Algebraic Identities: MCQ test, Set 8

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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.

$(m + 2n) \times 3m$ equals
A $3m^2 - 6mn$ B $3m + 6mn$ C $3m^2 + 6mn$ D $3m^2 + 2n$

Answer and solution

Answer: (C) $3m^2 + 6mn$

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

$(m + 2n) \times 3m = (m)(3m) + (2n)(3m) = 3m^2 + 6mn$
$(2a - 5)(2a + 6)$ equals
A $4a^2 - 22a - 30$ B $4a^2 - 30$ C $4a^2 + 2a - 30$ D $4a^2 + 2a + 30$

Answer and solution

Answer: (C) $4a^2 + 2a - 30$

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

$(2a - 5)(2a + 6) = (2a)(2a) + (2a)(6) - (5)(2a) - (5)(6)$

$= 4a^2 + 12a - 10a - 30 = 4a^2 + 2a - 30$
The product $(6m - 7n)(6m + 7n)$ is
A $6m^2 - 7n^2$ B $36m^2 + 49n^2$ C $36m^2 + 49n^2 - 84mn$ D $36m^2 - 49n^2$

Answer and solution

Answer: (D) $36m^2 - 49n^2$

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

$(6m - 7n)(6m + 7n) = (6m)^2 - (7n)^2$

$= 36m^2 - 49n^2$
The product $(4p^4 + 2q^3)(4p^4 - 2q^3)$ is
A $16p^6 - 4q^5$ B $16p^8 + 4q^6$ C $16p^8 - 4q^6$ D $4p^8 - 2q^6$

Answer and solution

Answer: (C) $16p^8 - 4q^6$

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

$(4p^4 + 2q^3)(4p^4 - 2q^3) = (4p^4)^2 - (2q^3)^2$ $= 16p^8 - 4q^6$
The expansion of $(4a + 6b)^2$ is
A $16a^2 + 36b^2 + 48ab$ B $4a^2 + 6b^2 + 48ab$ C $16a^2 + 36b^2 + 24ab$ D $16a^2 + 36b^2$

Answer and solution

Answer: (A) $16a^2 + 36b^2 + 48ab$

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

$(4a + 6b)^2 = (4a)^2 + (6b)^2 + 2(4a)(6b)$

$= 16a^2 + 36b^2 + 48ab$
$\displaystyle \left(5t + \frac{1}{7}\right)^2$ equals
A $\displaystyle 25t^2 + \frac{1}{7} + \frac{10t}{7}$ B $\displaystyle 25t^2 + \frac{1}{49} + \frac{10t}{7}$ C $\displaystyle 25t^2 + \frac{1}{49} + \frac{5t}{7}$ D $\displaystyle 25t^2 + \frac{1}{49}$

Answer and solution

Answer: (B) $\displaystyle 25t^2 + \frac{1}{49} + \frac{10t}{7}$

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

$\displaystyle \left(5t + \frac{1}{7}\right)^2 = (5t)^2 + \left(\frac{1}{7}\right)^2 + 2(5t)\left(\frac{1}{7}\right)$

$\displaystyle = 25t^2 + \frac{1}{49} + \frac{10t}{7}$
$(5p - 2q)^2$ equals
A $25p^2 - 4q^2 - 20pq$ B $25p^2 + 4q^2 - 20pq$ C $25p^2 + 4q^2 + 20pq$ D $25p^2 - 4q^2$

Answer and solution

Answer: (B) $25p^2 + 4q^2 - 20pq$

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

$(5p - 2q)^2 = (5p)^2 + (2q)^2 - 2(5p)(2q)$

$= 25p^2 + 4q^2 - 20pq$
$\displaystyle \left(\frac{2m}{3} - \frac{5n}{2}\right)^2$ equals
A $\displaystyle \frac{4m^2}{3} + \frac{25n^2}{2} - \frac{10mn}{3}$ B $\displaystyle \frac{4m^2}{9} + \frac{25n^2}{4} + \frac{10mn}{3}$ C $\displaystyle \frac{4m^2}{9} + \frac{25n^2}{4} - \frac{10mn}{3}$ D $\displaystyle \frac{4m^2}{9} + \frac{25n^2}{4} - \frac{5mn}{3}$

Answer and solution

Answer: (C) $\displaystyle \frac{4m^2}{9} + \frac{25n^2}{4} - \frac{10mn}{3}$

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

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

$\displaystyle = \frac{4m^2}{9} + \frac{25n^2}{4} - \frac{10mn}{3}$
$(3a + 6b)^2 + (3a - 6b)^2$ simplifies to
A $72ab$ B $18a^2 + 72b^2$ C $18a^2 + 72b^2 + 72ab$ D $9a^2 + 36b^2$

Answer and solution

Answer: (B) $18a^2 + 72b^2$

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

$(3a + 6b)^2 + (3a - 6b)^2 = 2[(3a)^2 + (6b)^2]$ $= 18a^2 + 72b^2$
$(9m + 5n)^2 - (9m - 5n)^2$ simplifies to
A $180mn$ B $162m^2 + 50n^2$ C $-180mn$ D $90mn$

Answer and solution

Answer: (A) $180mn$

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

$(9m + 5n)^2 - (9m - 5n)^2 = 4(9m)(5n) = 180mn$
If $a + b = 18$ and $a - b = 10$ , then $ab =$
A $106$ B $112$ C $224$ D $56$

Answer and solution

Answer: (D) $56$

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

$18^2 - 10^2 = 4ab$

$\displaystyle 324 - 100 = 4ab \Rightarrow ab = \frac{224}{4} = 56$
If $(2m + 6n)^2 - kmn = (2m - 6n)^2$ , then $k =$
A $24$ B $12$ C $48$ D $-48$

