M.C.Q

Question 11 Mark

Which of the following is rational?

$\sqrt{3}$
$\pi$
$\frac{4}{0}$
$\frac{0}{4}$

Answer

$\frac{0}{4}$

Solution:

$\sqrt{3}=1.732 \ ...=$ Non-terminating and non-repeating number, hence irrational

$\pi=3.14 \ ...$ also can not be terminated to $\frac{\text{p}}{\text{q}}$ form, and is non-terminating and non-repeating in nature. Hence, irrational.

$\frac{4}{0}$ is not a rational number because this is in the form $\frac{\text{p}}{\text{q}}$ where p and q are integers but q = 0

$\frac{0}{4}$ follows the defination of rational number.

Hence, correct option is (d).

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Question 21 Mark

An irrational number between 2 and 2.5 is:

$\sqrt{11}$
$\sqrt{5}$
$\sqrt{22.5}$
$\sqrt{12.5}$

Answer

$\sqrt{5}$

Solution:
$\sqrt{4}=2$ and $\sqrt{6.25}=2.5$
Option (a), (c) and (d): $\sqrt{11},\sqrt{22.5}$ and $\sqrt{12.5},$ all are greater than $\sqrt{6.25}$
⇒ Out of interval (2, 2.5)
Option (b): $\sqrt{4}<\sqrt{5}<\sqrt{6.25}\Rightarrow$ lies in the interval (2, 2.5)
Hence, option (b) is correct.

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Question 31 Mark

The number $0.\overline{32}$ when expressed in the form $\frac{\text{p}}{\text{q}}$ $\big($p, q are integers and $\text{q}\neq0\big),$ is:

$\frac{8}{25}$
$\frac{29}{90}$
$\frac{32}{99}$
$\frac{32}{199}$

Answer

$\frac{29}{90}$

Solution:
Let $\text{x}=0.\overline{32}=0.32222..(1)$
Now, $10\text{x}=3.2222=3.\overline{2}...(2),$
Subtracting equation (2) and (3), we get
$90\text{x}=29$
$\Rightarrow\text{x}=\frac{29}{90}$
Hence, option (b) is correct.

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Question 41 Mark

Which one of the following statements is true?

The sum of two irrational numbers is always an irrational number.
The sum of two irrational numbers is always a rational number.
The sum of two irrational numbers may be a rational number or an irrational number.
The sum of two irrational numbers is always an integer.

Answer

The sum of two irrational numbers may be a rational number or an irrational number.

Solution:
If two irrational numbers i.e. $\sqrt{2},\sqrt{5},2+\sqrt{3},2-\sqrt{3}$ etc. are added it is not necessary that sum comes out to be an irrational number always, or a rational nnumber always, or a rational number always...
Since $\sqrt{2}+\sqrt{5}=$ an irrational number
$2+\sqrt{\not\text{3}}+2-\sqrt{\not\text{3}}=4=$ a rational number
So we see that $\sqrt{2}$ and $\sqrt{5}$ are irrational numbers, and their sum is also irrational.
But $2+\sqrt{3}$ and $2-\sqrt{3}$ are also irrational numbers, and their sum is rational number '4'.
So sum of two irrational numbers can be either an irrational number or a rational number depending which numbers are being added.
So options (a) and (b) are totally wrong, because they are not 'always' true.
Option (c) is correcrt because sum can be either irrational or rational and option (c) is verifying this statement.
Option (d) - again it is not always true, if we add two irrational numbers like $2+\sqrt{3}$ and $2-\sqrt{3}.$
Sum is an integer = 4, but if we add $\sqrt{3}$ and $\sqrt{3},$ sum is $2\sqrt{3}$ which is not an integer but again an irrational number.
So option (d) is also incorrect.
Hence, correct option is (c).

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Question 51 Mark

The number $1.\overline{27}$ in the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ is:

$\frac{14}{9}$
$\frac{14}{11}$
$\frac{14}{13}$
$\frac{14}{15}$

Answer

$\frac{14}{11}$

Solution:
Let $\text{x}=1.\overline{27}=1.272727...(1)$
Now, $100\text{x}=127.272727=127.\overline{27}...(2)$
Subtracting equation (1) from (2), we get
$99\text{x}=126$
$\Rightarrow\text{x}=\frac{126}{99}=\frac{14}{11}$
Hence, option (b) is correct.

