A horsecarter has two wheels,frontwheel nd bachwheel...nd their circumference r in the ratio x:y. frontwheel take T rotations in one min nd backwheel takes T+x rotation in one minute.T,x,y r integers >0..so fing the second least value of y...when y>x?

• least integral value of y = 1
  second least value = 2
  Tx= (T+x)*y
  y= Tx/(T+x)
  so y=1, when T=x=2
  y=2, when T=x=4
• 11 years agoHelpfull: Yes(2) No(11)
• Tx=(T+x)y
  =>y=T/(1=T/x)
  as y>x and x>0
  so x=1,y=2
  =>T=2(T>0)
  =>Y=2
• 11 years agoHelpfull: Yes(1) No(2)
• but guys,,
  here given that y>x,,
  but you got y
• 11 years agoHelpfull: Yes(1) No(0)
• its given y>x,which implies that the ratio x:y is for backwheel:frontwheel.then (T+x)x=Ty.if Tx=(T+x)y,then:xy=T(x-y) which will be negative,as y>x,which is not possible.
  now if Tx+x^2=Ty.so x^2=T(y-x).so second least value comes out to be 3.then with x=2,T=4.y cannot be 1.so least value will be 2.
• 11 years agoHelpfull: Yes(0) No(5)
• @aman by the way u have done the question the 2nd least value of y is 2 only possible put T=3,x=6 i.e. 3*6/3+6 ie. 18/9=2
• 11 years agoHelpfull: Yes(0) No(1)