Question

The sum of two numbers is 12 and their product is 35. What is the sum of the reciprocals of these numbers?
(a) \[\dfrac{12}{35}\]
(b) \[\dfrac{1}{35}\]
(c) \[\dfrac{35}{8}\]
(d) \[\dfrac{7}{32}\]

Answer 1

Hint: Assume x and y as the two numbers. Form first relation in x and y by taking sum and equating with 12. Form a second relation in x and y by taking their product and equating with 35. Solve the two equations by forming a quadratic equation and using the middle term split method. Obtain the values of x and y and take the sum of their reciprocals, i.e. \[\left( \dfrac{1}{x}+\dfrac{1}{y} \right)\] to get the answer.

Complete step-by-step answer:
Here, we have to find two numbers such that their sum is 12 and their product is 35.
Now, let us assume the two numbers as x and y. So, according to the question, we have,
\[\Rightarrow x+y=12\]
\[\Rightarrow y=12-x\] - (1)
And \[xy=35\] - (2)
Substituting the value of y from equation (1) in equation (2), we get,
\[\begin{align}
  & \Rightarrow x\left( 12-x \right)=35 \\
 & \Rightarrow 12x-{{x}^{2}}=35 \\
 & \Rightarrow {{x}^{2}}-12x+35=0 \\
\end{align}\]
Using the middle term split method, we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}-7x-5x+35=0 \\
 & \Rightarrow x\left( x-7 \right)-5\left( x-7 \right)=0 \\
\end{align}\]
Taking \[\left( x-7 \right)\] common we get,
\[\Rightarrow \left( x-7 \right)\left( x-5 \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow x-7=0\] or x – 5 = 0
\[\Rightarrow x=7\] or x = 5
1. When x = 7: -
\[\Rightarrow y=12-x=12-7=5\]
2. When x = 5: -
\[\Rightarrow y=12-x=12-5=7\]
Now, we have been asked to find the sum of reciprocal of these numbers. So, let us consider both the cases one by one.
In first case we have x = 7 and y = 5, so taking their reciprocals, we get, \[\dfrac{1}{x}=\dfrac{1}{7}\] and \[\dfrac{1}{y}=\dfrac{1}{5}\]. So, we have,
\[\begin{align}
  & \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{7}+\dfrac{1}{5} \\
 & \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5+7}{7\times 5} \\
 & \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{12}{35} \\
\end{align}\]
Now, in second case we have x = 5 and y = 7, so taking their reciprocals, we get, \[\dfrac{1}{x}=\dfrac{1}{5}\] and \[\dfrac{1}{y}=\dfrac{1}{7}\]. So, we have,
\[\begin{align}
  & \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}+\dfrac{1}{7} \\
 & \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{12}{35} \\
\end{align}\]
Therefore, in both the cases we are getting the sum of the reciprocals of the obtained numbers equal to \[\dfrac{12}{35}\], which was necessary. If we would have got the value of sum different in both the cases then our answer would have been wrong.

So, the correct answer is “Option (a)”.

Note: One may note that we can also solve this question without finding the values of x and y. What we can so is we will take the sum of the reciprocal of these numbers, i.e. \[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}\] and equate the values of \[\left( x+y \right)\] and xy using equations (1) and (2). You may note that we have solved the obtained quadratic equation using the middle term split method. One can also use the discriminant method given as: - \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to get the values of x. Here, ‘a’ is the co – efficient of \[{{x}^{2}}\], ‘b’ is the co – efficient of x and ‘c’ is the constant term. As you can see that we have obtained two values of x after solving the quadratic equation and none of them is rejected. This is because both the values of x are valid.