Answer and solution

Answer: (C) $48$

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

$4ab = 4(2m)(6n) = 48mn$

$k = 48$
If $m - n = 5$ and $mn = 66$ , then $(m + n)^2 =$
A $269$ B $289$ C $157$ D $91$

Answer and solution

Answer: (B) $289$

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

$(m + n)^2 = (5)^2 + 4(66) = 25 + 264 = 289$
If $\displaystyle n - \frac{1}{n} = 10$ , then $\displaystyle n^2 + \frac{1}{n^2} =$
A $102$ B $98$ C $100$ D $104$

Answer and solution

Answer: (A) $102$

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

$= (10)^2 + 2 = 100 + 2 = 102$
If $m - n = 4$ and $mn = 77$ , then $m^2 + n^2 =$
A $324$ B $170$ C $93$ D $158$

Answer and solution

Answer: (B) $170$

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

$m^2 + n^2 = (4)^2 + 2(77) = 16 + 154 = 170$
The expansion of $(4a + 2b + c)^2$ is
A $16a^2 + 4b^2 + c^2 + 8ab + 2bc + 4ca$ B $16a^2 + 4b^2 + c^2 + 16ab + 4bc + 8ca$ C $4a^2 + 2b^2 + c^2 + 16ab + 4bc + 8ca$ D $16a^2 + 4b^2 + c^2$

Answer and solution

Answer: (B) $16a^2 + 4b^2 + c^2 + 16ab + 4bc + 8ca$

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

$(4a + 2b + c)^2 = (4a)^2 + (2b)^2 + (c)^2 + 2(4a)(2b) + 2(2b)(c) + 2(c)(4a)$
$(7p - 6q + 2r)^2$ equals
A $49p^2 + 36q^2 + 4r^2 - 84pq - 24qr + 28rp$ B $49p^2 + 36q^2 + 4r^2 - 84pq - 24qr - 28rp$ C $49p^2 + 36q^2 + 4r^2 - 84pq + 24qr + 28rp$ D $49p^2 + 36q^2 + 4r^2 + 84pq + 24qr + 28rp$

Answer and solution

Answer: (A) $49p^2 + 36q^2 + 4r^2 - 84pq - 24qr + 28rp$

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

$(7p - 6q + 2r)^2 = (7p)^2 + (6q)^2 + (2r)^2 - 2(7p)(6q) - 2(6q)(2r) + 2(2r)(7p)$
The coefficient of $ab$ in $(6a + 4b - 2c)^2$ is
A $-24$ B $24$ C $-48$ D $48$

Answer and solution

Answer: (D) $48$

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

The $ab$ term is $+ 2(6a)(4b) = 48ab$ , so the coefficient is $48$ .
The expansion of $(2n + 5)^3$ is
A $8n^3 + 125 + 20n^2 + 50n$ B $8n^3 + 125$ C $8n^3 + 125 + 30n$ D $8n^3 + 125 + 60n^2 + 150n$

Answer and solution

Answer: (D) $8n^3 + 125 + 60n^2 + 150n$

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

$(2n + 5)^3 = (2n)^3 + (5)^3 + 3(2n)(5)(2n + 5)$

$= 8n^3 + 125 + 60n^2 + 150n$
$(3x - y)^3$ equals
A $27x^3 - y^3 - 27x^2y + 9xy^2$ B $27x^3 - y^3 - 27x^2y - 9xy^2$ C $27x^3 - y^3$ D $27x^3 - y^3 + 27x^2y - 9xy^2$

Answer and solution

Answer: (A) $27x^3 - y^3 - 27x^2y + 9xy^2$

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

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

$= 27x^3 - y^3 - 27x^2y + 9xy^2$
The product $(p + 3)(p^2 - 3p + 9)$ is
A $p^3 - 27$ B $p^3 + 9p^2 + 27p + 27$ C $p^3 + 9$ D $p^3 + 27$

Answer and solution

Answer: (D) $p^3 + 27$

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

Here $a = p$ , $b = 3$ : $(p + 3)(p^2 - 3p + 9) = (p)^3 + (3)^3 = p^3 + 27$
If $m + n = 10$ and $mn = 16$ , then $m^3 + n^3 =$
A $1480$ B $952$ C $840$ D $520$

Answer and solution

Answer: (D) $520$

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

$= (10)^3 - 3(16)(10) = 1000 - 480 = 520$
If $p - q = 5$ and $pq = 50$ , then $p^3 - q^3 =$
A $-625$ B $275$ C $875$ D $375$

Answer and solution

Answer: (C) $875$

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

$= (5)^3 + 3(50)(5) = 125 + 750 = 875$
The product $(3m + 2n + 4)(9m^2 + 4n^2 + 16 - 6mn - 8n - 12m)$ is
A $27m^3 + 8n^3 + 64 - 72mn$ B $27m^3 + 8n^3 + 64 - 24mn$ C $27m^3 + 8n^3 + 64 + 72mn$ D $27m^3 + 8n^3 + 64$

Answer and solution

Answer: (A) $27m^3 + 8n^3 + 64 - 72mn$

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

Here $a = 3m$ , $b = 2n$ , $c = 4$

$= 27m^3 + 8n^3 + 64 - 72mn$
$(-13)^3 + 8^3 + 5^3 =$
A $-520$ B $1560$ C $-1560$ D $0$

Answer and solution

Answer: (C) $-1560$

$\text{Take } a = -13,\ b = 8,\ c = 5$

$a + b + c = -13 + 8 + 5 = 0$

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

$\therefore (-13)^3 + 8^3 + 5^3 = 3(-13)(8)(5) = -1560$

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MCQ test, Set 8