A horse-carter has two wheels,front-wheel and backwheel...and their circumference r in the ratio x:y. front-wheel take T rotations in one min and back-wheel takes T+x rotation in one minute.T,x,y r integers >0..so find the second least value of y...when y>x?

• C1:C2=x:y
  
  C1R1=C2R2 => x*T=y*(T+x) => x/y=(T+x)/T
  as T,x,y >0 & integrs so (T+x)/T is always > 1
  so, x/y=(T+x)/T > 1
  => x > y
  when y > x , no values of x,y .
• 10 years agoHelpfull: Yes(25) No(0)
• The question itself is show that it is not possible as back wheel is making more rotations than front means y < x, but in question it is given that y > x .. so it is not possible.
  
• 10 years agoHelpfull: Yes(6) No(0)
• no y for any value of x and t satisfying y>x. am i correct
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• y*T=x*(T+x)
  
• 10 years agoHelpfull: Yes(1) No(0)
• c1:c2=x:y
  
  c2 takes t rotations it meams total distace =ty
  c1 takes t+x rotations it meams total distace=(t+x)x
  ty=tx+x^2 for min value of we have to put min value of x and t
  PUT T=1,X=1 then y=2,
  put t=1, x=2 then y=6
  put t=2 x=1 the y=4
  hence 2nd least value of y=4m Ans
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• y=T*x/(T+x)
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