Which statement correctly compares 11/20 an 5/9

A:11/20<5/9
B:11/20>5/9
C:11/20=5/9

[tex]\frac{11}{20} =\frac{11 \times 9}{20 \times 9}=\frac{99}{180} \\ \\ \frac{5}{9}=\frac{5 \times 20}{9 \times 20}=\frac{100}{180} \\ \\ 99<100 \Rightarrow \frac{99}{180}<\frac{100}{180} \Rightarrow \boxed{\frac{11}{20} < \frac{5}{9}} \Leftarrow \hbox{answer A}[/tex]

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Isyllusid Isyllusid · Answer 2

Answer:

[tex]\dfrac{11}{20}<\dfrac{5}{9}[/tex]

A is correct.

Step-by-step explanation:

Given: [tex]\dfrac{11}{20}\text{ and }\dfrac{5}{9}[/tex]

We need to compare the two fraction.

We can compare either greater , less or equal.

If we compare two fraction. First we make their denominator same.

Denominators are 20 and 9

We will find the LCD of 20 and 9

20 = 2 x 2 x 5

9 = 3 x 3

LCD = 2 x 2 x 3 x 3 x 5 = 180

[tex]\text{First Fraction}\rightarrow \dfrac{11\times 9}{20\times 9}\Rightarrow \dfrac{99}{180}[/tex]

[tex]\text{Second Fraction}\rightarrow \dfrac{5\times 20}{9\times 20}\Rightarrow \dfrac{100}{180}[/tex]

Now, we can see their denominator are same.

We will compare the numerator.

99<100

Thus,

[tex]\dfrac{99}{180}<\dfrac{100}{180}[/tex]

[tex]\dfrac{11}{20}<\dfrac{5}{9}[/tex]

Hence, [tex]\dfrac{11}{20}<\dfrac{5}{9}[/tex] correct.