Question paper, Set 1

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

The integers between $-9$ and $-2$ are $-8, -7, -6, -5, -4, -3$.

Answer: $-8, -7, -6, -5, -4, -3$

Sol: (i) $1.\overline{6}$ is a non terminating recurring decimal, so it is rational.

(ii) $\sqrt{87} = 9.3273\ldots$ is a non terminating non recurring decimal, so it is irrational.

Answer: (i) Rational (ii) Irrational

Sol: $-\sqrt{100} = -10$

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

$-\sqrt{100}$ can be written in the form $\displaystyle \frac{p}{q}$, so the statement is true.

Answer: True

Sol: $-9 - 12\sqrt{21} + 1 + 3\sqrt{21}$

$= (-9 + 1) + (-12 + 3)\sqrt{21}$

$= -8 - 9\sqrt{21}$

Answer: $-8 - 9\sqrt{21}$

Sol: $(12 + 11\sqrt{14}) - (13 + 4\sqrt{14})$

$= 12 + 11\sqrt{14} - 13 - 4\sqrt{14}$

$= (12 - 13) + (11 - 4)\sqrt{14}$

$= -1 + 7\sqrt{14}$

Answer: $-1 + 7\sqrt{14}$

Sol: $\sqrt{11} \times \sqrt{13} = \sqrt{11 \times 13}$

$= \sqrt{143}$

Answer: $\sqrt{143}$

Sol: $11\sqrt{13} \times 6\sqrt{7} = 11 \times 6 \times \sqrt{13} \times \sqrt{7}$

$= 66 \times \sqrt{13 \times 7}$

$= 66\sqrt{91}$

Answer: $66\sqrt{91}$

Sol: (i) Positive terms: $300810\sqrt{6} + 40860\sqrt{6} = 341670\sqrt{6}$

Negative terms: $870\sqrt{6} + 7264\sqrt{6} = 8134\sqrt{6}$

$-870\sqrt{6} + 300810\sqrt{6} - 7264\sqrt{6} + 40860\sqrt{6} = 341670\sqrt{6} - 8134\sqrt{6}$

$= (341670 - 8134)\sqrt{6} = 333536\sqrt{6}$

(ii) Positive terms: $105\sqrt{14} + 800\sqrt{14} = 905\sqrt{14}$

Negative terms: $7640\sqrt{14}$

$105\sqrt{14} - 7640\sqrt{14} + 800\sqrt{14} = 905\sqrt{14} - 7640\sqrt{14}$

$= -(7640 - 905)\sqrt{14} = -6735\sqrt{14}$

Answer: (i) $333536\sqrt{6}$ (ii) $-6735\sqrt{14}$

Sol: (i) $9\sqrt{2} \times 8\sqrt{2} = 9 \times 8 \times \sqrt{2} \times \sqrt{2}$

$= 72 \times 2 = 144$ (since $\sqrt{a} \times \sqrt{a} = a$)

(ii) $7\sqrt{108} \times (-8\sqrt{3}) = 7 \times (-8) \times \sqrt{108 \times 3}$

$= -56 \times \sqrt{324}$

$\sqrt{324} = 18$

$= -56 \times 18 = -1008$

(iii) $-4\sqrt{12} \times 2\sqrt{192} = -4 \times 2 \times \sqrt{12 \times 192}$

$= -8 \times \sqrt{2304}$

Long division: $4 \times 4 = 16$, remainder $7$; bring down $04$ to get $704$; $88 \times 8 = 704$. So $\sqrt{2304} = 48$

$= -8 \times 48 = -384$

Answer: (i) $144$ (ii) $-1008$ (iii) $-384$

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

$\displaystyle \frac{\sqrt{180}}{\sqrt{5}} = \sqrt{\frac{180}{5}}$

$= \sqrt{36} = 6$

Answer: $6$

Sol: $\sqrt{320} = \sqrt{64 \times 5} = \sqrt{64} \times \sqrt{5} = 8\sqrt{5}$

Answer: $8\sqrt{5}$

Sol: $\sqrt{360} = \sqrt{36 \times 10} = \sqrt{36} \times \sqrt{10} = 6\sqrt{10}$

$\sqrt{160} = \sqrt{16 \times 10} = \sqrt{16} \times \sqrt{10} = 4\sqrt{10}$, so $2\sqrt{160} = 2 \times 4\sqrt{10} = 8\sqrt{10}$

