Tag: real number

Questions Related to real number


$\dfrac {17}{8}$ can be expressed as $.....$. It is a $........$ decimal.
maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. $ 2.125$, terminating

  2. $ 2.321321...$, non-terminating

  3. $1.125125124...$, recurring

  4. $2.125$, irrational

Show answer
Correct Option: A
Explanation:

$\dfrac {17}{8}=\dfrac{17\times125}{8\times 125}=  \dfrac{2125}{1000}=2.125$

When the division process does not end and the remainder is not equal to zero; then such decimal is known as ............... decimal

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. terminating

  2. non-terminating

  3. recurring

  4. irrational

Show answer
Correct Option: B
Explanation:

The division is completed when we get the remainder zero. In this division process we do not get a zero and it is never ending. This process of division is called a non- terminating decimal.
Therefore, $B$ is the correct answer.

Which of the following fractions will terminate when expressed as a decimal? (Choose all that apply.)

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. $\frac{1}{256}$

  2. $\frac{27}{100}$

  3. $\frac{100}{27}$

  4. $\frac{231}{660}$

  5. $\frac{7}{105}$

Show answer
Correct Option: A,B,D
Explanation:

Recall that in order for the decimal version of a fraction to terminate, the fraction's denominator in fully reduced form must have a prime factorization that consists of only 2's and/or 5's. 

The denominator in (A) is composed of only 2's $(256 = 2^8)$. 
The denominator in (B) is composed of only 2's and 5's $(100=2^2\times 5^2)$. 
In fully reduced form, the fraction in (D) is equal to $\frac{7}{20}$ and 20 is composed of only 2's and 5's $(20=2^2\times 5)$.
 By contrast, the denominator in (C) has prime factors other than Z's and 5's $(27 =3^3)$, and in fully reduced form, the fraction in (E) is equal to $\frac{1}{15}$, and 15 has a prime factor other than 2's and 5's $(15 = 3 \times 5)$.

Identify a non-terminating repeating decimal.

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. $\dfrac{24}{1600}$

  2. $\dfrac{171}{800}$

  3. $\dfrac{123}{2^2 \times 5^3}$

  4. $\dfrac{145}{2^3 \times 5^2 \times 7^2}$

Show answer
Correct Option: D
Explanation:

An fraction will be terminating decimal only if denominator will be in form of 

${2}^{m}\times{{5}^{n}}$. 
Now lets check denominator of each option 
Only option D does not satisfies this condition So it will be non terminating 
So correct answer will be option D

A rational number can be expressed as a terminating decimal if the denominator has factors _________.

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. $2$ or $5$

  2. $2$, $3$ or $5$

  3. $3$ or $5$

  4. Only $2$ and $3$

Show answer
Correct Option: A
Explanation:

Any rational number its denominator is in the form of ${ 2 }^{ m }\times { 5 }^{ n }$,where m,n are positive integer s are terminating decimals.

So correct answer will be option A

Which one of the following has a terminating decimal expansion?

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. $\displaystyle\frac{5}{32}$

  2. $\displaystyle\frac{7}{9}$

  3. $\displaystyle\frac{8}{15}$

  4. $\displaystyle\frac{1}{12}$

Show answer
Correct Option: A
Explanation:

We know that a terminating decimal is a decimal that ends. It's a decimal with a finite number of digits. For example $\dfrac {1}{4}=0.25$, it has only two decimal digits.


Now consider the fraction $\dfrac {5}{32}$ whose decimal form will be:

$\dfrac {5}{32}=0.15625$

The resulting decimal number ends with five decimal digits and therefore, it is terminating decimal.

Hence, $\dfrac {5}{32}$ has a terminating decimal expansion.

Which of the following numbers has the terminal decimal representation?

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

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

  2. $\dfrac{1}{3}$

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

  4. $\dfrac{17}{3}$

Show answer
Correct Option: C
Explanation:
Option A: $\dfrac {1}{7}=0.1428..$
Option B: $\dfrac {1}{3}=0.3333....$
Option C: $\dfrac{3}{5}=0.6\Rightarrow $ terminal decimal number
Lets check option D : $\dfrac {17}{3}=5.6666...$
Hence, option C is correct.

A real number $\displaystyle \frac{2^2 \times 3^2 \times 7^2}{2^5 \times 5^3 \times 3^2 \times 7}$ will have _________.

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. Terminating decimal

  2. Non-terminating decimal

  3. Non-terminating and non-repeating decimal

  4. Terminating repeating decimal

Show answer
Correct Option: A
Explanation:

$\dfrac{2^{2}\times3^{2}\times7^{2}}{2^{5}\times5^{3}\times3^{2}\times7} = 2^{2-5}\times3^{2-2}\times5^{-3}\times7^{2-1}$


                                 $= \dfrac{7}{2^{3}\times5^{3}}$

                                 $= \dfrac{7}{1000}$

                                 $= 0.007$
$0.007$ is a terminating non-repeating decimal

State the following statement is True or False

$\dfrac{987}{10500}$ will have terminating decimal expansion. 

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Terminating decimal expansion, because

$ \dfrac{987}{10500}=\dfrac{329}{3500}=\dfrac{329}{2^2.5^3.7}=\dfrac{47}{2^2.5^3}=.094. $

Given that $\dfrac {1}{7} = 0.\overline {142857}$, which is a repeating decimal having six different digits. If $x$ is the sum of such first three positive integers $n$ such that $\dfrac {1}{n} = 0.\overline {abcdef}$, where $a, b, c, d, e$ and $f$ are different digits, then the value of $x$ is

maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

  1. $20$

  2. $21$

  3. $41$

  4. $42$

Show answer
Correct Option: C
Explanation:

$1^{st}$ number
$x _{1} = 7$
$\Rightarrow \dfrac {1}{x _{1}} = 0.\overline {142857}$
such that,
$2^{nd}$ number
$x _{2} = 13$
$\Rightarrow \dfrac {1}{x _{2}} = 0.\overline {076923}$
$x _{3} = 21$
$\Rightarrow \dfrac {1}{x _{3}} = \dfrac {1}{21} = 0.\overline {047619}$
$x = x _{1} + x _{2} + x _{3}$
$\Rightarrow 7 + 13 + 21 = 41$.