Real Numbers Test

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Real Numbers Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

Every irrational number is:

A
a surd

B
a prime number

C
not a surd

D
none

Solution

An irrational number is a real number that cannot be represented as a ratio or a simple fraction.

By definition, a surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For eg, $$\sqrt2$$ is a surd since 2 is rational and $$\sqrt 2$$ is irrational.

Similarly, the cube root of 9 is also a surd since 9 is rational and the cube root of 9 is irrational.

On the other hand, $$\sqrtπ$$ is not a surd even though $$\sqrtπ$$ is irrational because π is not rational.

Thus, to answer the question, every surd is an irrational number, though an irrational number may or may not be a surd

The answer is Option C.
Question 2
1 / -0

$$20$$ is written as the product of primes as :

A
$${2\times 5 }$$

B
$${2\times 2\times 3\times 5}$$

C
$${2\times 2\times 5}$$

D
$${2\times 2\times 3}$$

Solution

To write a number as product of its primes, we divide it by various prime numbers $$ 2, 3, 5, 7 $$ etc one by one and check by which prime numbers it is divisible with and how many times.

Hence, $$ 20 = 2 \times 10 = 2 \times 2 \times 5 $$
Question 3
1 / -0

Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non -terminating decimal expansion
$$\displaystyle \frac{7}{210}$$

A
Terminating decimal expansion

B
Non -terminating decimal expansion

C
Cannot be determined

D
None

Solution

Simplify it by dividing nominator and denominator both by $$7$$ we get $$\dfrac{1}{30}$$
Factorize the denominator we get $$30=2 \times 3 \times 5$$
Denominator has $$3$$ also in denominator
So denominator is not in form of $$2^{n} \times 5^{n}$$
Hence it is non terminating.
Question 4
1 / -0

Prime factors of $$140$$ are :

A
$${2\times2\times 7}$$

B
$${2\times2\times5}$$

C
$${2\times2\times5\times7}$$

D
$${2\times2\times5\times7\times3}$$

Solution

Factors of $$140$$ are $$1,140,7,20,2,70,4,35,5, 28,10,14$$

Prime factors are $$2, 7$$ and $$5.$$

So, $$140$$ written as product of primes is $$2\times 2 \times 5\times 7.$$
Question 5
1 / -0

$$\sqrt3$$ is

A
rational number

B
irrational number

C
natural number

D
None

Solution

Let $$\sqrt3$$ is a rational number
$$\therefore \sqrt3 = \displaystyle \frac{a}{b}$$ [Where a & b are co-primes]
$$a^2=3b^2$$ .......(i)
$$\Rightarrow$$ $$3$$ divides $$a^2$$
$$\Rightarrow$$ $$3$$ also divides a
$$\Rightarrow$$ $$a=3c$$
[Where c is any non-zero positive integer]
$$\Rightarrow a^2 = 9c^2$$
From equation (i)
$$3b^2=9c^2$$
$$\Rightarrow b^2 = 3c^2 \Rightarrow$$ $$3$$ divides $$b^2$$
$$\Rightarrow$$ $$3$$ also divides b
So, $$3$$ is a common factor of a and b.
Our assumption is wrong, because a and b are not co - primes.
It means $$\sqrt3$$ is an irrational number.
Question 6
1 / -0

Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non -terminating decimal expansion
$$\displaystyle \frac{15}{1600}$$

A
Terminating decimal expansion

B
Non-terminating decimal expansion

C
Cannot be determined

D
None

Solution

Factorize the denominator we get $$1600=2 \times 2 \times 2 \times2 \times 2 \times 2 \times 5 \times 5 = 2^{6} \times 5^{2}$$
so denominator is in form of $$2^n \times 5^m$$
Hence $$\frac{15}{1600}$$ is terminating.
Question 7
1 / -0

$$(3 + \sqrt {5})$$ is ..............

A
whole number

B
an integer

C
rational

D
irrational

Solution

Sum or difference of a rational and irrational number is irrational. Therefore, $$D$$ is the correct answer.
Question 8
1 / -0

Euclids division lemma, the general equation can be represented as .......

A
$$a = b\times q + r$$

B
$$a = b\div q + r$$

C
$$a = b- q - r$$

D
$$a = b + q - r$$

Solution

Euclids division lemma states that for any two positive integers $$a$$ and $$b$$ we can find two whole numbers $$q$$ and $$r$$ such that $$a = b \times q + r$$ where $$0 \leq r < b.$$
Therefore, $$A$$ is the correct answer.
Question 9
1 / -0

The statement dividend $$=$$ divisor $$\times$$ quotient $$+$$ remainder is called

A
Euclid's multiplication Lemma

B
Euclid's addition Lemma

C
Euclid's subtraction Lemma

D
Euclid's division Lemma

Solution

Euclids division lemma states that $$a = b \times q + r$$ where $$0 \leq r < b.$$
$$a =$$ dividend $$,b =$$ divisor $$, q =$$ quotient and $$r =$$ remainder
Therefore, $$D$$ is the correct answer.
Question 10
1 / -0

Fundamental theorem of arithmetic is also called as ______ Factorization Theorem.

A
Algebra

B
Ambiguous

C
Unique

D
None of these

Solution

$$30 = 2 \times 3 \times 5$$, where $$2$$ and $$3$$ are prime numbers.
We cannot get the number $$30$$ by multiplying any other prime numbers.
Example : $$30 \neq 2 \times 5 \times 7.$$
It is only with one particular set of prime numbers.
Hence, it is called Unique Factorization Theorem.
Therefore, $$C$$ is the correct answer.

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Real Numbers Test - 22

Result

Real Numbers Test - 22

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Question 1

a surd

a prime number

not a surd

none

Solution

Question 2

$${2\times 5 }$$

$${2\times 2\times 3\times 5}$$

$${2\times 2\times 5}$$

$${2\times 2\times 3}$$

Solution

Question 3

Terminating decimal expansion

Non -terminating decimal expansion

Cannot be determined

None

Solution

Question 4

$${2\times2\times 7}$$

$${2\times2\times5}$$

$${2\times2\times5\times7}$$

$${2\times2\times5\times7\times3}$$

Solution

Question 5

rational number

irrational number

natural number

None

Solution

Question 6

Terminating decimal expansion

Non-terminating decimal expansion

Cannot be determined

None

Solution

Question 7

whole number

an integer

rational

irrational

Solution

Question 8

$$a = b\times q + r$$

$$a = b\div q + r$$

$$a = b- q - r$$

$$a = b + q - r$$

Solution

Question 9

Euclid's multiplication Lemma

Euclid's addition Lemma

Euclid's subtraction Lemma

Euclid's division Lemma

Solution

Question 10

Algebra

Ambiguous

Unique

None of these

Solution

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