Algebraic Identities: MCQ test, Set 10

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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.

$5x(2x - 7y) =$
A $10x^2 - 35xy$ B $10x^2 + 35xy$ C $10x^2 - 7y$ D $10x - 35xy$

Answer and solution

Answer: (A) $10x^2 - 35xy$

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

$5x(2x - 7y) = (5x)(2x) - (5x)(7y) = 10x^2 - 35xy$
$(3m - 6)(4m + 9) =$
A $12m^2 + 3m - 54$ B $12m^2 - 54$ C $12m^2 - 51m - 54$ D $12m^2 + 3m + 54$

Answer and solution

Answer: (A) $12m^2 + 3m - 54$

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

$(3m - 6)(4m + 9) = (3m)(4m) + (3m)(9) - (6)(4m) - (6)(9)$

$= 12m^2 + 27m - 24m - 54 = 12m^2 + 3m - 54$
$(8a + 5b)(8a - 5b) =$
A $64a^2 + 25b^2$ B $64a^2 + 25b^2 - 80ab$ C $64a^2 - 25b^2$ D $8a^2 - 5b^2$

Answer and solution

Answer: (C) $64a^2 - 25b^2$

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

$(8a + 5b)(8a - 5b) = (8a)^2 - (5b)^2$

$= 64a^2 - 25b^2$
The product $(6x^5 + 3y^3)(6x^5 - 3y^3)$ is
A $6x^{10} - 3y^6$ B $36x^{10} + 9y^6$ C $36x^7 - 9y^5$ D $36x^{10} - 9y^6$

Answer and solution

Answer: (D) $36x^{10} - 9y^6$

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

$(6x^5 + 3y^3)(6x^5 - 3y^3) = (6x^5)^2 - (3y^3)^2$ $= 36x^{10} - 9y^6$
$(3p + 9q)^2 =$
A $9p^2 + 81q^2 + 27pq$ B $9p^2 + 81q^2$ C $3p^2 + 9q^2 + 54pq$ D $9p^2 + 81q^2 + 54pq$

Answer and solution

Answer: (D) $9p^2 + 81q^2 + 54pq$

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

$(3p + 9q)^2 = (3p)^2 + (9q)^2 + 2(3p)(9q)$

$= 9p^2 + 81q^2 + 54pq$
$\displaystyle \left(6n + \frac{1}{7}\right)^2$ equals
A $\displaystyle 36n^2 + \frac{1}{7} + \frac{12n}{7}$ B $\displaystyle 36n^2 + \frac{1}{49}$ C $\displaystyle 36n^2 + \frac{1}{49} + \frac{12n}{7}$ D $\displaystyle 36n^2 + \frac{1}{49} + \frac{6n}{7}$

Answer and solution

Answer: (C) $\displaystyle 36n^2 + \frac{1}{49} + \frac{12n}{7}$

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

$\displaystyle \left(6n + \frac{1}{7}\right)^2 = (6n)^2 + \left(\frac{1}{7}\right)^2 + 2(6n)\left(\frac{1}{7}\right)$

$\displaystyle = 36n^2 + \frac{1}{49} + \frac{12n}{7}$
$(3m - 2n)^2 =$
A $9m^2 - 4n^2$ B $9m^2 + 4n^2 - 12mn$ C $9m^2 + 4n^2 + 12mn$ D $9m^2 - 4n^2 - 12mn$

Answer and solution

Answer: (B) $9m^2 + 4n^2 - 12mn$

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

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

$= 9m^2 + 4n^2 - 12mn$
$\displaystyle \left(\frac{3m}{2} - \frac{n}{5}\right)^2$ equals
A $\displaystyle \frac{9m^2}{2} + \frac{n^2}{5} - \frac{3mn}{5}$ B $\displaystyle \frac{9m^2}{4} + \frac{n^2}{25} - \frac{3mn}{10}$ C $\displaystyle \frac{9m^2}{4} + \frac{n^2}{25} - \frac{3mn}{5}$ D $\displaystyle \frac{9m^2}{4} + \frac{n^2}{25} + \frac{3mn}{5}$

Answer and solution

Answer: (C) $\displaystyle \frac{9m^2}{4} + \frac{n^2}{25} - \frac{3mn}{5}$

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

$\displaystyle \left(\frac{3m}{2} - \frac{n}{5}\right)^2 = \left(\frac{3m}{2}\right)^2 + \left(\frac{n}{5}\right)^2 - 2\left(\frac{3m}{2}\right)\left(\frac{n}{5}\right)$

$\displaystyle = \frac{9m^2}{4} + \frac{n^2}{25} - \frac{3mn}{5}$
$(7p + q)^2 + (7p - q)^2$ simplifies to
A $28pq$ B $49p^2 + q^2$ C $98p^2 + 2q^2$ D $98p^2 + 2q^2 + 28pq$

Answer and solution

Answer: (C) $98p^2 + 2q^2$

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

$(7p + q)^2 + (7p - q)^2 = 2[(7p)^2 + (q)^2]$ $= 98p^2 + 2q^2$
$(6x + 9y)^2 - (6x - 9y)^2$ simplifies to
A $-216xy$ B $108xy$ C $72x^2 + 162y^2$ D $216xy$

Answer and solution

Answer: (D) $216xy$

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

$(6x + 9y)^2 - (6x - 9y)^2 = 4(6x)(9y) = 216xy$
If $m + n = 14$ and $m - n = 12$ , then $mn =$
A $85$ B $52$ C $26$ D $13$

Answer and solution

Answer: (D) $13$

$(m + n)^2 - (m - n)^2 = 4mn$

$14^2 - 12^2 = 4mn$

$\displaystyle 196 - 144 = 4mn \Rightarrow mn = \frac{52}{4} = 13$
If $(4a + b)^2 - kab = (4a - b)^2$ , then $k =$
A $16$ B $4$ C $-16$ D $8$

