Algebraic Identities: MCQ test, Set 7

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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.

$2x(9x + 7y) =$
A $18x^2 - 14xy$ B $18x + 14xy$ C $18x^2 + 7y$ D $18x^2 + 14xy$

Answer and solution

Answer: (D) $18x^2 + 14xy$

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

$2x(9x + 7y) = (2x)(9x) + (2x)(7y) = 18x^2 + 14xy$
$(3m - 1)(m + 5) =$
A $3m^2 - 5$ B $3m^2 - 16m - 5$ C $3m^2 + 14m - 5$ D $3m^2 + 14m + 5$

Answer and solution

Answer: (C) $3m^2 + 14m - 5$

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

$(3m - 1)(m + 5) = (3m)(m) + (3m)(5) - (1)(m) - (1)(5)$

$= 3m^2 + 15m - m - 5 = 3m^2 + 14m - 5$
$(5x - 9y)(5x + 9y) =$
A $5x^2 - 9y^2$ B $25x^2 + 81y^2$ C $25x^2 - 81y^2$ D $25x^2 + 81y^2 - 90xy$

Answer and solution

Answer: (C) $25x^2 - 81y^2$

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

$(5x - 9y)(5x + 9y) = (5x)^2 - (9y)^2$

$= 25x^2 - 81y^2$
$(5m^4 + 2n^4)(5m^4 - 2n^4) =$
A $25m^8 - 4n^8$ B $5m^8 - 2n^8$ C $25m^6 - 4n^6$ D $25m^8 + 4n^8$

Answer and solution

Answer: (A) $25m^8 - 4n^8$

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

$(5m^4 + 2n^4)(5m^4 - 2n^4) = (5m^4)^2 - (2n^4)^2$ $= 25m^8 - 4n^8$
$(8x + 4y)^2 =$
A $8x^2 + 4y^2 + 64xy$ B $64x^2 + 16y^2 + 32xy$ C $64x^2 + 16y^2$ D $64x^2 + 16y^2 + 64xy$

Answer and solution

Answer: (D) $64x^2 + 16y^2 + 64xy$

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

$(8x + 4y)^2 = (8x)^2 + (4y)^2 + 2(8x)(4y)$

$= 64x^2 + 16y^2 + 64xy$
In the expansion of $(8a + 9b)^2$ , the coefficient of $ab$ is
A $288$ B $72$ C $34$ D $144$

Answer and solution

Answer: (D) $144$

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

$2ab = 2(8a)(9b) = 144ab$

So the coefficient of $ab$ is $144$ .
$(5m - n)^2 =$
A $25m^2 - n^2$ B $25m^2 + n^2 - 10mn$ C $25m^2 + n^2 + 10mn$ D $25m^2 - n^2 - 10mn$

Answer and solution

Answer: (B) $25m^2 + n^2 - 10mn$

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

$(5m - n)^2 = (5m)^2 + (n)^2 - 2(5m)(n)$

$= 25m^2 + n^2 - 10mn$
$\displaystyle \left(\frac{x}{4} - \frac{7y}{5}\right)^2 =$
A $\displaystyle \frac{x^2}{4} + \frac{49y^2}{5} - \frac{7xy}{10}$ B $\displaystyle \frac{x^2}{16} + \frac{49y^2}{25} + \frac{7xy}{10}$ C $\displaystyle \frac{x^2}{16} + \frac{49y^2}{25} - \frac{7xy}{20}$ D $\displaystyle \frac{x^2}{16} + \frac{49y^2}{25} - \frac{7xy}{10}$

Answer and solution

Answer: (D) $\displaystyle \frac{x^2}{16} + \frac{49y^2}{25} - \frac{7xy}{10}$

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

$\displaystyle \left(\frac{x}{4} - \frac{7y}{5}\right)^2 = \left(\frac{x}{4}\right)^2 + \left(\frac{7y}{5}\right)^2 - 2\left(\frac{x}{4}\right)\left(\frac{7y}{5}\right)$

$\displaystyle = \frac{x^2}{16} + \frac{49y^2}{25} - \frac{7xy}{10}$
$\displaystyle \left(5 + \frac{8}{q}\right)^2 + \left(5 - \frac{8}{q}\right)^2 =$
A $\displaystyle 25 + \frac{64}{q^2}$ B $\displaystyle 50 + \frac{128}{q^2} + \frac{160}{q}$ C $\displaystyle \frac{160}{q}$ D $\displaystyle 50 + \frac{128}{q^2}$

Answer and solution

Answer: (D) $\displaystyle 50 + \frac{128}{q^2}$

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

$\displaystyle \left(5 + \frac{8}{q}\right)^2 + \left(5 - \frac{8}{q}\right)^2 = 2[(5)^2 + \left(\frac{8}{q}\right)^2]$ $\displaystyle = 50 + \frac{128}{q^2}$
$(4m + 9n)^2 - (4m - 9n)^2 =$
A $32m^2 + 162n^2$ B $144mn$ C $-144mn$ D $72mn$

Answer and solution

Answer: (B) $144mn$

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

$(4m + 9n)^2 - (4m - 9n)^2 = 4(4m)(9n) = 144mn$
If $a + b = 16$ and $a - b = 2$ , then $ab =$
A $65$ B $63$ C $126$ D $252$

