Question

\[x%{ of y + y }\!\!%\!\!{ of }x\] is equal to:
a) \[3%\] of \[xy{ }\]
b) \[2%\] of \[xy{ }\]
c) \[5%\] of \[xy{ }\]
d) \[1%\] of \[xy{ }\]

Answer 1

Hint: The percentage form is given and we have to convert this into general form. First, we use the formula of percentage to convert this. Then after converting we add both of them and then assume the calculated result be \[t\] percent of \[xy\]and on comparing them and we will get the value of \[t\] that is the required answer.

Complete Step by Step Solution:
\[x%{ of y + y }\!\!%\!\!{ of }x\]
Since we know that,
\[x%{ }\]of \[{y}\] is equal to \[\dfrac{x}{100}\times y\]
And similarly, the other term \[{y }\!\!%\!\!{ }\] of \[x\] is equal to \[\dfrac{y}{100}\times x\]
Now according to question, we have to add these two terms
\[\Rightarrow \dfrac{x}{100}\times y+\dfrac{y}{100}\times x\]
\[\Rightarrow \dfrac{xy}{100}+\dfrac{yx}{100}\]
And by the algebraic law, we know that these two terms are like terms therefore we can add them
\[\Rightarrow \dfrac{2xy}{100}\]
Thus, we have solved and simplified the given question
\[\Rightarrow x%{ of y + y }\!\!%\!\!{ of }x=\dfrac{2xy}{100}\]
Now as the question asks this in terms of \[%{ of }xy\]
Therefore,
Let’s assume our calculated value be \[t%\] of \[xy\]
So, we can write this as
\[\Rightarrow \dfrac{t}{100}\times xy\]
\[\Rightarrow \dfrac{txy}{100}\]
Now, comparing value with our calculated one
\[\Rightarrow \dfrac{2xy}{100}=\dfrac{txy}{100}\]
On comparing clearly, \[t=2\]
Thus, the above form is \[2%\] of \[xy\]

Hence, the given percentage form \[x%{ of y + y }\!\!%\!\!{ of }x\] is equal to \[2%\] of \[xy\]
Option (b) is the correct answer.

Note:
First see carefully what is given, in this first convert the percentage form into general fractional form by just dividing the given percentage by \[100\] and multiply with the value whose percentage is given. After that calculate this for all terms and perform operations which are according to question and then we have to just simplify the answer as answer may be asked in another form.