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# SNAP Algebra Questions PDF [Important]

The Algebra is an important topic in the Quant section of the SNAP Exam. You can also download this Free Algebra Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Algebra questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practicing.

Question 1:Â If $a^{2}+b^{2}-c^{2}=0$, then the value of $\frac{2(a^{6}+b^{6}-c^{6})}{3a^{2}b^{2}c^{2}}$ is:

a)Â 3

b)Â 1

c)Â 0

d)Â 2

Solution:

If a + b + cÂ  = 0 thenÂ $a^3 + b^3 + c^3 = 3abc$ so,

$a^{6}+b^{6}-c^{6} = 3a^2b^2c^2$

$\frac{2(a^{6}+b^{6}-c^{6})}{3a^{2}b^{2}c^{2}}$

=Â $\frac{2(3a^{2}b^{2}c^{2})}{3a^{2}b^{2}c^{2}}$ = 2

Question 2:Â If $a+\frac{1}{a}=5$ then $a^{3} + \frac{1}{a^{3}}$ is:

a)Â 110

b)Â 10

c)Â 80

d)Â 140

$a^{3} + \frac{1}{a^{3}}$
= $($a+\frac{1}{a})^3 – 3(a+\frac{1}{a})$($\because (a + b)^3 = a^3 + b^3 + 3ab(a + b))$=$5^3 – 3(5)$= 110 Question 3:Â The coefficient of x in$(x – 3y)^{3}$is : a)Â$3 y^{2}$b)Â$27 y^{2}$c)Â$-27 y^{2}$d)Â$-3 y^{2}$3)Â AnswerÂ (B) Solution:$(x – 3y)^{3} = x^3 – (3y)^3 – 3x.3y(x – 3y)$($(a – b)^3 = a^3 – b^3 – 3ab(a – b)$) =$x^3 – 27y^3 – 9xy(x – 3y)$=$x^3 – 27y^3 – 9x^2y – 27xy^2$The coefficient of x = 27$y^2$Question 4:Â Expand$\left(\frac{x}{3} + \frac{y}{5}\right)^3$a)Â$\frac{x^3}{27} + \frac{x^2y}{25} + \frac{xy^2}{25} + \frac{y^3}{125}$b)Â$\frac{x^3}{25} + \frac{x^2y}{15} + \frac{xy^2}{25} + \frac{y^3}{125}$c)Â$\frac{x^3}{27} + \frac{xy}{15} + \frac{xy^2}{25} + \frac{y^3}{125}$d)Â$\frac{x^3}{27} + \frac{x^2y}{15} + \frac{xy^2}{25} + \frac{y^3}{125}$4)Â AnswerÂ (D) Solution:$\left(\frac{x}{3} + \frac{y}{5}\right)^3$($\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$) =Â$(\frac{x}{3})^3 + (\frac{y}{5})^3 + 3(\frac{x}{3})(\frac{y}{5})(\frac{x}{3} + \frac{y}{5})Â $=$\frac{x^3}{27} + \frac{y^3}{125} + \frac{xy}{5} (\frac{x}{3} + \frac{y}{5}) $=$\frac{x^3}{27} + \frac{y^3}{125} + \frac{x^2y}{15} + \frac{xy^2}{25} $Question 5:Â If$a^2 + b^2 + c^2 = 300$and$ab + bc + ca = 50$, then what is the value of$a + b + c$? (Given that a, b and c are all positive.) a)Â 22 b)Â 20 c)Â 25 d)Â 15 5)Â AnswerÂ (B) Solution:$(a + b + c)^2 =Â a^2 + b^2 + c^2 +Â 2(ab + bc + ca)(a + b + c)^2 = 300 + 2(50)(a + b + c)^2 = 400$a + b + c = 20 TakeÂ SNAP mock tests here Question 6:Â If x + y + z = 10 and xy + yz + zx = 15, then find the value of$x^3 + y^3 + z^3 â€” 3xyz$. a)Â 660 b)Â 525 c)Â 550 d)Â 575 6)Â AnswerÂ (C) Solution:$x^3 + y^3 + z^3 â€” 3xyz = (x + y + z)(x^2 + y^2 +Â z^2 – xy – yz – xz)$x + y + z = 10 Taking square on both sides,$(x + y + z)^2 = 100x^2 + y^2 + z^2 + 2(xy + yz + xz) = 100x^2 + y^2 + z^2 = 100 – 2\times 15 = 00 – 30 = 70x^3 + y^3 + z^3 â€” 3xyz = (10)(70 – 15) = 10 \times 55 = 550$Question 7:Â If$x^{2} – 4x+4=0 $, then the value of 16$(x^{4} – \frac{1}{x^{4}})$is a)Â 127 b)Â 255 c)Â$\frac{127}{16}$d)Â$\frac{255}{16}$7)Â AnswerÂ (B) Solution:$x^{2} â€”4x+4=0 x^{2} â€”2x – 2x+4=0 x(x â€” 2) – 2(x â€” 2)=0 (x â€” 2)(x â€” 2)=0 $x = 2 now, 16$(x^{4}-\frac{1}{x^{4}})$=Â 16$(2^{4}-\frac{1}{2^{4}})$= 16$(16-\frac{1}{16})$=$16^2 – 1$= 255 Question 8:Â If$a^{3}+\frac{1}{a^{3}} = 52$then the value of$2\left(a + \frac{1}{a}\right)$is : a)Â 8 b)Â 2 c)Â 6 d)Â 4 8)Â AnswerÂ (A) Solution:$a^{3}+\frac{1}{a^{3}} = 52(a + \frac{1}{a})^3 – 3.a.\frac{1}{a}(a + \frac{1}{a}) = 52(\because a^3 + b^3 = (a + b)^3 – 3ab(a + b))(a + \frac{1}{a})^3 – 3(a + \frac{1}{a}) = 52$From the option A) – Put the value of$2(a + \frac{1}{a}) = 8$,$(a + \frac{1}{a}) = 4$L.H.S.,$4^3 – 3 \times 4$= 52 = R.H.S.$\therefore$The value of$2\left(a + \frac{1}{a}\right)$is 8. Question 9:Â If$b + c = ax, c + a = by, a + b = cz$, then the value$\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$is: a)Â$\frac{1}{9}$b)Â 1 c)Â 0 d)Â$\frac{1}{3}$9)Â AnswerÂ (A) Solution:$b + c = ax, c + a = by, a + b = cz$x =$\frac{b + c}{a}$y =$\frac{c + a}{b}$z =$\frac{a + b}{c}$Now,$\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$x + 1 =Â$\frac{b + c}{a}$+ 1 =$\frac{a + b + c}{a}$y + 1 =$\frac{c + a}{b}$+ 1 =$\frac{a + b + c}{b}$z + 1 =$\frac{a + b}{c}$+ 1 =$\frac{a + b + c}{c}$=Â$\frac{1}{9}\left[\frac{1}{\frac{a + b + c}{a}}+\frac{1}{\frac{a + b + c}{b}}+\frac{1}{\frac{a + b + c}{c}}\right]\frac{1}{9}\left[\frac{a}{a + b + c}+\frac{b}{a + b + c}+\frac{c}{a + b + c}\right]\frac{1}{9}[\frac{a + b + c}{a + b + c}]$= 1/9 Question 10:Â Find the product of$(a + b + 2c)(a^{2} + b^{2} + 4c^{2} – ab – 2bc – 2ca)$a)Â$a^{3} + b^{3} + 8c^{3} – 2abc$b)Â$a^{3} + b^{3} + 8c^{3} – abc$c)Â$a^{3} + b^{3} + 6c^{3} – 6abc$d)Â$a^{3} + b^{3} + 8c^{3} – 6abc$10)Â AnswerÂ (D) Solution:$(a + b + 2c)(a^{2} + b^{2} + 4c^{2} – ab – 2bc – 2ca)= a^3 + b^3 + (2c)^3 – 3 \times a \times b \times 2c(\becauseÂ a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca))=Â a^3 + b^3 + 8c^3 – 6abc$Question 11:Â$25a^{2}-9$is factored as a)Â$(5a + 3)(5aÂ –Â 3)$b)Â$(5a + 1)(5a –Â 9)$c)Â$(5a – 3)^{2}$d)Â$(25a +Â 1)(a -9)$11)Â AnswerÂ (A) Solution:$25a^{2}-9$=$(5a)^2 – (3)^2$= ($\because a^2 – b^2 = (a + b)(a – b)$) = (5a + 3)(5a – 3) Question 12:Â If$a^{4} + \frac{1}{a^{4}} = 50$, then find the value of$a^{3} + \frac{1}{a^{3}}$a)Â$\sqrt{2(1+\sqrt{3})}+(-1+2\sqrt{13})$b)Â$\sqrt{2(1+\sqrt{3})}(3-2\sqrt{13})$c)Â$\sqrt{2(\sqrt{13}+1)}(3+2\sqrt{13})$d)Â$\sqrt{2(1-\sqrt{3})}(-1+2\sqrt{13})$12)Â AnswerÂ (C) Solution:$a^{4} + \frac{1}{a^{4}} = 50a^{4} + \frac{1}{a^{4}} + 2= 50 + 2(a^2+\frac{1}{a^2})^2=52(a^2+\frac{1}{a^2})=\sqrt{52}a^2+\frac{1}{a^2} + 2Â = \sqrt{52} + 2(a + \frac{1}{a})^2 =Â \sqrt{52} + 2(a + \frac{1}{a})Â = \sqrt{\sqrt{52} + 2}a^{3} + \frac{1}{a^{3}} = (a + b)^3 + 3ab(a + b)$=$(\sqrt{\sqrt{52} + 2})^3 +Â \sqrt{\sqrt{52} + 2}$=$(\sqrt{2\sqrt{13} + 2})^3 + \sqrt{2\sqrt{13} + 2}$=$\sqrt{2\sqrt{13} + 2}(1 +Â (\sqrt{2\sqrt{13} + 2})^2)$=$\sqrt{2\sqrt{13} + 2}(1 + {2\sqrt{13} + 2})$=$\sqrt{2(\sqrt{13} + 1})(3 + {2\sqrt{13}})$Question 13:Â$(a + b – c + d)^2 – (a – b + c – d)^2 = ?$a)Â$4a(b + d – c)$b)Â$2a(a + b – c)$c)Â$2a(b + c – d)$d)Â$4a(b – d + c)$13)Â AnswerÂ (A) Solution:$(a + b – c + d)^2 – (a – b + c – d)^2$= [(a + b – c + d) +Â (a – b + c – d)][(a + b – c + d) – (a – b + c – d)] ($\because a^2 – b^2 = (a + b)(a – b)$) = (2a)(2b-2c + 2d) = 4a(b – c + d) Question 14:Â The value of$27a^3 – 2\sqrt{2}b^3$is equal to: a)Â$(3a – \sqrt{2}b)(9a^2 – 2b^2 + 6\sqrt{2}ab)$b)Â$(3a – \sqrt{2}b)(9a^2 + 2b^2 + 6\sqrt{2}ab)$c)Â$(3a – \sqrt{2}b)(9a^2 + 2b^2 + 3\sqrt{2}ab)$d)Â$(3a – \sqrt{2}b)(9a^2 – 2b^2 – 3\sqrt{2}ab)$14)Â AnswerÂ (C) Solution:$27a^3 – 2\sqrt{2}b^3 = (3a – \sqrt{2}b)(9a^2 + 2b^2 + 6\sqrt{2}ab)$($\because a^3 – b^3 = (a – b)(a^2 +Â ab + b^2)$) here, a = 3a b =$\sqrt{2}b$Question 15:Â If$x+3y+2=0$then value of$x^{3}+27y^{3}+8-18xy$is: a)Â -2 b)Â 2 c)Â 1 d)Â 0 15)Â AnswerÂ (D) Solution:$x+3y+2=0$x + 3y = -2 Taking cube both sides,$(x + 3y)^3 = -8x^3 + 27y^3 + 3x.3y(x +Â 3y) = -8x^3 + 27y^3 + 9xy(-2) = -8 x^{3}+27y^{3} -18xy = -8x^{3}+27y^{3}+8-18xy$= 0 Question 16:Â If$p+q=7$and$pq=5$, then the value of$p^{3}+q^{3}$is: a)Â 34 b)Â 238 c)Â 448 d)Â 64 16)Â AnswerÂ (B) Solution:$p^{3}+q^{3} = (p + q)^3 – 3pq(pÂ + q)$=$7^3 – 3 \times 5(7)$= 343 – 105 = 238 Question 17:Â If$30x^2 – 15x + 1 = 0$, then what is the value of$25x^2 + (36x^2)^{-1}$? a)Â$\frac{9}{2}$b)Â$6\frac{1}{4}$c)Â$\frac{65}{12}$d)Â$\frac{55}{12}$17)Â AnswerÂ (D) Solution:$30x^2 – 15x + 1 = 0$Dividing by x,$30x – 15 + \frac{1}{x} = 05x – 15/6 +Â \frac{1}{6x} = 