Real Numbers and Their Properties: MCQ test, Set 3

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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.

Which of the following is not an integer?
A $-8$ B $15$ C $0$ D $\displaystyle \frac{3}{7}$

Answer and solution

Answer: (D) $\displaystyle \frac{3}{7}$

$\displaystyle \frac{3}{7}$ is a fraction, not an integer.

$Z = \{\ldots, -2, -1, 0, 1, 2, 3, \ldots\}$
The integers between $-12$ and $-8$ are
A $-11, -10, -9, -8$ B $-12, -11, -10, -9, -8$ C $-10, -9$ D $-11, -10, -9$

Answer and solution

Answer: (D) $-11, -10, -9$

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

So the integers between $-12$ and $-8$ are $-11, -10, -9$ .
Which of these numbers is irrational?
A $3.\overline{3}$ B $\displaystyle \frac{10}{9}$ C $\sqrt{40}$ D $7.8$

Answer and solution

Answer: (C) $\sqrt{40}$

$\sqrt{40} = 6.3245\ldots$ 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.
$\sqrt{39} = 6.2449\ldots$ is
A an integer B a non terminating non recurring decimal, so irrational C a non terminating recurring decimal, so rational D a terminating decimal, so rational

Answer and solution

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

$\sqrt{39} = 6.2449\ldots$ is a non terminating non recurring decimal, so it is irrational.
Which of the following is false?
A $0 \in R$ B $-32 \in N$ C $40 \in Q$ D $\sqrt{67} \in R$

Answer and solution

Answer: (B) $-32 \in N$

$N = \{1, 2, 3, 4, \ldots\}$

$-32$ is not in this set, so the statement is false.

The other three statements are true.
$\sqrt{66}$ belongs to which of these sets?
A $Q^*$ (irrational numbers) B $Q$ (rational numbers) C $N$ (natural numbers) D $W$ (whole numbers)

Answer and solution

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

$\sqrt{66} = 8.1240\ldots$ is a non terminating non recurring decimal, so it is irrational.

So $\sqrt{66}$ belongs to $Q^*$ ; it does not belong to $N$ , $W$ or $Q$ .
Add $16\sqrt{3}$ and $-6\sqrt{3}$ .
A $10\sqrt{3}$ B $-96\sqrt{3}$ C $22\sqrt{3}$ D $-22\sqrt{3}$

Answer and solution

Answer: (A) $10\sqrt{3}$

$16\sqrt{3} - 6\sqrt{3} = (16 - 6)\sqrt{3}$

$= 10\sqrt{3}$
Add $-13 - 10\sqrt{17}$ and $7 + 3\sqrt{17}$ .
A $-20 - 7\sqrt{17}$ B $-6 - 13\sqrt{17}$ C $-13\sqrt{17}$ D $-6 - 7\sqrt{17}$

Answer and solution

Answer: (D) $-6 - 7\sqrt{17}$

$-13 - 10\sqrt{17} + 7 + 3\sqrt{17} = (-13 + 7) + (-10 + 3)\sqrt{17}$

$= -6 - 7\sqrt{17}$

Only like terms are added: numbers with numbers, $\sqrt{17}$ terms with $\sqrt{17}$ terms.
Subtract $-14\sqrt{6}$ from $22\sqrt{6}$ .
A $-36\sqrt{6}$ B $-8\sqrt{6}$ C $36\sqrt{6}$ D $8\sqrt{6}$

Answer and solution

Answer: (C) $36\sqrt{6}$

$22\sqrt{6} - (-14\sqrt{6}) = 22\sqrt{6} + 14\sqrt{6}$

$= 36\sqrt{6}$
Subtract $(12 + \sqrt{6})$ from $(8 - 7\sqrt{6})$ .
A $-4 - 6\sqrt{6}$ B $-4 - 8\sqrt{6}$ C $20 - 8\sqrt{6}$ D $4 + 8\sqrt{6}$

Answer and solution

Answer: (B) $-4 - 8\sqrt{6}$

$(8 - 7\sqrt{6}) - (12 + \sqrt{6}) = 8 - 7\sqrt{6} - 12 - \sqrt{6}$

$= -4 - 8\sqrt{6}$

The sign of every term of the bracket being subtracted changes.
$5\sqrt{5} \times 2\sqrt{13} =$
A $10\sqrt{18}$ B $7\sqrt{65}$ C $10\sqrt{65}$ D $-10\sqrt{65}$

Answer and solution

Answer: (C) $10\sqrt{65}$

$5\sqrt{5} \times 2\sqrt{13} = 5 \times 2 \times \sqrt{5 \times 13}$

$= 10\sqrt{65}$
$\sqrt{7} \times \sqrt{19} =$
A $\sqrt{26}$ B $133$ C $\sqrt{133}$ D $7\sqrt{19}$

Answer and solution

Answer: (C) $\sqrt{133}$

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

$\sqrt{7} \times \sqrt{19} = \sqrt{133}$
$25\sqrt{5} + 27\sqrt{5} - 11\sqrt{5} =$
A $40\sqrt{5}$ B $41\sqrt{5}$ C $63\sqrt{5}$ D $-41\sqrt{5}$

