Number Systems Test

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Number Systems Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

If A : The quotient of two integers is always a rational number, and
R : $$\displaystyle \frac{1}{0}$$ is not rational, then which of the following statements is true ?

A
A is True and R is the correct explanation of A

B
A is False and R is the correct explanation of A

C
A is True and R is False

D
Both A and R are False

Solution

Since $$\displaystyle \frac{1}{0}$$ is not rational, the quotient of two integers is not rational.
Question 2
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Choose the rational number which does not lie between rational numbers $$\displaystyle\frac{3}{5}$$ and $$\displaystyle\frac{2}{3}$$

A
$$\displaystyle\frac{46}{75}$$

B
$$\displaystyle\frac{47}{75}$$

C
$$\displaystyle\frac{49}{75}$$

D
$$\displaystyle\frac{50}{75}$$

Solution

Consider the given rational numbers,

$$\dfrac{3}{5}$$ and $$\dfrac{2}{3}$$

Now,

$$\dfrac{3\times 15}{5\times 15}$$ and $$\dfrac{2\times 25}{3\times 25}$$

$$\dfrac{45}{75}$$ and $$\dfrac{50}{75}$$

Now, rational number which does not lie between thes rational numbers is,

$$\dfrac{50}{75}$$

Hence, this is the answer.

.
Question 3
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Which of the following rational numbers lies between $$\dfrac {-4}{5}$$ and $$\dfrac {-7}{5}$$?

A
$$\dfrac {-36}{45}$$

B
$$\dfrac {-35}{45}$$

C
$$\dfrac {-39}{45}$$

D
$$\dfrac {-63}{45}$$

Solution

As per the given options, let us multiply both numbers with $$\dfrac {9}{9}$$.

$$\therefore \dfrac {-4}{5}\times \dfrac {9}{9} = \dfrac {-36}{45}$$ and $$\dfrac {-7}{5}\times \dfrac {9}{9} = \dfrac {-63}{45}$$

$$\therefore$$ Clearly, $$\dfrac {-39}{45}$$ lies between $$\dfrac {-4}{5}$$ and $$\dfrac {-7}{5}$$.
Question 4
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Find five rational numbers between $$1$$ and $$2.$$

A
$$\dfrac {1}{10}, \dfrac {2}{10}, \dfrac {3}{10}, \dfrac {4}{10}, \dfrac {5}{10}$$

B
$$\dfrac {1}{5}, \dfrac {2}{5}, \dfrac {3}{5}, \dfrac {4}{5}, \dfrac {5}{5}$$

C
$$\dfrac {1}{2}, \dfrac {1}{3}, \dfrac {1}{4}, \dfrac {1}{5}, \dfrac {1}{6}$$

D
$$\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$$

Solution

$$\textbf{Step -1: Find required rational numbers.}$$
$$\text{We need five rational numbers between }1\text{ and }2.$$
$$\text{So, multiply and divide the numbers with a natural number greater or equal to } 5+1=6.$$
$$\text{Lets take }7.$$
$$1=1\times\dfrac{7}{7}=\dfrac{7}{7}$$
$$\text{and }2=2\times\dfrac{7}{7}=\dfrac{14}{7}$$
$$\text{Thus, }1\text{ and }2\text{ becomes }\dfrac{7}{7}\text{ and }\dfrac{14}{7}\text{ respectively.}$$
$$\text{Since, }7<8<9<10<11<12<13<14$$
$$\therefore 1=\dfrac{7}{7}<\dfrac{8}{7}<\dfrac{9}{7}<\dfrac{10}{7}<\dfrac{11}{7}<\dfrac{12}{7}<\dfrac{13}{7}<\dfrac{14}{7}=2$$
$$\text{Hence, five rational numbers between }1\text{ and }2\text{ are }\dfrac{8}{7},\dfrac{9}{7},\dfrac{10}{7},\dfrac{11}{7},\text{ and }\dfrac{12}{7}.$$
$$\textbf{Hence , the correct option is D.}$$
Question 5
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The value of $$ \displaystyle 2\sqrt{3}-3\sqrt{12}+5\sqrt{75} $$ is equal to:

A
$$ \displaystyle 4\sqrt{3} $$

B
$$ \displaystyle 7\sqrt{5} $$

C
$$ \displaystyle 3\sqrt{5} $$

D
$$ \displaystyle 21\sqrt{3} $$

Solution

$$\displaystyle 2\sqrt{3}-3\sqrt{12}+5\sqrt{75}$$
$$=\displaystyle 2\sqrt{3}-3\sqrt{3\times 2\times 2}+5\sqrt{5\times 5\times 3}$$
$$=\displaystyle 2\sqrt{3}-6\sqrt{3}+25\sqrt{3}$$
$$=\displaystyle \sqrt{3}\left ( 2-6+25 \right )$$
$$=\displaystyle \sqrt{3}(21)=21\sqrt{3}$$.
Therefore, option $$D$$ is correct.
Question 6
1 / -0

The square root of any prime number is

A
rational

B
irrational

C
co-prime

D
composite

Solution

The square root of any prime number is irrational.
Example: $$\sqrt{2}$$ is a irrational number.
Question 7
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Simplify : $$(6^{-1} - 8^{-1})^{-1} + (2^{-1} - 3^{-1})^{-1}$$ is-

