Real Numbers Test

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Real Numbers Test - 32

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Question 1
1 / -0

$$\dfrac { 317 } { 3125 }$$ represents ______.

A
A terminating decimal

B
A non-recurring decimal

C
A recurring decimal

D
An Integer

Solution

If the denominator can be in the powers of either $$2$$ or $$5$$ or both, then it will be a terminating decimal, otherwise non-terminating.
Now, the denominator $$3125=5^5$$, i.e. is in the form $$2^m\times5^n$$; where $$n,m$$ are non-negative.
Hence, the rational number is terminating.
Therefore, $$\dfrac{317}{3125} $$represents a terminating decimal.
Question 2
1 / -0

The HCF of $$420$$ and $$130$$ is

A
$$10$$

B
$$15$$

C
$$5$$

D
$$3$$
Question 3
1 / -0

Find the H.C.F. of :

A
$$8$$ and $$40$$

B
$$15$$ and $$25$$

C
$$16$$ and $$60$$

D
$$14$$ and $$35$$

E
$$15,$$ $$25$$ and $$60$$

Solution

$${\textbf{Step - 1: Solving A}}$$

$${\text{Prime factorizing 8 and 40,}}$$

   $$ \Rightarrow {\text{ 8 = 2 }} \times {\text{ 2 }} \times {\text{ 2}}$$

   $$ \Rightarrow {\text{ 40 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 5}}$$

   $${\text{Multiplying common prime factors to get HCF, }}$$

   $${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 }} \times {\text{ 2 = 8}}$$

$${\textbf{Step - 2: Solving B}}$$

   $${\text{Prime factorizing 15 and 25,}}$$

   $$ \Rightarrow {\text{ 15 = 3 }} \times {\text{ 5}}$$

   $$ \Rightarrow {\text{ 25 = 5 }} \times {\text{ 5}}$$

   $${\text{Multiplying common prime factors to get HCF, }}$$

   $${\text{H}}{\text{.C}}{\text{.F = 5}}$$

$${\textbf{Step - 3: Solving C}}$$

  $${\text{Prime factorizing 16 and 60,}}$$

  $$ \Rightarrow {\text{ 16 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2}}$$

   $$ \Rightarrow {\text{ 60 = 2 }} \times {\text{ 2 }} \times {\text{ 3 }} \times {\text{ 5}}$$

   $${\text{Multiplying common prime factors to get HCF, }}$$

   $${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 = 4}}$$

$${\textbf{Step - 4: Solving D}}$$

   $${\text{Prime factorizing 14 and 35,}}$$

   $$ \Rightarrow {\text{ 14 = 2 }} \times {\text{ 7}}$$

   $$ \Rightarrow {\text{ 35 = 5 }} \times {\text{ 7}}$$

   $${\text{Multiplying common prime factors to get HCF, }}$$

   $${\text{H}}{\text{.C}}{\text{.F = 7}}$$

$${\textbf{Step - 5: Solving E}}$$

   $${\text{Prime factorizing 15, 25 and 60,}}$$

   $$ \Rightarrow {\text{ 15 = 3 }} \times {\text{ 5}}$$

   $$ \Rightarrow {\text{ 25 = }}5{\text{ }} \times {\text{ 5}}$$

   $$ \Rightarrow {\text{ 60 = 2 }} \times {\text{ 2 }} \times {\text{ 3 }} \times {\text{ 5}}$$

   $${\text{Multiplying common prime factors to get HCF, }}$$

   $${\text{H}}{\text{.C}}{\text{.F = }}5$$

$${\textbf{Step - 6: Solving F}}$$

   $${\text{Prime factorizing 16, 48 and 80,}}$$

   $$ \Rightarrow {\text{ 16 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2}}$$

   $$ \Rightarrow {\text{ 48 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 3}}$$

   $$ \Rightarrow {\text{ 80 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 5}}$$

   $${\text{Multiplying common prime factors to get HCF, }}$$

   $${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 = 16}}$$

$${\textbf{Hence, we have found H.C.F of different pair of numbers using prime factorization method}}$$
Question 4
1 / -0

Mark the correct alternative of the following.
The HCF of $$100$$ and $$101$$ is _________.

A
$$1$$

B
$$7$$

C
$$37$$

D
None of these

Solution

The given numbers $$100$$ and $$101$$ are consecutive numbers and respectively even and odd numbers.
$$\implies$$ Numbers are co-prime
So the HCF is $$1$$.
Question 5
1 / -0

The HCF of 144 and 198 is

A
$$9$$

B
$$12$$

C
$$6$$

D
$$18$$

Solution

$$\textbf{Step-1: Find the prime factors of given numbers & apply the concept of H.C.F.}$$

$$\text{Given numbers are}$$ $$144$$ $$\text{and}$$ $$198$$

$$\text{Representing the given numbers as the product of their prime factors:}$$

$$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$$

$$ = 2^4 \times 3^2$$

$$ 198 = 2 \times 3 \times 3 \times 11$$

$$ = 2 \times 3^2 \times 11$$

$$\text{We know that}$$ $$HCF$$ $$\text{of two or more numbers is the product of all}$$
$$\text{ prime factors common between them.}$$

$$\text{So,}$$ $$HCF(144, 198)=2\times 3^2$$

$$=2\times 9$$

$$=18$$

$$\text{So, the}$$ $$HCF$$ $$\text{of}$$ $$144$$ $$\text{and}$$ $$198$$ $$\text{is}$$ $$18$$.