05xÂ  +Â \frac{1}{6x}Â = 5/2$taking square both side,$(5x + \frac{1}{6x})^2 = 25/425x^2 + \frac{1}{36x^2} + 2 \times 5x \timesÂ \frac{1}{6x} = 25/425x^2 + \frac{1}{36x^2} = 25/4 – 5/325x^2 + \frac{1}{36x^2} = \frac{55}{12} $Question 18:Â If a + b + c = 7 and ab + bc + ca = -6, then the value of$a^3 + b^3 + c^3 – 3abc$is: a)Â 469 b)Â 472 c)Â 463 d)Â 479 18)Â AnswerÂ (A) Solution: We know that,$a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 +Â b^2 +Â c^2 – (ab + bc + ac))$a + b + c = 7 Squaring both sides,$(a + b + c)^2 = 49a^2 + b^2 + c^2 + 2(ab + bc + ac) = 49a^2 + b^2 + c^2 = 49 + 12 = 61a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – (ab + bc + ac))$= 7(61 – (-6)) = 7$\times$67 = 469 Question 19:Â The given table represents the revenue (in â‚¹ crores) of a company from the sale of four products A, B, C and D in 6 years. Study the table carefully and answer the question that follows. By what percentage is the total revenue of the company from the sale of products A, B and D in 2012 and 2013 more than the total revenue from the sale of product B in 2013 to 2016?(Correct to one decimal place) a)Â 44.5 b)Â 31.2 c)Â 43.6 d)Â 45.4 19)Â AnswerÂ (D) Solution: Total revenue of the company from the sale of products A, B and D in 2012 and 2013 = 98 + 74 + 74 + 94 + 96 + 102 = 538 Total revenue from the sale of product B in 2013 to 2016 = 96 + 92 + 84 + 98 = 370 Required percentage =$\frac{538 – 370}{370} \times 100$= 45.4% Question 20:Â If$P = \frac{x^4 – 8x}{x^3 – x^2 – 2x}, Q = \frac{x^2 + 2x + 1}{x^2 – 4x – 5}$and$R = \frac{2x^2 + 4x + 8}{x – 5}$, then$(P \times Q) \div R$is equal to: a)Â$\frac{1}{2}$b)Â 1 c)Â 2 d)Â 4 20)Â AnswerÂ (A) Solution:$P = \frac{x^4 – 8x}{x^3 – x^2 – 2x}Q = \frac{x^2 + 2x + 1}{x^2 – 4x – 5}$=Â$\frac{(x +Â 1)^2}{x^2 – 4x – 5 + 9 – 9}(P \times Q) \div R$=$(\frac{x^4 – 8x}{x^3 – x^2 – 2x} \times \frac{x^2 + 2x + 1}{x^2 – 4x – 5}) \div \frac{2x^2 + 4x + 8}{x – 5}$=$\frac{x^4 – 8x}{x^3 – x^2 – 2x} \times \frac{x^2 + 2x + 1}{x^2 – 4x – 5} \times \frac{x – 5}{2x^2 + 4x + 8}$=$\frac{x(x^3 – 8)}{x^3 – x^2 – 2x} \times \frac{x^2 + 2x + 1}{x^2 – 4x – 5} \times \frac{x – 5}{2(x^2 + 2x + 4)}$=$\frac{x(x – 2)(x^2 + 2x – 4)}{x(x^2 – x – 2)} \times \frac{(x + 1)^2}{x^2 – 5x + x – 5} \times \frac{x – 5}{2(x^2 + 2x + 4)}$=$\frac{(x – 2)}{(x^2 – 2x + x- 2)} \times \frac{(x + 1)^2}{(x +1)(x – 5)} \times \frac{x – 5}{2}$=$\frac{(x – 2)}{(x – 2)(x + 1)} \times \frac{(x + 1)}{2}$=$\frac{1}{2}\$