Question

If L.C.M of two numbers is 180, H.C.F is 15 and their sum is 105, what is the difference between those two numbers?

Answer 1

Hint:We will use the property of L.C.M and H.C.F to find the required value of a in this question. We should remember the concept that the product of L.C.M and H.C.F or G.C.D of any two given natural numbers is always equivalent to the product of the given numbers. For example, if we have two numbers, a and b, then, $ a\times b=H.C.F\left( a,b \right)\times L.C.M\left( a,b \right) $ .

Complete step-by-step answer:
We have been given that the L.C.M of two numbers is 180, H.C.F is 15 and their sum is 105 a.nd we have been asked to find the difference between the two numbers. Let us assume the two numbers as, a and b. So, according to the question, we can write as,
L.C.M (a, b) = 180
H.C.F (a, b) = 15
And a + b = 105
So, we can say that, b = 105 - a ………… (i)
We know that the property of L.C.M and H.C.F that the product of L.C.M and H.C.F of any two natural numbers is equal to the product of the numbers. So, we can write as,
 $ \begin{align}
  & a\times b=L.C.M\left( a,b \right)\times H.C.F\left( a,b \right) \\
 & \Rightarrow a\times b=180\times 15 \\
 & \Rightarrow a\times b=2700 \\
\end{align} $
Now, we will use the value of b in terms of a from equation (i) in the above equality. So, we get,
 $ \begin{align}
  & a\times \left( 105-a \right)=2700 \\
 & \Rightarrow 105a-{{a}^{2}}=2700 \\
\end{align} $
On rearranging the terms, we will get,
 $ {{a}^{2}}-105a+2700=0 $
We know the formula to find the roots of a quadratic equation is as, follows: if $ a{{x}^{2}}+bx+c=0\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ . So, we will apply this in the above equation. So, for a = 1, b = 105, c= 2700 and x = a, we get,
 $ \begin{align}
  & a=\dfrac{105\pm \sqrt{{{\left( 105 \right)}^{2}}-4\times 2700}}{2} \\
 & \Rightarrow a=\dfrac{105\pm \sqrt{11025-10800}}{2} \\
 & \Rightarrow a=\dfrac{105\pm \sqrt{225}}{2} \\
 & \Rightarrow a=\dfrac{105\pm 15}{2} \\
 & \Rightarrow a=\dfrac{105+15}{2}\text{ and a}=\dfrac{105-15}{2} \\
 & \Rightarrow a=\dfrac{120}{2}\text{ and a}=\dfrac{90}{2} \\
 & \Rightarrow a=60\text{ and a}=45 \\
\end{align} $
If we take the value of a = 60 and substitute this value in equation (i), we will get,
b = 105 - 60 = 45.
So, the difference between the numbers will be,
 $ \begin{align}
  & \left| a-b \right|=60-45 \\
 & \Rightarrow \left| a-b \right|=15 \\
\end{align} $
And if we take the value of a = 45, in equation (i), we will get,
b = 105 - 45 = 60
So, the difference between the numbers will be,
 $ \begin{align}
  & \left| a-b \right|=\left| 45-60 \right| \\
 & \Rightarrow \left| a-b \right|=15 \\
\end{align} $
Therefore, we get the difference between the two numbers as 15.

Note: We should be careful while finding the roots of the quadratic equation and use the formula to find the roots, otherwise it would be difficult to find it by another method. We should also remember that the difference is always positive, so we will take the modulus value of the difference.