For this, we follow the given steps:
When we divide, we get two types of decimals, which are as follows:
(1) Terminating decimals: The decimals that end with a finite (limited) number of digits are known as terminating decimals.
For example: \(0.8,\;0.25,\;0.46374\) etc.
NOTE : The fractions having the denominators that can be written in the form of powers of \(2\;\text{[ example: 4, 8, 16....]}\) or \(5\;\text{[ example: 5, 25, 125...]}\) result in terminating decimals.
(2) Repeating decimals: The decimals which do not end but one or more digits keeps on repeating are known as repeating decimals or non-terminating but repeating decimals.
Here, 'bar' represents the repetition.
For example: \(0.6262...=0.\overline{62}\)
NOTE : We can only put the bar over the digits which repeat.
For example: \(0.888=0.\overline8\)
\(2.1575757...=2.1\overline{57}\)
(3) Non-terminating and non-repeating decimals:
The decimals which neither end with a finite number of digits nor do they repeat are called non-terminating and non-repeating decimals.
For example: \(2.42152434...\)
NOTE: Those square roots which are not perfect square numbers, always give non-terminating and non-repeating decimals.
For example: \(\sqrt 3=1.732.....\)
A \(0.\overline2\)
B \(0.\overline1\)
C \(0.\overline4\)
D \(0.\overline5\)
For example: \(1.2323.....\)
\(0.\overline4\)
For example: Converting \(0.\overline{09}\) into a fraction.
Step 1. Consider the decimal as a variable.
Let \(x=0.\overline{09}\) \(.......(1)\)
Step 2. Multiply both the sides of the equation by \(10\) if one digit repeats, by \(100\) if two digits repeat and so on.
In \(0.\overline{09}\), two digits are repeating every time.
\(\therefore\) We multiply both the sides of equation \((1)\) by \(100\).
\(x=0.\overline{09}\)
\(\Rightarrow x=0.090909....\)
\(\Rightarrow100×x=100×(0.0909....)\)
\(\Rightarrow100x=09.0909.....\)
\(\Rightarrow100x=9.0909.....\)
We can write \(9.0909.....=9+0.0909......\)
\(\therefore\;\;100x=9+0.0909.....\)
\(\Rightarrow100x=9+0.\overline{09}\;\;\;\;\;\;\;\;..........(2)\)
Putting \(x\) in place of \(0.\overline{09}\) in equation \((2)\) from equation \((1)\).
\(100x=9+x\)
Step 4. Solving the equation to find the value of \(x\).
\(100x=9+x\)
Subtracting \(x\) from both the sides,
\(\Rightarrow100x-x=9+x-x\)
\(\Rightarrow(100-1)x=9\) \((x-x=0)\)
\(99x=9\)
Dividing both the sides by \(99\)
\(\Rightarrow\dfrac{99x}{99}=\dfrac{9}{99}\)
\(\Rightarrow x=\dfrac{1}{11}\) \(\left( \dfrac{9\div9}{99\div9}=\dfrac{1}{11} \right)\)
Therefore, the repeating decimal \(0.\overline{09}=\dfrac{1}{11}\)
A \(\dfrac{3}{5}\)
B \(\dfrac{3}{8}\)
C \(\dfrac{7}{9}\)
D \(\dfrac{8}{5}\)
Consider two fractions as two points on the number line.
On a number line, a fraction which is placed to the left is smaller than the fraction which is placed to the right side.
For example: Consider \(\dfrac{1}{4}\) and \(\dfrac{5}{4}\).
\(\dfrac{1}{4}\) is placed to the left of \(\dfrac{5}{4}\).
\(\therefore\) \(\dfrac{1}{4}<\dfrac{5}{4}\)
Note: While comparing two or more fractions, make sure that each one of them has the same denominator.
A \(P\)= \(\dfrac{9}{6}\), \(R\)= \(\dfrac{5}{6}\), \(P>R\)
B \(P\)= \(\dfrac{5}{6}\), \(R\)= \(\dfrac{9}{6}\), \(P>R\)
C \(P\)= \(\dfrac{9}{6}\), \(R\)= \(\dfrac{5}{6}\), \(R>P\)
D \(P\)= \(\dfrac{5}{6}\), \(R\)= \(\dfrac{9}{6}\), \(R>P\)
Convert \(\dfrac{11}{2}\) into a mixed number.
Step 1: Rewrite the given fraction as a division problem and solve it.
