Question

Divide 27 into two parts such that the sum of their reciprocals is \[\dfrac{3}{20}\].

Answer 1

Hint: Assume x and y as the two parts in which 27 must be divided. Form first relation in x and y by taking their sum and equating with 27. Form a second relation by taking the reciprocals of x and y and equating their sum with \[\dfrac{3}{20}\]. Solve the two equations by forming a quadratic equation and using the middle term split method.

Complete step-by-step answer:
Here, we have been provided with a number 27 and we have to decide it in two parts such that the sum of the reciprocals of these parts in \[\dfrac{3}{20}\].
Now, let us assume the two parts in which 27 must be divided as x and y. So, we have,
\[\Rightarrow x+y=27\]
\[\Rightarrow y=27-x\] - (1)
Taking the reciprocals of x and y we get \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] respectively. So, according to the question, we have,
\[\begin{align}
  & \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{30} \\
 & \Rightarrow \dfrac{x+y}{xy}=\dfrac{3}{30} \\
\end{align}\]
Substituting the value of \[x+y=27\] in the above relation, we get,
\[\begin{align}
  & \Rightarrow \dfrac{27}{xy}=\dfrac{3}{20} \\
 & \Rightarrow xy=\dfrac{20\times 27}{3} \\
 & \Rightarrow xy=20\times 9 \\
 & \Rightarrow xy=180 \\
\end{align}\]
Now, using equation (1), we get,
\[\begin{align}
  & \Rightarrow x\left( 27-x \right)=180 \\
 & \Rightarrow 27x-{{x}^{2}}=180 \\
\end{align}\]
\[\Rightarrow {{x}^{2}}-27x+180=0\]
Using the middle term split method, we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}-15x-12x+180=0 \\
 & \Rightarrow x\left( x-15 \right)-12\left( x-15 \right)=0 \\
\end{align}\]
Taking \[\left( x-15 \right)\] common we get,
\[\Rightarrow \left( x-15 \right)\left( x-12 \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow x-15=0\] or x – 12 = 0
\[\Rightarrow x=15\] or x = 12
1. When x = 15: -
\[\Rightarrow y=27-15=12\]
2. When x = 12: -
\[\Rightarrow y=27-12=15\]
So, there can be two ways to divide 27 into two parts so that the given conditions might get fulfilled. They are: - if x = 15 then y = 12 and if x = 12 then y = 15.

Note: One may note that we can also solve this question without using the second variable y. We can directly assume the two numbers as x and (27 – x) because we can clearly see that their sum will be 27. Now, 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 answer. 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.