Question

The HCF of two numbers is 28 and their LCM is 336. If one number is 112 then the other number is
A. 64
B. 84
C. 34
D. none

Answer 1

Hint: We will be using the concepts of number theory to solve the problem and using the concept of H.C.F. and L.C.M. to find the number then form the equation with the help of data given to get the desired answer.

Complete step-by-step answer:
Now, we have been given that the H.C.F. of two numbers is 28. So, we know by definition that H.C.F. is the highest common factor. Therefore, we let the two numbers be 28a, 28b where a and b are co- primes.
Now, we have been given that the L.C.M. of two numbers is 336. Now, we know by definition of L.C.M. that it is the least common multiple of two numbers. So, therefore the numbers we have are 28a, 28b their L.C.M. will be 28ab as it is the least number that is divisible by both 28a and 28 b.
Now, we have been given this equal to 336. So, we have,
$\begin{align}
  & 28ab=336 \\
 & ab=\dfrac{336}{28} \\
 & ab=12.............\left( 1 \right) \\
\end{align}$
Now, we have been given one number 112 so let,
$\begin{align}
  & 28a=112 \\
 & a=\dfrac{112}{28} \\
 & a=4................\left( 2 \right) \\
\end{align}$
Using (1) and (2) we will find the values of b,
$\begin{align}
  & 4\times b=12 \\
 & b=3 \\
\end{align}$
So, the other number 28b,
$\begin{align}
  & =28\times 3 \\
 & =84 \\
\end{align}$

Note: Alternative way:
We know that the product of two numbers is equal to the product of their L.C.M. and H.C.F. Therefore, we have,
Product of two numbers = Product of their L.C.M and H.C.F
\[\begin{align}
  & 112\times b=336\times 28 \\
 & b=\dfrac{336\times 28}{112} \\
 & b=28\times 3 \\
 & b=84 \\
\end{align}\]