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Quizzes (518) Questions (9324)

Questions Related to maths

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

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

  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 $\displaystyle \frac{8}{11}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  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. $\displaystyle -\frac{4}{9},\frac{5}{6},\frac{1}{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $-\dfrac { 1 }{ 2 } ,-\dfrac { 4 }{ 9 } ,\dfrac { 5 }{ 6 }$

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

Which of the following statements is true ?

  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. $\displaystyle {\frac{5}{7}\, <\, \frac{9}{11}\, <\, \frac{11}{13}\, <\, \frac{7}{9}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $\dfrac{5}{7} < \dfrac{7}{9} <  \dfrac{9}{11} <  \dfrac{11}{13}$

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

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

  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. $\displaystyle {\frac{3}{4}\, >\, \frac{4}{-5}\, >\, \frac{-11}{20}\, >\, -2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\dfrac { 3 }{ 4 } >\dfrac { -11 }{ 20 } >\dfrac { 4 }{ -5 } >-2$.

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

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:

  1. $\displaystyle \frac{86}{90}$
  2. $\displaystyle \frac{86}{45}$
  3. $\displaystyle \frac{43}{45}$
  4. $\displaystyle \frac{43}{90}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 = $\displaystyle {\frac{2/5\, +\, 5/9}{2}\, =\, \frac{43}{90}}$

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

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:

  1. $\displaystyle \frac{1}{2}$
  2. $\displaystyle \frac{-7}{8}$
  3. $\displaystyle \frac{4}{-5}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\dfrac {4}{-5}$.

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

What is the percentage of least number in the greatest number if $\displaystyle \frac{3}{5},\, \displaystyle \frac{9}{5},\, \displaystyle \frac{1}{5},\, \displaystyle \frac{7}{5}$ are arranged ascending or descending order?

  1. $11\, \displaystyle \frac{1}{9}\, \%$
  2. $10\, \%$
  3. $20\, \%$
  4. $25\, \%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given numbers can be arranged in the ascending order as $\displaystyle \frac{1}{5}\, <\, \displaystyle \frac{3}{5}\, <\, \displaystyle \frac{7}{5}\, <\, \displaystyle \frac{9}{5}$
Greatest number $=\, \displaystyle \frac{9}{5}$;
Least number $=\, \displaystyle \frac{1}{5}$.
We have, $\displaystyle \frac{9}{5}\, \times\, \displaystyle \frac{x}{100}\, =\, \displaystyle \frac{1}{5}$
$x\, =\, \displaystyle \frac{100}{9}\, =\, 11\, \displaystyle \frac{1}{9}\, \%$

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

Which of the following statements is true ?

  1. $\displaystyle\frac{-2}{3}\, <\, \frac{4}{-9}\,<\,\frac{-5}{12}\, <\, \frac{7}{-18}$
  2. $\displaystyle\frac{7}{-18}\, <\, \frac{-5}{12}\,<\,\frac{4}{9}\, <\, \frac{-2}{3}$
  3. $\displaystyle\frac{4}{-9}\, <\, \frac{7}{-18}\,<\,\frac{-5}{12}\, <\, \frac{2}{-3}$
  4. $\displaystyle\frac{-5}{12}\, <\, \frac{-2}{3}\,<\,\frac{4}{-9}\, <\, \frac{7}{-18}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This question is very easy if we solve it by verification
process.

Take (A) i.e. $\displaystyle\frac{-2}{3}\,<\,\frac{4}{-9}\, \frac{-5}{12}\, <\,\frac{7}{-18}$

First take \displaystyle\frac{-2}{3},\,\frac{-4}{9}$

$-2\,\times\,9,\, 4\,\times\,3$

-18, -12

$\because\, -12\,>\,-18$

So, $\displaystyle\frac{-4}{9}, \frac{-5}{12}$

$- 4\,\times\, 12, \, - 5\,\times\, 9$

- 48, - 45

$\because\, -45\, >\, -48$

So, $\displaystyle\frac{-5}{12}\,>\, \frac{-4}{9},\,i.e.,\frac{-4}{9}\, <\, \frac{-5}{12}$

Finally, $\displaystyle\frac{-5}{12}, \frac{-7}{18}$

$5\,\times\, 18, \, -7\,\times\, 12$

- 90, - 84

$\because\, -84\, >\, -90$

So, $\displaystyle\frac{-7}{18}\, >\, \frac{-5}{12}, i.e., \frac{-5}{12}\, <\, \frac{-7}{18}$

$\therefore\, \displaystyle\frac{-2}{3}\, <\, \frac{-4}{9}\,<\,  \frac{-5}{12}\,<\, \frac{-7}{18}$

You can identify the answer by observing the question by practicing this method.