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Question 61 Mark

Every point on a number line represents:

a unique real number.
a natural number.
a rational number.
an irrational number.

Answer

a unique real number.

Solution:
On number line, we have $-\infty$ to $\infty$ numbers, consisting $-\infty...-4,-3,-2,-1,0,1,2,3,4...\infty,$ 1.12,1.14 and 1.41406532, 3.146201286295... etc. That means on number line, there are natural nmbers (1, 2, 3, 4 ...), integers, rational numbers $\frac{1}{2},\frac{1}{3},1.3333,$ irrational numbers 1.4148625385...
But if we see every number as a complete family, it becomes
Real numbers (any number which can be represent on Real axes)
So, every point on the number line reoresent a unique real number which contains every type.
Hence, option (a) is correct.

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Question 71 Mark

The number $1.\overline{3}$ in the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ is:

$\frac{33}{100}$
$\frac{3}{10}$
$\frac{1}{3}$
$\frac{3}{100}$

Answer

$\frac{1}{3}$

Solution:
Let $\text{x}=1.\overline{3}=0.3333..(1)$
Now, $10\text{x}=3.3333=3.\overline{3}...(2)$
Subtracting equation (1) from (2), we get
$9\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{9}=\frac{1}{3}$
$\Rightarrow0.\overline{3}=\frac{1}{3}$
Hence, option (c) is correct.

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Question 81 Mark

Which of the following is irrational?

0.14
$0.14\overline{16}$
$0.\overline{1416}$
0.1014001400014...

Answer

0.1014001400014...

Solution:

$0.14=\frac{14}{100},$ which is a Rational number

$0.14\overline{16}$ is non-terminating but repeating, hence a rational number

$0.\overline{1416}$ is non-terminating but repeating, hence a rational number

0.1014001400014... is non-terminating as well as non-repeating number, which is irrational in nature.

Hence, correct option is (d).

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Question 91 Mark

Which of the following statements is true?

Product of two irrational numbers is always irrational.
Product of a rational and an irrational number is always irrational.
Sum of two irrational numbers can never be irrational.
Sum of an integer and a rational number can never be an integer.

Answer

Product of a rational and an irrational number is always irrational.

Solution:

Is incorrect, Product of two irrational numbers is not always irrational, it can be also rational sometimes.
when an irrational number is multiplied to itself, or multiplied by another irrational, that product becomes a perfect square.
Example:
$\sqrt{2}\times\sqrt{2}=2$ (Rational)
$\sqrt{2}\times\sqrt{8}=\sqrt{16}=\pm4$ (Rational)

Is correct, because when a rational number is multiplied to an irrational number, it can not make an irrational number terminating or Non-terminating Repeating. Product again becomes a Non-terminating Non-Repeating number.
as: $2\times\sqrt{3}=2\sqrt{3}$
$\frac{2}{3}\times\sqrt{3}=\frac{2}{\sqrt{3}}$
So, product of a rational number and an irrational number is always an irrational, because irrational number is just changed in magnitude not in properties.

Is incorrect, Sum of two irrational numbers can be an irrational number. i.e. if we add $\sqrt{2}$ and $\sqrt{3},$ we will get $\sqrt{2}+\sqrt{3}$ which is also an irrational.

Is incorrect, Sum of an integer and a rational number can be a integer. Because all integers are rational numbers and also we can say some rational numbers are integers. So their sum with integer would be a integer

i.e. 2 + 3 = 5
Hence, correct option is (b).

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Question 101 Mark

Which one of the following is a correct statement?

Decimal expansion of a rational number is terminating.
Decimal expansion of a rational number is non-terminating.
Decimal expansion of an irrational number is terminating.
Decimal expansion of an irrational number is non-terminating and non-repeating.

Answer

Decimal expansion of an irrational number is non-terminating and non-repeating.

Solution:
Decimal Expansion of a Rational number is not only terminating,
It can be either terminating like $\frac{1}{2}=0.5$ or non-terminating Repeating like $\frac{1}{3}=0.3333333......$ So option (a) is not true alone.
Now we know that Non-Terminating numbers are of two types:
One is Non-Terminating Repeating and other is Non-Terminating Non-Repeating.
The Decimal expansion of a Rational number matches one of it's kind i.e Non-Terminating Repeating of Non-Terminating numbers.
So Rational number does not consist both the kinds of Non-Terminating numbers.
Hence, they are not Non-Terminating numbers.
An irrational number is always Non-Terminating in nature, but again not of both of it's kinds.
The decimal Expansion of an irrational number is Non-Terminating Non-Repeating in Nature.
So from all above points and theory we can conclude an Irrational number is Non-Terminating but Non-Repeating in nature
i.e. $\sqrt{2}=1.4142135623730...$
So, option (d) is correct.

