Algebraic Identities: MCQ test, Set 1

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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.

$2p(2p + 5q) =$
A $4p + 10pq$ B $4p^2 + 10pq$ C $4p^2 + 5q$ D $4p^2 - 10pq$

Answer and solution

Answer: (B) $4p^2 + 10pq$

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

$2p(2p + 5q) = (2p)(2p) + (2p)(5q) = 4p^2 + 10pq$
$(a + 1)(2a - 3) =$
A $2a^2 + 5a - 3$ B $2a^2 - a + 3$ C $2a^2 - a - 3$ D $2a^2 - 3$

Answer and solution

Answer: (C) $2a^2 - a - 3$

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

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

$= 2a^2 - 3a + 2a - 3 = 2a^2 - a - 3$
$(4m + 6n)(4m - 6n) =$
A $4m^2 - 6n^2$ B $16m^2 + 36n^2$ C $16m^2 - 36n^2$ D $16m^2 + 36n^2 - 48mn$

Answer and solution

Answer: (C) $16m^2 - 36n^2$

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

$(4m + 6n)(4m - 6n) = (4m)^2 - (6n)^2$

$= 16m^2 - 36n^2$
$(3a^3 + 2b^3)(3a^3 - 2b^3) =$
A $9a^6 + 4b^6$ B $9a^6 - 4b^6$ C $3a^6 - 2b^6$ D $9a^5 - 4b^5$

Answer and solution

Answer: (B) $9a^6 - 4b^6$

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

$(3a^3 + 2b^3)(3a^3 - 2b^3) = (3a^3)^2 - (2b^3)^2$ $= 9a^6 - 4b^6$
$(8p + 6q)^2 =$
A $64p^2 + 36q^2 + 96pq$ B $8p^2 + 6q^2 + 96pq$ C $64p^2 + 36q^2 + 48pq$ D $64p^2 + 36q^2$

Answer and solution

Answer: (A) $64p^2 + 36q^2 + 96pq$

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

$(8p + 6q)^2 = (8p)^2 + (6q)^2 + 2(8p)(6q)$

$= 64p^2 + 36q^2 + 96pq$
In the expansion of $(7a + 4b)^2$ , the coefficient of $ab$ is
A $22$ B $112$ C $28$ D $56$

Answer and solution

Answer: (D) $56$

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

$2ab = 2(7a)(4b) = 56ab$

So the coefficient of $ab$ is $56$ .
$(8p - q)^2 =$
A $64p^2 + q^2 - 16pq$ B $64p^2 - q^2 - 16pq$ C $64p^2 + q^2 + 16pq$ D $64p^2 - q^2$

Answer and solution

Answer: (A) $64p^2 + q^2 - 16pq$

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

$(8p - q)^2 = (8p)^2 + (q)^2 - 2(8p)(q)$

$= 64p^2 + q^2 - 16pq$
$\displaystyle \left(\frac{3a}{4} - \frac{3b}{2}\right)^2 =$
A $\displaystyle \frac{9a^2}{4} + \frac{9b^2}{2} - \frac{9ab}{4}$ B $\displaystyle \frac{9a^2}{16} + \frac{9b^2}{4} - \frac{9ab}{8}$ C $\displaystyle \frac{9a^2}{16} + \frac{9b^2}{4} - \frac{9ab}{4}$ D $\displaystyle \frac{9a^2}{16} + \frac{9b^2}{4} + \frac{9ab}{4}$

Answer and solution

Answer: (C) $\displaystyle \frac{9a^2}{16} + \frac{9b^2}{4} - \frac{9ab}{4}$

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

$\displaystyle \left(\frac{3a}{4} - \frac{3b}{2}\right)^2 = \left(\frac{3a}{4}\right)^2 + \left(\frac{3b}{2}\right)^2 - 2\left(\frac{3a}{4}\right)\left(\frac{3b}{2}\right)$

$\displaystyle = \frac{9a^2}{16} + \frac{9b^2}{4} - \frac{9ab}{4}$
$\displaystyle \left(3 + \frac{7}{x}\right)^2 + \left(3 - \frac{7}{x}\right)^2 =$
A $\displaystyle 9 + \frac{49}{x^2}$ B $\displaystyle 18 + \frac{98}{x^2}$ C $\displaystyle 18 + \frac{98}{x^2} + \frac{84}{x}$ D $\displaystyle \frac{84}{x}$

Answer and solution

Answer: (B) $\displaystyle 18 + \frac{98}{x^2}$

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

$\displaystyle \left(3 + \frac{7}{x}\right)^2 + \left(3 - \frac{7}{x}\right)^2 = 2[(3)^2 + \left(\frac{7}{x}\right)^2]$ $\displaystyle = 18 + \frac{98}{x^2}$
$(8x + 7y)^2 - (8x - 7y)^2 =$
A $112xy$ B $128x^2 + 98y^2$ C $-224xy$ D $224xy$

Answer and solution

Answer: (D) $224xy$

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

$(8x + 7y)^2 - (8x - 7y)^2 = 4(8x)(7y) = 224xy$
If $m + n = 18$ and $m - n = 2$ , then $mn =$
A $80$ B $82$ C $320$ D $160$

Answer and solution

Answer: (A) $80$

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

$18^2 - 2^2 = 4mn$

$\displaystyle 324 - 4 = 4mn \Rightarrow mn = \frac{320}{4} = 80$
If $(3p - 9q)^2 + kpq = (3p + 9q)^2$ , then $k =$
A $27$ B $-108$ C $54$ D $108$

