Method 1: In an inverse variation as one variable increases the other decreases. #color(blue)(yprop1/x#

We now insert a constant #k# into the equation:

#y = k * 1/x#

#color(blue)(y = k /x#

We now substitute the value of variables as provided:

#x=10 , y =50#

#50 = k /10#

#k = 50 * 10 = color(green)(500#

Now we use the value of this constant to find #y:# As per data provided #x =20# #color(blue)(y = k /x#

#y = 500 /20#

#color(purple)(y = 25#

Method 2:

As per data provided when #color(green)(x =10)# , #y =50#

So, when #x =20#, the value is doubled from the initial value of #x#, (#20 = color(green)(10) * 2)#

Since the value of #x# is doubled , the value of #y# will have to be halved from the initial value, since there is an inverse variation.

So, #y = 50/2, color(purple)(y = 25#