Question paper, Set 9

A constant has a fixed value. A variable can take various numerical values.

$q$, $z$ and $p$ can take various values, so they are variables.

$-24$ has a fixed value, so it is the constant.

Answer: $-24$

Variables can take various numerical values: $y$, $c$ and $t$.

Constants have a fixed value: $\displaystyle \frac{7}{8}$, $\displaystyle -\frac{1}{7}$ and $\displaystyle -\frac{1}{3}$.

Answer: (i) $y$, $c$, $t$ (ii) $\displaystyle \frac{7}{8}$, $\displaystyle -\frac{1}{7}$, $\displaystyle -\frac{1}{3}$

Multiply: $19 \times m = 19m$

Answer: $19m$

(i) $z$ divided by $8$ is $\displaystyle \frac{z}{8}$

(ii) Take $4z$ away from $13$. $13 - 4z$

Answer: (i) $\displaystyle \frac{z}{8}$, (ii) $13 - 4z$

(i) $43$ more than $p$: add $43$ to $p$. $p + 43$

(ii) Take $60$ away from $m$. $m - 60$

(iii) $p$ divided by $7$ is $\displaystyle \frac{p}{7}$

(iv) $16$ divided by $q$ is $\displaystyle \frac{16}{q}$

(v) $10$ times $t$ is $10t$; $11$ times $a$ is $11a$. Add $10t$ to $11a$: $11a + 10t$

Answer: (i) $p + 43$, (ii) $m - 60$, (iii) $\displaystyle \frac{p}{7}$, (iv) $\displaystyle \frac{16}{q}$, (v) $11a + 10t$

(i) One third of $c$ is $\displaystyle \frac{c}{3}$. $8$ more than it: add $8$. $\displaystyle \frac{c}{3} + 8$

(ii) One third of $m$ is $\displaystyle \frac{m}{3}$. $13$ less than it: take away $13$. $\displaystyle \frac{m}{3} - 13$

Answer: (i) $\displaystyle \frac{c}{3} + 8$, (ii) $\displaystyle \frac{m}{3} - 13$

Rule: $1 \times a = a$

$1 \times xyz = xyz$

Answer: $xyz$

(i) Rule $\displaystyle \frac{a}{-1} = -a$, so $\displaystyle \frac{pqt}{-1} = -pqt$

(ii) Rule $0 \times a = 0$, so $0 \times bc = 0$

(iii) Rule $\displaystyle \frac{a}{1} = a$, so $\displaystyle \frac{p}{1} = p$

Answer: (i) $-pqt$, (ii) $0$, (iii) $p$

The terms are separated by $+$ and $-$ signs; each term keeps the sign in front of it.

The terms are $3c$, $-19$, $12qr$ and $4yz$.

Answer: $3c$, $-19$, $12qr$, $4yz$

$24ac = 24 \times a \times c$

Each numerical (or) literal value in the product is a factor: $24, a, c$.

Answer: $24ac = 24 \times a \times c$; factors $24, a, c$

(i) $-19mnq = -19 \times mnq$, so the coefficient of $mnq$ is $-19$.

(ii) $-19mnq = -19mn \times q$, so the coefficient of $q$ is $-19mn$.

(iii) $-19mnq = -19q \times mn$, so the coefficient of $mn$ is $-19q$.

Answer: (i) $-19$, (ii) $-19mn$, (iii) $-19q$

$\displaystyle \frac{xz}{12} = \frac{1}{12} \times xz$

So the coefficient of $xz$ is $\displaystyle \frac{1}{12}$.

Answer: $\displaystyle \frac{1}{12}$

(i) $15$ has no literal factor, $15pqt$ has literal factors $pqt$: different, so unlike terms.

(ii) $\displaystyle -\frac{9}{4}pq$, $\displaystyle -\frac{3}{2}qp$: the same literal factors $pq$ ($qp = pq$), so like terms.

Answer: (i) Unlike, (ii) Like

(i) $10n$, $-6n$: the same literal factors $n$, so like terms.

(ii) $-5n$ has literal factors $n$, $-5$ has no literal factor: different, so unlike terms.

(iii) $7pr$, $\displaystyle -\frac{3}{2}pr$: the same literal factors $pr$, so like terms.

(iv) $\displaystyle -\frac{12}{5}yz$, $15yz$: the same literal factors $yz$, so like terms.

(v) $-15bq$ has literal factors $bq$, $-9b$ has literal factors $b$: different, so unlike terms.

Answer: (i) Like, (ii) Unlike, (iii) Like, (iv) Like, (v) Unlike

$19ab + 5ab = (19 + 5)ab$

$= 24ab$

Answer: $24ab$

$7y + 23y + 14y = (7 + 23 + 14)y$

$= 44y$

Answer: $44y$

Bring the like terms together and add.

$14b + 8t + 7y + 9t + 12y + 14b = (14b + 14b) + (8t + 9t) + (7y + 12y)$

$= (14 + 14)b + (8 + 9)t + (7 + 12)y$

$= 28b + 17t + 19y$

Answer: $28b + 17t + 19y$

(i) $-13n + 11n = (-13 + 11)n = -2n$

(ii) $25n - 14n = (25 - 14)n = 11n$

(iii) $-27n - 26n = (-27 - 26)n = -53n$

Answer: (i) $-2n$, (ii) $11n$, (iii) $-53n$

(i) $26xy + 5xy = (26 + 5)xy = 31xy$

(ii) $-26xy - 5xy = (-26 - 5)xy = -31xy$

(iii) $-26xy + 5xy = (-26 + 5)xy = -21xy$

(iv) $26xy - 5xy = (26 - 5)xy = 21xy$

(v) $-25xy + 25xy = (-25 + 25)xy = 0 \times xy = 0$

Answer: (i) $31xy$, (ii) $-31xy$, (iii) $-21xy$, (iv) $21xy$, (v) $0$

(i) $6mp - 20y + 7y + 3mp = (6mp + 3mp) + (-20y + 7y)$

$= (6 + 3)mp + (-20 + 7)y$

$= 9mp - 13y$

(ii) $14xy - 16ac - 13 - 10xy + 13ac - 8 = (14xy - 10xy) + (-16ac + 13ac) + (-13 - 8)$

$= (14 - 10)xy + (-16 + 13)ac + (-13 - 8)$

$= 4xy - 3ac - 21$

Answer: (i) $9mp - 13y$, (ii) $4xy - 3ac - 21$