The value of $\frac{99}{100} \times 99$ is
This solution details the calculation steps to find the value of the mathematical expression $\frac{99}{100} \times 99$. The goal is to simplify the expression and present it as a mixed number.
Start with the given expression:
$ \frac{99}{100} \times 99 $
Rewrite the fraction $\frac{99}{100}$ to make the calculation easier. Notice that $\frac{99}{100}$ is equal to $1 - \frac{1}{100}$.
$ \left(1 - \frac{1}{100}\right) \times 99 $
Distribute the $99$ to both terms inside the parentheses:
$ (1 \times 99) - \left(\frac{1}{100} \times 99\right) $
Simplify the terms:
$ 99 - \frac{99}{100} $
To subtract the fraction, express $99$ as an equivalent fraction with a denominator of $100$.
$ 99 = \frac{99 \times 100}{100} = \frac{9900}{100} $
Perform the subtraction:
$ \frac{9900}{100} - \frac{99}{100} = \frac{9900 - 99}{100} $
Calculate the numerator:
$ 9900 - 99 = 9801 $
The result is the improper fraction $\frac{9801}{100}$.
Convert the improper fraction $\frac{9801}{100}$ to a mixed number. Divide $9801$ by $100$.
$9801 \div 100 = 98$ with a remainder of $1$.
Therefore, the mixed number is $98 \frac{1}{100}$.
The value of $\frac{99}{100} \times 99$ is $98 \frac{1}{100}$.
The value of 0.18÷0.015 is:
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Simplify: \((x^{\frac{m}{n}})^{m+n} \times (x^{\frac{n}{p}})^{n+p} \times (x^{p} \times x^{m})^{p-m}\)
Which of the following is correct for divisibility?
(A) A number is divisible by 6 if it is divisible by 3 or 2.
(B) A number is divisible by 5 if its unit digit is 0 or 5.
(C) A number is divisible by 3 if its unit digit is divisible by 3.
(D) A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
Choose the correct answer from the options given below: