Tag: comparing fractions

Questions Related to comparing fractions

A student was asked to solve the fraction $\cfrac { \cfrac { 7 }{ 3 } +\left( 1\cfrac { 1 }{ 2 }  \times\cfrac { 5 }{ 3 } \right) }{ 2+1\cfrac { 2 }{ 3 }  } $ and his answer was $\cfrac{1}{4}$. By how much was his answer wrong?

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. $1$

  2. $\cfrac{1}{55}$

  3. $\cfrac{1}{220}$

  4. None of these

Show answer
Correct Option: D
Explanation:
$ \dfrac{\dfrac{7}{3}+(1\dfrac{1}{2}\times \dfrac{5}{3})}{2+1\dfrac{2}{3}}$

$ = \dfrac{\dfrac{7}{3}+(\dfrac{3}{2}\times \dfrac{5}{3})}{(2+\dfrac{5}{3})}$

$  = \dfrac{\dfrac{7}{3}+\dfrac{5}{2}}{\dfrac{11}{3}} = \dfrac{14+15}{6}\times \dfrac{3}{11} = \dfrac{29}{22}$

$ \Rightarrow $ His answer was wrong by

$ \dfrac{29}{22}-\dfrac{1}{4} = \dfrac{116-22}{88} = \dfrac{94}{88} = \dfrac{47}{44}$


Which of the following fraction is the smallest? $\dfrac{7}{6}, \dfrac{7}{9}, \dfrac{4}{5}, \dfrac{5}{7}$

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. $\dfrac{7}{6}$

  2. $\dfrac{7}{9}$

  3. $\dfrac{4}{5}$

  4. $\dfrac{5}{7}$

Show answer
Correct Option: D
Explanation:
let the fractions be a, b, c, d
$ \dfrac{a}{b} = \dfrac{7}{6}\times \dfrac{9}{7} = \dfrac{9}{6}> 1\Rightarrow a> b$
a is not smallest
$ \dfrac{b}{c} = \dfrac{7}{9}\times \dfrac{4}{5} = \dfrac{28}{45}< 1\Rightarrow c> b$
 c is not smallest
$ \dfrac{b}{d} = \dfrac{7}{9}\times \dfrac{7}{5} = \dfrac{49}{45}> 1\Rightarrow b> d$
$ \Rightarrow $ d  is smallest $\Rightarrow (D)$

Write the following as fractions in their simplest form.

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. 0.4

  2. 1.5

  3. 25.75

  4. 0.072

  5. 1.248

Show answer
Correct Option: A
Explanation:

$0.4=\dfrac 4{10}=\dfrac 25\1.5=\dfrac{15}{10}=\dfrac 32\25.75=\dfrac{2575}{100}=\dfrac{103}{4}\0.072=\dfrac{72}{1000}=\dfrac{9}{125}\1.248=\dfrac{1248}{1000}=\dfrac{156}{125}$

$\dfrac{2}{3}$ is equal to $\dfrac{4}{6}$.

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Now,

$\dfrac{4}{6}$

$=\dfrac{2\times 2}{2\times 3}$

$=\dfrac{2}{3}$.

Hence the given statement is correct.

The fraction $\displaystyle \frac{3}{5}$ is found between which pair of fractions on a number line?

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. $\displaystyle \frac{7}{10}$ and $\displaystyle \frac{3}{4}$

  2. $\displaystyle \frac{2}{5}$ and $\displaystyle \frac{1}{2}$

  3. $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{5}{13}$

  4. $\displaystyle \frac{2}{7}$ and $\displaystyle \frac{8}{11}$

Show answer
Correct Option: D
Explanation:

(a) Let us consider the first set of fraction $\dfrac { 7 }{ 10 } ,\dfrac { 3 }{ 4 }$ and another given fraction $\dfrac { 3 }{ 5 }$ 


Taking the LCM to make the denominators same of the above fractions, we have

$\dfrac { 7\times 2 }{ 10\times 2 } ,\dfrac { 3\times 4 }{ 5\times 4 } ,\dfrac { 3\times 5 }{ 4\times 5 } \\ =\dfrac { 14 }{ 20 } ,\dfrac { 12 }{ 20 } ,\dfrac { 15 }{ 20 } \\ \Rightarrow \dfrac { 12 }{ 20 } <\dfrac { 14 }{ 20 } <\dfrac { 15 }{ 20 } \\ \Rightarrow \dfrac { 3 }{ 5 } <\dfrac { 7 }{ 10 } <\dfrac { 3 }{ 4 }$   

Therefore, $\dfrac { 3 }{ 5 }$ does not lie between the first set of fraction $\dfrac { 7 }{ 10 } ,\dfrac { 3 }{ 4 }$.

(b) Now, consider the set of fraction $\dfrac { 2 }{ 5 } ,\dfrac { 1 }{ 2 }$ and another given fraction $\dfrac { 3 }{ 5 }$

Taking the LCM to make the denominators same of the above fractions, we have

$\dfrac { 2\times 2 }{ 5\times 2 } ,\dfrac { 3\times 2 }{ 5\times 2 } ,\dfrac { 1\times 5 }{ 2\times 5 } \\ =\dfrac { 4 }{ 10 } ,\dfrac { 6 }{ 10 } ,\dfrac { 5 }{ 10 } \\ \Rightarrow \dfrac { 4 }{ 10 } <\dfrac { 5 }{ 10 } <\dfrac { 6 }{ 10 } \\ \Rightarrow \dfrac { 2 }{ 5 } <\dfrac { 1 }{ 2 } <\dfrac { 3 }{ 5 }$     

Therefore, $\dfrac { 3 }{ 5 }$ does not lie between the set of fraction $\dfrac { 2 }{ 5 } ,\dfrac { 1 }{ 2 }$.

(c) Now, consider the set of fraction $\dfrac { 1 }{ 3 } ,\dfrac { 5 }{ 13 }$ and another given fraction $\dfrac { 3 }{ 5 }$


Taking the LCM to make the denominators same of the above fractions, we have

$\dfrac { 1\times 65 }{ 3\times 2 } ,\dfrac { 3\times 39 }{ 5\times 39 } ,\dfrac { 5\times 15 }{ 13\times 15 } \\ =\dfrac { 65 }{ 195 } ,\dfrac { 108 }{ 195 } ,\dfrac { 75 }{ 195 } \\ \Rightarrow \dfrac { 65 }{ 195 } <\dfrac { 75 }{ 195 } <\dfrac { 108 }{ 195 } \\ \Rightarrow \dfrac { 1 }{ 3 } <\dfrac { 5 }{ 13 } <\dfrac { 3 }{ 5 }$     

Therefore, $\dfrac { 3 }{ 5 }$ does not lie between the set of fraction $\dfrac { 1 }{ 3 } ,\dfrac { 5 }{ 13 }$.

(d) Now, consider the set of fraction $\dfrac { 2 }{ 7 } ,\dfrac { 8 }{ 11 }$ and another given fraction $\dfrac { 3 }{ 5 }$

Taking the LCM to make the denominators same of the above fractions, we have

$\dfrac { 2\times 55 }{ 7\times 55 } ,\dfrac { 3\times 77 }{ 5\times 77 } ,\dfrac { 8\times 35 }{ 11\times 35 } \\ =\dfrac { 110 }{ 385 } ,\dfrac { 221 }{ 385 } ,\dfrac { 280 }{ 385 } \\ \Rightarrow \dfrac { 110 }{ 385 } <\dfrac { 221 }{ 385 } <\dfrac { 280 }{ 385 } \\ \Rightarrow \dfrac { 2 }{ 7 } <\dfrac { 3 }{ 5 } <\dfrac { 8 }{ 11 }$     

Therefore, $\dfrac { 3 }{ 5 }$ lies between the set of fraction $\dfrac { 2 }{ 7 } ,\dfrac { 8 }{ 11 }$.

Hence, the fraction $\dfrac { 3 }{ 5 }$ is found between $\dfrac { 2 }{ 7 }$ and $\dfrac { 8 }{ 11 }$ on a number line.

Which one of the following sets of fractions is in the correct sequence of ascending order of their values ?

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. $\displaystyle -\frac{1}{2},\frac{5}{6},\frac{-4}{9}$

  2. $\displaystyle -\frac{3}{7},\frac{-5}{6},\frac{3}{5}$

  3. $\displaystyle -\frac{1}{2},-\frac{4}{9},\frac{5}{6}$

  4. $\displaystyle -\frac{4}{9},\frac{5}{6},\frac{1}{6}$

Show answer
Correct Option: C
Explanation:

(a) Let us consider the first set of fraction $-\dfrac { 1 }{ 2 } ,\dfrac { 5 }{ 6 } ,-\dfrac { 4 }{ 9 }$ 


Taking the LCM to make the denominators same of the above fractions, we have

$-\dfrac { 1\times 9 }{ 2\times 9 } ,\dfrac { 5\times 3 }{ 6\times 3 } ,-\dfrac { 4\times 2 }{ 9\times 2 } \\ =-\dfrac { 9 }{ 18 } ,\dfrac { 15 }{ 18 } ,-\dfrac { 8 }{ 18 } \\ \Rightarrow -\dfrac { 9 }{ 18 } <-\dfrac { 8 }{ 18 } <\dfrac { 15 }{ 18 } \\ \Rightarrow -\dfrac { 1 }{ 2 } <-\dfrac { 4 }{ 9 } <\dfrac { 5 }{ 6 }$   

Therefore, the first set of fraction $-\dfrac { 1 }{ 2 } ,\dfrac { 5 }{ 6 } ,-\dfrac { 4 }{ 9 }$ is not in ascending order.

(b) Now, consider the set of fraction $-\dfrac { 3 }{ 7 } ,-\dfrac { 5 }{ 6 } ,\dfrac { 3 }{ 5 }$ 

Taking the LCM to make the denominators same of the above fractions, we have

$-\dfrac { 3\times 30 }{ 7\times 30 } ,-\dfrac { 5\times 15 }{ 6\times 15 } ,\dfrac { 3\times 42 }{ 5\times 42 } \\ =-\dfrac { 90 }{ 210 } ,-\dfrac { 175 }{ 210 } ,\dfrac { 126 }{ 210 } \\ \Rightarrow -\dfrac { 175 }{ 210 } <-\dfrac { 90 }{ 210 } <\dfrac { 126 }{ 210 } \\ \Rightarrow -\dfrac { 5 }{ 6 } <-\dfrac { 3 }{ 7 } <\dfrac { 3 }{ 5 }$    

Therefore, the set of fraction $-\dfrac { 3 }{ 7 } ,-\dfrac { 5 }{ 6 } ,\dfrac { 3 }{ 5 }$ is not in ascending order.


(c) Now, consider the set of fraction $-\dfrac { 1 }{ 2 } ,-\dfrac { 4 }{ 9 } ,\dfrac { 5 }{ 6 }$ 


Taking the LCM to make the denominators same of the above fractions, we have

$-\dfrac { 1\times 9 }{ 2\times 9 } ,-\dfrac { 4\times 2 }{ 9\times 2 } ,\dfrac { 5\times 3 }{ 6\times 3 } \\ =-\dfrac { 9 }{ 18 } ,-\dfrac { 8 }{ 18 } ,\dfrac { 5 }{ 18 } \\ \Rightarrow -\dfrac { 9 }{ 18 } <-\dfrac { 8 }{ 18 } <\dfrac { 5 }{ 18 } \\ \Rightarrow -\dfrac { 1 }{ 2 } <-\dfrac { 4 }{ 9 } <\dfrac { 5 }{ 6 }$    

Therefore, the set of fraction $-\dfrac { 1 }{ 2 } ,-\dfrac { 4 }{ 9 } ,\dfrac { 5 }{ 6 }$ is in ascending order.

(d) Now, consider the set of fraction $-\dfrac { 4 }{ 9 } ,\dfrac { 5 }{ 6 } ,\dfrac { 1 }{ 6 }$ 

Taking the LCM to make the denominators same of the above fractions, we have

$-\dfrac { 4\times 2 }{ 9\times 2 } ,\dfrac { 5\times 3 }{ 6\times 3 } ,\dfrac { 1\times 3 }{ 6\times 3 } \\ =-\dfrac { 8 }{ 18 } ,\dfrac { 15 }{ 18 } ,\dfrac { 3 }{ 18 } \\ \Rightarrow -\dfrac { 8 }{ 18 } <\dfrac { 3 }{ 18 } <\dfrac { 15 }{ 18 } \\ \Rightarrow -\dfrac { 4 }{ 9 } <\dfrac { 1 }{ 6 } <\dfrac { 5 }{ 6 }$    

Therefore, the set of fraction $-\dfrac { 4 }{ 9 } ,\dfrac { 5 }{ 6 } ,\dfrac { 1 }{ 6 }$ is not in ascending order.

Hence, the only set of fraction in ascending order is $-\dfrac { 1 }{ 2 } ,-\dfrac { 4 }{ 9 } ,\dfrac { 5 }{ 6 }$

Which of the following statements is true ?

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. $\displaystyle {\frac{5}{7}\, <\, \frac{7}{9}\, <\, \frac{9}{11}\, <\, \frac{11}{13}}$

  2. $\displaystyle {\frac{11}{13}\, <\, \frac{9}{11}\, <\, \frac{7}{9}\, <\, \frac{5}{7}}$

  3. $\displaystyle {\frac{5}{7}\, <\, \frac{11}{13}\, <\, \frac{7}{9}\, <\, \frac{9}{11}}$

  4. $\displaystyle {\frac{5}{7}\, <\, \frac{9}{11}\, <\, \frac{11}{13}\, <\, \frac{7}{9}}$

Show answer
Correct Option: A
Explanation:
Here we have four factors $\dfrac{5}{7},  \dfrac{7}{9},   \dfrac{9}{11},   \dfrac{11}{13}$
LCM of 7, 9, 11 and 13 is 9009
So, 
$\dfrac{5}{7} \times\dfrac{1287}{1287}$ = $\dfrac{6435}{9009}$

$\dfrac{7}{9} \times\dfrac{1001}{1001}$ = $\dfrac{7007}{9009}$

$\dfrac{9}{11} \times\dfrac{819}{819}$ = $\dfrac{7371}{9009}$

$\dfrac{11}{13} \times\dfrac{693}{693}$ = $\dfrac{7623}{9009}$
As, 
6435 < 7007 < 7371 < 7623
So, $\dfrac{5}{7} < \dfrac{7}{9} <  \dfrac{9}{11} <  \dfrac{11}{13}$

Arrange the following numbers in descending order.
$-2,\, \displaystyle {\frac{4}{-5},\, \frac{-11}{20},\, \frac{3}{4}}$

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. $\displaystyle {\frac{3}{4}\, >\, -2\, >\, \frac{-11}{20}\, >\, \frac{4}{-5}}$

  2. $\displaystyle {\frac{3}{4}\, >\, \frac{-11}{20}\, >\, \frac{4}{-5}\, >\, -2}$

  3. $\displaystyle {\frac{3}{4}\, >\, \frac{4}{-5}\, >\, -2\, >\, \frac{-11}{20}}$

  4. $\displaystyle {\frac{3}{4}\, >\, \frac{4}{-5}\, >\, \frac{-11}{20}\, >\, -2}$

Show answer
Correct Option: B
Explanation:

The rational number $\dfrac {4}{-5}$ is same as $\dfrac {-4}{5}$.


Now consider the given rational numbers $-2,\dfrac {-11}{20},\dfrac {-4}{5}$ and $\dfrac {3}{4}$ and make their denominator same by taking the LCM of the denominators as follows:

LCM$(5,20,4)=20$

The given fractions now with denominator $20$ can be written as:

$\dfrac { -2\times 20 }{ 1\times 20 } =\dfrac { -40 }{ 20 } \ \dfrac { -4\times 4 }{ 5\times 4 } =\dfrac { -16 }{ 20 } \ \dfrac { -11\times 1 }{ 20\times 1 } =\dfrac { -11 }{ 20 } \ \dfrac { 3\times 5 }{ 4\times 5 } =\dfrac { 15 }{ 20 }$ 

The descending order of the rational numbers is:

$\dfrac { 15 }{ 20 } >\dfrac { -11 }{ 20 } >\dfrac { -16 }{ 20 } >\dfrac { -40 }{ 20 } \ \Rightarrow \dfrac { 3 }{ 4 } >\dfrac { -11 }{ 20 } >\dfrac { 4 }{ -5 } >-2$ 

Hence, the descending order is $\dfrac { 3 }{ 4 } >\dfrac { -11 }{ 20 } >\dfrac { 4 }{ -5 } >-2$.

The average of the middle two rational numbers if $\displaystyle {\frac{4}{7},\, \frac{1}{3},\, \frac{2}{5},\, \frac{5}{9}}$ are arranged in ascending order is:

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. $\displaystyle \frac{86}{90}$

  2. $\displaystyle \frac{86}{45}$

  3. $\displaystyle \frac{43}{45}$

  4. $\displaystyle \frac{43}{90}$

Show answer
Correct Option: D
Explanation:

$\displaystyle {\frac{4}{7},\, \frac{1}{3},\, \frac{2}{5},\, \frac{5}{9}}$
The above numbers in ascending order are
$\displaystyle {\frac{1}{3}\, <\, \frac{2}{5}\, <\, \frac{5}{9}\, <\, \frac{4}{7}}$
Middle two numbers are $\displaystyle \frac{2}{5}$ and $\displaystyle \frac{5}{9}$
$\therefore$ Average = $\displaystyle {\frac{2/5\, +\, 5/9}{2}\, =\, \frac{43}{90}}$

The given rational numbers are $\displaystyle {\frac{1}{2},\, \frac{4}{-5},\, \frac{-7}{8}}.$ If these numbers are arranged in the ascending order or descending order, then the middle number is:

maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

  1. $\displaystyle \frac{1}{2}$

  2. $\displaystyle \frac{-7}{8}$

  3. $\displaystyle \frac{4}{-5}$

  4. None of these

Show answer
Correct Option: C
Explanation:

The rational number $\dfrac {4}{-5}$ is same as $\dfrac {-4}{5}$.


Now consider the given rational numbers $\dfrac {1}{2},\dfrac {-4}{5}$ and $\dfrac {-7}{8}$ and make their denominator same by taking the LCM of the denominators as follows:

LCM$(2,5,8)=40$

The given fractions now with denominator $40$ can be written as:

$\dfrac { 1\times 20 }{ 2\times 20 } =\dfrac { 20 }{ 40 } \ \dfrac { -4\times 8 }{ 5\times 8 } =\dfrac { -32 }{ 40 } \ \dfrac { -7\times 5 }{ 8\times 5 } =\dfrac { -35 }{ 40 }$ 

The ascending order of the rational numbers is:

$\dfrac { -35 }{ 40 } ,\dfrac { -32 }{ 40 } ,\dfrac { 20 }{ 40 } \ \Rightarrow \dfrac { -7 }{ 8 } ,\dfrac { 4 }{ -5 } ,\dfrac { 1 }{ 2 } .......(1)$

The descending order of the rational numbers is:

$\dfrac { 20 }{ 40 } ,\dfrac { -32 }{ 40 } ,\dfrac { -35 }{ 40 } \ \Rightarrow \dfrac { 1 }{ 2 } ,\dfrac { 4 }{ -5 } ,\dfrac { -7 }{ 8 } .......(2)$ 

From equations 1 and 2, we conclude that the middle rational number in both ascending and descending order is same that is $\dfrac {4}{-5}$.

Hence, the middle number is $\dfrac {4}{-5}$.