Algebraic Identities: MCQ test, Set 2

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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.

$(6p + 8q) \times 3p$ equals
A $18p^2 + 8q$ B $18p^2 + 24pq$ C $18p + 24pq$ D $18p^2 - 24pq$

Answer and solution

Answer: (B) $18p^2 + 24pq$

$(a + b)c = ac + bc$

$(6p + 8q) \times 3p = (6p)(3p) + (8q)(3p) = 18p^2 + 24pq$
$(4p - 8)(3p + 2)$ equals
A $12p^2 - 16p - 16$ B $12p^2 - 16$ C $12p^2 - 16p + 16$ D $12p^2 - 32p - 16$

Answer and solution

Answer: (A) $12p^2 - 16p - 16$

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

$(4p - 8)(3p + 2) = (4p)(3p) + (4p)(2) - (8)(3p) - (8)(2)$

$= 12p^2 + 8p - 24p - 16 = 12p^2 - 16p - 16$
The product $(p + 9q)(p - 9q)$ is
A $p^2 - 9q^2$ B $p^2 - 81q^2$ C $p^2 + 81q^2 - 18pq$ D $p^2 + 81q^2$

Answer and solution

Answer: (B) $p^2 - 81q^2$

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

$(p + 9q)(p - 9q) = (p)^2 - (9q)^2$

$= p^2 - 81q^2$
The product $(3m^3 + 5n^4)(3m^3 - 5n^4)$ is
A $3m^6 - 5n^8$ B $9m^5 - 25n^6$ C $9m^6 - 25n^8$ D $9m^6 + 25n^8$

Answer and solution

Answer: (C) $9m^6 - 25n^8$

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

$(3m^3 + 5n^4)(3m^3 - 5n^4) = (3m^3)^2 - (5n^4)^2$ $= 9m^6 - 25n^8$
The expansion of $(2a + 9b)^2$ is
A $2a^2 + 9b^2 + 36ab$ B $4a^2 + 81b^2 + 18ab$ C $4a^2 + 81b^2$ D $4a^2 + 81b^2 + 36ab$

Answer and solution

Answer: (D) $4a^2 + 81b^2 + 36ab$

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

$(2a + 9b)^2 = (2a)^2 + (9b)^2 + 2(2a)(9b)$

$= 4a^2 + 81b^2 + 36ab$
$\displaystyle \left(3q + \frac{1}{9}\right)^2$ equals
A $\displaystyle 9q^2 + \frac{1}{81}$ B $\displaystyle 9q^2 + \frac{1}{9} + \frac{2q}{3}$ C $\displaystyle 9q^2 + \frac{1}{81} + \frac{q}{3}$ D $\displaystyle 9q^2 + \frac{1}{81} + \frac{2q}{3}$

Answer and solution

Answer: (D) $\displaystyle 9q^2 + \frac{1}{81} + \frac{2q}{3}$

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

$\displaystyle \left(3q + \frac{1}{9}\right)^2 = (3q)^2 + \left(\frac{1}{9}\right)^2 + 2(3q)\left(\frac{1}{9}\right)$

$\displaystyle = 9q^2 + \frac{1}{81} + \frac{2q}{3}$
$(7x - 2y)^2$ equals
A $49x^2 - 4y^2 - 28xy$ B $49x^2 + 4y^2 + 28xy$ C $49x^2 + 4y^2 - 28xy$ D $49x^2 - 4y^2$

Answer and solution

Answer: (C) $49x^2 + 4y^2 - 28xy$

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

$(7x - 2y)^2 = (7x)^2 + (2y)^2 - 2(7x)(2y)$

$= 49x^2 + 4y^2 - 28xy$
$\displaystyle \left(\frac{3a}{2} - \frac{7b}{5}\right)^2$ equals
A $\displaystyle \frac{9a^2}{4} + \frac{49b^2}{25} - \frac{21ab}{5}$ B $\displaystyle \frac{9a^2}{4} + \frac{49b^2}{25} + \frac{21ab}{5}$ C $\displaystyle \frac{9a^2}{2} + \frac{49b^2}{5} - \frac{21ab}{5}$ D $\displaystyle \frac{9a^2}{4} + \frac{49b^2}{25} - \frac{21ab}{10}$

