Algebraic Expressions and Their Uses: MCQ test, Set 3

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

A symbol which has a fixed value is called a
A variable B term C co-efficient D constant

Answer and solution

Answer: (D) constant

A symbol with a fixed value is a constant, e.g. $15$ , $-9$ , $0$ .
" $10$ less than half of $b$ " in symbolic form is
A $\displaystyle \frac{b - 10}{2}$ B $\displaystyle \frac{b}{2} - 10$ C $\displaystyle 10 - \frac{b}{2}$ D $2b - 10$

Answer and solution

Answer: (B) $\displaystyle \frac{b}{2} - 10$

Half of $b$ is $\displaystyle \frac{b}{2}$ ; $10$ less means subtract $10$ .

So it is $\displaystyle \frac{b}{2} - 10$ .
The number of terms in $\displaystyle 2m^2n + \frac{7n}{6} - 5m^2 + 9m + 1$ is
A $6$ B $5$ C $7$ D $4$

Answer and solution

Answer: (B) $5$

Terms are separated by $+$ and $-$ signs.

The terms are $2m^2n$ , $\displaystyle \frac{7n}{6}$ , $-5m^2$ , $9m$ , $1$ .

So there are $5$ terms.
$10mt^3$ written as a product of its factors is
A $10 \times m \times t \times t \times t$ B $10 + m + t + t + t$ C $10 \times m \times t \times 3$ D $10 \times m \times t$

Answer and solution

Answer: (A) $10 \times m \times t \times t \times t$

$t^3$ means $t$ multiplied $3$ times.

$10mt^3 = 10 \times m \times t \times t \times t$
The co-efficient of $xy^3$ in $\displaystyle -\frac{5}{4}xy^3z^2$ is
A $\displaystyle -\frac{5}{4}$ B $\displaystyle -\frac{5}{4}z^2$ C $z^2$ D $\displaystyle \frac{5}{4}z^2$

Answer and solution

Answer: (B) $\displaystyle -\frac{5}{4}z^2$

$\displaystyle -\frac{5}{4}xy^3z^2 = \left(-\frac{5}{4}z^2\right) \times xy^3$

So the co-efficient of $xy^3$ is $\displaystyle -\frac{5}{4}z^2$ .
Which of the following is a like term of $5yz^3$ ?
A $5y$ B $5rz^3$ C $9z^3y$ D $5y^3z$

Answer and solution

Answer: (C) $9z^3y$

Like terms have the same literal factors.

$9z^3y$ has the literal factors $z^3y = yz^3$ , the same as $5yz^3$ .
$\displaystyle \frac{p}{5} + \frac{p}{4} =$
A $\displaystyle \frac{9p}{20}$ B $\displaystyle \frac{9p^2}{20}$ C $\displaystyle \frac{2p}{9}$ D $\displaystyle \frac{p}{20}$

Answer and solution

Answer: (A) $\displaystyle \frac{9p}{20}$

$\displaystyle \frac{p}{5} + \frac{p}{4}$

$\displaystyle = (\frac{1}{5} + \frac{1}{4})p$

For $p$ : $\displaystyle \frac{1}{5} + \frac{1}{4} = \frac{4 + 5}{20} = \frac{9}{20}$

$\displaystyle = \frac{9p}{20}$
$5t - 6t =$
A $t$ B $-t$ C $-11t$ D $11t$

Answer and solution

Answer: (B) $-t$

$5t - 6t$

$= (5 - 6)t$

$= -t$
Which of the following is a binomial?
A $2y^2 + 3x^2y$ B $\displaystyle -\frac{5}{3}x^3$ C $6y^2 + 8x^2y^2 + 7y + 8x^2$ D $7x^2y + 6x^2y^2 - 3y$

Answer and solution

Answer: (A) $2y^2 + 3x^2y$

A binomial has two monomials.

$2y^2 + 3x^2y$ has two terms.
Which of the following statements is true?
A Every polynomial is a multinomial. B $\displaystyle z + \frac{9}{z}$ is a polynomial. C Every multinomial is a polynomial. D $4\sqrt{z}$ is a monomial.

Answer and solution

Answer: (A) Every polynomial is a multinomial.

A polynomial has one or more monomials; a multinomial has one or more terms.

So all polynomials are multinomials, but not the other way round.
The degree of $12a^4b^3m^2$ is
A $12$ B $3$ C $9$ D $4$

Answer and solution

Answer: (C) $9$

Add the powers of the variables:

$4 + 3 + 2 = 9$
The degree of $-5mn^4 - 6 - 7m^4n^3 + 9mn^3$ is
A $5$ B $7$ C $16$ D $4$

Answer and solution

Answer: (B) $7$

Degrees of the terms are $5, 0, 7, 4$ .

The greatest is $7$ .
The value of $q^2 - 2qr + r^2$ when $q = 2$ and $r = -1$ is
A $7$ B $6$ C $-9$ D $9$

Answer and solution

Answer: (D) $9$

Put $q = 2$ and $r = -1$ :

$(2)^2 - 2(2)(-1) + (-1)^2$

$= 4 + 4 + 1$

$= 9$
The value of $p^3 + p$ when $p = -1$ is
A $-4$ B $-2$ C $2$ D $0$

Answer and solution

Answer: (B) $-2$

Put $p = -1$ :

$(-1)^3 + (-1)$

$= -1 - 1$

$= -2$
The sum of $5x^2 + 5z^2 + 8xz$ and $-6x^2 - 2z^2 + 6xz$ is
A $-11x^2 + 3z^2 + 14xz$ B $x^2 + 3z^2 + 14xz$ C $-x^2 + 3z^2 + 14xz$ D $11x^2 + 7z^2 + 2xz$

