Question

A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number.

Answer 1

Hint: The given problem is a word problem. Consider the number to be x. Using the information given in the question, form an equation. Solve the equation to get the value of x, which will be the required value of the number.

Complete step-by-step answer:
Let the number be x. In the question, it is given that when x is increased by 12, becomes equal to 160 times its reciprocal of the x.
So, we will add 12 to x and equate it with reciprocal of x multiplied by 160 . \[\Rightarrow x+12=160\left( \dfrac{1}{x} \right)\ldots \ldots .\left( 1 \right)\]
Now, we will solve equation 1 to find the value of x.
\[x+12=160\left( \dfrac{1}{x} \right)\]
\[\Rightarrow {{x}^{2}}+12x=160\]
Now subtract 160 on both sides to convert the equation in the form \[a{{x}^{2}}+bx+c=0\] .
\[\Rightarrow {{x}^{2}}+12x-160=0\]
Now, the equation is in the form \[a{{x}^{2}}+bx+c=0\]. We know that, the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\] is given by:
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
On comparing \[{{x}^{2}}+12x-160=0\] with \[a{{x}^{2}}+bx+c=0\] , we can say that \[a=1,b=12\] and \[c=-160\] .
Now, we will substitute the values of a, b and c in \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] .
\[\Rightarrow x=\dfrac{-12\pm \sqrt{{{\left( 12 \right)}^{2}}-4(1)(-160)}}{2(1)}\]
\[\Rightarrow x=\dfrac{-12\pm \sqrt{144+640}}{2}\]
\[\Rightarrow x=\dfrac{-12\pm \sqrt{784}}{2(1)}\]
We know that \[{{28}^{2}}=784\] .
\[\Rightarrow \sqrt{784}=28\]
\[\Rightarrow x=\dfrac{-12\pm 28}{2}\]


\[\Rightarrow x=\dfrac{16}{2},\dfrac{-40}{2}\]
\[\Rightarrow x=8,-20\]
Since, x is a natural number. The value of a natural number can never be negative.
Therefore, the natural number is 8.

Note: We can solve equation 1 by using the factor method.
\[x+12=160\left( \dfrac{1}{x} \right)\ldots \ldots .\left( 1 \right)\]
Multiply with x on both the sides.
\[\Rightarrow {{x}^{2}}+12x-160=0\]
Here the 12x can be written as 20x-8x.
\[\Rightarrow {{x}^{2}}+20x-8x-160=0\]
\[\Rightarrow x\left( x+20 \right)-8\left( x+20 \right)=0\]
Here take x+20 common
\[\Rightarrow \left( x+20 \right)\left( x-8 \right)=0\]
\[\Rightarrow x=8,-20\]
But x is a natural number. So, it cannot be negative.
\[\Rightarrow x=8\]
Therefore, the number is 8.