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Question: If the sum of the two numbers is \[9\] and the sum of their reciprocals is\[\dfrac{1}{2}\] . Find th...

If the sum of the two numbers is 99 and the sum of their reciprocals is12\dfrac{1}{2} . Find the numbers.

Explanation

Solution

If xx is a number then its reciprocal is 1x\dfrac{1}{x} .
Then frame the equations according to the given data.
If you get the equation of the general form ax2+bx+c=0a{x^2} + bx + c = 0 use the below formula to find x, x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Stepwise Solution:
Given: Sum of the two numbers is 99 .
The sum of their reciprocals is12\dfrac{1}{2} .
Let the number be xx and therefore its reciprocal becomes 1x\dfrac{1}{x} .
Let the other number be yy and therefore its reciprocal becomes 1y\dfrac{1}{y} .
According to the question,
The sum of the two numbers is 99
x+y=9\Rightarrow x + y = 9
y=9x\Rightarrow y = 9 - x …………….. Equation 1
The sum of their reciprocals is12\dfrac{1}{2}
1x+1y=12\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{2} …………… Equation 2
On substituting equation 1 in equation 2 we get,
1x+19x=12 9x+xx(9x)=12 99xx2=12 18=9xx2 x29x+18=0  \dfrac{1}{x} + \dfrac{1}{{9 - x}} = \dfrac{1}{2} \\\ \Rightarrow \dfrac{{9 - x + x}}{{x\left( {9 - x} \right)}} = \dfrac{1}{2} \\\ \Rightarrow \dfrac{9}{{9x - {x^2}}} = \dfrac{1}{2} \\\ \Rightarrow 18 = 9x - {x^2} \\\ \Rightarrow {x^2} - 9x + 18 = 0 \\\
Now the above equation is in the form of general quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 ,
solve for x using x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Comparing given equation with the general form of quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 we get,
From the above equation a is 11 , b is 9 - 9 and c is 1818

x=(9)±(9)24×1×182×1 x=9±81722 x=9±92 x=9±32  x = \dfrac{{ - ( - 9) \pm \sqrt {{{( - 9)}^2} - 4 \times 1 \times 18} }}{{2 \times 1}} \\\ \Rightarrow x = \dfrac{{9 \pm \sqrt {81 - 72} }}{2} \\\ \Rightarrow x = \dfrac{{9 \pm \sqrt 9 }}{2} \\\ \Rightarrow x = \dfrac{{9 \pm 3}}{2} \\\

Therefore,

x=9+32 x=122 x=6  x = \dfrac{{9 + 3}}{2} \\\ \Rightarrow x = \dfrac{{12}}{2} \\\ \Rightarrow x = 6 \\\

0r

x=932 x=62 x=3  x = \dfrac{{9 - 3}}{2} \\\ \Rightarrow x = \dfrac{6}{2} \\\ \Rightarrow x = 3 \\\

If x=6x = 6 then y=96=3y = 9 - 6 = 3
If x=3x = 3 then y=93=6y = 9 - 3 = 6
Therefore the required numbers are 6,36,3

Note: In such types of questions which involve wordy information they might ask you to frame equations and so we will need to have knowledge about solving equations. Most of the time solving equations leads to quadratic equation forms and so we need to have knowledge about the formula to find the value of x. As calculations play a critical role in these kinds of questions we will need to be vigilant about that while solving.