Real Numbers and Their Properties: MCQ test, Set 10

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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 $8$ ?
A $7$ B $8$ C $5$ D $6$

Answer and solution

Answer: (D) $6$

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

The integers between $1$ and $8$ are $2, 3, 4, 5, 6, 7$ , that is $6$ numbers.
The integers between $-9$ and $-4$ are
A $-9, -8, -7, -6, -5, -4$ B $-8, -7, -6, -5, -4$ C $-7, -6, -5$ D $-8, -7, -6, -5$

Answer and solution

Answer: (D) $-8, -7, -6, -5$

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

So the integers between $-9$ and $-4$ are $-8, -7, -6, -5$ .
Which of the following belongs to $Q$ ?
A $\sqrt{50}$ B $\pi$ C $\displaystyle \frac{25}{11}$ D $\sqrt{12}$

Answer and solution

Answer: (C) $\displaystyle \frac{25}{11}$

$\displaystyle \frac{25}{11}$ is in the form $\displaystyle \frac{p}{q}$ with $q \ne 0$ , so it is rational.

The other numbers are non terminating non recurring decimals, so they are irrational.
$\sqrt{47} = 6.8556\ldots$ is
A a non terminating non recurring decimal, so irrational B a non terminating recurring decimal, so rational C a terminating decimal, so rational D an integer

Answer and solution

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

$\sqrt{47} = 6.8556\ldots$ is a non terminating non recurring decimal, so it is irrational.
$e$ belongs to which of these sets?
A $N$ (natural numbers) B $W$ (whole numbers) C $Q^*$ (irrational numbers) D $Z$ (integers)

Answer and solution

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

$e = 2.71828\ldots$ is a non terminating non recurring decimal, so it is irrational.

So $e$ belongs to $Q^*$ ; it does not belong to $N$ , $W$ or $Z$ .
$\sqrt{94}$ belongs to which of these sets?
A $Z$ (integers) B $Q$ (rational numbers) C $N$ (natural numbers) D $Q^*$ (irrational numbers)

Answer and solution

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

$\sqrt{94} = 9.6953\ldots$ is a non terminating non recurring decimal, so it is irrational.

So $\sqrt{94}$ belongs to $Q^*$ ; it does not belong to $Z$ , $Q$ or $N$ .
Add $-13\sqrt{3}$ and $-15\sqrt{3}$ .
A $-28\sqrt{3}$ B $-28\sqrt{6}$ C $2\sqrt{3}$ D $28\sqrt{3}$

Answer and solution

Answer: (A) $-28\sqrt{3}$

$-13\sqrt{3} - 15\sqrt{3} = (-13 - 15)\sqrt{3}$

$= -28\sqrt{3}$
Add $8\sqrt{7} - 8\sqrt{13}$ and $-10\sqrt{7} + \sqrt{13}$ .
A $18\sqrt{7} - 7\sqrt{13}$ B $-2\sqrt{7} - 9\sqrt{13}$ C $-2\sqrt{7} - 7\sqrt{13}$ D $-2\sqrt{7} + 7\sqrt{13}$

Answer and solution

Answer: (C) $-2\sqrt{7} - 7\sqrt{13}$

$8\sqrt{7} - 8\sqrt{13} - 10\sqrt{7} + \sqrt{13} = (8 - 10)\sqrt{7} + (-8 + 1)\sqrt{13}$

$= -2\sqrt{7} - 7\sqrt{13}$
Subtract $(-12 + 7\sqrt{13})$ from $(-12 - 7\sqrt{13})$ .
A $-14\sqrt{13}$ B $-24$ C $0$ D $14\sqrt{13}$

Answer and solution

Answer: (A) $-14\sqrt{13}$

$(-12 - 7\sqrt{13}) - (-12 + 7\sqrt{13}) = -12 - 7\sqrt{13} + 12 - 7\sqrt{13}$

$= -14\sqrt{13}$
Subtract $-23\sqrt{5}$ from $25\sqrt{5}$ .
A $-48\sqrt{5}$ B $-2\sqrt{5}$ C $2\sqrt{5}$ D $48\sqrt{5}$

Answer and solution

Answer: (D) $48\sqrt{5}$

$25\sqrt{5} - (-23\sqrt{5}) = 25\sqrt{5} + 23\sqrt{5}$

$= 48\sqrt{5}$
$3\sqrt{21} \times 5\sqrt{21} =$
A $15\sqrt{21}$ B $-315$ C $168$ D $315$

Answer and solution

Answer: (D) $315$

$3\sqrt{21} \times 5\sqrt{21} = 3 \times 5 \times \sqrt{21} \times \sqrt{21}$

$= 15 \times 21 = 315$ (since $\sqrt{a} \times \sqrt{a} = a$ )
$6\sqrt{2} \times (-5\sqrt{11}) =$
A $\sqrt{22}$ B $30\sqrt{22}$ C $-30\sqrt{22}$ D $-30\sqrt{13}$

Answer and solution

Answer: (C) $-30\sqrt{22}$

$6\sqrt{2} \times (-5\sqrt{11}) = 6 \times (-5) \times \sqrt{2 \times 11}$

$= -30\sqrt{22}$
$-7\sqrt{13} + 12\sqrt{13} + 24\sqrt{13} =$
A $28\sqrt{13}$ B $29\sqrt{13}$ C $43\sqrt{13}$ D $-29\sqrt{13}$