Answer and solution

Answer: (A) $\displaystyle \frac{9a^2}{4} + \frac{49b^2}{25} - \frac{21ab}{5}$

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

$\displaystyle \left(\frac{3a}{2} - \frac{7b}{5}\right)^2 = \left(\frac{3a}{2}\right)^2 + \left(\frac{7b}{5}\right)^2 - 2\left(\frac{3a}{2}\right)\left(\frac{7b}{5}\right)$

$\displaystyle = \frac{9a^2}{4} + \frac{49b^2}{25} - \frac{21ab}{5}$
$(5a + 2b)^2 + (5a - 2b)^2$ simplifies to
A $50a^2 + 8b^2 + 40ab$ B $40ab$ C $25a^2 + 4b^2$ D $50a^2 + 8b^2$

Answer and solution

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

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

$(5a + 2b)^2 + (5a - 2b)^2 = 2[(5a)^2 + (2b)^2]$ $= 50a^2 + 8b^2$
$(m + 3n)^2 - (m - 3n)^2$ simplifies to
A $-12mn$ B $12mn$ C $6mn$ D $2m^2 + 18n^2$

Answer and solution

Answer: (B) $12mn$

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

$(m + 3n)^2 - (m - 3n)^2 = 4(m)(3n) = 12mn$
If $x + y = 22$ and $x - y = 2$ , then $xy =$
A $122$ B $240$ C $120$ D $480$

Answer and solution

Answer: (C) $120$

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

$22^2 - 2^2 = 4xy$

$\displaystyle 484 - 4 = 4xy \Rightarrow xy = \frac{480}{4} = 120$
If $(8m + 9n)^2 - kmn = (8m - 9n)^2$ , then $k =$
A $288$ B $72$ C $144$ D $-288$

Answer and solution

Answer: (A) $288$

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

$4ab = 4(8m)(9n) = 288mn$

$k = 288$
If $m - n = 7$ and $mn = 60$ , then $(m + n)^2 =$
A $247$ B $169$ C $109$ D $289$

Answer and solution

Answer: (D) $289$

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

$(m + n)^2 = (7)^2 + 4(60) = 49 + 240 = 289$
If $\displaystyle y - \frac{1}{y} = 7$ , then $\displaystyle y^2 + \frac{1}{y^2} =$
A $47$ B $51$ C $49$ D $53$

Answer and solution

Answer: (B) $51$

$a^2 + b^2 = (a - b)^2 + 2ab$ , with $\displaystyle ab = y \cdot \frac{1}{y} = 1$

$= (7)^2 + 2 = 49 + 2 = 51$
If $x - y = 2$ and $xy = 35$ , then $x^2 + y^2 =$
A $74$ B $144$ C $72$ D $39$

Answer and solution

Answer: (A) $74$

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

$x^2 + y^2 = (2)^2 + 2(35) = 4 + 70 = 74$
The expansion of $(6x + y + 4z)^2$ is
A $36x^2 + y^2 + 16z^2 + 12xy + 8yz + 48zx$ B $6x^2 + y^2 + 4z^2 + 12xy + 8yz + 48zx$ C $36x^2 + y^2 + 16z^2$ D $36x^2 + y^2 + 16z^2 + 6xy + 4yz + 24zx$

