Real Numbers and Their Properties: MCQ test, Set 1

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

How many integers are there between $-1$ and $3$ ?
A $5$ B $3$ C $2$ D $4$

Answer and solution

Answer: (B) $3$

$Z = \{\ldots, -2, -1, 0, 1, 2, 3, \ldots\}$

The integers between $-1$ and $3$ are $0, 1, 2$ , that is $3$ numbers.
The integers between $-7$ and $-2$ are
A $-7, -6, -5, -4, -3, -2$ B $-6, -5, -4, -3, -2$ C $-5, -4, -3$ D $-6, -5, -4, -3$

Answer and solution

Answer: (D) $-6, -5, -4, -3$

$Z = \{\ldots, -2, -1, 0, 1, 2, 3, \ldots\}$

So the integers between $-7$ and $-2$ are $-6, -5, -4, -3$ .
Which of the following is an irrational number?
A $5.8$ B $\sqrt{7}$ C $5.\overline{8}$ D $\displaystyle \frac{8}{13}$

Answer and solution

Answer: (B) $\sqrt{7}$

$\sqrt{7} = 2.6457\ldots$ is a non terminating non recurring decimal, so it is irrational.

The other numbers can be written in the form $\displaystyle \frac{p}{q}$ , so they are rational.
$\sqrt{24} = 4.8989\ldots$ is
A an integer B a non terminating non recurring decimal, so irrational C a terminating decimal, so rational D a non terminating recurring decimal, so rational

Answer and solution

Answer: (B) a non terminating non recurring decimal, so irrational

$\sqrt{24} = 4.8989\ldots$ is a non terminating non recurring decimal, so it is irrational.
$\sqrt{13}$ belongs to which of these sets?
A $N$ (natural numbers) B $W$ (whole numbers) C $Q$ (rational numbers) D $Q^*$ (irrational numbers)

Answer and solution

Answer: (D) $Q^*$ (irrational numbers)

$\sqrt{13} = 3.6055\ldots$ is a non terminating non recurring decimal, so it is irrational.

So $\sqrt{13}$ belongs to $Q^*$ ; it does not belong to $N$ , $W$ or $Q$ .
Which of the following is true?
A $e \in Z$ B $0 \in W$ C $\sqrt{54} \in Z$ D $\sqrt{43} \in Q$

Answer and solution

Answer: (B) $0 \in W$

$W = \{0, 1, 2, 3, 4, \ldots\}$

$0$ is in this set, so the statement is true.

The other three statements are false.
Add $-10\sqrt{19}$ and $25\sqrt{19}$ .
A $15\sqrt{38}$ B $-35\sqrt{19}$ C $-15\sqrt{19}$ D $15\sqrt{19}$

Answer and solution

Answer: (D) $15\sqrt{19}$

$-10\sqrt{19} + 25\sqrt{19} = (-10 + 25)\sqrt{19}$

$= 15\sqrt{19}$
Add $-4 - 10\sqrt{19}$ and $-11 + 12\sqrt{19}$ .
A $-13\sqrt{19}$ B $7 + 2\sqrt{19}$ C $-15 + 2\sqrt{19}$ D $-15 - 22\sqrt{19}$

Answer and solution

Answer: (C) $-15 + 2\sqrt{19}$

$-4 - 10\sqrt{19} - 11 + 12\sqrt{19} = (-4 - 11) + (-10 + 12)\sqrt{19}$

$= -15 + 2\sqrt{19}$

Only like terms are added: numbers with numbers, $\sqrt{19}$ terms with $\sqrt{19}$ terms.
Subtract $-25\sqrt{10}$ from $-9\sqrt{10}$ .
A $16\sqrt{10}$ B $34\sqrt{10}$ C $-34\sqrt{10}$ D $-16\sqrt{10}$

Answer and solution

Answer: (A) $16\sqrt{10}$

$-9\sqrt{10} - (-25\sqrt{10}) = -9\sqrt{10} + 25\sqrt{10}$

$= 16\sqrt{10}$
Subtract $(8 + 2\sqrt{21})$ from $(9 - 5\sqrt{21})$ .
A $17 - 7\sqrt{21}$ B $1 - 3\sqrt{21}$ C $1 - 7\sqrt{21}$ D $-1 + 7\sqrt{21}$

Answer and solution

Answer: (C) $1 - 7\sqrt{21}$

$(9 - 5\sqrt{21}) - (8 + 2\sqrt{21}) = 9 - 5\sqrt{21} - 8 - 2\sqrt{21}$

$= 1 - 7\sqrt{21}$

The sign of every term of the bracket being subtracted changes.
$2\sqrt{3} \times (-6\sqrt{10}) =$
A $-12\sqrt{13}$ B $-12\sqrt{30}$ C $-4\sqrt{30}$ D $12\sqrt{30}$

Answer and solution

Answer: (B) $-12\sqrt{30}$

$2\sqrt{3} \times (-6\sqrt{10}) = 2 \times (-6) \times \sqrt{3 \times 10}$

$= -12\sqrt{30}$
$\sqrt{11} \times \sqrt{3} =$
A $33$ B $11\sqrt{3}$ C $\sqrt{33}$ D $\sqrt{14}$

Answer and solution

Answer: (C) $\sqrt{33}$

$\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$

$\sqrt{11} \times \sqrt{3} = \sqrt{33}$
$-25\sqrt{17} + 14\sqrt{17} + 20\sqrt{17} =$
A $9\sqrt{17}$ B $8\sqrt{17}$ C $-9\sqrt{17}$ D $59\sqrt{17}$