A
$$20$$

B
$$30$$

C
$$10$$

D
$$40$$

Solution

Law of exponents,
Given - $$(6^{-1}-8^{-1})^{-1}+(2^{-1}-3^{-1})^{-1}$$
$$=\left(\dfrac{1}{6}-\dfrac{1}{8}\right)^{-1}+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)^{-1}$$
$$=\left(\dfrac{8-6}{6\times 8}\right)^{-1}+\left(\dfrac{3-2}{2\times 3}\right)^{-1}$$
$$=\left(\dfrac{2}{36\times 8}\right)^{-1}+\left(\dfrac{1}{2\times 3}\right)^{-1}$$
$$=\left(\dfrac{1}{24}\right)^{-1}+\left(\dfrac{1}{6}\right)^{-1}$$
$$=24+6$$
$$=30$$
Hence, option B is correct.
Question 8
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The value of $$ \displaystyle \frac{9}{\sqrt{11}+\sqrt{2}} $$ is equal to

A
$$ \displaystyle 9\left ( \sqrt{11}+\sqrt{2} \right ) $$

B
$$9 \displaystyle \left ( \sqrt{11}-\sqrt{2} \right ) $$

C
$$ \displaystyle \sqrt{11}-\sqrt{2} $$

D
$$\dfrac{9}{ \displaystyle \left ( \sqrt{11}-\sqrt{2} \right ) }$$

Solution

$$\displaystyle \frac{9}{\sqrt{11}+\sqrt{2}}=\frac{9}{\sqrt{11}+\sqrt{2}}\times \frac{\sqrt{11}-\sqrt{2}}{\sqrt{11}-\sqrt{2}}$$ ..... [By rationalizing]

$$=\displaystyle \frac{9\left ( \sqrt{11}-\sqrt{2} \right )}{11-2}$$

$$=\displaystyle \frac{9\left ( \sqrt{11}-\sqrt{2} \right )}{9}=\sqrt{11}-\sqrt{2}$$
Question 9
1 / -0

Choose the rational number which does not lie between rational numbers $$-\cfrac {2}{5}$$ and $$-\cfrac {1}{5}$$

A
$$-\dfrac {1}{4}$$

B
$$-\dfrac {3}{10}$$

C
$$\dfrac {3}{10}$$

D
$$-\dfrac {7}{20}$$

Solution

Since the given rational numbers $$-\dfrac {2}{5}$$ and $$-\dfrac {1}{5}$$ are negative rational numbers, therefore, none of the positive rational number can lie between them.

Hence, the rational number $$\dfrac {3}{10}$$ does not lie between the rational numbers $$-\dfrac {2}{5}$$ and $$-\dfrac {1}{5}$$
Question 10
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Which of the following numbers is different from others?

A
$$\sqrt 2$$

B
$$\sqrt 3$$

C
$$\sqrt 4$$

D
$$\sqrt 5$$

Solution

$$\sqrt{2}, \sqrt{3}, \sqrt{5}$$ are irrational numbers.
But $$\sqrt{4}=2$$ is a rational number.
So, $$\sqrt{4}$$ is different from others.

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Number Systems Test - 31

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Number Systems Test - 31

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Question 1

A is True and R is the correct explanation of A

A is False and R is the correct explanation of A

A is True and R is False

Both A and R are False

Solution

Question 2

$$\displaystyle\frac{46}{75}$$

$$\displaystyle\frac{47}{75}$$

$$\displaystyle\frac{49}{75}$$

$$\displaystyle\frac{50}{75}$$

Solution

Question 3

$$\dfrac {-36}{45}$$

$$\dfrac {-35}{45}$$

$$\dfrac {-39}{45}$$

$$\dfrac {-63}{45}$$

Solution

Question 4

$$\dfrac {1}{10}, \dfrac {2}{10}, \dfrac {3}{10}, \dfrac {4}{10}, \dfrac {5}{10}$$

$$\dfrac {1}{5}, \dfrac {2}{5}, \dfrac {3}{5}, \dfrac {4}{5}, \dfrac {5}{5}$$

$$\dfrac {1}{2}, \dfrac {1}{3}, \dfrac {1}{4}, \dfrac {1}{5}, \dfrac {1}{6}$$

$$\dfrac {8}{7}, \dfrac {9}{7}, \dfrac {10}{7}, \dfrac {11}{7}, \dfrac {12}{7}$$

Solution

Question 5

$$ \displaystyle 4\sqrt{3} $$

$$ \displaystyle 7\sqrt{5} $$

$$ \displaystyle 3\sqrt{5} $$

$$ \displaystyle 21\sqrt{3} $$

Solution

Question 6

rational

irrational

co-prime

composite

Solution

Question 7

$$20$$

$$30$$

$$10$$

$$40$$

Solution

Question 8

$$ \displaystyle 9\left ( \sqrt{11}+\sqrt{2} \right ) $$

$$9 \displaystyle \left ( \sqrt{11}-\sqrt{2} \right ) $$

$$ \displaystyle \sqrt{11}-\sqrt{2} $$

$$\dfrac{9}{ \displaystyle \left ( \sqrt{11}-\sqrt{2} \right ) }$$

Solution

Question 9

$$-\dfrac {1}{4}$$

$$-\dfrac {3}{10}$$

$$\dfrac {3}{10}$$

$$-\dfrac {7}{20}$$

Solution

Question 10

$$\sqrt 2$$

$$\sqrt 3$$

$$\sqrt 4$$

$$\sqrt 5$$

Solution

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