Tag: operations on rational numbers

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Questions Related to operations on rational numbers

Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers
State true or false:

Ten rational numbers between $\dfrac{3}{5}$ and $\dfrac {3}{4}$ are 

$\displaystyle\frac{97}{160},\frac{98}{160},\frac{99}{160},\frac{100}{160},\frac{101}{160},\frac{102}{160},\frac{103}{160},\frac{104}{160},\frac{105}{160},\frac{106}{160}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To get the rational numbers between $\displaystyle\frac{3}{5}$ and $\displaystyle\frac{3}{4}$

Take an LCM of these two numbers: $\displaystyle\frac{12}{20}$ and $\displaystyle\frac{15}{20}$


So now to make the denominator $(160)$ as per the question we need to
multiply numerator and denominator by $8$: $\displaystyle\frac{96}{160}$ and $\displaystyle\frac{120}{160}$

All the numbers between $\displaystyle\frac{96}{160}$ and $\displaystyle\frac{120}{160}$ form the answer

Some of these numbers are $\displaystyle\frac{97}{160}$, $\displaystyle\frac{98}{160}$, $\displaystyle\frac{99}{160}$, $\displaystyle\frac{100}{160}$, $\displaystyle\frac{101}{160}$, $\displaystyle\frac{102}{160}$, $\displaystyle\frac{103}{160}$, $\displaystyle\frac{104}{160}$, $\displaystyle\frac{105}{160}$, $\displaystyle\frac{106}{160}$

Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

Two rational numbers between $\dfrac{1}{5}$ and $\dfrac{4}{5}$ are :

  1. 1 and $\dfrac{3}{5}$
  2. $\dfrac{2}{5}$ and $\dfrac{3}{5}$
  3. $\dfrac{1}{2}$ and $\dfrac{2}{1}$
  4. $\dfrac{3}{5}$ and $\dfrac{6}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the denominator of both rational numbers are same. So, for getting the rational numbers between the given rational numbers, we only have to consider the numerators of the rational numbers.

Two numbers between 1 & 4 are 2 and 3.
So, two rational numbers between the given rational numbers will be $\dfrac { 2 }{ 5 }$ and $ \dfrac { 3 }{ 5 } $
So, correct answer is option B.

Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

A rational number lying between $\sqrt{2}$ and $\sqrt{3}$ is :

  1. $\dfrac{\sqrt{2}+\sqrt{3}}{2}$
  2. $\sqrt{6}$

  3. 1.6

  4. 1.9
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\sqrt {2  } \cong $ 1.41....and $\sqrt {3  } \cong $ 1.73..

Now we see that option A is irrational number so it is incorrect , Option B is also irrational number so it is also incorrect ,Option C is rational number and lie between the given number so it is correct , Option D is rational number but it does not lie between the given number so it is incorrect.

Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

Which of the rational number lies between $-\dfrac { 2}{ 3} $ and $\dfrac {1 }{4}$

  1. $ {\dfrac{ - 5} {24}}$
  2. ${\dfrac {25} {12}}$
  3. ${\dfrac {51} {24}}$
  4. ${\dfrac {5} {12}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the fractions $-\dfrac { 2}{ 3} $ and $\dfrac {1 }{4}$ can be written as $-\dfrac{8}{12}$ and $\dfrac{3}{12}$ or, $-\dfrac{16}{24}$ and $\dfrac{6}{24}$.

Now it is clear that $-\dfrac{5}{24}$ lies between the given fractions.

Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

What is the sum of the addictive inverse of $\frac{2}{3}$ and the reciprocal of $\frac{9}{8}$?

  1. $\frac{3}{8}$
  2. -$\frac{3}{8}$
  3. $\frac{2}{9}$
  4. -$\frac{2}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Additive inverse of 2/3 is -2/3. Reciprocal of 9/8 is 8/9. Sum = -2/3 + 8/9 = -6/9 + 8/9 = 2/9.

Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

Find five rational numbers between $\displaystyle\frac{-3}{2}$ and $\displaystyle\frac{5}{3}$.

  1. $\displaystyle\frac{-8}{6},\,\displaystyle\frac{-13}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{2}{6}$
  2. $\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{13}{6}$
  3. $\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{11}{6}$
  4. $\displaystyle\frac{-8}{6},\,\displaystyle\frac{-7}{6},\,\displaystyle\frac{0}{6},\,\displaystyle\frac{1}{6}$ and $\displaystyle\frac{2}{6}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Converting the given rational numbers with the same denominators
$\cfrac{2}{3}=\cfrac{2\times5}{3\times5}=\cfrac{10}{15}$ and $\cfrac{4}{5}=\cfrac{4\times3}{5\times3}=\cfrac{12}{15}$

Also, $\cfrac{2}{3}=\cfrac{10}{15}=\cfrac{10\times4}{15\times4}=\cfrac{40}{60}$ and $\cfrac{4}{5}=\cfrac{12}{15}=\cfrac{12\times4}{15\times4}=\cfrac{48}{60}$

We know that $40,\,<\,41\,<\,42\,<\,43\,<44\,<45\,<\,46\,<\,47\,<\,48$
$\Rightarrow\cfrac{40}{60}\,<\,\cfrac{41}{60}\,<\,\cfrac{42}{60}\,<\,\dots\,<\,\cfrac{47}{60}\,<\,\cfrac{48}{60}$
Thus, we have the following five rational numbers between $\cfrac{2}{3}$ and $\cfrac{4}{5}$
$\cfrac{41}{60},\,\cfrac{43}{60},\,\cfrac{43}{60}\,\cfrac{44}{60}$ and $\cfrac{45}{60}$.

Converting the given rational numbers with the same denominators 
$\cfrac{-3}{2}=\cfrac{-3\times3}{2\times3}=\cfrac{-9}{6}$ and $\cfrac{5}{3}=\cfrac{5\times2}{3\times2}=\cfrac{10}{6}$

We know that $-9\,<\,-8\,<\,-7\,<\,-6\,<\,\dots\,<\,0\,<\,1\,<\,2\,<\,8\,<\,9\,<\,10$
$\Rightarrow\cfrac{-9}{6}\,<\,\cfrac{-8}{6}\,<\,\cfrac{-7}{6}\,<\,\cfrac{-6}{6}\,<\,\dots\,<\,\cfrac{0}{6}\,<\,\cfrac{1}{6}\,<\,\cfrac{2}{6}\,<\,\dots\,<\,\cfrac{8}{6}\,<\,\cfrac{9}{6}\,<\,\cfrac{10}{6}$.

Thus, we have the following five rational numbers between $\cfrac{-3}{2}$ and $\cfrac{5}{3}$ 
$\Rightarrow \cfrac{-8}{6},\,\cfrac{-7}{6},\,\cfrac{0}{6},\,\cfrac{1}{6}and\cfrac{2}{6}$
Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

State True or False.

The five rational numbers between $\dfrac{3}{5}$ and $\dfrac{4}{5}$ are $ \displaystyle \frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since we want five  numbers we write $\frac{3}{5}$ and $\frac{4}{5}$
So multiply in numerator and denominator by 5+1 = 6 we get
$\Rightarrow \frac{3}{5}=\frac{3\times 6}{5\times 6}=\frac{18}{30}$
$\Rightarrow \frac{4}{5}=\frac{4\times 6}{5\times 6}=\frac{24}{30}$
We know that $18<19<20<21<22<23<24$
$\Rightarrow \frac{18}{30}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{24}{30}$
Hence 5 rational number between $\frac{3}{5} and  \frac{4}{5}$are
$\frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30},$