Question paper, Set 6

Sol: A rational number is a number of the form $\displaystyle \frac{p}{q}$.

$p$ and $q$ are integers and $q \ne 0$ ($q$ not equal to zero).

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

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

Sol: (i) $\displaystyle \sqrt{169} = 13 = \frac{13}{1}$, so it is rational.

(ii) $\sqrt{52} = 7.2111\ldots$ is a non terminating non recurring decimal, so it is irrational.

(iii) $0.\overline{2}$ is a non terminating recurring decimal, so it is rational.

(iv) $\sqrt{67} = 8.1853\ldots$ is a non terminating non recurring decimal, so it is irrational.

Answer: (i) rational (ii) irrational (iii) rational (iv) irrational

Sol: $\displaystyle \frac{24}{7}$ is in the form $\displaystyle \frac{p}{q}$ with $q \ne 0$, so it is rational.

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

$\displaystyle \frac{24}{7}$ can be written in the form $\displaystyle \frac{p}{q}$, so the statement is true.

Answer: True

Sol: $7\sqrt{14} + 10\sqrt{11} + 9\sqrt{14} - \sqrt{11}$

$= (7 + 9)\sqrt{14} + (10 - 1)\sqrt{11}$

$= 16\sqrt{14} + 9\sqrt{11}$

Answer: $16\sqrt{14} + 9\sqrt{11}$

Sol: $(-4\sqrt{11} + \sqrt{5}) - (-8\sqrt{11} - 9\sqrt{5})$

$= -4\sqrt{11} + \sqrt{5} + 8\sqrt{11} + 9\sqrt{5}$

$= (-4 + 8)\sqrt{11} + (1 + 9)\sqrt{5}$

$= 4\sqrt{11} + 10\sqrt{5}$

Answer: $4\sqrt{11} + 10\sqrt{5}$

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

$= \sqrt{42}$

Answer: $\sqrt{42}$

Sol: $9\sqrt{14} \times 2\sqrt{14} = 9 \times 2 \times \sqrt{14} \times \sqrt{14}$

$= 18 \times 14$

$= 252$

Answer: $252$

Sol: (i) Positive terms: $7952\sqrt{6}$

Negative terms: $516989\sqrt{6} + 341\sqrt{6} + 4296\sqrt{6} = 521626\sqrt{6}$

$-516989\sqrt{6} + 7952\sqrt{6} - 341\sqrt{6} - 4296\sqrt{6} = 7952\sqrt{6} - 521626\sqrt{6}$

$= -(521626 - 7952)\sqrt{6} = -513674\sqrt{6}$

(ii) Positive terms: $1042\sqrt{13} + 7546\sqrt{13} = 8588\sqrt{13}$

Negative terms: $66153\sqrt{13} + 54975\sqrt{13} = 121128\sqrt{13}$

$1042\sqrt{13} - 66153\sqrt{13} - 54975\sqrt{13} + 7546\sqrt{13} = 8588\sqrt{13} - 121128\sqrt{13}$

$= -(121128 - 8588)\sqrt{13} = -112540\sqrt{13}$

Answer: (i) $-513674\sqrt{6}$ (ii) $-112540\sqrt{13}$

Sol: (i) $6\sqrt{19} \times (-5\sqrt{19}) = 6 \times (-5) \times \sqrt{19} \times \sqrt{19}$

$= -30 \times 19 = -570$ (since $\sqrt{a} \times \sqrt{a} = a$)

(ii) $2\sqrt{5} \times (-7\sqrt{20}) = 2 \times (-7) \times \sqrt{5 \times 20}$

$= -14 \times \sqrt{100}$

$\sqrt{100} = 10$

$= -14 \times 10 = -140$

(iii) $6\sqrt{539} \times 4\sqrt{11} = 6 \times 4 \times \sqrt{539 \times 11}$

$= 24 \times \sqrt{5929}$

Long division: $7 \times 7 = 49$, remainder $10$; bring down $29$ to get $1029$; $147 \times 7 = 1029$. So $\sqrt{5929} = 77$

$= 24 \times 77 = 1848$

Answer: (i) $-570$ (ii) $-140$ (iii) $1848$

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

$\displaystyle \frac{\sqrt{50}}{\sqrt{2}} = \sqrt{\frac{50}{2}}$

$= \sqrt{25} = 5$

Answer: $5$

Sol: $\sqrt{56} = \sqrt{4 \times 14} = \sqrt{4} \times \sqrt{14} = 2\sqrt{14}$

Answer: $2\sqrt{14}$

Sol: $\sqrt{160} = \sqrt{16 \times 10} = \sqrt{16} \times \sqrt{10} = 4\sqrt{10}$

