
What is \[20\% \] of 90 ?
Answer
510.3k+ views
Hint: The given question requires us to find a specific percentage of a given number. To solve this problem, we need to know how to write a number in percentage. We can solve this problem by two methods. We can solve either by normal method or by finding the percentage of the whole number. Such questions require accuracy in arithmetic.
Complete step-by-step solutions:
Let’s consider the given problem. We have to find \[20\% \] of \[90\].
So, let \[20\% \] of \[90\] be x.
Then, \[x = 20\% \] of \[90\].
Then, \[x = \dfrac{{20}}{{100}} \times 90\].
So, cancelling the common factors in numerator and denominator, we get,
\[x = \dfrac{2}{{10}} \times 90\]
\[x = 2\times 9\]
Simplifying the calculations further, we get,
\[ \Rightarrow x = 18\]
So, \[20\% \] of 90 is 18.
This is our required answer to the given problem.
Thus the final answer is 18.
Note: Solving problems by the unitary method needs better understanding in the concept of percentage. While solving such a problem using a unitary method, we consider the given number as \[100\% \] of itself and then then find out the desired percentage of the number. Work as many problems as possible to crack these types of problems in a limited time period.
Complete step-by-step solutions:
Let’s consider the given problem. We have to find \[20\% \] of \[90\].
So, let \[20\% \] of \[90\] be x.
Then, \[x = 20\% \] of \[90\].
Then, \[x = \dfrac{{20}}{{100}} \times 90\].
So, cancelling the common factors in numerator and denominator, we get,
\[x = \dfrac{2}{{10}} \times 90\]
\[x = 2\times 9\]
Simplifying the calculations further, we get,
\[ \Rightarrow x = 18\]
So, \[20\% \] of 90 is 18.
This is our required answer to the given problem.
Thus the final answer is 18.
Note: Solving problems by the unitary method needs better understanding in the concept of percentage. While solving such a problem using a unitary method, we consider the given number as \[100\% \] of itself and then then find out the desired percentage of the number. Work as many problems as possible to crack these types of problems in a limited time period.
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