Answer and solution

Answer: (B) $29\sqrt{13}$

$-7\sqrt{13} + 12\sqrt{13} + 24\sqrt{13} = (-7 + 12 + 24)\sqrt{13}$

$= 29\sqrt{13}$
$-49147\sqrt{5} + 8715\sqrt{5} - 60911\sqrt{5} + 60073\sqrt{5} =$
A $41270\sqrt{5}$ B $-41370\sqrt{5}$ C $-41270\sqrt{5}$ D $-178846\sqrt{5}$

Answer and solution

Answer: (C) $-41270\sqrt{5}$

Positive terms: $8715\sqrt{5} + 60073\sqrt{5} = 68788\sqrt{5}$

Negative terms: $49147\sqrt{5} + 60911\sqrt{5} = 110058\sqrt{5}$

$-49147\sqrt{5} + 8715\sqrt{5} - 60911\sqrt{5} + 60073\sqrt{5} = 68788\sqrt{5} - 110058\sqrt{5}$

$= -(110058 - 68788)\sqrt{5} = -41270\sqrt{5}$
$-7\sqrt{7} \times (-2\sqrt{63}) =$
A $6174$ B $294$ C $-294$ D $-189$

Answer and solution

Answer: (B) $294$

$-7\sqrt{7} \times (-2\sqrt{63}) = -7 \times (-2) \times \sqrt{7 \times 63}$

$= 14 \times \sqrt{441}$

Long division: $2 \times 2 = 4$ , remainder $0$ ; bring down $41$ to get $41$ ; $41 \times 1 = 41$ . So $\sqrt{441} = 21$

$= 14 \times 21 = 294$
$-4\sqrt{28} \times (-5\sqrt{7}) =$
A $-126$ B $-280$ C $280$ D $3920$

Answer and solution

Answer: (C) $280$

$-4\sqrt{28} \times (-5\sqrt{7}) = -4 \times (-5) \times \sqrt{28 \times 7}$

$= 20 \times \sqrt{196}$

$\sqrt{196} = 14$

$= 20 \times 14 = 280$
$\displaystyle \frac{\sqrt{720}}{6\sqrt{5}} =$
A $2$ B $6$ C $12$ D $24$

Answer and solution

Answer: (A) $2$

$\sqrt{720} = \sqrt{144} \times \sqrt{5} = 12\sqrt{5}$

$\displaystyle \frac{12\sqrt{5}}{6\sqrt{5}} = 2$
$\displaystyle \frac{5 \times \sqrt{169}}{\sqrt{13}} =$
A $5\sqrt{13}$ B $\sqrt{169}$ C $\sqrt{13}$ D $65$

Answer and solution

Answer: (A) $5\sqrt{13}$

$\displaystyle \frac{\sqrt{169}}{\sqrt{13}} = \sqrt{\frac{169}{13}} = \sqrt{13}$

$\therefore 5 \times \sqrt{13} = 5\sqrt{13}$
$\sqrt{112} =$
A $16\sqrt{7}$ B $4\sqrt{7}$ C $4\sqrt{28}$ D $28$

Answer and solution

Answer: (B) $4\sqrt{7}$

$\sqrt{112} = \sqrt{16 \times 7} = \sqrt{16} \times \sqrt{7} = 4\sqrt{7}$
$\sqrt{45} - \sqrt{180} =$
A $-27\sqrt{5}$ B $3\sqrt{5}$ C $-3\sqrt{10}$ D $-3\sqrt{5}$

Answer and solution

Answer: (D) $-3\sqrt{5}$

$\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$

$\sqrt{180} = \sqrt{36 \times 5} = \sqrt{36} \times \sqrt{5} = 6\sqrt{5}$

$3\sqrt{5} - 6\sqrt{5} = -3\sqrt{5}$
If $a = 7\sqrt{10}$ and $b = -8\sqrt{10}$ , then $a + b$ is
A $-1$ B $-\sqrt{10}$ C $-\sqrt{20}$ D $15\sqrt{10}$

Answer and solution

Answer: (B) $-\sqrt{10}$

$a + b = -\sqrt{10} \in R$
If $66 \div 11 = 6$ , then $11 \div 66 =$
A $-6$ B $\displaystyle \frac{1}{6}$ C $\displaystyle -\frac{1}{6}$ D $6$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{6}$

$\displaystyle 11 \div 66 = \frac{11}{66} = \frac{1}{6}$

Division is not commutative.
$32 - (38 - 4) =$
A $-2$ B $66$ C $74$ D $-10$

Answer and solution

Answer: (A) $-2$

$32 - (38 - 4) = 32 - 34 = -2$

Note $(32 - 38) - 4 = -10$ : subtraction is not associative.
$26 + \underline{\quad\quad} = 26$ . The number in the blank is
A $1$ B $0$ C $26$ D $-26$

Answer and solution

Answer: (B) $0$

$a + 0 = a$ : $0$ is the additive identity.
The additive inverse of $-6$ is
A $0$ B $\displaystyle -\frac{1}{6}$ C $\displaystyle \frac{1}{6}$ D $6$

Answer and solution

Answer: (D) $6$

Additive inverse of $a$ is $-a$ , since $a + (-a) = 0$ .

$-6 + 6 = 0$

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MCQ test, Set 10