Solution:

The equation for the given problem of joint variation is

x = Kyz where K is the constant.

For the given data

16 = K × 4 × 6

or, K = 46$\frac{4}{6}$.

So substituting the value of K the equation becomes

x = 4yz6$\frac{4yz}{6}$

Now for the required condition

x = 4×8×126$\frac{4\times 8\times 12}{6}$

   = 64

Hence the value of x will be 64.

Solution:

From the given problem equation for the joint variation is

A = KBC²

From the given data value of the constant K is

K = BC2A$\frac{B{C}^2}{A}$

K = 4×32144$\frac{4\times 3^2}{144}$

   = 36144$\frac{36}{144}$

    = 14$\frac{1}{4}$.

Substituting the value of K in the equation

A = BC24$\frac{B{C}^2}{4}$

A = 6×424$\frac{6\times 4^2}{4}$

   = 24

Solution:

We know the area of triangle is half the product of base and height. So the joint variation equation for area of triangle is A = bh2$\frac{bh}{2}$ where A is the area, b is the base and h is the height.

Here 12$\frac{1}{2}$ is the constant for the equation.

Base is increased by 10%, so it will be b x 110100$\frac{110}{100}$ = 11b10$\frac{11b}{10}$.

Height is decreased by 10%, so it will be h x 90100$\frac{90}{100}$ = 9h10$\frac{9h}{10}$.

So the new area after the changes of base and height is

11b10×9h102$\frac{\frac{11b}{10}\times \frac{9h}{10}}{2}$

= (99100$\frac{99}{100}$)bh2$\frac{bh}{2}$ = 99100$\frac{99}{100}$A.

So the area of the triangle is decreased by 1%.