Answer and solution

Answer: (A) $9\sqrt{17}$

$-25\sqrt{17} + 14\sqrt{17} + 20\sqrt{17} = (-25 + 14 + 20)\sqrt{17}$

$= 9\sqrt{17}$
$46690\sqrt{3} - 9009\sqrt{3} - 30057\sqrt{3} + 1350\sqrt{3} =$
A $8974\sqrt{3}$ B $-8974\sqrt{3}$ C $87106\sqrt{3}$ D $8984\sqrt{3}$

Answer and solution

Answer: (A) $8974\sqrt{3}$

Positive terms: $46690\sqrt{3} + 1350\sqrt{3} = 48040\sqrt{3}$

Negative terms: $9009\sqrt{3} + 30057\sqrt{3} = 39066\sqrt{3}$

$46690\sqrt{3} - 9009\sqrt{3} - 30057\sqrt{3} + 1350\sqrt{3} = 48040\sqrt{3} - 39066\sqrt{3}$

$= (48040 - 39066)\sqrt{3} = 8974\sqrt{3}$
$5\sqrt{162} \times 5\sqrt{32} =$
A $129600$ B $720$ C $-1800$ D $1800$

Answer and solution

Answer: (D) $1800$

$5\sqrt{162} \times 5\sqrt{32} = 5 \times 5 \times \sqrt{162 \times 32}$

$= 25 \times \sqrt{5184}$

Long division: $7 \times 7 = 49$ , remainder $2$ ; bring down $84$ to get $284$ ; $142 \times 2 = 284$ . So $\sqrt{5184} = 72$

$= 25 \times 72 = 1800$
$-7\sqrt{3} \times 5\sqrt{27} =$
A $315$ B $-2835$ C $-315$ D $-18$

Answer and solution

Answer: (C) $-315$

$-7\sqrt{3} \times 5\sqrt{27} = -7 \times 5 \times \sqrt{3 \times 27}$

$= -35 \times \sqrt{81}$

$\sqrt{81} = 9$

$= -35 \times 9 = -315$
$\displaystyle \frac{\sqrt{70}}{\sqrt{10}} =$
A $\sqrt{60}$ B $\sqrt{7}$ C $7$ D $\sqrt{700}$

Answer and solution

Answer: (B) $\sqrt{7}$

$\displaystyle \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$

$\displaystyle \frac{\sqrt{70}}{\sqrt{10}} = \sqrt{7}$
$\displaystyle \frac{42}{\sqrt{42}} =$
A $\sqrt{42}$ B $42$ C $1$ D $42\sqrt{42}$

Answer and solution

Answer: (A) $\sqrt{42}$

$\displaystyle \frac{a}{\sqrt{a}} = \frac{\sqrt{a} \times \sqrt{a}}{\sqrt{a}} = \sqrt{a}$

$\displaystyle \frac{42}{\sqrt{42}} = \sqrt{42}$
$\sqrt{52} =$
A $2\sqrt{26}$ B $4\sqrt{13}$ C $13\sqrt{2}$ D $2\sqrt{13}$

Answer and solution

Answer: (D) $2\sqrt{13}$

$\sqrt{52} = \sqrt{4 \times 13} = \sqrt{4} \times \sqrt{13} = 2\sqrt{13}$
$\sqrt{448} + \sqrt{175} =$
A $13\sqrt{7}$ B $89\sqrt{7}$ C $\sqrt{623}$ D $13\sqrt{14}$

Answer and solution

Answer: (A) $13\sqrt{7}$

$\sqrt{448} = \sqrt{64 \times 7} = \sqrt{64} \times \sqrt{7} = 8\sqrt{7}$

$\sqrt{175} = \sqrt{25 \times 7} = \sqrt{25} \times \sqrt{7} = 5\sqrt{7}$

$8\sqrt{7} + 5\sqrt{7} = 13\sqrt{7}$
If $a = 0$ and $b = -22$ , which of the following does not belong to $R$ ?
A $\displaystyle \frac{b}{a}$ B $a \times b$ C $\displaystyle \frac{a}{b}$ D $a + b$

Answer and solution

Answer: (A) $\displaystyle \frac{b}{a}$

$\displaystyle \frac{b}{a} = \frac{-22}{0}$ is not defined, so it does not belong to $R$ .

$\displaystyle \frac{a}{b} = \frac{0}{-22} = 0 \in R$ .
For $a = 13$ and $b = 17$ , which of the following is NOT true?
A $a + b = b + a$ B $a \times 1 = a$ C $a - b = b - a$ D $a + 0 = a$

Answer and solution

Answer: (C) $a - b = b - a$

$a - b = 13 - 17 = -4$ , $b - a = 17 - 13 = 4$

So $a - b \ne b - a$ .
$(-12 \div 6) \div 3 =$
A $-6$ B $\displaystyle -\frac{2}{3}$ C $-4$ D $-2$

Answer and solution

Answer: (B) $\displaystyle -\frac{2}{3}$

$\displaystyle (-12 \div 6) \div 3 = -2 \div 3 = -\frac{2}{3}$

Do the bracket first. Division is not associative, so $-12 \div (6 \div 3) = -6$ is different.
$22 + \underline{\quad\quad} = 22$ . The number in the blank is
A $-22$ B $22$ C $0$ D $1$

Answer and solution

Answer: (C) $0$

$a + 0 = a$ : $0$ is the additive identity.
The multiplicative inverse of $-14$ is
A $14$ B $\displaystyle \frac{1}{14}$ C $1$ D $\displaystyle -\frac{1}{14}$

Answer and solution

Answer: (D) $\displaystyle -\frac{1}{14}$

Multiplicative inverse of $a$ is $\displaystyle \frac{1}{a}$ , since $\displaystyle a \times \frac{1}{a} = 1$ .

$\displaystyle -14 \times (-\frac{1}{14}) = 1$

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MCQ test, Set 1