Real Numbers Test

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Real Numbers Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is an irrational number?

A
$\sqrt{41616}$

B
$23.232323$

C
$\displaystyle\frac{(1+\sqrt{3})^3-(1-\sqrt{3})^3}{\sqrt 3}$

D
$23.10100100010000...$

Solution

Op-(A)
$\sqrt{41616}=204$
$\sqrt{41616}$ is a perfect square, so it is a rational number.

Op-(B)
$23.232323....$
It is recurring decimal, so it is a rational number.

Op-(C)
$\dfrac{(1+\sqrt{3})^3-(1-\sqrt{3})^3}{\sqrt{3}}=\dfrac{2\sqrt{3}(1+2\sqrt{3}+3-2+1-2\sqrt{3}+3)}{\sqrt{3}}=\dfrac{2\sqrt{3}(6)}{\sqrt{3}}=\dfrac{12\sqrt{3}}{\sqrt{3}}=12$ .
Only, the number $23.10100100010000.......$ is not terminating.
So, D is the irrational number.
Question 2
1 / -0

Let $x=\dfrac { p }{ q }$ be a rational number, such that the prime factorization of $q$ is of the form $2^n 5^m$ , where $n, m$ are non-negative integers. Then $x$ has a decimal expansion which terminates.

A
True

B
False

C
Neither

D
Either

Solution

The form of q is $2^n*5^m$
q can be $1,2,5,10,20,40....$
Any integer divided by these numbers will always give a terminating decimal number.
Question 3
1 / -0

Which of the following will have a terminating decimal expansion?

A
$\displaystyle \frac{77}{210}$

B
$\displaystyle \frac{23}{30}$

C
$\displaystyle \frac{125}{441}$

D
$\displaystyle \frac{23}{8}$

Solution

We know that the divisor of the forms $2^n5^m$ always form a terminating decimal number. Let us simplify the expressions and find if the expansion have a terminating decimal expansion.

Option A :
$\dfrac{77}{210}$
$=\dfrac{7\times11}{2\times 3\times 5 \times 7 }$
$=\dfrac{11}{2\times 3 \times 5}$
Since there is a factor of 3 in the denominator, the decimal expansion will not be terminating.

Option B :
$\dfrac{23}{30}$
$=\dfrac{23}{2\times 3\times 5}$
Since, the denominator contains a power of 3 , it is non-terminating.

Option C :
$\dfrac{125}{441}$
$=\dfrac{5\times5\times5}{3\times3\times7\times7}$
This is also non-terminating.

Option D :
$\dfrac{23}{8}$
$=\dfrac{23}{2\times2\times2}$
This contains power of 2 in the denominator. Hence, the decimal expansion is terminating.
Question 4
1 / -0

If $p$ is prime, then $\sqrt{p}$ is irrational. So
$\sqrt{7}$ is:

A
a rational number

B
an irrational number

C
not a real number

D
terminating decimal

Solution

A rational number can be represented in the form of $\dfrac pq,$ where $p$ and $q$ are integers and $q$ is a non-zero integer.

Here, $\sqrt 7$ is not a perfect square and thus cannot be expressed in the form of $\dfrac pq,$ thus it is an irrational number.
Question 5
1 / -0

The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

A
1

B
2

C
3

D
4

Solution

$Given\quad that\quad product\quad of\quad the\quad number\quad is\quad 5400=30\times 3\times 2\times 30.\\ \therefore \quad Possible\quad pairs\quad as\quad per\quad the\quad requirment\quad are-\\ (1)\quad 30\times (3\times 2\times 30)=30\times 180\\ (2)\quad (30\times 3)\times (2\times 30)=90\times 60\\ \therefore \quad Total\quad number\quad of\quad pairs=2\quad \quad (Ans)$
Question 6
1 / -0

According to Euclid's division algorithm, HCF of any two positive integers $a$ and $b$ with $a > b$ is obtained by applying Euclid's division lemma to $a$ and $b$ to find $q$ and $r$ such that $a = bq + r$ , where $r$ must satisfy

A
$1< r< b$

B
$0< r< b$

C
$0\leq r< b$

D
$0< r\leq b$

Solution

According to Euclid's division algorithm, HCF of any two positive integers $a$ and $b$ with $a > b$ is obtained by applying Euclid's division lemma to $a$ and $b$ to find $q$ and $r$ such that $a = bq + r$
The remainder $r$ is either equal to or greater than $0$ but it is always smaller than divisor $b$ .
i.e $0\le r<b$
Hence, option C is correct.
Question 7
1 / -0

The decimal representation of $\dfrac {93}{1500}$ will be:

A
terminating

B
non-terminating

C
non-terminating, repeating

D
non-terminating, non-repeating

Solution

$\dfrac {93}{1500} =\dfrac{3\times 31}{3\times 5\times 5\times 5\times 2\times 2 } =0.062$

Also, in fully reduced form, $\dfrac {93}{1500}$ has a denominator composed of $2$ 's and $5$ 's only ( $500=2^2\times 5^3$ ).