Answer and solution

Answer: (C) $-x^2 + 3z^2 + 14xz$

$5x^2 + 5z^2 + 8xz - 6x^2 - 2z^2 + 6xz$

$= (5 - 6)x^2 + (5 - 2)z^2 + (8 + 6)xz$

$= -x^2 + 3z^2 + 14xz$
The sum of $-x + 2y$ , $9y + 9t$ and $2x - 8t$ is
A $3x + 11y + t$ B $-3x + 11y + 17t$ C $-x + 11y + t$ D $x + 11y + t$

Answer and solution

Answer: (D) $x + 11y + t$

Add the like terms column by column:

$\begin{array}{rrr} -x & +2y & \\ & 9y & +9t \\ 2x & & -8t \\ \hline x & +11y & +t \end{array}$
Subtract $-8a$ from $11a$ .
A $-19a$ B $-3a$ C $3a$ D $19a$

Answer and solution

Answer: (D) $19a$

$(11a) - (-8a)$

$= 11a + 8a$

$= (11 + 8)a$

$= 19a$
If $A = 4b^2 - 5b + 5$ and $B = 6b^2 - 7b + 1$ , then $2A - B =$
A $-2b^2 + 2b + 4$ B $14b^2 - 17b + 11$ C $2b^2 - 3b + 9$ D $2b^2 - 17b + 11$

Answer and solution

Answer: (C) $2b^2 - 3b + 9$

$2(4b^2 - 5b + 5) - (6b^2 - 7b + 1)$

$= 8b^2 - 10b + 10 - 6b^2 + 7b - 1$

$= (8 - 6)b^2 + (-10 + 7)b + (10 - 1)$

$= 2b^2 - 3b + 9$
Subtract $6a^2 + 3ab - 6$ from $3a^2 + 4$ .
A $3a^2 + 3ab - 10$ B $-3a^2 - 2 + 3ab$ C $9a^2 - 2 + 3ab$ D $-3a^2 + 10 - 3ab$

Answer and solution

Answer: (D) $-3a^2 + 10 - 3ab$

Change the signs of $6a^2 + 3ab - 6$ and add:

$\begin{array}{rrr} 3a^2 & +4 & \\ 6a^2 & -6 & +3ab \\ (-) & (+) & (-) \\ \hline -3a^2 & +10 & -3ab \end{array}$

$= -3a^2 + 10 - 3ab$
$6x^2y \times 3xy^3 \times 3xy^2 =$
A $12x^4y^6$ B $54x^4y^3$ C $54x^4y^6$ D $54x^2y^6$

Answer and solution

Answer: (C) $54x^4y^6$

Multiply the numbers and add the powers of the same letters:

$= (6 \times 3 \times 3)(x^2 \times x \times x)(y \times y^3 \times y^2)$

$= 54x^4y^6$
$-2b^2c(-8bc + 6b^2c + 2) =$
A $16b^3c^2 + 6b^2c + 2$ B $16b^3c^2 - 12b^4c^2 - 4b^2c$ C $16b^2c - 12b^4c^2 - 4b^2c$ D $16b^3c^2 - 12b^4c^2 + 4b^2c$

Answer and solution

Answer: (B) $16b^3c^2 - 12b^4c^2 - 4b^2c$

Multiply $-2b^2c$ with every term inside the bracket:

$= (-2b^2c)(-8bc) + (-2b^2c)(6b^2c) + (-2b^2c)(2)$

$= 16b^3c^2 - 12b^4c^2 - 4b^2c$
$(4a - 9)(a + 7) =$
A $4a^2 - 63$ B $4a^2 + 19a + 63$ C $4a^2 - 19a - 63$ D $4a^2 + 19a - 63$

Answer and solution

Answer: (D) $4a^2 + 19a - 63$

$(a + b)(c + d) = ac + ad + bc + bd$

$= 4a^2 + 28a - 9a - 63$

$= 4a^2 + 19a - 63$
$(a - 6c)(a + 2c) =$
A $a^2 - 4ac - 12c^2$ B $a^2 + 4ac - 12c^2$ C $a^2 - 12c^2$ D $a^2 - 4ac + 12c^2$

Answer and solution

Answer: (A) $a^2 - 4ac - 12c^2$

$(a + b)(c + d) = ac + ad + bc + bd$

$= a^2 + 2ac - 6ac - 12c^2$

$= a^2 - 4ac - 12c^2$
$2x(2x + 2y) - 2y(3x - 4y) =$
A $4x^2 - 2xy + 8y^2$ B $4x^2 - 2xy - 8y^2$ C $-4x^2 + 2xy - 8y^2$ D $4x^2 + 10xy - 8y^2$

Answer and solution

Answer: (A) $4x^2 - 2xy + 8y^2$

Multiply out each bracket and add like terms:

$= 4x^2 + 4xy - 6xy + 8y^2$

$= 4x^2 + (4 - 6)xy + 8y^2$

$= 4x^2 - 2xy + 8y^2$
$2a + \{5c - (3c - 4a)\} =$
A $-2a + 2c$ B $-2a + 8c$ C $6a + 2c$ D $-2a - 2c$

Answer and solution

Answer: (C) $6a + 2c$

Remove the round brackets first, then the curly brackets:

$= 2a + \{5c - 3c + 4a\}$

$= 2a + \{2c + 4a\}$

$= 2a + 2c + 4a$

$= 6a + 2c$

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MCQ test, Set 3