Question 75

The average of a number and its reciprocal is 4. The average of its cube and its reciprocal is equal to:

Solution

Let the number be x and its reciprocal is $$\frac{1}{x}$$

Average of x and $$\frac{1}{x}$$ = 4

$$\frac{x+\frac{1}{x}}{2}=4$$

$$x + \frac{1}{x}=8$$

Cubing both side, we get

$$x^{3} + \frac{1}{x^{3}}+3(x+\frac{1}{x} = 512)$$

$$x^{3} + \frac{1}{x^{3}}+24 = 512$$ $$(x + \frac{1}{x}=8)$$

$$x^{3} + \frac{1}{x^{3}}=512-24=488$$

Required average = $$\frac{488}{2}=244$$

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