• The sum of two numbers is 10. Their product is 20. Find the sum of the reciprocals of the two numbers.
 A) 1 B) $$\Large \frac{3}{5}$$ C) $$\Large \frac{1}{2}$$ D) $$\Large \frac{11}{6}$$

 C) $$\Large \frac{1}{2}$$

Let the numbers be x and y, Then,

x + y = 10 ...(i)

and xy = 20 ...(ii)

Therefore, $$\Large \left(x-y\right)^{2}= \left(x+y\right)^{2}-4xy$$

$$\Large \left(x-y\right)^{2}= \left(10\right)^{2}-4 \times 20=20$$

$$\Large x - y = \sqrt{20} = 2\sqrt{5}$$ ...(iii)

On adding Eqs. (i) and (iii), we get

x + y = 10

$$\Large x - y = 2\sqrt{5}$$

$$\Large 2x = 10 + 2\sqrt{5}$$

Therefore, $$\Large x = 5 + \sqrt{5}$$

On putting the value of x in Eq. (i), we

x + y = 10

=> $$\Large 5 + \sqrt{5} + y = 10$$

Therefore, $$\Large y = 5 - \sqrt{5}$$

Sum of reciprocals of x and y = $$\Large \frac{1}{x}$$ + $$\Large \frac{1}{y}$$

$$\Large =\frac{1}{5+\sqrt{5}}+\frac{1}{5-\sqrt{5}}= \frac{5-\sqrt{5}+5+\sqrt{5}}{ \left(5+\sqrt{5}\right) \left(5-\sqrt{5}\right) }$$

$$\Large = \frac{10}{ \left(5\right)^{2} - \left(\sqrt{5}\right)^{2} } = \frac{10}{20} = \frac{1}{2}$$

##### Similar Questions
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