Answer and solution

Answer: (A) $16$

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

$4ab = 4(4a)(b) = 16ab$

$k = 16$
If $m - n = 9$ and $mn = 10$ , then $(m + n)^2 =$
A $101$ B $121$ C $49$ D $41$

Answer and solution

Answer: (B) $121$

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

$(m + n)^2 = (9)^2 + 4(10) = 81 + 40 = 121$
If $\displaystyle z + \frac{1}{z} = 16$ , then $\displaystyle z^2 + \frac{1}{z^2} =$
A $252$ B $254$ C $256$ D $258$

Answer and solution

Answer: (B) $254$

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

$= (16)^2 - 2 = 256 - 2 = 254$
If $a - b = 4$ and $ab = 32$ , then $a^2 + b^2 =$
A $144$ B $48$ C $80$ D $68$

Answer and solution

Answer: (C) $80$

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

$a^2 + b^2 = (4)^2 + 2(32) = 16 + 64 = 80$
The expansion of $(6a + 4b + 5c)^2$ is
A $36a^2 + 16b^2 + 25c^2 + 24ab + 20bc + 30ca$ B $6a^2 + 4b^2 + 5c^2 + 48ab + 40bc + 60ca$ C $36a^2 + 16b^2 + 25c^2$ D $36a^2 + 16b^2 + 25c^2 + 48ab + 40bc + 60ca$

Answer and solution

Answer: (D) $36a^2 + 16b^2 + 25c^2 + 48ab + 40bc + 60ca$

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

$(6a + 4b + 5c)^2 = (6a)^2 + (4b)^2 + (5c)^2 + 2(6a)(4b) + 2(4b)(5c) + 2(5c)(6a)$
$(2a + 4b - 5c)^2 =$
A $4a^2 + 16b^2 + 25c^2 + 16ab - 40bc - 20ca$ B $4a^2 + 16b^2 + 25c^2 + 16ab + 40bc + 20ca$ C $4a^2 + 16b^2 + 25c^2 + 16ab - 40bc + 20ca$ D $4a^2 + 16b^2 + 25c^2 + 16ab + 40bc - 20ca$

Answer and solution

Answer: (A) $4a^2 + 16b^2 + 25c^2 + 16ab - 40bc - 20ca$

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

$(2a + 4b - 5c)^2 = (2a)^2 + (4b)^2 + (5c)^2 + 2(2a)(4b) - 2(4b)(5c) - 2(5c)(2a)$
The coefficient of $zx$ in $(5x - 6y - z)^2$ is
A $-10$ B $5$ C $10$ D $-5$

Answer and solution

Answer: (A) $-10$

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

The $zx$ term is $- 2(z)(5x) = -10zx$ , so the coefficient is $-10$ .
The expansion of $(3z + 2)^3$ is
A $27z^3 + 8$ B $27z^3 + 8 + 54z^2 + 36z$ C $27z^3 + 8 + 18z^2 + 12z$ D $27z^3 + 8 + 18z$

Answer and solution

Answer: (B) $27z^3 + 8 + 54z^2 + 36z$

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

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

$= 27z^3 + 8 + 54z^2 + 36z$
$(2t - 5)^3$ equals
A $8t^3 - 125 - 60t^2 + 150t$ B $8t^3 - 125 - 60t^2 - 150t$ C $8t^3 - 125$ D $8t^3 - 125 + 60t^2 - 150t$

Answer and solution

Answer: (A) $8t^3 - 125 - 60t^2 + 150t$

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

$(2t - 5)^3 = (2t)^3 - (5)^3 - 3(2t)(5)(2t - 5)$

$= 8t^3 - 125 - 60t^2 + 150t$
The product $(3m + 4)(9m^2 - 12m + 16)$ is
A $27m^3 + 108m^2 + 144m + 64$ B $27m^3 + 16$ C $27m^3 + 64$ D $27m^3 - 64$

Answer and solution

Answer: (C) $27m^3 + 64$

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

Here $a = 3m$ , $b = 4$ : $(3m + 4)(9m^2 - 12m + 16) = (3m)^3 + (4)^3 = 27m^3 + 64$
If $x + y = 13$ and $xy = 36$ , then $x^3 + y^3 =$
A $1729$ B $793$ C $2089$ D $3601$

Answer and solution

Answer: (B) $793$

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

$= (13)^3 - 3(36)(13) = 2197 - 1404 = 793$
If $x - y = 6$ and $xy = 16$ , then $x^3 - y^3 =$
A $312$ B $264$ C $-72$ D $504$

Answer and solution

Answer: (D) $504$

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

$= (6)^3 + 3(16)(6) = 216 + 288 = 504$
The product $(2p + 3q + 5)(4p^2 + 9q^2 + 25 - 6pq - 15q - 10p)$ is
A $8p^3 + 27q^3 + 125 + 90pq$ B $8p^3 + 27q^3 + 125 - 90pq$ C $8p^3 + 27q^3 + 125 - 30pq$ D $8p^3 + 27q^3 + 125$

Answer and solution

Answer: (B) $8p^3 + 27q^3 + 125 - 90pq$

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

Here $a = 2p$ , $b = 3q$ , $c = 5$

$= 8p^3 + 27q^3 + 125 - 90pq$
$(-5)^3 + 2^3 + 3^3 =$
A $0$ B $-30$ C $-90$ D $90$

Answer and solution

Answer: (C) $-90$

$\text{Take } a = -5,\ b = 2,\ c = 3$

$a + b + c = -5 + 2 + 3 = 0$

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

$\therefore (-5)^3 + 2^3 + 3^3 = 3(-5)(2)(3) = -90$

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MCQ test, Set 10