y varies directly as x and inversely as z y100 when x5 and z

y varies directly as x and inversely as z. y=100 when x=5 and z=10. Find y when x=3 and z=60.

Solution

y varies directly as x and inversely as z

Let the relation between x ,y and z be : y = k1(x/z) where k1 is constant we need to find it.

y = 100 ; x= 5 and z=10

100 = k1*5/10 ----> k1 =200

So, y = 200x/z

Now we have x= 3 and z=60 , find y

y = 200*(3/60)

=200/20 = 10

y=10