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The sum of two numbers is 10. Their product is 20. Find the sum of the reciprocals of the two numbers.

Correct Answer:

A) 1 |
B) \( \Large \frac{3}{5} \) |

C) \( \Large \frac{1}{2} \) |
D) \( \Large \frac{11}{6} \) |

Correct Answer:

C) \( \Large \frac{1}{2} \) |

Description for Correct answer:

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 = \( \Large \frac{1}{x} \) + \( \Large \frac{1}{y} \)

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

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

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 = \( \Large \frac{1}{x} \) + \( \Large \frac{1}{y} \)

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

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

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