MCQ test, Set 4

Answer: (D) $5$

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

The integers between $-12$ and $-6$ are $-11, -10, -9, -8, -7$, that is $5$ numbers.

Answer: (A) $1, 2, 3, 4$

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

So the natural numbers between $-1$ and $5$ are $1, 2, 3, 4$.

Answer: (C) $\sqrt{169}$

$\displaystyle \sqrt{169} = 13 = \frac{13}{1}$, so it is rational.

The other numbers are non terminating non recurring decimals, so they are irrational.

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

$\sqrt{76} = 8.7177\ldots$ is a non terminating non recurring decimal, so it is irrational.

Answer: (C) $Z$ (integers)

So $-28$ belongs to $Z$; it does not belong to $N$, $Q^*$ or $W$.

Answer: (B) $3.\overline{2} \in Q$

$3.\overline{2}$ is a non terminating recurring decimal, so it is rational.

$\displaystyle Q = \{\frac{p}{q}$ where $p, q$ are integers and $q \ne 0\}$

$3.\overline{2}$ can be written in the form $\displaystyle \frac{p}{q}$, so the statement is true.

The other three statements are false.

Answer: (B) $-42\sqrt{10}$

$-18\sqrt{10} - 24\sqrt{10} = (-18 - 24)\sqrt{10}$

$= -42\sqrt{10}$

Answer: (C) $14\sqrt{10} + 2\sqrt{15}$

$4\sqrt{10} - \sqrt{15} + 10\sqrt{10} + 3\sqrt{15} = (4 + 10)\sqrt{10} + (-1 + 3)\sqrt{15}$

$= 14\sqrt{10} + 2\sqrt{15}$

Answer: (D) $2\sqrt{14}$

$(11 + \sqrt{14}) - (11 - \sqrt{14}) = 11 + \sqrt{14} - 11 + \sqrt{14}$

$= 2\sqrt{14}$

Answer: (A) $-14\sqrt{15}$

$-2\sqrt{15} - (12\sqrt{15}) = -2\sqrt{15} - 12\sqrt{15}$

$= -14\sqrt{15}$

Answer: (C) $-120$

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

$= -12 \times 10 = -120$ (since $\sqrt{a} \times \sqrt{a} = a$)

Answer: (D) $-16\sqrt{39}$

$2\sqrt{13} \times (-8\sqrt{3}) = 2 \times (-8) \times \sqrt{13 \times 3}$

$= -16\sqrt{39}$

Answer: (D) $15\sqrt{10}$

$26\sqrt{10} - 8\sqrt{10} - 3\sqrt{10} = (26 - 8 - 3)\sqrt{10}$

$= 15\sqrt{10}$

Answer: (C) $73756\sqrt{14}$

Positive terms: $95809\sqrt{14} + 10639\sqrt{14} = 106448\sqrt{14}$

Negative terms: $8377\sqrt{14} + 24315\sqrt{14} = 32692\sqrt{14}$

$-8377\sqrt{14} + 95809\sqrt{14} - 24315\sqrt{14} + 10639\sqrt{14} = 106448\sqrt{14} - 32692\sqrt{14}$

$= (106448 - 32692)\sqrt{14} = 73756\sqrt{14}$

Answer: (B) $864$

$-6\sqrt{108} \times (-4\sqrt{12}) = -6 \times (-4) \times \sqrt{108 \times 12}$

$= 24 \times \sqrt{1296}$

Long division: $3 \times 3 = 9$, remainder $3$; bring down $96$ to get $396$; $66 \times 6 = 396$. So $\sqrt{1296} = 36$

$= 24 \times 36 = 864$

Answer: (B) $1764$

$7\sqrt{216} \times 7\sqrt{6} = 7 \times 7 \times \sqrt{216 \times 6}$

$= 49 \times \sqrt{1296}$

Long division: $3 \times 3 = 9$, remainder $3$; bring down $96$ to get $396$; $66 \times 6 = 396$. So $\sqrt{1296} = 36$

$= 49 \times 36 = 1764$

Answer: (A) $2$

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

$\displaystyle \frac{6\sqrt{3}}{3\sqrt{3}} = 2$

Answer: (D) $3\sqrt{19}$

$\displaystyle \frac{\sqrt{361}}{\sqrt{19}} = \sqrt{\frac{361}{19}} = \sqrt{19}$

$\therefore 3 \times \sqrt{19} = 3\sqrt{19}$

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

$\sqrt{325} = \sqrt{25 \times 13} = \sqrt{25} \times \sqrt{13} = 5\sqrt{13}$

Answer: (A) $10\sqrt{7}$

$\sqrt{252} = \sqrt{36 \times 7} = \sqrt{36} \times \sqrt{7} = 6\sqrt{7}$

$\sqrt{112} = \sqrt{16 \times 7} = \sqrt{16} \times \sqrt{7} = 4\sqrt{7}$

$6\sqrt{7} + 4\sqrt{7} = 10\sqrt{7}$

Answer: (B) division

Example: $7, 0 \in R$ but $\displaystyle \frac{7}{0}$ is not defined, so $\displaystyle \frac{7}{0} \notin R$.

Sum, difference and product of two real numbers are always real numbers.

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

$a - b = -19 - (-7) = -12$, $b - a = -7 - (-19) = 12$

So $a - b \ne b - a$.

Answer: (C) associative property under multiplication

The numbers stay in the same order; only the grouping (brackets) changes.

$a \times (b \times c) = (a \times b) \times c$

Answer: (B) $0$

$a + 0 = 0 + a = a$, so $0$ is the additive identity.

Answer: (B) $-29$

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

$29 + (-29) = 0$