Fraction \(=\dfrac{11}{2}\)
Step 2: To get a mixed number from division:
Quotient \(=5\longleftarrow\) Whole number
Remainder \(=1\longleftarrow\) Numerator
Divisor \(=2\longleftarrow\) Denominator
Steps 3: Write the required mixed number.
Whole number \(=5\)
Fraction \(=\dfrac{1}{2}\)
Mixed number \(=5\dfrac{1}{2}\)
Step 4: Simplify the fraction part of the mixed number.
\(1\) and \(2\) do not have any common factor other than \(1\).
\(\therefore\;\dfrac{1}{2}\) is in its simplest form.
Thus, \(5\dfrac{1}{2}\) is the answer.
Convert \(5\dfrac{1}{3}\) into an improper fraction.
Step 1: Multiply the whole number by the denominator and add the numerator,
\(=5×3+1\)
\(=15+1\)
\(=16\)
Step 2: Put the result over the original denominator,
\(=\dfrac{16}{3}\)
Step 3: The fraction obtained is an improper fraction.
As \(16>3\)
\(\therefore\;\dfrac{16}{3}\) is an improper fraction.
A \(3\dfrac{1}{2}\)
B \(3\dfrac{1}{3}\)
C \(2\dfrac{1}{3}\)
D \(5\dfrac{1}{2}\)
Arrange \(2\dfrac{1}{3}\), \(\dfrac{3}{4}\) and \(2\) from greatest to least.
First, make sure that all the numbers are in the same form.
So, we convert them into fraction form.
Here, \(2\dfrac{1}{3}=\dfrac{2×3+1}{3}=\dfrac{7}{3}\) and \(2=\dfrac{2}{1}\)
Step 1: Make sure that the denominators are equal, if they are not, convert them into like fractions.
Here, the denominators are not equal.
So, we convert them into like fractions.
L.C.M. of \(4\), \(3\) and \(1=12\)
L.C.M. becomes the least common denominator.
\(\dfrac{3}{4}×\dfrac{3}{3}=\dfrac{9}{12}\)
\(\dfrac{7}{3}×\dfrac{4}{4}=\dfrac{28}{12}\) and \(\dfrac{2}{1}×\dfrac{12}{12}=\dfrac{24}{12}\)
Step 2: Compare the numerators.
The fraction with larger numerator has the higher value than the others.
Here, \(28>24>9\)
So, \(\dfrac{28}{12}>\dfrac{24}{12}>\dfrac{9}{12}\)
Step 3. Arranging the original numbers from greatest to least.
\(2\dfrac{1}{3},\;2,\;\dfrac{3}{4}\)
A \(\text{Carl, Aron, Sam }\)
B \(\text{Sam, Carl, Aron}\)
C \(\text{Aron, Carl, Sam}\)
D \(\text{Carl, Sam, Aron}\)
Which one is greater between \(\dfrac{2}{5}\) and \(\dfrac{4}{5}\)?
Step 1: If the denominators are same, compare the numerators.
Numerator of \(\dfrac{4}{5}=4\)
Step 2: The fraction with larger numerator is greater than the fraction with smaller numerator.
\(4\) is greater than \(2\).
Thus, \(\dfrac{4}{5}\) is greater than \(\dfrac{2}{5}\).
Step 3: Write the result using greater than (>), less than (<) or equal to (=) signs.
\(4>2\)
\(\therefore\;\dfrac{4}{5}>\dfrac{2}{5}\)
Compare \(\dfrac{3}{4}\) and \(\dfrac{2}{3}\).
Step 1: Find the least common multiple (L.C.M) of denominators.
L.C.M of \(4\) and \(3,\)
Multiples of \(4=4,\;8,\;12\,...\)
Multiples of \(3=3,\;6,\;9,\;12\,...\)
L.C.M \(=12\)
Step 2: L.C.M becomes the lowest common denominator.
L.C.D \(=12\)
Step 3: Find the equivalent fractions having L.C.D as denominator.
Equivalent fraction of \(\dfrac{3}{4}=\dfrac{3×3}{4×3}=\dfrac{9}{12}\)
Equivalent fraction of \(\dfrac{2}{3}=\dfrac{2×4}{3×4}=\dfrac{8}{12}\)
Step 4: Compare the fractions having same denominators.
Compare \(\dfrac{9}{12}\) and \(\dfrac{8}{12}\).
\(9>8\)
\(\therefore\;\dfrac{9}{12}>\dfrac{8}{12}\)
Step 5: Rewrite the original fractions for the answer.
\(\dfrac{3}{4}>\dfrac{2}{3}\)
A \(\dfrac{1}{2}\)
B \(\dfrac{3}{2}\)
C Both are equal
D