Answer and solution

Answer: (B) $63$

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

$16^2 - 2^2 = 4ab$

$\displaystyle 256 - 4 = 4ab \Rightarrow ab = \frac{252}{4} = 63$
If $(7p - 4q)^2 + kpq = (7p + 4q)^2$ , then $k =$
A $56$ B $-112$ C $112$ D $28$

Answer and solution

Answer: (C) $112$

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

$4ab = 4(7p)(4q) = 112pq$

$k = 112$
If $x + y = 8$ and $xy = 7$ , then $(x - y)^2 =$
A $36$ B $50$ C $78$ D $92$

Answer and solution

Answer: (A) $36$

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

$(x - y)^2 = (8)^2 - 4(7) = 64 - 28 = 36$
If $\displaystyle q + \frac{1}{q} = 15$ , then $\displaystyle q^2 + \frac{1}{q^2} =$
A $225$ B $227$ C $221$ D $223$

Answer and solution

Answer: (D) $223$

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

$= (15)^2 - 2 = 225 - 2 = 223$
If $m + n = 6$ and $mn = 5$ , then $m^2 + n^2 =$
A $31$ B $46$ C $26$ D $16$

Answer and solution

Answer: (C) $26$

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

$m^2 + n^2 = (6)^2 - 2(5) = 36 - 10 = 26$
In the expansion of $(3a + 2b + 6c)^2$ , the coefficient of $ca$ is
A $72$ B $36$ C $18$ D $45$

Answer and solution

Answer: (B) $36$

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

The $ca$ term is $2(6c)(3a) = 36ca$ , so the coefficient is $36$ .
$(7a + b - 6c)^2 =$
A $49a^2 + b^2 + 36c^2 + 14ab + 12bc + 84ca$ B $49a^2 + b^2 + 36c^2 + 14ab + 12bc - 84ca$ C $49a^2 + b^2 + 36c^2 + 14ab - 12bc + 84ca$ D $49a^2 + b^2 + 36c^2 + 14ab - 12bc - 84ca$

Answer and solution

Answer: (D) $49a^2 + b^2 + 36c^2 + 14ab - 12bc - 84ca$

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

$(7a + b - 6c)^2 = (7a)^2 + (b)^2 + (6c)^2 + 2(7a)(b) - 2(b)(6c) - 2(6c)(7a)$
The coefficient of $qr$ in $(3p - q - 7r)^2$ is
A $14$ B $-14$ C $-7$ D $7$

Answer and solution

Answer: (A) $14$

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

The $qr$ term is $+ 2(q)(7r) = 14qr$ , so the coefficient is $14$ .
$(3m + 2n)^3 =$
A $27m^3 + 8n^3$ B $27m^3 + 8n^3 + 18m^2n + 12mn^2$ C $27m^3 + 8n^3 + 54m^2n + 36mn^2$ D $27m^3 + 8n^3 + 18mn$

Answer and solution

Answer: (C) $27m^3 + 8n^3 + 54m^2n + 36mn^2$

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

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

$= 27m^3 + 8n^3 + 54m^2n + 36mn^2$
$(r - 4)^3 =$
A $r^3 - 64$ B $r^3 - 64 - 12r^2 + 48r$ C $r^3 - 64 + 12r^2 - 48r$ D $r^3 - 64 - 12r^2 - 48r$

Answer and solution

Answer: (B) $r^3 - 64 - 12r^2 + 48r$

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

$(r - 4)^3 = (r)^3 - (4)^3 - 3(r)(4)(r - 4)$

$= r^3 - 64 - 12r^2 + 48r$
$(3y + 2)(9y^2 - 6y + 4) =$
A $27y^3 + 8$ B $27y^3 + 54y^2 + 36y + 8$ C $27y^3 + 4$ D $27y^3 - 8$

Answer and solution

Answer: (A) $27y^3 + 8$

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

Here $a = 3y$ , $b = 2$ : $(3y + 2)(9y^2 - 6y + 4) = (3y)^3 + (2)^3 = 27y^3 + 8$
If $x + y = 5$ and $xy = 6$ , then $x^3 + y^3 =$
A $35$ B $107$ C $215$ D $95$

Answer and solution

Answer: (A) $35$

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

$= (5)^3 - 3(6)(5) = 125 - 90 = 35$
$(4z - 1)(16z^2 + 4z + 1) =$
A $64z^3 - 1 - 48z^2 + 12z$ B $64z^3 + 1$ C $64z^3 - 1$ D $4z^3 - 1$

Answer and solution

Answer: (C) $64z^3 - 1$

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

$(4z - 1)(16z^2 + 4z + 1) = (4z)^3 - (1)^3 = 64z^3 - 1$
$(2m + n + 3)(4m^2 + n^2 + 9 - 2mn - 3n - 6m) =$
A $8m^3 + n^3 + 27 - 18mn$ B $8m^3 + n^3 + 27 - 6mn$ C $8m^3 + n^3 + 27 + 18mn$ D $8m^3 + n^3 + 27$

Answer and solution

Answer: (A) $8m^3 + n^3 + 27 - 18mn$

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

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

$= 8m^3 + n^3 + 27 - 18mn$
The value of $5^3 - 3^3 - 2^3$ is
A $30$ B $90$ C $0$ D $-90$

Answer and solution

Answer: (B) $90$

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

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

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

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

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MCQ test, Set 7