Question

If a varies jointly as b and c and inversely as the square of d, and a = 120 when b = 5, c = 2, and d = 9, find a when b = 12, c = 9 and d = 9.

Answer 1

$a=k\frac{bc}{d^2}.$

$120=k\frac{5*2}{9^2} \to k=\frac{120*81}{10}=972.$

When b=12, c=9, d=9: $a=972\frac{12*9}{9^2}=1296.$