced by 5kmph
  s1=80 kmph
  d=s.t
  d=80*(3/4)= 60 km
  
• 9 years agoHelpfull: Yes(3) No(0)
• S * (45/60) = (S-5) * (48/60)
  on simplifying s=80 kmphr i.e speed
  distance = 80*(45/60) or 75*(48/60) = 60km
• 9 years agoHelpfull: Yes(2) No(0)
• (1) 25 days
  (2) 80km
• 8 years agoHelpfull: Yes(1) No(1)
• 2) s=d/t
  => (s1/s2)=(d1/d2)*(t2/t1)
  =>( x / (x-5) )=(d / d) * ( 48 / 45 )
  =>( 4x-20 ) *4 = 15x
  =>x=80
• 7 years agoHelpfull: Yes(1) No(0)
• speed(s)=distance(d)/time(t)...............here distance constant
  s*t=constant
  s1*t1=s2*t2
  speed time
  x 45
  x-5 48
  x*45=(x-5)*48
  
  3x=240
  
  x=80km
• 9 years agoHelpfull: Yes(0) No(5)
• 25 DAYS WOULD BE THE ANSWER
• 7 years agoHelpfull: Yes(0) No(0)
• 1. A+B+C's 1 day work 1/10
  B+C's 1 day work=3/50
  A's 1 day work=1/10-3/50=1/25
  A can complete work in 25 days
  2.
  60x/45=60x/48+5
  4x/3=5x/4+5
  16x=15x+60
  x=60 km
• 7 years agoHelpfull: Yes(0) No(0)
• given 1/a+1/b+1/c=1/10
  so 1/b+1/c=1/10-1/a
  now a work for 4 days and b nd c work for 14 days now
  4/a+14/b+14/c=1
  4/a+14(1/10-1/a) =1
  by solving it we got a=25 days
• 7 years agoHelpfull: Yes(0) No(0)
• 1
  A+B+C CAN COMPLETE IN 10 DAYS
  LET'S TAKE WORK TO BE 10 UNIT THEN THEY TOGHTER COMPLETE 1 UNIT A DAY
  IN 4 DAYS THEY COMPLETE 4 UNIT I.E 6 UNIT IS REMAINING . NOW B+C CAN COMPLETE THIS 6 UNIT IN 10 DAYS i.e 6/10 UNIT A DAY
  A'S 1 DAY WORK IS 1-6/10=4/10 unit
  HENCE A WILL COMPLETE WORK IN = TOTAL WORK/ONE DAY WORK
  =10unit/4/10unit/DAY
  =10*10/4
  =25 DAYS
• 7 years agoHelpfull: Yes(0) No(0)
• 2. speed=distance/time
  lets assume distance=x;
  change the time in hrs=t1=(45/60)=3/4,t2=(48/60)=4/5
  speed of train reduced=distance/t1-distance/t2=(how much speed reduced)=5
  x/t1-x/t2=5
  4x/3-5x/4=5 do LCM
  16x-15x=60
  x=60
• 4 years agoHelpfull: Yes(0) No(0)