$\sqrt{162} = \sqrt{81 \times 2} = \sqrt{81} \times \sqrt{2} = 9\sqrt{2}$, so $3\sqrt{162} = 3 \times 9\sqrt{2} = 27\sqrt{2}$

$\sqrt{128} = \sqrt{64 \times 2} = \sqrt{64} \times \sqrt{2} = 8\sqrt{2}$, so $2\sqrt{128} = 2 \times 8\sqrt{2} = 16\sqrt{2}$

$\therefore \sqrt{360} - 2\sqrt{160} + 3\sqrt{162} + 2\sqrt{128} = 6\sqrt{10} - 8\sqrt{10} + 27\sqrt{2} + 16\sqrt{2}$

$= (6 - 8)\sqrt{10} + (27 + 16)\sqrt{2}$

$= -2\sqrt{10} + 43\sqrt{2}$

Answer: $-2\sqrt{10} + 43\sqrt{2}$

Sol: $a + b = 10 + 22 = 32$

$a - b = 10 - 22 = -12$

$a \times b = 10 \times 22 = 220$

All three are real numbers: real numbers are closed under addition, subtraction and multiplication.

Answer: $a + b = 32$, $a - b = -12$, $a \times b = 220$; all belong to $R$

Sol: $a + b = -3\sqrt{21} + 8\sqrt{21} = 5\sqrt{21}$

$a - b = -3\sqrt{21} - 8\sqrt{21} = -11\sqrt{21}$

$a \times b = -3\sqrt{21} \times 8\sqrt{21} = -24 \times 21 = -504$

All three are real numbers, so $a + b, a - b, a \times b \in R$.

Answer: $a + b = 5\sqrt{21}$, $a - b = -11\sqrt{21}$, $a \times b = -504$; each $\in R$

Sol: Real numbers are commutative under addition: $a + b = b + a$.

$\displaystyle \therefore -3\sqrt{10} + \frac{3}{7} = \frac{3}{7} + (-3\sqrt{10})$

Answer: $-3\sqrt{10}$

Sol: $a - b = 12 - 4 = 8$

$b - a = 4 - 12 = -8$

$\therefore a - b \ne b - a$

$a \div b = 12 \div 4 = 3$

$\displaystyle b \div a = 4 \div 12 = \frac{1}{3}$

$\therefore a \div b \ne b \div a$

Answer: $a - b = 8 \ne b - a = -8$; $\displaystyle a \div b = 3 \ne b \div a = \frac{1}{3}$

Sol: (i) LHS $= 36 + (12 + 3) = 36 + 15 = 51$, RHS $= (36 + 12) + 3 = 48 + 3 = 51$

LHS $=$ RHS

(ii) LHS $= 36 - (12 - 3) = 36 - 9 = 27$, RHS $= (36 - 12) - 3 = 24 - 3 = 21$

LHS $\ne$ RHS

(iii) LHS $= 36 \times (12 \times 3) = 36 \times 36 = 1296$, RHS $= (36 \times 12) \times 3 = 432 \times 3 = 1296$

LHS $=$ RHS

(iv) LHS $= (36 \div 12) \div 3 = 3 \div 3 = 1$, RHS $= 36 \div (12 \div 3) = 36 \div 4 = 9$

LHS $\ne$ RHS

Answer: (i) $51 = 51$ (ii) $27 \ne 21$ (iii) $1296 = 1296$ (iv) $1 \ne 9$

Sol: $a + 0 = a$, where $0$ is the additive identity.

$\therefore 9\sqrt{6} + 0 = 9\sqrt{6}$

Answer: $0$

Sol: (i) Additive inverse of $a$ is $-a$, since $a + (-a) = 0$. So it is $7\sqrt{6}$.

(ii) Multiplicative inverse of $a$ is $\displaystyle \frac{1}{a}$, since $\displaystyle a \times \frac{1}{a} = 1$. So it is $\displaystyle -\frac{11}{16}$.

Answer: (i) $7\sqrt{6}$ (ii) $\displaystyle -\frac{11}{16}$

Sol: Additive inverse of $\displaystyle -\frac{23}{11}$ is $\displaystyle \frac{23}{11}$.

Multiplicative inverse of $\displaystyle -\frac{23}{11}$ is $\displaystyle -\frac{11}{23}$.

Check: $\displaystyle -\frac{23}{11} + \frac{23}{11} = 0$

Check: $\displaystyle -\frac{23}{11} \times (-\frac{11}{23}) = 1$

Answer: Additive inverse $\displaystyle \frac{23}{11}$, multiplicative inverse $\displaystyle -\frac{11}{23}$