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

$$\sqrt{41616}$$

$$23.232323$$

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

$$23.10100100010000...$$

Solution

Question 2

True

False

Neither

Either

Solution

Question 3

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

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

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

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

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

$$1< r< b$$

$$0< r< b$$

$$0\leq r< b$$

$$0< r\leq b$$

Solution

Question 7

terminating

non-terminating

non-terminating, repeating

non-terminating, non-repeating

Solution

Question 8

$$2$$

$$14$$

$$142$$

None of these

Solution

Question 9

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

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

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

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

Solution

Question 10

Irrational number

Rational number

Less Data

None of the above

Solution

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