Question paper, Set 3

Sol: Whole numbers are $0$ and all the natural numbers.

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

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

Sol: (i) $8.\overline{4}$ is a non terminating recurring decimal, so it is rational.

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

Answer: (i) Rational (ii) Irrational

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

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

$1.\overline{1}$ can be written in the form $\displaystyle \frac{p}{q}$, so the statement is true.

Answer: True

Sol: $2 - 6\sqrt{21} + 15 - 4\sqrt{21}$

$= (2 + 15) + (-6 - 4)\sqrt{21}$

$= 17 - 10\sqrt{21}$

Answer: $17 - 10\sqrt{21}$

Sol: $(4 + 11\sqrt{10}) - (-15 - 6\sqrt{10})$

$= 4 + 11\sqrt{10} + 15 + 6\sqrt{10}$

$= (4 + 15) + (11 + 6)\sqrt{10}$

$= 19 + 17\sqrt{10}$

Answer: $19 + 17\sqrt{10}$

Sol: $\sqrt{5} \times \sqrt{22} = \sqrt{5 \times 22}$

$= \sqrt{110}$

Answer: $\sqrt{110}$

Sol: $11\sqrt{6} \times 8\sqrt{7} = 11 \times 8 \times \sqrt{6} \times \sqrt{7}$

$= 88 \times \sqrt{6 \times 7}$

$= 88\sqrt{42}$

Answer: $88\sqrt{42}$

Sol: (i) Positive terms: $23278\sqrt{13} + 526637\sqrt{13} = 549915\sqrt{13}$

Negative terms: $1453\sqrt{13} + 4004\sqrt{13} = 5457\sqrt{13}$

$-1453\sqrt{13} - 4004\sqrt{13} + 23278\sqrt{13} + 526637\sqrt{13} = 549915\sqrt{13} - 5457\sqrt{13}$

$= (549915 - 5457)\sqrt{13} = 544458\sqrt{13}$

(ii) Positive terms: $8404\sqrt{11}$

Negative terms: $553\sqrt{11} + 902\sqrt{11} + 8531\sqrt{11} = 9986\sqrt{11}$

$-553\sqrt{11} - 902\sqrt{11} - 8531\sqrt{11} + 8404\sqrt{11} = 8404\sqrt{11} - 9986\sqrt{11}$

$= -(9986 - 8404)\sqrt{11} = -1582\sqrt{11}$

Answer: (i) $544458\sqrt{13}$ (ii) $-1582\sqrt{11}$

Sol: (i) $6\sqrt{2} \times (-6\sqrt{2}) = 6 \times (-6) \times \sqrt{2} \times \sqrt{2}$

$= -36 \times 2 = -72$ (since $\sqrt{a} \times \sqrt{a} = a$)

(ii) $4\sqrt{98} \times (-5\sqrt{2}) = 4 \times (-5) \times \sqrt{98 \times 2}$

$= -20 \times \sqrt{196}$

$\sqrt{196} = 14$

$= -20 \times 14 = -280$

(iii) $-2\sqrt{5} \times 4\sqrt{245} = -2 \times 4 \times \sqrt{5 \times 245}$

$= -8 \times \sqrt{1225}$

Long division: $3 \times 3 = 9$, remainder $3$; bring down $25$ to get $325$; $65 \times 5 = 325$. So $\sqrt{1225} = 35$

$= -8 \times 35 = -280$

Answer: (i) $-72$ (ii) $-280$ (iii) $-280$

Sol: $\displaystyle \frac{7 \times \sqrt{25}}{\sqrt{5}} = 7 \times \frac{\sqrt{25}}{\sqrt{5}}$

$\displaystyle = 7 \times \sqrt{\frac{25}{5}} = 7 \times \sqrt{5}$

$= 7\sqrt{5}$

Answer: $7\sqrt{5}$

Sol: $\sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \times \sqrt{6} = 4\sqrt{6}$

Answer: $4\sqrt{6}$

Sol: $\sqrt{243} = \sqrt{81 \times 3} = \sqrt{81} \times \sqrt{3} = 9\sqrt{3}$, so $2\sqrt{243} = 2 \times 9\sqrt{3} = 18\sqrt{3}$

