Question

If A varies as C, and B varies as C, then $A\pm B$ and $\sqrt{AB}$ will each vary as C.

Answer 1

Hint: Now we know that if A varies C we have $A=kC$ where k is constant. Similarly we have $B=lC$ where l is constant. Now we will substitute the values of A and B in $A\pm B$ and $\sqrt{AB}$ and then check the proportionality of the two terms with C.

Complete step-by-step answer:
Now let us first understand the concept of proportionality.
Now the objects can be directly proportional or inversely proportional.
Now let us first understand the concept of directly proportional.
If the two quantities A and B are directly proportional then the ratio of A and B is constant. Hence if A increases by some amount then B also increases by the same amount.
Now this is written as $A\prec B$ and is read as A varies with B.
Here let us say that k is constant of proportionality then we have $A=kB$.
Now let us understand the concept of inversely proportional
If two quantities are inversely proportional then the multiplication of two terms is constant.
Now suppose we have A and B inversely proportional then we have A varies $\dfrac{1}{B}$ . Which means if A is increased by a certain amount then B is decreased by the same amount. Hence we have,
$\begin{align}
  & \Rightarrow A\prec \dfrac{1}{B} \\
 & \Rightarrow A=\dfrac{k}{B} \\
\end{align}$
Now consider the given problem we have A varies as C, and B varies as C.
Consider A varies as C. Let k be the constant of proportionality.
$A=kC$
Now consider B varies as C.
Let l be the constant of proportionality. Hence we have,
$B=lC$
Now consider $A\pm B$,
$\begin{align}
  & \Rightarrow A\pm B=kC\pm lC \\
 & \Rightarrow A\pm B=\left( k\pm l \right)C \\
\end{align}$
Now we have k and l as constant. Hence $k\pm l$ is also a constant. Let this constant be $k'$ .
Hence we get $A\pm B=k'C$ . Hence we can say that $A\pm B$ varies as C.
Now consider $\sqrt{AB}$
$\begin{align}
  & \Rightarrow \sqrt{AB}=\sqrt{kClC} \\
 & \Rightarrow \sqrt{AB}=\sqrt{kl}\times C \\
\end{align}$
Now again we know that $\sqrt{kl}$ is a constant as k and m both are constant. Let us say this constant is $k''$ . Hence we get, $\sqrt{AB}=k''C$.
Hence we get $\sqrt{AB}$ varies as C.

Note: Now note that a variable can also be proportional to the powers of another variable. Suppose we have x is proportional to ${{y}^{2}}$ then we have $x\prec {{y}^{2}}$ . Now if k is the constant of proportionality then we have $x=k{{y}^{2}}$.