$$\textbf{Hence, the correct option is D}$$
Question 6
1 / -0

The HCF of $$144, 180$$ and $$192$$ is

A
$$12$$

B
$$16$$

C
$$18$$

D
$$8$$

Solution

Hence $$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$$
$$180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5$$
$$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3$$
Hence $$2^2 \times 3=12$$ is the highest common factor of all the three numbers
Option (a) is the correct answer
Question 7
1 / -0

Choose the correct alternative answer for the question given below.
Decimal expansion of which of the following is non-terminating recurring?

A
$$\dfrac{2}{5}$$

B
$$\dfrac{3}{16}$$

C
$$\dfrac{3}{11}$$

D
$$\dfrac{137}{25}$$

Solution

1) Since, $$5 = 2^0 \times 5^1$$, which is in the form of $$2^m \times 5^n$$, where m and n are no - negative integers.
So, the decimal expansion of $$\dfrac{2}{5}$$ is terminating.
2) Since, $$16 = 2^4 \times 5^0$$, which in the form of $$2^m \times 5^n$$, where m and n are no - negative integers.
So, the decimal expansion of $$\dfrac{3}{16}$$ is terminating.
3) Since, $$11 = 2^0 \times 5^0 \times 11^1$$, which is not in the form of $$2^m \times 5^n$$, where m and n are non - negative integers.
So, the decimal expansion of $$\dfrac{3}{11}$$ is non - terminating recurring.
4) Since, $$25 = 2^0 x 5^2$$, which is in the form of $$2^m \times 5^n$$, where m and n are non - negative integers.
So, the decimal expansion of $$\dfrac{137}{25}$$ is terminating.
Question 8
1 / -0

In the decimal expansion of rational number $$ \dfrac{43}{{2}^{2} \times {5}^{3}} $$, after how many digits decimal will end ?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Given, $$ \dfrac{43}{2^2 \times 5^3} $$ .
Multiply by $$2$$ in numerator and denominator of number,
$$ \dfrac{43}{2^2 \times 5^3} $$ $$ =\dfrac{43\times2}{2^3 \times 5^3} = \dfrac{86}{10^3} = 0.086$$.

Therefore, after three digits of decimal, the decimal expansion of the given rational number will terminate.
Question 9
1 / -0

Decimal expansion of number $$ \dfrac{441}{2^{2} \times 5^{7} \times 7^{2}} $$ will be:

A
Terminating

B
Non-terminating repeating

C
Terminating and non-terminating both

D
Non-rational

Solution

$$\textbf{Step 1 : Simplify the given fraction. }$$

$$\text{The given fraction is :}$$$$\dfrac{441}{2^2\times5^7\times7^2}$$.

$$\text{It can be reduced to}$$ $$\dfrac{9}{2^2\times5^7}$$

$$\text{Since the new denominator is of the form}$$ $$2^{m} \times 5^{n} $$,
$$\text{the given number will have a terminating decimal expansion.}$$

$$\textbf{Hence, correct option is A.}$$
Question 10
1 / -0

If a rational number $$ x = \dfrac{p}{q} $$ such as a prime factor of $$q$$ is not the form of $$2^{n} 5^{m}$$ where $$ n , m $$ is a zero integer, then decimal expansion of $$x$$ is

A
terminating decimal expansion

B
non-terminating repeating decimal

C
choice (A) and (B) both correct

D
none of these

Solution

A rational number $$x$$ expressed as $$ x = \dfrac{p}{q} $$ where the prime factorization of $$q$$ in the form of $$2^n5^m$$ is having the terminating decimal expansion.
So, if a rational number $$ x = \dfrac{p}{q} $$ such as a prime factor of $$q$$ is not the form of $$2^{n} 5^{m}$$ where $$ n , m $$ is a zero integer, then decimal expansion of $$x$$ is non terminating repeating decimal

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Real Numbers Test - 32

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Real Numbers Test - 32

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Question 1

A terminating decimal

A non-recurring decimal

A recurring decimal

An Integer

Solution

Question 2

$$10$$

$$15$$

$$5$$

$$3$$

Question 3

$$8$$ and $$40$$

$$15$$ and $$25$$

$$16$$ and $$60$$

$$14$$ and $$35$$

$$15,$$ $$25$$ and $$60$$

Solution

Question 4

$$1$$

$$7$$

$$37$$

None of these

Solution

Question 5

$$9$$

$$12$$

$$6$$

$$18$$

Solution

Question 6

$$12$$

$$16$$

$$18$$

$$8$$

Solution

Question 7

$$\dfrac{2}{5}$$

$$\dfrac{3}{16}$$

$$\dfrac{3}{11}$$

$$\dfrac{137}{25}$$

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

Terminating

Non-terminating repeating

Terminating and non-terminating both

Non-rational

Solution

Question 10

terminating decimal expansion

non-terminating repeating decimal

choice (A) and (B) both correct

none of these

Solution

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