Number Systems Test

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Number Systems Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$[1^{-2} + 2^{-2}+3^{-2}] \times 6^2$$ is ______.

A
$$\dfrac{49}{36}$$

B
$$\dfrac{36}{49}$$

C
$$36$$

D
$$49$$

Solution

$$\left[\dfrac{36}{1} + \dfrac{36}{4} + \dfrac{36}{9} \right]$$
$$[1^{-2} + 2^{-2}+3^{-2}] \times 6^2 = \left[\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} \right] \times 6^2$$
$$=\left[\dfrac{1}{1} + \dfrac{1}{4} + \dfrac{1}{9} \right] \times 36$$
$$\left[\dfrac{36+9+4}{36}\right] \times 36$$
$$\left[\dfrac{49}{36}\right] \times 36=49$$
Question 2
1 / -0

A rational number between $$\sqrt {2}$$ and $$\sqrt {3}$$ is

A
$$\dfrac {\sqrt {2} + \sqrt {3}}{2}$$

B
$$\dfrac {\sqrt {2} \cdot \sqrt {3}}{2}$$

C
$$1.5$$

D
$$1.8$$

Solution

We know that
$$\sqrt {2} = 1.4142135....$$ and $$\sqrt {3} = 1.732050807 ...$$
So, we can clearly see that $$1.5$$ is a rational number and it lies between $$1.4142135...$$ and $$1.732050807...$$

Hence, $$(C)$$ is the correct answer.
Question 3
1 / -0

After rationalizing the denominator of $$\dfrac {7}{3\sqrt {3} - 2\sqrt {2}}$$, we get the denominator as

A
$$13$$

B
$$19$$

C
$$5$$

D
$$35$$

Solution

$$\dfrac {7}{3\sqrt {3} - 2\sqrt {2}} = \dfrac {7}{3\sqrt {3} - 2\sqrt {2}} \times \dfrac {3\sqrt {3} + 2\sqrt {2}}{3\sqrt {3} + 2\sqrt {2}}$$

$$= \dfrac {7(3\sqrt {3} + 2\sqrt {2})}{(3\sqrt {3})^{2} - (2\sqrt {2})^{2}} = \dfrac {7(3\sqrt {3} + 2\sqrt {2})}{27 - 8}$$

$$= \dfrac {7(3\sqrt {3} + 2\sqrt {2})}{19}$$
Therefore, we get the denominator as $$19$$.
Hence, $$(B)$$ is the correct answer.
Question 4
1 / -0

$$\dfrac {1}{\sqrt {9} - \sqrt {8}}$$ is equal to

A
$$\dfrac {1}{2} (3 - 2\sqrt {2})$$

B
$$\dfrac {1}{2} (3 + 2\sqrt {2})$$

C
$$3 - 2\sqrt {2}$$

D
$$3 + 2\sqrt {2}$$

Solution

$$\dfrac {1}{\sqrt {9} - \sqrt {8}} = \dfrac {1}{\sqrt {3\times 3} - \sqrt {4\times 2}} = \dfrac {1}{3 - 2\sqrt {2}}$$

$$= \dfrac {1}{3 - 2\sqrt {2}} \times \dfrac {3 + 2\sqrt {2}}{3 + 2\sqrt {2}} $$

$$= \dfrac {3 + 2\sqrt {2}}{(3)^{2} - (2\sqrt {2})^{2}}$$

$$= \dfrac {3 + 2\sqrt {2}}{9 - 8} = \dfrac {3 + 2\sqrt {2}}{1} = 3 + 2\sqrt {2}$$
Hence, $$(D)$$ is the correct answer.
Question 5
1 / -0

Choose the correct alternative answer for the question given below.
Which one of the following is an irrational number?

A
$$\sqrt{\dfrac{16}{25}}$$

B
$$\sqrt{5}$$

C
$$\dfrac{3}{9}$$

D
$$\sqrt{196}$$

Solution

Since,
$$\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}$$ is a rational number;
$$\dfrac{3}{9}$$ is a rational number,
$$\sqrt{196}=14$$ is a rational number; and
$$\sqrt{5}$$ is an irrational number.
Question 6
1 / -0

Which of the following is not an irrational number?

A
$$\sqrt{2}$$

B
$$\sqrt{3}$$

C
$$\sqrt{9}$$

D
$$\sqrt{5}$$

Solution
Question 7
1 / -0

Choose the correct alternative answer for the question given below.
What is $$\sqrt{n}$$, if n is not a perfect square number?

