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The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.

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Hint: - Use the formula ${\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy$
Let the numbers be $x$ and $y$.
According to the question , the sum of two numbers is 8.
$ \Rightarrow x + y = 8...............\left( 1 \right)$
Reciprocals of the numbers be$\dfrac{1}{x}$and $\dfrac{1}{y}$
According to question 15 times the sum of reciprocals is also 8
$ \Rightarrow 15\left( {\dfrac{1}{x} + \dfrac{1}{y}} \right) = 8$
$ \Rightarrow 15\left( {\dfrac{{x + y}}{{xy}}} \right) = 8$
From equation 1
$
   \Rightarrow 15\left( {\dfrac{8}{{xy}}} \right) = 8 \\
   \Rightarrow xy = 15.........\left( 2 \right) \\
$
Now it is known fact that ${\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy$
Now from equation (1) and (2)
$
  {\left( {x - y} \right)^2} = {8^2} - 4 \times 15 = 64 - 60 = 4 = {2^2} \\
   \Rightarrow x - y = 2...........\left( 3 \right) \\
$
Now add equation (1) and (3)
$
   \Rightarrow x + y + x - y = 8 + 2 \\
   \Rightarrow 2x = 10 \\
   \Rightarrow x = 5 \\
$
From equation 1
$
   \Rightarrow x + y = 8 \\
   \Rightarrow 5 + y = 8 \\
   \Rightarrow y = 3 \\
$
So, the required numbers are 5 and 3.

Note: - Whenever we face such types of problems the key concept we have to remember is that always remember the formula ${\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy$. It will give us a simple approach to solve this kind of problem, then simplify we will get the required answer.

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