Algebraic Identities: MCQ test, Set 6

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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.

The product $2a(3a - 8b)$ is
A $6a - 16ab$ B $6a^2 - 8b$ C $6a^2 - 16ab$ D $6a^2 + 16ab$

Answer and solution

Answer: (C) $6a^2 - 16ab$

$a(b - c) = ab - ac$

$2a(3a - 8b) = (2a)(3a) - (2a)(8b) = 6a^2 - 16ab$
The product $(2a - 1)(3a + 6)$ is
A $6a^2 + 9a + 6$ B $6a^2 - 15a - 6$ C $6a^2 - 6$ D $6a^2 + 9a - 6$

Answer and solution

Answer: (D) $6a^2 + 9a - 6$

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

$(2a - 1)(3a + 6) = (2a)(3a) + (2a)(6) - (1)(3a) - (1)(6)$

$= 6a^2 + 12a - 3a - 6 = 6a^2 + 9a - 6$
$(2p + 6q)(2p - 6q)$ equals
A $2p^2 - 6q^2$ B $4p^2 + 36q^2$ C $4p^2 + 36q^2 - 24pq$ D $4p^2 - 36q^2$

Answer and solution

Answer: (D) $4p^2 - 36q^2$

$(a + b)(a - b) = a^2 - b^2$

$(2p + 6q)(2p - 6q) = (2p)^2 - (6q)^2$

$= 4p^2 - 36q^2$
The product $(4p^5 + 5q^5)(4p^5 - 5q^5)$ is
A $4p^{10} - 5q^{10}$ B $16p^{10} - 25q^{10}$ C $16p^7 - 25q^7$ D $16p^{10} + 25q^{10}$

Answer and solution

Answer: (B) $16p^{10} - 25q^{10}$

$(a + b)(a - b) = a^2 - b^2$

$(4p^5 + 5q^5)(4p^5 - 5q^5) = (4p^5)^2 - (5q^5)^2$ $= 16p^{10} - 25q^{10}$
$(2p + 3q)^2$ equals
A $4p^2 + 9q^2 + 6pq$ B $4p^2 + 9q^2 + 12pq$ C $4p^2 + 9q^2$ D $2p^2 + 3q^2 + 12pq$

Answer and solution

Answer: (B) $4p^2 + 9q^2 + 12pq$

$(a + b)^2 = a^2 + b^2 + 2ab$

$(2p + 3q)^2 = (2p)^2 + (3q)^2 + 2(2p)(3q)$

$= 4p^2 + 9q^2 + 12pq$
$\displaystyle \left(5q + \frac{1}{3}\right)^2$ equals
A $\displaystyle 25q^2 + \frac{1}{9} + \frac{10q}{3}$ B $\displaystyle 25q^2 + \frac{1}{3} + \frac{10q}{3}$ C $\displaystyle 25q^2 + \frac{1}{9}$ D $\displaystyle 25q^2 + \frac{1}{9} + \frac{5q}{3}$

Answer and solution

Answer: (A) $\displaystyle 25q^2 + \frac{1}{9} + \frac{10q}{3}$

$(a + b)^2 = a^2 + b^2 + 2ab$

$\displaystyle \left(5q + \frac{1}{3}\right)^2 = (5q)^2 + \left(\frac{1}{3}\right)^2 + 2(5q)\left(\frac{1}{3}\right)$

$\displaystyle = 25q^2 + \frac{1}{9} + \frac{10q}{3}$
The expansion of $(3x - 6y)^2$ is
A $9x^2 + 36y^2 + 36xy$ B $9x^2 - 36y^2$ C $9x^2 - 36y^2 - 36xy$ D $9x^2 + 36y^2 - 36xy$

Answer and solution

Answer: (D) $9x^2 + 36y^2 - 36xy$

$(a - b)^2 = a^2 + b^2 - 2ab$

$(3x - 6y)^2 = (3x)^2 + (6y)^2 - 2(3x)(6y)$

$= 9x^2 + 36y^2 - 36xy$
$\displaystyle \left(\frac{7x}{3} - \frac{6y}{5}\right)^2$ equals
A $\displaystyle \frac{49x^2}{3} + \frac{36y^2}{5} - \frac{28xy}{5}$ B $\displaystyle \frac{49x^2}{9} + \frac{36y^2}{25} + \frac{28xy}{5}$ C $\displaystyle \frac{49x^2}{9} + \frac{36y^2}{25} - \frac{28xy}{5}$ D $\displaystyle \frac{49x^2}{9} + \frac{36y^2}{25} - \frac{14xy}{5}$