Answer and solution

Answer: (D) $108$

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

$4ab = 4(3p)(9q) = 108pq$

$k = 108$
If $a + b = 18$ and $ab = 72$ , then $(a - b)^2 =$
A $468$ B $612$ C $180$ D $36$

Answer and solution

Answer: (D) $36$

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

$(a - b)^2 = (18)^2 - 4(72) = 324 - 288 = 36$
If $\displaystyle r + \frac{1}{r} = 13$ , then $\displaystyle r^2 + \frac{1}{r^2} =$
A $169$ B $165$ C $167$ D $171$

Answer and solution

Answer: (C) $167$

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

$= (13)^2 - 2 = 169 - 2 = 167$
If $x + y = 9$ and $xy = 14$ , then $x^2 + y^2 =$
A $25$ B $53$ C $109$ D $67$

Answer and solution

Answer: (B) $53$

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

$x^2 + y^2 = (9)^2 - 2(14) = 81 - 28 = 53$
In the expansion of $(4a + 6b + 3c)^2$ , the coefficient of $ca$ is
A $48$ B $12$ C $14$ D $24$

Answer and solution

Answer: (D) $24$

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

The $ca$ term is $2(3c)(4a) = 24ca$ , so the coefficient is $24$ .
$(5p + 2q - 3r)^2 =$
A $25p^2 + 4q^2 + 9r^2 + 20pq + 12qr + 30rp$ B $25p^2 + 4q^2 + 9r^2 + 20pq - 12qr - 30rp$ C $25p^2 + 4q^2 + 9r^2 + 20pq + 12qr - 30rp$ D $25p^2 + 4q^2 + 9r^2 + 20pq - 12qr + 30rp$

Answer and solution

Answer: (B) $25p^2 + 4q^2 + 9r^2 + 20pq - 12qr - 30rp$

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

$(5p + 2q - 3r)^2 = (5p)^2 + (2q)^2 + (3r)^2 + 2(5p)(2q) - 2(2q)(3r) - 2(3r)(5p)$
The coefficient of $rp$ in $(4p - 6q - 3r)^2$ is
A $-24$ B $12$ C $24$ D $-12$

Answer and solution

Answer: (A) $-24$

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

The $rp$ term is $- 2(3r)(4p) = -24rp$ , so the coefficient is $-24$ .
$(4x + 3y)^3 =$
A $64x^3 + 27y^3$ B $64x^3 + 27y^3 + 36xy$ C $64x^3 + 27y^3 + 48x^2y + 36xy^2$ D $64x^3 + 27y^3 + 144x^2y + 108xy^2$

Answer and solution

Answer: (D) $64x^3 + 27y^3 + 144x^2y + 108xy^2$

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

$(4x + 3y)^3 = (4x)^3 + (3y)^3 + 3(4x)(3y)(4x + 3y)$

$= 64x^3 + 27y^3 + 144x^2y + 108xy^2$
$(6t - 1)^3 =$
A $216t^3 - 1 + 108t^2 - 18t$ B $216t^3 - 1$ C $216t^3 - 1 - 108t^2 + 18t$ D $216t^3 - 1 - 108t^2 - 18t$

Answer and solution

Answer: (C) $216t^3 - 1 - 108t^2 + 18t$

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

$(6t - 1)^3 = (6t)^3 - (1)^3 - 3(6t)(1)(6t - 1)$

$= 216t^3 - 1 - 108t^2 + 18t$
$(3q + 6)(9q^2 - 18q + 36) =$
A $27q^3 + 216$ B $27q^3 + 36$ C $27q^3 - 216$ D $27q^3 + 162q^2 + 324q + 216$

Answer and solution

Answer: (A) $27q^3 + 216$

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

Here $a = 3q$ , $b = 6$ : $(3q + 6)(9q^2 - 18q + 36) = (3q)^3 + (6)^3 = 27q^3 + 216$
If $p + q = 13$ and $pq = 42$ , then $p^3 + q^3 =$
A $1651$ B $559$ C $3835$ D $2071$

Answer and solution

Answer: (B) $559$

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

$= (13)^3 - 3(42)(13) = 2197 - 1638 = 559$
$(3x - 6)(9x^2 + 18x + 36) =$
A $27x^3 - 216$ B $3x^3 - 216$ C $27x^3 + 216$ D $27x^3 - 216 - 162x^2 + 324x$

Answer and solution

Answer: (A) $27x^3 - 216$

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

$(3x - 6)(9x^2 + 18x + 36) = (3x)^3 - (6)^3 = 27x^3 - 216$
$(m + 2n + 5)(m^2 + 4n^2 + 25 - 2mn - 10n - 5m) =$
A $m^3 + 8n^3 + 125 - 10mn$ B $m^3 + 8n^3 + 125$ C $m^3 + 8n^3 + 125 - 30mn$ D $m^3 + 8n^3 + 125 + 30mn$

Answer and solution

Answer: (C) $m^3 + 8n^3 + 125 - 30mn$

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

Here $a = m$ , $b = 2n$ , $c = 5$

$= m^3 + 8n^3 + 125 - 30mn$
The value of $15^3 - 8^3 - 7^3$ is
A $0$ B $840$ C $2520$ D $-2520$

Answer and solution

Answer: (C) $2520$

$\text{Take } a = 15,\ b = -8,\ c = -7$

$a + b + c = 15 - 8 - 7 = 0$

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

$\therefore 15^3 - 8^3 - 7^3 = 3(15)(-8)(-7) = 2520$

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MCQ test, Set 1