$\sqrt{108} = \sqrt{36 \times 3} = \sqrt{36} \times \sqrt{3} = 6\sqrt{3}$, so $3\sqrt{108} = 3 \times 6\sqrt{3} = 18\sqrt{3}$

$\sqrt{147} = \sqrt{49 \times 3} = \sqrt{49} \times \sqrt{3} = 7\sqrt{3}$

$\sqrt{192} = \sqrt{64 \times 3} = \sqrt{64} \times \sqrt{3} = 8\sqrt{3}$

$\therefore 2\sqrt{243} + 3\sqrt{108} - \sqrt{147} - \sqrt{192} = 18\sqrt{3} + 18\sqrt{3} - 7\sqrt{3} - 8\sqrt{3}$

$= 21\sqrt{3}$

Answer: $21\sqrt{3}$

Sol: (i) $\displaystyle \frac{0}{18} = 0 \in R$, so it is true.

(ii) $\displaystyle \frac{-39}{0}$ is not defined, so $\displaystyle \frac{-39}{0} \notin R$; it is false.

Answer: (i) True (ii) False

Sol: $a + b = 2\sqrt{14} + (-8\sqrt{14}) = -6\sqrt{14}$

$a - b = 2\sqrt{14} - (-8\sqrt{14}) = 10\sqrt{14}$

$a \times b = 2\sqrt{14} \times (-8\sqrt{14}) = -16 \times 14 = -224$

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

Answer: $a + b = -6\sqrt{14}$, $a - b = 10\sqrt{14}$, $a \times b = -224$; each $\in R$

Sol: The order of the two numbers is changed and the answer is the same.

This is the commutative property under multiplication.

Answer: Commutative property under multiplication

Sol: $a - b = 25 - 5 = 20$

$b - a = 5 - 25 = -20$

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

$a \div b = 25 \div 5 = 5$

$\displaystyle b \div a = 5 \div 25 = \frac{1}{5}$

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

Answer: $a - b = 20 \ne b - a = -20$; $\displaystyle a \div b = 5 \ne b \div a = \frac{1}{5}$

Sol: (i) LHS $= 60 + (10 + 5) = 60 + 15 = 75$, RHS $= (60 + 10) + 5 = 70 + 5 = 75$

LHS $=$ RHS

(ii) LHS $= 60 - (10 - 5) = 60 - 5 = 55$, RHS $= (60 - 10) - 5 = 50 - 5 = 45$

LHS $\ne$ RHS

(iii) LHS $= 60 \times (10 \times 5) = 60 \times 50 = 3000$, RHS $= (60 \times 10) \times 5 = 600 \times 5 = 3000$

LHS $=$ RHS

(iv) LHS $\displaystyle = (60 \div 10) \div 5 = 6 \div 5 = \frac{6}{5}$, RHS $= 60 \div (10 \div 5) = 60 \div 2 = 30$

LHS $\ne$ RHS

Answer: (i) $75 = 75$ (ii) $55 \ne 45$ (iii) $3000 = 3000$ (iv) $\displaystyle \frac{6}{5} \ne 30$

Sol: $1 \times a = a$, where $1$ is the multiplicative identity.

$\therefore 1 \times 4 = 4$

Answer: $1$

Sol: (i) Additive inverse of $a$ is $-a$, since $a + (-a) = 0$. So it is $-\sqrt{22}$.

(ii) Multiplicative inverse of $a$ is $\displaystyle \frac{1}{a}$, since $\displaystyle a \times \frac{1}{a} = 1$. So it is $\displaystyle -\frac{1}{4\sqrt{19}}$.

Answer: (i) $-\sqrt{22}$ (ii) $\displaystyle -\frac{1}{4\sqrt{19}}$

Sol: Additive inverse of $\displaystyle -\frac{12}{5}$ is $\displaystyle \frac{12}{5}$.

Multiplicative inverse of $\displaystyle -\frac{12}{5}$ is $\displaystyle -\frac{5}{12}$.

Check: $\displaystyle -\frac{12}{5} + \frac{12}{5} = 0$

Check: $\displaystyle -\frac{12}{5} \times (-\frac{5}{12}) = 1$

Answer: Additive inverse $\displaystyle \frac{12}{5}$, multiplicative inverse $\displaystyle -\frac{5}{12}$