Answer and solution

Answer: (A) $36x^2 + y^2 + 16z^2 + 12xy + 8yz + 48zx$

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

$(6x + y + 4z)^2 = (6x)^2 + (y)^2 + (4z)^2 + 2(6x)(y) + 2(y)(4z) + 2(4z)(6x)$
$(6a - 7b + 5c)^2$ equals
A $36a^2 + 49b^2 + 25c^2 - 84ab + 70bc + 60ca$ B $36a^2 + 49b^2 + 25c^2 - 84ab - 70bc + 60ca$ C $36a^2 + 49b^2 + 25c^2 - 84ab - 70bc - 60ca$ D $36a^2 + 49b^2 + 25c^2 + 84ab + 70bc + 60ca$

Answer and solution

Answer: (B) $36a^2 + 49b^2 + 25c^2 - 84ab - 70bc + 60ca$

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

$(6a - 7b + 5c)^2 = (6a)^2 + (7b)^2 + (5c)^2 - 2(6a)(7b) - 2(7b)(5c) + 2(5c)(6a)$
The coefficient of $xy$ in $(5x + 3y - z)^2$ is
A $-15$ B $15$ C $30$ D $-30$

Answer and solution

Answer: (C) $30$

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

The $xy$ term is $+ 2(5x)(3y) = 30xy$ , so the coefficient is $30$ .
The expansion of $(2m + 3)^3$ is
A $8m^3 + 27 + 18m$ B $8m^3 + 27 + 12m^2 + 18m$ C $8m^3 + 27$ D $8m^3 + 27 + 36m^2 + 54m$

Answer and solution

Answer: (D) $8m^3 + 27 + 36m^2 + 54m$

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

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

$= 8m^3 + 27 + 36m^2 + 54m$
$(z - 7)^3$ equals
A $z^3 - 343 + 21z^2 - 147z$ B $z^3 - 343 - 21z^2 - 147z$ C $z^3 - 343$ D $z^3 - 343 - 21z^2 + 147z$

Answer and solution

Answer: (D) $z^3 - 343 - 21z^2 + 147z$

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

$(z - 7)^3 = (z)^3 - (7)^3 - 3(z)(7)(z - 7)$

$= z^3 - 343 - 21z^2 + 147z$
The product $(x + 6)(x^2 - 6x + 36)$ is
A $x^3 + 36$ B $x^3 + 216$ C $x^3 - 216$ D $x^3 + 18x^2 + 108x + 216$

Answer and solution

Answer: (B) $x^3 + 216$

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

Here $a = x$ , $b = 6$ : $(x + 6)(x^2 - 6x + 36) = (x)^3 + (6)^3 = x^3 + 216$
If $p + q = 12$ and $pq = 32$ , then $p^3 + q^3 =$
A $2880$ B $1632$ C $576$ D $1344$

Answer and solution

Answer: (C) $576$

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

$= (12)^3 - 3(32)(12) = 1728 - 1152 = 576$
If $m - n = 3$ and $mn = 40$ , then $m^3 - n^3 =$
A $-333$ B $387$ C $147$ D $27$

Answer and solution

Answer: (B) $387$

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

$= (3)^3 + 3(40)(3) = 27 + 360 = 387$
The product $(3x + y + 2)(9x^2 + y^2 + 4 - 3xy - 2y - 6x)$ is
A $27x^3 + y^3 + 8 - 18xy$ B $27x^3 + y^3 + 8 + 18xy$ C $27x^3 + y^3 + 8$ D $27x^3 + y^3 + 8 - 6xy$

Answer and solution

Answer: (A) $27x^3 + y^3 + 8 - 18xy$

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

Here $a = 3x$ , $b = y$ , $c = 2$

$= 27x^3 + y^3 + 8 - 18xy$
$11^3 - 4^3 - 7^3 =$
A $-924$ B $0$ C $924$ D $308$

Answer and solution

Answer: (C) $924$

$\text{Take } a = 11,\ b = -4,\ c = -7$

$a + b + c = 11 - 4 - 7 = 0$

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

$\therefore 11^3 - 4^3 - 7^3 = 3(11)(-4)(-7) = 924$

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MCQ test, Set 2