Answer and solution

Answer: (B) $41\sqrt{5}$

$25\sqrt{5} + 27\sqrt{5} - 11\sqrt{5} = (25 + 27 - 11)\sqrt{5}$

$= 41\sqrt{5}$
$34070\sqrt{6} - 250\sqrt{6} + 321713\sqrt{6} - 5009\sqrt{6} =$
A $361042\sqrt{6}$ B $-350524\sqrt{6}$ C $350534\sqrt{6}$ D $350524\sqrt{6}$

Answer and solution

Answer: (D) $350524\sqrt{6}$

Positive terms: $34070\sqrt{6} + 321713\sqrt{6} = 355783\sqrt{6}$

Negative terms: $250\sqrt{6} + 5009\sqrt{6} = 5259\sqrt{6}$

$34070\sqrt{6} - 250\sqrt{6} + 321713\sqrt{6} - 5009\sqrt{6} = 355783\sqrt{6} - 5259\sqrt{6}$

$= (355783 - 5259)\sqrt{6} = 350524\sqrt{6}$
$6\sqrt{45} \times 4\sqrt{5} =$
A $5400$ B $150$ C $360$ D $-360$

Answer and solution

Answer: (C) $360$

$6\sqrt{45} \times 4\sqrt{5} = 6 \times 4 \times \sqrt{45 \times 5}$

$= 24 \times \sqrt{225}$

$\sqrt{225} = 15$

$= 24 \times 15 = 360$
$7\sqrt{13} \times (-5\sqrt{117}) =$
A $-1365$ B $1365$ C $-53235$ D $78$

Answer and solution

Answer: (A) $-1365$

$7\sqrt{13} \times (-5\sqrt{117}) = 7 \times (-5) \times \sqrt{13 \times 117}$

$= -35 \times \sqrt{1521}$

Long division: $3 \times 3 = 9$ , remainder $6$ ; bring down $21$ to get $621$ ; $69 \times 9 = 621$ . So $\sqrt{1521} = 39$

$= -35 \times 39 = -1365$
$\displaystyle \frac{\sqrt{10}}{\sqrt{5}} =$
A $\sqrt{2}$ B $\sqrt{50}$ C $\sqrt{5}$ D $2$

Answer and solution

Answer: (A) $\sqrt{2}$

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

$\displaystyle \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}$
$\displaystyle \frac{45}{\sqrt{45}} =$
A $45\sqrt{45}$ B $\sqrt{45}$ C $45$ D $1$

Answer and solution

Answer: (B) $\sqrt{45}$

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

$\displaystyle \frac{45}{\sqrt{45}} = \sqrt{45}$
$\sqrt{243} - \sqrt{108} =$
A $\sqrt{135}$ B $3\sqrt{6}$ C $45\sqrt{3}$ D $3\sqrt{3}$

Answer and solution

Answer: (D) $3\sqrt{3}$

$\sqrt{243} = \sqrt{81 \times 3} = \sqrt{81} \times \sqrt{3} = 9\sqrt{3}$

$\sqrt{108} = \sqrt{36 \times 3} = \sqrt{36} \times \sqrt{3} = 6\sqrt{3}$

$9\sqrt{3} - 6\sqrt{3} = 3\sqrt{3}$
$\sqrt{375}$ in the simplest form is
A $25\sqrt{15}$ B $5\sqrt{15}$ C $5\sqrt{75}$ D $15\sqrt{5}$

Answer and solution

Answer: (B) $5\sqrt{15}$

$\sqrt{375} = \sqrt{25 \times 15} = \sqrt{25} \times \sqrt{15} = 5\sqrt{15}$
If $a = 14$ and $b = 0$ , which of the following does not belong to $R$ ?
A $\displaystyle \frac{a}{b}$ B $a + b$ C $a - b$ D $a \times b$

Answer and solution

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

$\displaystyle \frac{a}{b} = \frac{14}{0}$ is not defined, so $\displaystyle \frac{14}{0} \notin R$ .

$a + b = 14, a - b = 14, a \times b = 0$ all belong to $R$ .
$\displaystyle -\frac{12}{11} + 5\sqrt{14} = 5\sqrt{14} + (-\frac{12}{11})$ is an example of
A the associative property under addition B the closure under addition C the commutative property under addition D the additive identity

Answer and solution

Answer: (C) the commutative property under addition

The two numbers change places and the result is the same: $a + b = b + a$ .
$18 - (5 - 4) =$
A $9$ B $19$ C $27$ D $17$

Answer and solution

Answer: (D) $17$

$18 - (5 - 4) = 18 - 1 = 17$

Note $(18 - 5) - 4 = 9$ : subtraction is not associative.
$\sqrt{7} + \underline{\quad\quad} = \sqrt{7}$ . The number in the blank is
A $0$ B $\sqrt{7}$ C $1$ D $-\sqrt{7}$

Answer and solution

Answer: (A) $0$

$a + 0 = a$ : $0$ is the additive identity.
The multiplicative inverse of $-20$ is
A $\displaystyle \frac{1}{20}$ B $1$ C $20$ D $\displaystyle -\frac{1}{20}$

Answer and solution

Answer: (D) $\displaystyle -\frac{1}{20}$

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

$\displaystyle -20 \times (-\frac{1}{20}) = 1$

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MCQ test, Set 3