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Here you can download CMAT 2022 – important MAH – CET Algebra Questions PDF by Cracku. Very Important MAH – CET 2022 and These questions will help your MAH – CET preparation. So kindly download the PDF for reference and do more practice.

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

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

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 $-3 y^{2}$

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 $\frac{x^3}{27} + \frac{x^2y}{15} + \frac{xy^2}{25} + \frac{y^3}{125}$

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

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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

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 $\frac{255}{16}$

Question 8: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 $\frac{1}{3}$

Question 9: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 $a^{3} + b^{3} + 8c^{3} – 6abc$

Question 10: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

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 $(25a +Â 1)(a -9)$

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 $\sqrt{2(1-\sqrt{3})}(-1+2\sqrt{13})$

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 $4a(b – d + c)$

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 $(3a – \sqrt{2}b)(9a^2 – 2b^2 – 3\sqrt{2}ab)$

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

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

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 $\frac{55}{12}$

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

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

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

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

$a^{3} + \frac{1}{a^{3}}$
= $($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 3)Â AnswerÂ (B)$(x – 3y)^{3} = x^3 – (3y)^3 – 3x.3y(x – 3y)$($(a – b)^3 = a^3 – b^3 – 3ab(a – b)$) =$x^3 – 27y^3 – 9xy(x – 3y)$=$x^3 – 27y^3 – 9x^2y – 27xy^2$The coefficient of x = 27$y^2$4)Â AnswerÂ (D)$\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} $5)Â AnswerÂ (B)$(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$a + b + c = 20 6)Â AnswerÂ (C)$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$7)Â AnswerÂ (B)$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 8)Â AnswerÂ (A)$b + c = ax, c + a = by, a + b = cz$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 9)Â AnswerÂ (D)$(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$10)Â AnswerÂ (A)$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. 11)Â AnswerÂ (A)$25a^{2}-9$=$(5a)^2 – (3)^2$= ($\because a^2 – b^2 = (a + b)(a – b)$) = (5a + 3)(5a – 3) 12)Â AnswerÂ (C)$a^{4} + \frac{1}{a^{4}} = 50a^{4} + \frac{1}{a^{4}} + 2= 50 + 2(a^2+\frac{1}{a^2})^2=52(a^2+\frac{1}{a^2})=\sqrt{52}a^2+\frac{1}{a^2} + 2Â = \sqrt{52} + 2(a + \frac{1}{a})^2 =Â \sqrt{52} + 2(a + \frac{1}{a})Â = \sqrt{\sqrt{52} + 2}a^{3} + \frac{1}{a^{3}} = (a + b)^3 + 3ab(a + b)$=$(\sqrt{\sqrt{52} + 2})^3 +Â \sqrt{\sqrt{52} + 2}$=$(\sqrt{2\sqrt{13} + 2})^3 + \sqrt{2\sqrt{13} + 2}$=$\sqrt{2\sqrt{13} + 2}(1 +Â (\sqrt{2\sqrt{13} + 2})^2)$=$\sqrt{2\sqrt{13} + 2}(1 + {2\sqrt{13} + 2})$=$\sqrt{2(\sqrt{13} + 1})(3 + {2\sqrt{13}})$13)Â AnswerÂ (A)$(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) 14)Â AnswerÂ (C)$27a^3 – 2\sqrt{2}b^3 = (3a – \sqrt{2}b)(9a^2 + 2b^2 + 6\sqrt{2}ab)$($\because a^3 – b^3 = (a – b)(a^2 +Â ab + b^2)$) here, a = 3a b =$\sqrt{2}b$15)Â AnswerÂ (D)$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$= 0 16)Â AnswerÂ (B)$p^{3}+q^{3} = (p + q)^3 – 3pq(pÂ + q)$=$7^3 – 3 \times 5(7)$= 343 – 105 = 238 17)Â AnswerÂ (D)$30x^2 – 15x + 1 = 0$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} $18)Â AnswerÂ (A) We know that,$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$67 = 469 19)Â AnswerÂ (D) Total revenue of the company from the sale of products A, B and D in 2012 and 2013 = 98 + 74 + 74 + 94 + 96 + 102 = 538 Total revenue from the sale of product B in 2013 to 2016 = 96 + 92 + 84 + 98 = 370 Required percentage =$\frac{538 – 370}{370} \times 100$= 45.4% 20)Â AnswerÂ (A)$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}\$