A man travels three-fifths of distance AB at a speed of 3a, and the remaining at a speed of 2b, if he goes from A to B and back at a speed of 5c in the same time, then prove that 1/a + 1/b = 1/c

• Let the distance between A and B is X
  Three-fifths of distance AB =3X/5
  Remaining distance=2X/5
  If he goes from B to A and back i.e X+X=2X
  
  Now from the question we have
  
  (3X/(5*3a))+(2x/(5*2b))=2X/5c
  Or 1/a+1/b=2/c
  
• 13 years agoHelpfull: Yes(15) No(4)
• this que shoud be like this...............
  
  A man travels three fifths of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and returns at a speed 5C in the same time then :
  @opt a
  1/a + 1/b =1/c
  
  @opt b
  a+b=c
  @opt-ans
  1/a + 1/b =2/c
  @opt
  None of these
• 13 years agoHelpfull: Yes(6) No(4)
• (3/5 * d / 3a)+(2/5 * d / 2b) = d/5c
  then
  
  (1/a)+(1/b)=(1/c)
• 13 years agoHelpfull: Yes(3) No(3)
• its ans will be 1/a +1/b =2/c
• 13 years agoHelpfull: Yes(2) No(5)
• X be the distance AB
  X/(3a+2b) = X/5c
  
  5c = 3a + 2b
• 13 years agoHelpfull: Yes(1) No(4)
• Let the distance between A and B is X
  Three-fifths of distance AB =3X/5
  time taken is (3X/5)/(3a)= 3X/15a
  Remaining distance=2X/5
  time taken is (2X/5)/(2a)= 3X/10a.
  and on return journey X/5c.
  
  so eq is
  
  3X/15a + 2X/10b = X/5c
  
  X/a + X/b = X/c
  
• 11 years agoHelpfull: Yes(1) No(0)
• data in sufficient
• 13 years agoHelpfull: Yes(0) No(4)
• Let the distance between A and B is X
  Three-fifths of distance AB =3X/5
  Remaining distance=1-3x/5=2X/5
  If he goes from A to B and back then total distance X+X=2X
  and 2*[((3X/(5*3a))+(2x/(5*2b))]
  Now acc. to question
  
  2*[(3X/(5*3a))+(2x/(5*2b))]=2X/5c
  solve it and you get 1/a+1/b=1/c
• 12 years agoHelpfull: Yes(0) No(0)
• speed=dist./time
  3a=(3/5AB)/t1
  2b=(2/5AB)/t2
  and also given
  t1+t2=AB/5c
  now, ((3/5AB)/3a)+((2/5AB)2b)=AB/5c
  it becomes,3AB/15a + 2AB/5b=AB/5c
  take 1/5AB common from both side and hence it is proved..
• 8 years agoHelpfull: Yes(0) No(0)
• A....................C...............................B
  AB = x => AC = 3x/5 and CB = 2x/5.
  Distance = Speed * Time
  Time taken to travel the distance between A and C = 3x/5 = 3a * T1 => T1 = x/5a
  Time taken to travel the distance between C and B = 2x/5 = 2b * T2 => T2 = x/5b
  Time taken to travel between A and B back and fourth = x= 5c * T => T = x/5c
  Given that, T = T1 + T2
  => x/5a + x/5b = x/5c
  => 1/a + 1/b = 1/c.
• 8 years agoHelpfull: Yes(0) No(0)