Algebraic Identities: MCQ test, Set 5

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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.

$3x(8x - 2y)$ equals
A $24x - 6xy$ B $24x^2 - 2y$ C $24x^2 - 6xy$ D $24x^2 + 6xy$

Answer and solution

Answer: (C) $24x^2 - 6xy$

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

$3x(8x - 2y) = (3x)(8x) - (3x)(2y) = 24x^2 - 6xy$
$(3p - 7)(p + 6)$ equals
A $3p^2 - 25p - 42$ B $3p^2 + 11p + 42$ C $3p^2 + 11p - 42$ D $3p^2 - 42$

Answer and solution

Answer: (C) $3p^2 + 11p - 42$

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

$(3p - 7)(p + 6) = (3p)(p) + (3p)(6) - (7)(p) - (7)(6)$

$= 3p^2 + 18p - 7p - 42 = 3p^2 + 11p - 42$
The product $(a + 2b)(a - 2b)$ is
A $a^2 + 4b^2 - 4ab$ B $a^2 - 4b^2$ C $a^2 + 4b^2$ D $a^2 - 2b^2$

Answer and solution

Answer: (B) $a^2 - 4b^2$

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

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

$= a^2 - 4b^2$
$(4m^4 + 5n^3)(4m^4 - 5n^3) =$
A $16m^8 - 25n^6$ B $4m^8 - 5n^6$ C $16m^8 + 25n^6$ D $16m^6 - 25n^5$

Answer and solution

Answer: (A) $16m^8 - 25n^6$

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

$(4m^4 + 5n^3)(4m^4 - 5n^3) = (4m^4)^2 - (5n^3)^2$ $= 16m^8 - 25n^6$
The expansion of $(9m + 3n)^2$ is
A $81m^2 + 9n^2 + 54mn$ B $9m^2 + 3n^2 + 54mn$ C $81m^2 + 9n^2 + 27mn$ D $81m^2 + 9n^2$

Answer and solution

Answer: (A) $81m^2 + 9n^2 + 54mn$

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

$(9m + 3n)^2 = (9m)^2 + (3n)^2 + 2(9m)(3n)$

$= 81m^2 + 9n^2 + 54mn$
In the expansion of $(3m + 8n)^2$ , the coefficient of $mn$ is
A $22$ B $24$ C $96$ D $48$

Answer and solution

Answer: (D) $48$

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

$2ab = 2(3m)(8n) = 48mn$

So the coefficient of $mn$ is $48$ .
$(7p - 3q)^2$ equals
A $49p^2 - 9q^2$ B $49p^2 + 9q^2 - 42pq$ C $49p^2 - 9q^2 - 42pq$ D $49p^2 + 9q^2 + 42pq$

Answer and solution

Answer: (B) $49p^2 + 9q^2 - 42pq$

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

$(7p - 3q)^2 = (7p)^2 + (3q)^2 - 2(7p)(3q)$

$= 49p^2 + 9q^2 - 42pq$
$\displaystyle \left(\frac{3p}{2} - \frac{5q}{4}\right)^2 =$
A $\displaystyle \frac{9p^2}{4} + \frac{25q^2}{16} - \frac{15pq}{8}$ B $\displaystyle \frac{9p^2}{2} + \frac{25q^2}{4} - \frac{15pq}{4}$ C $\displaystyle \frac{9p^2}{4} + \frac{25q^2}{16} + \frac{15pq}{4}$ D $\displaystyle \frac{9p^2}{4} + \frac{25q^2}{16} - \frac{15pq}{4}$

Answer and solution

Answer: (D) $\displaystyle \frac{9p^2}{4} + \frac{25q^2}{16} - \frac{15pq}{4}$

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

$\displaystyle \left(\frac{3p}{2} - \frac{5q}{4}\right)^2 = \left(\frac{3p}{2}\right)^2 + \left(\frac{5q}{4}\right)^2 - 2\left(\frac{3p}{2}\right)\left(\frac{5q}{4}\right)$

$\displaystyle = \frac{9p^2}{4} + \frac{25q^2}{16} - \frac{15pq}{4}$
$\displaystyle \left(2 + \frac{6}{y}\right)^2 + \left(2 - \frac{6}{y}\right)^2 =$
A $\displaystyle 4 + \frac{36}{y^2}$ B $\displaystyle 8 + \frac{72}{y^2} + \frac{48}{y}$ C $\displaystyle \frac{48}{y}$ D $\displaystyle 8 + \frac{72}{y^2}$

Answer and solution

Answer: (D) $\displaystyle 8 + \frac{72}{y^2}$

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

$\displaystyle \left(2 + \frac{6}{y}\right)^2 + \left(2 - \frac{6}{y}\right)^2 = 2[(2)^2 + \left(\frac{6}{y}\right)^2]$ $\displaystyle = 8 + \frac{72}{y^2}$
$(7x + 5y)^2 - (7x - 5y)^2 =$
A $70xy$ B $140xy$ C $-140xy$ D $98x^2 + 50y^2$

Answer and solution

Answer: (B) $140xy$

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

$(7x + 5y)^2 - (7x - 5y)^2 = 4(7x)(5y) = 140xy$
If $x + y = 6$ and $x - y = 4$ , then $xy =$
A $13$ B $20$ C $5$ D $10$