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Question 111 Mark

The number 0.318564318564318564 ........ is:

a natural number
an integer
a rational number
an irrational number

Answer

a rational number

Solution:
$0.318564318564318564 \ ...=0.\overline{318564}$ is a Non-terminating repeating Number.
Hence, it is a rational number.
So, correct option is (c).

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Question 121 Mark

$0.\overline{001}$ when expressed in the form $\frac{\text{p}}{\text{q}}$ $\big($p, q are integers and $\text{q}\neq0\big),$ is:

$\frac{1}{1000}$
$\frac{1}{100}$
$\frac{1}{1999}$
$\frac{1}{999}$

Answer

$\frac{1}{999}$

Solution:
Let $\text{x}=0.\overline{001}=0.001001001...(1)$
Now, $1000\text{x}=001.001001001...(2)$
Subtracting equation (1) from (2), we get
$999\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{999}$
Hence, option (d) is correct.

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Question 131 Mark

Which of the following is a correct statement?

Sum of two irrational numbers is always irrational.
Sum of a rational and irrational number is always an irrational number.
Square of an irrational number is always a rational number.
Sum of two rational numbers can never be an integer.

Answer

Sum of a rational and irrational number is always an irrational number.

Solution:

Is incorrect, because sum of two irrational numbers is not an irrational number always. It can also be a rational number

i.e. if we add $2+\sqrt{3}$ and $2-\sqrt{3},$ sum comes out to be $2+\sqrt{\not\text{3}}+2-\sqrt{\not\text{3}}=4,$ which is a rational number.

Is correct, if a rational number is added to an irrational number means to a Non- terminating-repeating number, the sum will also be non-terminating and Non-repeating number, i.e an irrational number.

Example: a rational number '2' and an irrational no $'\sqrt{3}'$ is added, sum $=2+\sqrt{3}$ which is again a non-terminating and non-repeating number, hence an irrational number always.

Is incorrect, Square of an irrational number is not necessarily a rational number. Again it can be either a rational or irrational.

i.e $(\sqrt{2})^2=2$ (Rational)
$(2+\sqrt{3})^2=4+3+2\times2\sqrt{3}=7+4\sqrt{3}$ (irrational)

Is incorrect, Sum of two rational numbers can be an integer and a rational number both.

i.e $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$ (Rational number)
Hence, correct option is (b).

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Question 141 Mark

The value of $0.\overline{23}+0.\overline{22}$ is:

$0.\overline{45}$
$0.\overline{43}$
$0.\overline{45}$
$0.45$

Answer

$0.\overline{45}$

Solution:
Let $\text{x}=0.\overline{23}=0.232323...(1)$
Now, $\text{y}=0.\overline{22}=0.22222...(2)$
Adding equation (1) and (2), we get
$\text{x}+\text{y}=0.454545=0.\overline{45}$
$\Rightarrow0.\overline{23}+0.\overline{22}=0.\overline{45}$
Hence, option (a) is correct.

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Question 151 Mark

Which of the following is irrational?

$\sqrt{\frac{4}{9}}$
$\frac{4}{5}$
$\sqrt{7}$
$\sqrt{81}$

Answer

$\sqrt{7}$

Solution:

Is incorrect, because $\sqrt{\frac{4}{9}}=\frac{\sqrt{4}}{\sqrt{9}}=\pm\frac{2}{3}$ (Rational)

Is also incorrect, as $\frac{4}{5}$ is in the form of $\frac{\text{P}}{\text{Q}}(\text{Q}\neq0),$ (Rational)

Is incorrect, because $\sqrt{7}$ is a non-terminating and Non-Repeating number.

Is incorrect, because $\sqrt{81}=\pm9$ (Rational)

Hence, correct option is (c).

