Question 155220
y varies jointly as a and b and inversely as the square root of c
then 
{{{y=k(ab)/(sqrt(c))}}}
y=48 when a=4, b=8 and c=36.

48=k(32)/6
then k=48(6)/32=9

then
{{{y=9(ab)/(sqrt(c))}}}
a=2, b=7 and c=16.

then

{{{y=126/4=31.5}}}

correct option is b)