Answer and solution

Answer: (C) $5$

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

$6^2 - 4^2 = 4xy$

$\displaystyle 36 - 16 = 4xy \Rightarrow xy = \frac{20}{4} = 5$
If $(6x - 4y)^2 + kxy = (6x + 4y)^2$ , then $k =$
A $48$ B $-96$ C $96$ D $24$

Answer and solution

Answer: (C) $96$

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

$4ab = 4(6x)(4y) = 96xy$

$k = 96$
If $m + n = 12$ and $mn = 35$ , then $(m - n)^2 =$
A $284$ B $4$ C $74$ D $214$

Answer and solution

Answer: (B) $4$

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

$(m - n)^2 = (12)^2 - 4(35) = 144 - 140 = 4$
If $\displaystyle m - \frac{1}{m} = 9$ , then $\displaystyle m^2 + \frac{1}{m^2} =$
A $83$ B $79$ C $81$ D $85$

Answer and solution

Answer: (A) $83$

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

$= (9)^2 + 2 = 81 + 2 = 83$
If $a + b = 18$ and $ab = 80$ , then $a^2 + b^2 =$
A $164$ B $484$ C $4$ D $244$

Answer and solution

Answer: (A) $164$

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

$a^2 + b^2 = (18)^2 - 2(80) = 324 - 160 = 164$
In the expansion of $(7x + 4y + z)^2$ , the coefficient of $zx$ is
A $7$ B $28$ C $14$ D $16$

Answer and solution

Answer: (C) $14$

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

The $zx$ term is $2(z)(7x) = 14zx$ , so the coefficient is $14$ .
$(p - 5q + 7r)^2$ equals
A $p^2 + 25q^2 + 49r^2 - 10pq - 70qr - 14rp$ B $p^2 + 25q^2 + 49r^2 - 10pq - 70qr + 14rp$ C $p^2 + 25q^2 + 49r^2 - 10pq + 70qr + 14rp$ D $p^2 + 25q^2 + 49r^2 + 10pq + 70qr + 14rp$

Answer and solution

Answer: (B) $p^2 + 25q^2 + 49r^2 - 10pq - 70qr + 14rp$

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

$(p - 5q + 7r)^2 = (p)^2 + (5q)^2 + (7r)^2 - 2(p)(5q) - 2(5q)(7r) + 2(7r)(p)$
The coefficient of $qr$ in $(p + 7q - 3r)^2$ is
A $-21$ B $21$ C $42$ D $-42$

Answer and solution

Answer: (D) $-42$

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

The $qr$ term is $- 2(7q)(3r) = -42qr$ , so the coefficient is $-42$ .
$(2x + 4)^3 =$
A $8x^3 + 64 + 48x^2 + 96x$ B $8x^3 + 64 + 24x$ C $8x^3 + 64 + 16x^2 + 32x$ D $8x^3 + 64$

Answer and solution

Answer: (A) $8x^3 + 64 + 48x^2 + 96x$

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

$(2x + 4)^3 = (2x)^3 + (4)^3 + 3(2x)(4)(2x + 4)$

$= 8x^3 + 64 + 48x^2 + 96x$
$(2m - n)^3 =$
A $8m^3 - n^3 - 12m^2n + 6mn^2$ B $8m^3 - n^3$ C $8m^3 - n^3 + 12m^2n - 6mn^2$ D $8m^3 - n^3 - 12m^2n - 6mn^2$

Answer and solution

Answer: (A) $8m^3 - n^3 - 12m^2n + 6mn^2$

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

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

$= 8m^3 - n^3 - 12m^2n + 6mn^2$
$(3t + 5)(9t^2 - 15t + 25) =$
A $27t^3 + 25$ B $27t^3 - 125$ C $27t^3 + 125$ D $27t^3 + 135t^2 + 225t + 125$

Answer and solution

Answer: (C) $27t^3 + 125$

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

Here $a = 3t$ , $b = 5$ : $(3t + 5)(9t^2 - 15t + 25) = (3t)^3 + (5)^3 = 27t^3 + 125$
If $x + y = 14$ and $xy = 48$ , then $x^3 + y^3 =$
A $2600$ B $728$ C $2072$ D $4760$

Answer and solution

Answer: (B) $728$

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

$= (14)^3 - 3(48)(14) = 2744 - 2016 = 728$
$(4t - 3)(16t^2 + 12t + 9) =$
A $64t^3 + 27$ B $4t^3 - 27$ C $64t^3 - 27$ D $64t^3 - 27 - 144t^2 + 108t$

Answer and solution

Answer: (C) $64t^3 - 27$

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

$(4t - 3)(16t^2 + 12t + 9) = (4t)^3 - (3)^3 = 64t^3 - 27$
$(4m + 3n + 2)(16m^2 + 9n^2 + 4 - 12mn - 6n - 8m) =$
A $64m^3 + 27n^3 + 8 - 24mn$ B $64m^3 + 27n^3 + 8$ C $64m^3 + 27n^3 + 8 + 72mn$ D $64m^3 + 27n^3 + 8 - 72mn$

Answer and solution

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

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

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

$= 64m^3 + 27n^3 + 8 - 72mn$
The value of $4^3 + (-10)^3 + 6^3$ is
A $-240$ B $0$ C $720$ D $-720$

Answer and solution

Answer: (D) $-720$

$\text{Take } a = 4,\ b = -10,\ c = 6$

$a + b + c = 4 - 10 + 6 = 0$

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

$\therefore 4^3 + (-10)^3 + 6^3 = 3(4)(-10)(6) = -720$

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MCQ test, Set 5