A
Natural number

B
Rational number

C
Irrational number

D
Options A, B, C all are correct

Solution

If n is not a perfect square number, then $$\sqrt{n}$$ is an irrational number.
Question 8
1 / -0

Choose the correct alternative answer for the question given below.
Which one of the following is an irrational number?

A
$$0.17$$

B
$$1.\overline{513}$$

C
$$0.\overline{2746}$$

D
$$0.101001000.....$$

Solution

Since,
$$0.17$$ has a terminating decimal expansion, so, it is rational number;
$$1.\overline{513}$$ has non-terminating recurring decimal expansion, so ,it is rational number;
$$0.\overline{2746}$$ has non-terminating recurring decimal expansion, so, it is rational number;
$$0.101001000....$$ has non-terminating non-recurring decimal expansion, so, it is irrational number;
Question 9
1 / -0

$$\sqrt{2}$$ is

A
a rational number

B
an irrational number

C
an integer

D
a natural number

Solution

$$ {\textbf{Step - 1: Formation}} $$

$$ \sqrt 2 {\text{ = }}\dfrac{{\text{p}}}{{\text{q}}} $$

$$ {{\text{Here p and q are coprime numbers and q}} \ne {\text{0}}} $$

$$ {\sqrt 2 {\text{ = }}\dfrac{{\text{p}}}{{\text{q}}}} $$

$$ {{\text{On squaring both the sides we get,}}} $$

$$ {{\Rightarrow \text{ 2 = }}{{\left( {\dfrac{{\text{p}}}{{\text{q}}}} \right)}^{\text{2}}}} $$

$$ {{\Rightarrow \text{ 2}}{{\text{q}}^{\text{2}}}{\text{ = }}{{\text{p}}^{\text{2}}}{{ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots }}..\left( {\text{1}} \right)} $$
$$ {\Rightarrow} \dfrac{{{p^2}}}{2} = q $$
$$ {\textbf{Step - 2: Calculation}} $$
$$ {\text{So 2 divides p and p is multiple of 2}} $$
$$   \Rightarrow {\text{p = 2m}} $$

$$   \Rightarrow {{\text{p}}^2} = 4{m^2} ..........................(2) $$

$$ {\text{From equation (1) and (2), we get}} $$

$$   \Rightarrow 2{q^2} = 4{m^2} $$

$$   \Rightarrow {q^2} = 2{m^2} $$

$$ {q^2}\;{\text{is amultiple of 2}} $$

$$ {\text{So, q is multiple of 2}} $$

$$ {\text{Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. }}$$
$${\text{Therefore, p/q is not a rational number}} $$
  $$ \sqrt 2{\text{ is an irrational number.}}$$
$${\textbf{Hence, the correct answer is option B.}}$$
Question 10
1 / -0

Multiply $$10^4$$ by $$10^2$$

A
$$10^8$$

B
$$10^2$$

C
$$10^6$$

D
$$10^{-2}$$

E
$$10^3$$

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Re-Attempt Weekly Quiz Competition

Number Systems Test - 37

Result

Number Systems Test - 37

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SHARING IS CARING

Question 1

$$\dfrac{49}{36}$$

$$\dfrac{36}{49}$$

$$36$$

$$49$$

Solution

Question 2

$$\dfrac {\sqrt {2} + \sqrt {3}}{2}$$

$$\dfrac {\sqrt {2} \cdot \sqrt {3}}{2}$$

$$1.5$$

$$1.8$$

Solution

Question 3

$$13$$

$$19$$

$$5$$

$$35$$

Solution

Question 4

$$\dfrac {1}{2} (3 - 2\sqrt {2})$$

$$\dfrac {1}{2} (3 + 2\sqrt {2})$$

$$3 - 2\sqrt {2}$$

$$3 + 2\sqrt {2}$$

Solution

Question 5

$$\sqrt{\dfrac{16}{25}}$$

$$\sqrt{5}$$

$$\dfrac{3}{9}$$

$$\sqrt{196}$$

Solution

Question 6

$$\sqrt{2}$$

$$\sqrt{3}$$

$$\sqrt{9}$$

$$\sqrt{5}$$

Solution

Question 7

Natural number

Rational number

Irrational number

Options A, B, C all are correct

Solution

Question 8

$$0.17$$

$$1.\overline{513}$$

$$0.\overline{2746}$$

$$0.101001000.....$$

Solution

Question 9

a rational number

an irrational number

an integer

a natural number

Solution

Question 10

$$10^8$$

$$10^2$$

$$10^6$$

$$10^{-2}$$

$$10^3$$

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