$\sqrt{384} = \sqrt{64 \times 6} = \sqrt{64} \times \sqrt{6} = 8\sqrt{6}$

$\sqrt{216} = \sqrt{36 \times 6} = \sqrt{36} \times \sqrt{6} = 6\sqrt{6}$

$\sqrt{490} = \sqrt{49 \times 10} = \sqrt{49} \times \sqrt{10} = 7\sqrt{10}$

$\therefore \sqrt{160} + \sqrt{384} + \sqrt{216} - \sqrt{490} = 4\sqrt{10} + 8\sqrt{6} + 6\sqrt{6} - 7\sqrt{10}$

$= (4 - 7)\sqrt{10} + (8 + 6)\sqrt{6}$

$= -3\sqrt{10} + 14\sqrt{6}$

Answer: $-3\sqrt{10} + 14\sqrt{6}$

Sol: (i) $\displaystyle \frac{0}{6} = 0 \in R$, so it is true.

(ii) $\displaystyle \frac{-3}{0}$ is not defined, so $\displaystyle \frac{-3}{0} \notin R$; it is false.

Answer: (i) True (ii) False

Sol: $a + b = 3\sqrt{11} + (-12\sqrt{11}) = -9\sqrt{11}$

$a - b = 3\sqrt{11} - (-12\sqrt{11}) = 15\sqrt{11}$

$a \times b = 3\sqrt{11} \times (-12\sqrt{11}) = -36 \times 11 = -396$

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

Answer: $a + b = -9\sqrt{11}$, $a - b = 15\sqrt{11}$, $a \times b = -396$; each $\in R$

Sol: Real numbers are commutative under multiplication: $a \times b = b \times a$.

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

Answer: $-4$

Sol: $a + b = 5\sqrt{11} + 4\sqrt{11} = 9\sqrt{11}$

$b + a = 4\sqrt{11} + 5\sqrt{11} = 9\sqrt{11}$

$a \times b = 5\sqrt{11} \times 4\sqrt{11} = 20 \times 11 = 220$

$b \times a = 4\sqrt{11} \times 5\sqrt{11} = 20 \times 11 = 220$

So $a + b = b + a$ and $a \times b = b \times a$.

Answer: $a + b = b + a = 9\sqrt{11}$, $a \times b = b \times a = 220$

Sol: (i) LHS $= -45 + (9 + 3) = -45 + 12 = -33$, RHS $= (-45 + 9) + 3 = -36 + 3 = -33$

LHS $=$ RHS

(ii) LHS $= -45 - (9 - 3) = -45 - 6 = -51$, RHS $= (-45 - 9) - 3 = -54 - 3 = -57$

LHS $\ne$ RHS

(iii) LHS $= -45 \times (9 \times 3) = -45 \times 27 = -1215$, RHS $= (-45 \times 9) \times 3 = -405 \times 3 = -1215$

LHS $=$ RHS

(iv) LHS $\displaystyle = (-45 \div 9) \div 3 = -5 \div 3 = -\frac{5}{3}$, RHS $= -45 \div (9 \div 3) = -45 \div 3 = -15$

LHS $\ne$ RHS

Answer: (i) $-33 = -33$ (ii) $-51 \ne -57$ (iii) $-1215 = -1215$ (iv) $\displaystyle -\frac{5}{3} \ne -15$

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

$\therefore 7\sqrt{13} + 0 = 7\sqrt{13}$

Answer: $0$

Sol: (i) Additive inverse of $a$ is $-a$, since $a + (-a) = 0$. So it is $\displaystyle -\frac{20}{11}$.

(ii) Multiplicative inverse of $a$ is $\displaystyle \frac{1}{a}$, since $\displaystyle a \times \frac{1}{a} = 1$. So it is $\displaystyle -\frac{1}{3\sqrt{19}}$.

Answer: (i) $\displaystyle -\frac{20}{11}$ (ii) $\displaystyle -\frac{1}{3\sqrt{19}}$

Sol: Additive inverse of $-23$ is $23$.

Multiplicative inverse of $-23$ is $\displaystyle -\frac{1}{23}$.

Check: $-23 + 23 = 0$

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

Answer: Additive inverse $23$, multiplicative inverse $\displaystyle -\frac{1}{23}$