Question

The LCM of two numbers is $495$ and their HCF is $5$. If the sum of the numbers is $100$ then their difference is?
$A)10$
$B)46$
$C)70$
$D)90$

Answer 1

Hint: First to solve the given question we need to know about the concept of the LCM and HCF. Since is the least common multiples of the given two or more than two numbers, the process can be obtained by taking the common multiples of the given number and then to find least among them or we can also able to use the prime factorization method, which is to simplify the number into set of prime number.
HCF is the highest common factor or also known as the process of greatest common divisor, which is to find the common divisors first and then choose the greatest among them

Complete step-by-step solution:
Given that the LCM of the two numbers is expressed as $495$, which means the least common multiple of any two numbers will make the number $495$. Also, the same two numbers will make the HCF as $5$
Also given that the sum of the number as $100$ which means the number one and two addition is $100$
Hence let as assume the first number as $x$ (unknown) and then the second number as $100-x$ so that the sum of the two numbers will make the number $100$
We know that the LCM of any numbers and HCF of same numbers product will make the product of the given numbers, which means $a,b$ are given numbers then we have $a\times b=LCM(a,b)\times HCF(a,b)$
Hence making use of the result we have $x\times (100-x)=LCM\times HCF$ and since the LCM of two numbers is $495$ and their HCF is $5$. Thus, we get $x\times (100-x)=495\times 5$
Further solving we get $$x\times (100-x)=495\times 5\Rightarrow 100x-x^2=2475$$ and hence we get the quadratic equation as in the form of $x^2-100x+2475=0$
The roots of a quadratic equation can be found by using the formula $${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$ where $$a$$ - coefficient of the term $$x^2$$, $$b$$ - coefficient of the term $$x$$, and $$c$$ - constant term.
From the given equation $x^2-100x+2475=0$, we have $$a=1$$, $$b=-100$$, and $$c=2475$$.
On substituting these terms in the formula, we get
$$x={\displaystyle \frac{-\left(-100\right)\pm \sqrt{{\left(-100\right)}^2-4\left(1\right)\left(2475\right)}}{2\left(1\right)}}$$
On simplifying this we get
$$x={\displaystyle \frac{100\pm \sqrt{10000-9900}}{2}}$$
On further simplification we get
$$x={\displaystyle \frac{100\pm \sqrt{100}}{2}}$$ and thus, we get $$x={\displaystyle \frac{100\pm 10}{2}}$$
$$x={\displaystyle \frac{100+10}{2}}$$ and $$x={\displaystyle \frac{100-10}{2}}$$
On solving the above, we get
$$x=55$$ and $$x=45$$
Thus, we got the roots of the given quadratic equation that is $$x=55$$ and $$x=45$$
Suppose if the $$x=55$$ and we get another number as $100-55=45$
Suppose if the $$x=45$$ and we get another number as $100-45=55$
Therefore, the difference of the two numbers is $55-45=10$
Hence the option $A)10$ is correct.

Note: Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the formula.
The word quadratic means second-degree values of the given variables.
The formula that we need to know before solving the problem:
Let $$a{x}^2+bx+c=0$$ be a quadratic equation then the roots of this equation are given by $${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$.
The roots of the equation are nothing but the possible value of the unknown variable in that equation. Also, the number of roots of an equation depends on its degree. The degree of an equation is the highest power of the unknown variable in that equation.
Also, in the quadratic equation, the $a=0$ is never possible, because then it will be a linear equation.