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Question 161 Mark

The smallest rational number by which $\frac{1}{3}$ should be multiplied so that its decimal expansion terminates after one place of decimal, is:

$\frac{1}{10}$
$\frac{3}{10}$
$3$
$30$

Answer

$\frac{3}{10}$

Solution:
$\frac{1}{3}=0.33333...$ (a Non - termnating number)
Now, if we remove 3 from denominator it will terminate.
So, if we multiply by $\frac{3}{10}$
i.e. $\frac{1}{3}\times\frac{1}{3}=\frac{1}{10}=0.1$ ( terminates after one place of decimal)
By multiplying by $\frac{1}{10},$ 3 does not replaces.
By multiplying by 3, we get 1, which is not terminating after one place of decimal
And, by multiplying by 30, we get 10, again not terminating after one palce of decimal.
Hence, option (b) is correct.

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Question 171 Mark

Which of the following is irrational?

0.15
0.01516
$0.\overline{1516}$
0.5015001500015.

Answer

0.5015001500015.

Solution:

$0.15=\frac{15}{100}=$ Rational number
$0.1516=\frac{1516}{100000}=$ Rational number
$0.\overline{1516}$ is a Non-terminating Repeating number = Rational number.
0.5015001500015. is a Non-terminating, Non-Repeating decimal number, So is a irrational number.

Hence, option (d) is correct.

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Question 181 Mark

If n is a natural number, then $\sqrt{\text{n}}$ is:

always a natural number.
always an irrational number.
always an irrational number.
sometimes a natural number and sometimes an irrational number.

Answer

sometimes a natural number and sometimes an irrational number.

Solution:

Is incorrect, because $\sqrt{\text{n}}$ can not be always a natural number

i.e. if $\text{n}=2, \ \sqrt{\text{n}}=\sqrt{2}$ (not a natural no.)

Is incorrect, similiarly, if n = 2, 5, …. Or any odd no. or not perfect square, $\sqrt{\text{n}}=\sqrt{2},\sqrt{5},\sqrt{7}$ are Non-terminating and non-repeating, So irrational in nature, So, not always a rational number.
Is also incorrect, $\sqrt{\text{n}}$ can aslo be rational or say a natural number. If n = 4, 9, 16... or any perfect square number then $\sqrt{\text{n}}=2,3,4...$ natural numbers.
Is fully correct because if n is any odd number or non-perfect square number then $\sqrt{\text{n}}$ would be irrational, but if n is a perfect square number $\sqrt{\text{n}}$ then will be a natural number.

If n = 2, 3, 5, 8 ... $\sqrt{\text{n}}=\sqrt{2},\sqrt{3},\sqrt{8}...$ (irrational)
If n = 4, 9, 16 ... = 2, 3, 4 ... (Natural number)
So, correct option is (d).

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Question 191 Mark

Which of the following numbers can be represented as non-terminating, repeating decimals?

$\frac{39}{24}$
$\frac{3}{16}$
$\frac{3}{11}$
$\frac{137}{25}$

Answer

$\frac{3}{11}$

Solution:

$\frac{39}{24}=1.625=$ Terminating Decimal
$\frac{3}{16}=0.1875=$ Terminating Decimal
$\frac{3}{11}=0.27272727 \ ...=$ Non-Terminating Decimal
$\frac{137}{25}=5.48=$ Terminating Decimal

Hence, option (c) is correct.

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Question 201 Mark

$23.\overline{43}$ when expressed in the form $\frac{\text{p}}{\text{q}}$ $\big($p, q are integers and $\text{q}\neq0\big),$ is:

$\frac{2320}{99}$
$\frac{2343}{100}$
$\frac{2343}{999}$
$\frac{2320}{199}$

Answer

$\frac{2320}{99}$

Solution:
Let $\text{x}=23.\overline{43}=23.434343...(1)$
Now, $100\text{x}=2343.43333...(2)$
Subtracting equation (1) from (2), we get
$99\text{x}=2320$
$\Rightarrow\text{x}=\frac{2320}{99}$
Hence, option (a) is correct.

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MCQ 211 Mark

The number of consecutive zeros in $2^3 \times 3^4 \times 5^4 \times 7,$ is:

✓
$0$
B
$2$
C
$4$
D
$5$

Answer

Correct option: A.

$0$

$5 \times 2 = 10$
$\Rightarrow$ one $5$ and one $2$ make one zero, so $5 \times 2 \times 5 \times 2 = 100$
Numbers of pairs of $5$ and $2$ will be equal to the number of consecutive zeros in the given number.
In the given number, there are three $2\ 's$ and four $5\ 's.$
So number of pairs of $5$ and $2$ are only three.
So there will be three consecutive zeros in the given number

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