Answer and solution

Answer: (C) $\displaystyle \frac{49x^2}{9} + \frac{36y^2}{25} - \frac{28xy}{5}$

$(a - b)^2 = a^2 + b^2 - 2ab$

$\displaystyle \left(\frac{7x}{3} - \frac{6y}{5}\right)^2 = \left(\frac{7x}{3}\right)^2 + \left(\frac{6y}{5}\right)^2 - 2\left(\frac{7x}{3}\right)\left(\frac{6y}{5}\right)$

$\displaystyle = \frac{49x^2}{9} + \frac{36y^2}{25} - \frac{28xy}{5}$
$(2a + 5b)^2 + (2a - 5b)^2$ simplifies to
A $8a^2 + 50b^2$ B $8a^2 + 50b^2 + 40ab$ C $40ab$ D $4a^2 + 25b^2$

Answer and solution

Answer: (A) $8a^2 + 50b^2$

$(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$

$(2a + 5b)^2 + (2a - 5b)^2 = 2[(2a)^2 + (5b)^2]$ $= 8a^2 + 50b^2$
$(7m + 4n)^2 - (7m - 4n)^2$ simplifies to
A $56mn$ B $-112mn$ C $112mn$ D $98m^2 + 32n^2$

Answer and solution

Answer: (C) $112mn$

$(a + b)^2 - (a - b)^2 = 4ab$

$(7m + 4n)^2 - (7m - 4n)^2 = 4(7m)(4n) = 112mn$
If $p + q = 12$ and $p - q = 2$ , then $pq =$
A $35$ B $140$ C $37$ D $70$

Answer and solution

Answer: (A) $35$

$(p + q)^2 - (p - q)^2 = 4pq$

$12^2 - 2^2 = 4pq$

$\displaystyle 144 - 4 = 4pq \Rightarrow pq = \frac{140}{4} = 35$
If $(8p + 2q)^2 - kpq = (8p - 2q)^2$ , then $k =$
A $-64$ B $16$ C $64$ D $32$

Answer and solution

Answer: (C) $64$

$(a - b)^2 = (a + b)^2 - 4ab$

$4ab = 4(8p)(2q) = 64pq$

$k = 64$
If $a - b = 11$ and $ab = 12$ , then $(a + b)^2 =$
A $169$ B $73$ C $59$ D $145$

Answer and solution

Answer: (A) $169$

$(a + b)^2 = (a - b)^2 + 4ab$

$(a + b)^2 = (11)^2 + 4(12) = 121 + 48 = 169$
If $\displaystyle z^2 + \frac{1}{z^2} = 23$ , then $\displaystyle \left(z + \frac{1}{z}\right)^2 =$
A $23$ B $25$ C $21$ D $27$

Answer and solution

Answer: (B) $25$

$(a + b)^2 = a^2 + b^2 + 2ab$

$\displaystyle \left(z + \frac{1}{z}\right)^2 = z^2 + \frac{1}{z^2} + 2 \cdot z \cdot \frac{1}{z} = 23 + 2 = 25$
If $a - b = 8$ and $ab = 48$ , then $a^2 + b^2 =$
A $160$ B $112$ C $256$ D $104$

Answer and solution

Answer: (A) $160$

$a^2 + b^2 = (a - b)^2 + 2ab$

$a^2 + b^2 = (8)^2 + 2(48) = 64 + 96 = 160$
The expansion of $(p + 2q + 7r)^2$ is
A $p^2 + 4q^2 + 49r^2$ B $p^2 + 4q^2 + 49r^2 + 4pq + 28qr + 14rp$ C $p^2 + 2q^2 + 7r^2 + 4pq + 28qr + 14rp$ D $p^2 + 4q^2 + 49r^2 + 2pq + 14qr + 7rp$

Answer and solution

Answer: (B) $p^2 + 4q^2 + 49r^2 + 4pq + 28qr + 14rp$

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

$(p + 2q + 7r)^2 = (p)^2 + (2q)^2 + (7r)^2 + 2(p)(2q) + 2(2q)(7r) + 2(7r)(p)$
The expansion of $(3x - 6y - 7z)^2$ is
A $9x^2 + 36y^2 + 49z^2 + 36xy + 84yz + 42zx$ B $9x^2 + 36y^2 + 49z^2 - 36xy + 84yz + 42zx$ C $9x^2 + 36y^2 + 49z^2 - 36xy - 84yz - 42zx$ D $9x^2 + 36y^2 + 49z^2 - 36xy + 84yz - 42zx$

