Question

The HCF of two numbers is 12 and their difference is also 12. The numbers are
\[\begin{array}{*{35}{l}}
   \left( \text{a} \right)\text{ 12, 84} \\
   \left( \text{b} \right)\text{ 84, 96} \\
   \left( \text{c} \right)\text{ 64, 76} \\
   \left( \text{d} \right)\text{ 100, 112} \\
\end{array}\]

Answer 1

Hint: In this, we will use the HCF of two numbers to find the two numbers. HCF (highest common factor) of the two numbers is the largest positive number which divided both the numbers.

Complete step-by-step solution:
Given that the HCF of two numbers is 12 and their difference is 12.
Let $x$ and $y$ be the two numbers
Since the HCF of x and y is 12
$\Rightarrow$ $x=12a$ and $y=12b$ where a and bare prime to each other
Also, the difference between the two numbers is 12.
$\Rightarrow $ $x-y=12$
$\Rightarrow $$12a-12b=12$
$\Rightarrow $$12\left( a-b \right)=12$
By dividing both sides by 12, we get
$\Rightarrow $$a-b=1\text{ }...\text{(1)}$
$\Rightarrow $$a$ and $b$ are consecutive to each other and also are prime to each other.
Now, we will check options one by one and then eliminate it.
Option (a), let $x=8\text{4 and }y=12.$
$\begin{align}
  & \Rightarrow x-y=84-12. \\
 & \Rightarrow x-y=72. \\
\end{align}$
Since, difference between to numbers is 12
$\Rightarrow $ Option (a) is not correct.
Option (b), let $x=96\text{ and }y=84.$
$\begin{align}
  & \Rightarrow x-y=96-84. \\
 & \Rightarrow x-y=12. \\
\end{align}$
Hence, the difference of two numbers is 12.
Since HCF of two numbers is 12. We will check whether both number satisfies the equation (1) or not.
By dividing $x\text{ and }y$ by 12 and then subtracting $\dfrac{y}{12}\text{ from }\dfrac{x}{12}$, we get
 $\begin{align}
  & \Rightarrow \dfrac{x}{12}-\dfrac{y}{12}=\dfrac{96}{12}-\dfrac{84}{12}=8-7=1 \\
 & \Rightarrow \dfrac{x}{12}-\dfrac{y}{12}=1. \\
\end{align}$
This show $x\text{ and }y$ satisfies equation (1) when dividing 12.
Option (b) is correct.
Option (c), let $x=76\text{ and }y=64.$
Since, 12 does not divides $x=76$
$\Rightarrow $12 is not HCF of $x=76\text{ and }y=64.$
Option (c) is not correct.
Option (d), let $x=100\text{ and }y=112.$
Since, 12 does not divides both $x=100\text{ and }y=112.$
$\Rightarrow $12 is not HCF of $x=100\text{ and }y=112.$
Option (d) is not correct.
Hence by eliminating options we get that (b) is the correct option.

Note: In this problem, it should know that numbers are said to be consecutive if they are continuously followed by each other from smallest to largest in order and two numbers are said to be co-prime or primes of each other if the HCF of two numbers is 1.