Asked in Bangalore Elitmus Test (29/10/2017)

A can do a piece of work in a days, B can do a piece of work in b days, C can do a piece of work in c days.If A+B can do a piece of work in 6 days B+C can do a piece of work in 7.5 days, C+A can do a piece of work in 10 days find the increasing order of a,b,c.

• A+B = 1/6 ......(1)
  B+C = 1/7.5......(2)
  C+A = 1/10 .........(3)
  
  From this equation
  C = 1/30
  B = 1/10
  A = 1/15
  
  so,.. c = 30, b = 10, a = 15;
  Hence,.. c>a>b
• 6 years agoHelpfull: Yes(18) No(10)
• Given,
  A+B = 6 ....(1)
  B+C = 7.5 ....(2)
  C+A = 10 ....(3)
  Adding equation 1+2+3 , we get
  A+B+C = 47/4
  6+C = 47/4 ( A+B=6)
  C= 13/4
  So, Work done by C = (13/4)t=1
  t=4/13 ( for c)
  Putting the value of C in eqn (2) we get B= 17/4.
  So work done by B = (17/4)t= 1
  t=4/17 ( for B)
  Putting the value of B=17/4 We get A= 27/4
  So , Work done by A= (27/4)t=1
  t= 4/27( for A)
  Hence A= 4/27 , B= 4/17 , C=4/13
  C>B>A
• 6 years agoHelpfull: Yes(13) No(17)
• A+B=6days
  B+C=7.5days
  C+A=10days
  2(a+b+c)=23.5
  So a+b+c=11.75 days
  A=4.25 days
  B=1.75 days
  C=5.75 days
  
  According to efficiency
  A+B=25/day
  B+C=20/day
  C+A=15/day
  So 2(a+b+c)=60
  A+B+C= 30
  So A=10
  B=15
  C=5
  In question they ask increasing order according to days so ans is 1.75 , 4.25 ,5.75
  Means b-->a-->c
• 6 years agoHelpfull: Yes(10) No(2)
• ans is ... B>A>C
• 6 years agoHelpfull: Yes(9) No(12)
• Total work=150
  (A+B)'1DAY=25WORK
  (B+C)'1DAY=20WORK
  (C+A)'1DAY=15WORK
  (A+B+C)'1DAY WORK=30
  A'1DAY=10 WORK
  SO, 150/10=15DAYS
  
  likewise all.....
  B=10 days
  C=30 days
  C>A>B
• 6 years agoHelpfull: Yes(6) No(1)
• A+B = 6 ....(1)
  B+C = 7.5 ....(2)
  C+A = 10 ....(3)
  
  by eq. 1 and 2 "B" is comman in both but in eq 1 it takes 6 days which is less than eq. 2 (7.5 days )
  so A>C (A is faster than C)
  by eq. 1 and 3 "A" is comman in both but in eq 1 it takes 6 days which is less than eq. 3 (10 days )
  so B>C (B is faster than C)
  by eq. 2 and 3 "C" is comman in both but in eq 2 it takes 7.5 days which is less than eq. 3 (10 days )
  so B>A (B is faster than A)
  
  order will be B>A>C
• 6 years agoHelpfull: Yes(2) No(2)
• A+B=6 days, B+C=7.5 days, C+A=10 days. L.C.M of the above three values will be= 30, which is the total work to be done. then A+B=30/6=5 units, B+C=30/7.5=4 units, C+A=30/10=3 units. So, A+B+C=6 units in one day. Therefore, A=2units in 1 day, B=3units in 1 day, C=1 unit in 1 day. Now, a=30/A, which is 30/2= 15 days. And similarly, b=10 days. and c=30 days. therefore answer is c>a>b.
• 5 years agoHelpfull: Yes(2) No(0)
• (1/a+1/b)6=7.5(1/b+1/c)=10(1/a+1/c)
  now equating the equations we will get c = 3b and a = 3b/2
  So the order will be b
• 5 years agoHelpfull: Yes(1) No(1)
• A+B=6
  B+C=7.5
  C+A=10
  
  So,we have to select that value of a,b,c which would satisfy the above value
  
  A=10
  B=15
  C=25
• 6 years agoHelpfull: Yes(0) No(9)
• b=10
  a=15
  c=30
  b
• 6 years agoHelpfull: Yes(0) No(3)
• (A+B+C)=47/4
  THEN,
  A=47/4-7.5
  B=47/4-10
  C=47/4-6
  THUS C WILL TAKE LEAST DAY FOLLOWED BY A THEN B
• 5 years agoHelpfull: Yes(0) No(0)
• A+B= 6 =5
  B+C =7.5 =4
  C+A= 10 = 3 LCM OF(6,7.5,10)=30
  
  COMBINING THESE 3 ,WE GET 2(A+B+C)= 12
  => (A+B+C)= 6
  FROM THIS WE GET, A=2, B=3, C=1
  SO, c>a>b
• 5 years agoHelpfull: Yes(0) No(1)
• A+B =6
  B+C =7.5
  C+A =10
  BY TAKING LCM WE GATE =150
  then, we gate the value of A+B+C=30
  THEN FROM here we get A=10,B=15,C=5.
  THEN order of increase is b>a>c
• 3 years agoHelpfull: Yes(0) No(0)
• b c a
  in the ratio of efficiencies
• 3 years agoHelpfull: Yes(0) No(0)