Therefore, the decimal representation will be terminating.

Thus, option $A$ is correct.
Question 8
1 / -0

The HCF of $256,442$ and $940$ is

A
$2$

B
$14$

C
$142$

D
None of these

Solution

Prime factors of numbers are
$256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\\ 442=2\times 13\times 17\\ 940=2\times 2\times 5\times 47$
HCF is the highest common factor among the numbers.
Thus among the given numbers $2$ is the common factor. Hence HCF of given numbers is $2$ .
So, correct answer is option A.
Question 9
1 / -0

The rational number which can be expressed as a terminating decimal is:

A
$\displaystyle \frac{1}{6}$

B
$\displaystyle \frac{1}{12}$

C
$\displaystyle \frac{1}{15}$

D
$\displaystyle \frac{1}{20}$

Solution

A terminating decimal is a decimal that ends after a finite number of steps. It's a decimal with a finite number of digits.
Also, a decimal is terminating if in fully reducible form, the denominator is composed of $2$ 's and $5$ 's only.
Here, only option $D$ has a denominator composed of $2$ 's and $5$ 's, i.e. $20=2^2\times5$
and $\dfrac {1}{20}= 0.05$ .
In all the other options, the decimal does not end with a finite number of digits and the denominator has factors other than $2$ 's and $5$ 's.
Therefore, option $D$ is correct.
Question 10
1 / -0

Classify the following numbers as rational or irrational : $2-\sqrt{5}$

A
Irrational number

B
Rational number

C
Less Data

D
None of the above

Solution

$2$ is rational
$\sqrt 5 =2.035.........$ which is non terminating and non repeating hence irrational number.
We know that, $\text{rational}$ $-$ $\text{irrational}= \text{irrational number.}$
Hence $2-\sqrt 5= irrational \, number$
Hence, option A is the correct answer.

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Real Numbers Test - 21

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Real Numbers Test - 21

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Question 1

41616\sqrt{41616}41616​

23.23232323.23232323.232323

(1+3)3−(1−3)33\displaystyle\frac{(1+\sqrt{3})^3-(1-\sqrt{3})^3}{\sqrt 3}3​(1+3​)3−(1−3​)3​

23.10100100010000...23.10100100010000...23.10100100010000...

Solution

Question 2

True

False

Neither

Either

Solution

Question 3

77210\displaystyle \frac{77}{210}21077​

2330\displaystyle \frac{23}{30}3023​

125441\displaystyle \frac{125}{441}441125​

238\displaystyle \frac{23}{8}823​

Solution

Question 4

a rational number

an irrational number

not a real number

terminating decimal

Solution

Question 5

1

2

3

4

Solution

Question 6

1<r<b1< r< b1<r<b

0<r<b0< r< b0<r<b

0≤r<b0\leq r< b0≤r<b

0<r≤b0< r\leq b0<r≤b

Solution

Question 7

terminating

non-terminating

non-terminating, repeating

non-terminating, non-repeating

Solution

Question 8

222

141414

142142142

None of these

Solution

Question 9

16\displaystyle \frac{1}{6}61​

112\displaystyle \frac{1}{12}121​

115\displaystyle \frac{1}{15}151​

120\displaystyle \frac{1}{20}201​

Solution

Question 10

Irrational number

Rational number

Less Data

None of the above

Solution

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$\sqrt{41616}$

$23.232323$

$\displaystyle\frac{(1+\sqrt{3})^3-(1-\sqrt{3})^3}{\sqrt 3}$

$23.10100100010000...$

$\displaystyle \frac{77}{210}$

$\displaystyle \frac{23}{30}$

$\displaystyle \frac{125}{441}$

$\displaystyle \frac{23}{8}$

$1< r< b$

$0< r< b$

$0\leq r< b$

$0< r\leq b$

$2$

$14$

$142$

$\displaystyle \frac{1}{6}$

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

$\displaystyle \frac{1}{15}$

$\displaystyle \frac{1}{20}$