Latest Elitmus Aptitude Question SOLUTION: 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 o

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(#M40033095) ELITMUS Numerical Ability question apti Keep an EYE

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?

Asked In Elitmus ankit chauhan (10 years ago)

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RAKESH  (10 years ago)

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 .

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manivikas  (10 years ago)

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.

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Sridhar Reddy  (10 years ago)

no y for any value of x and t satisfying y>x. am i correct

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ashish bharti  (10 years ago)

y*T=x*(T+x)

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deepak kumar  (10 years ago)

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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sk sohel ali  (10 years ago)

y=T*x/(T+x)

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