Answer and solution

Answer: (D) $9x^2 + 36y^2 + 49z^2 - 36xy + 84yz - 42zx$

$(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ca$

$(3x - 6y - 7z)^2 = (3x)^2 + (6y)^2 + (7z)^2 - 2(3x)(6y) + 2(6y)(7z) - 2(7z)(3x)$
The coefficient of $qr$ in $(6p - 4q + r)^2$ is
A $-4$ B $8$ C $4$ D $-8$

Answer and solution

Answer: (D) $-8$

$(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca$

The $qr$ term is $- 2(4q)(r) = -8qr$ , so the coefficient is $-8$ .
The expansion of $(2m + 3n)^3$ is
A $8m^3 + 27n^3$ B $8m^3 + 27n^3 + 36m^2n + 54mn^2$ C $8m^3 + 27n^3 + 12m^2n + 18mn^2$ D $8m^3 + 27n^3 + 18mn$

Answer and solution

Answer: (B) $8m^3 + 27n^3 + 36m^2n + 54mn^2$

$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$

$(2m + 3n)^3 = (2m)^3 + (3n)^3 + 3(2m)(3n)(2m + 3n)$

$= 8m^3 + 27n^3 + 36m^2n + 54mn^2$
$(4x - 2y)^3$ equals
A $64x^3 - 8y^3 - 96x^2y - 48xy^2$ B $64x^3 - 8y^3 + 96x^2y - 48xy^2$ C $64x^3 - 8y^3 - 96x^2y + 48xy^2$ D $64x^3 - 8y^3$

Answer and solution

Answer: (C) $64x^3 - 8y^3 - 96x^2y + 48xy^2$

$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$

$(4x - 2y)^3 = (4x)^3 - (2y)^3 - 3(4x)(2y)(4x - 2y)$

$= 64x^3 - 8y^3 - 96x^2y + 48xy^2$
The product $(2q + 4)(4q^2 - 8q + 16)$ is
A $8q^3 - 64$ B $8q^3 + 48q^2 + 96q + 64$ C $8q^3 + 64$ D $8q^3 + 16$

Answer and solution

Answer: (C) $8q^3 + 64$

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Here $a = 2q$ , $b = 4$ : $(2q + 4)(4q^2 - 8q + 16) = (2q)^3 + (4)^3 = 8q^3 + 64$
If $a + b = 15$ and $ab = 54$ , then $a^3 + b^3 =$
A $945$ B $2565$ C $3213$ D $5805$

Answer and solution

Answer: (A) $945$

$a^3 + b^3 = (a + b)^3 - 3ab(a + b)$

$= (15)^3 - 3(54)(15) = 3375 - 2430 = 945$
If $x - y = 4$ and $xy = 12$ , then $x^3 - y^3 =$
A $-80$ B $208$ C $112$ D $100$

Answer and solution

Answer: (B) $208$

$a^3 - b^3 = (a - b)^3 + 3ab(a - b)$

$= (4)^3 + 3(12)(4) = 64 + 144 = 208$
The product $(p + 2q + 3)(p^2 + 4q^2 + 9 - 2pq - 6q - 3p)$ is
A $p^3 + 8q^3 + 27 - 6pq$ B $p^3 + 8q^3 + 27 + 18pq$ C $p^3 + 8q^3 + 27 - 18pq$ D $p^3 + 8q^3 + 27$

Answer and solution

Answer: (C) $p^3 + 8q^3 + 27 - 18pq$

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Here $a = p$ , $b = 2q$ , $c = 3$

$= p^3 + 8q^3 + 27 - 18pq$
$14^3 - 6^3 - 8^3 =$
A $672$ B $-2016$ C $0$ D $2016$

Answer and solution

Answer: (D) $2016$

$\text{Take } a = 14,\ b = -6,\ c = -8$

$a + b + c = 14 - 6 - 8 = 0$

If $a + b + c = 0$ , then $a^3 + b^3 + c^3 = 3abc$ .

$\therefore 14^3 - 6^3 - 8^3 = 3(14)(-6)(-8) = 2016$

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MCQ test, Set 6