Question 192

What will come in place of both the question marks (?) in the following equation ?
$$\frac{(?)^{2}}{108}=\frac{16}{(?)}$$

Solution

Given, $$\frac{(x)^{2}}{108}=\frac{16}{(x)}$$

Rearranging them gives, $$x^{3}$$ = 16*108

$$x^{3}$$ = (4*4)*(9*12)

= (4*4)*(3*3*4*3)

= $$4^{3}$$*$$3^{3}$$

= $$12^{3}$$

Hence, x = 12


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