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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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