Algebra Questions for CMAT 2022 – Download PDF

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Algebra Questions for CMAT
Algebra Questions for CMAT

Algebra Questions for CMAT 2022 – Download PDF

Download CMAT 2022 Algebra Questions pdf by Cracku. Very Important Algebra Questions for CMAT 2022 based on asked questions in previous exam papers. These questions will help your CMAT preparation. So kindly download the PDF for reference and do more practice.

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Question 1: If x=35, then the value of x216x+6 is:

a) 4

b) 0

c) 2

d) -2

Question 2: If x=32, then the value of 1+x+1x1+x1x is equal to:

a) 2

b) 3

c) 3

d) 2

Question 3: If a3+b3=62 and a + b = 2, then the value of ab is:

a) -6

b) 9

c) 6

d) -9

Question 4: If ab=18 and a3b3=324, then find ab.

a) 105

b) -102

c) -104

d) 103

Question 5: If a2+2a2=16, then find the value of 72a2a4+2+8a2

a) 2

b) 4

c) 1

d) 3

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Question 6: If a + 3b = 12 and ab = 9, then the value of (a – 3b) is:

a) 9

b) 8

c) 6

d) 4

Question 7: If x36x2+ax+b is divisible by (x23x+2), then the values of a and b are:

a) a = -6 and b = -11

b) a = -11 and b = 6

c) a = 11 and b = -6

d) a = 6 and b = 11

Question 8: If a + b + c + d = 2, then the maximum value of (1 + a)(1 + b)(1 + c)(1 + d) is

a) 919

b) 6322

c) 5413

d) 8116

Question 9: If x+1x=4, then the value of x4+1x4 is :

a) 16

b) 196

c) 194

d) 14

Question 10: If x – y = 13 and xy = 25, then the value of x2y2 = ?

a) 13240

b) 13229

c) 13269

d) 13210

Question 11: 2x — 3y is a factor of:

a) 4x2+2x3y+9y212xy

b) 4x2+9y2+12xy

c) 8x3+27y3

d) 4x2+2x3y+36y2+12xy

Question 12: If a and b are two positive real numbers such that a + b = 20 and ab = 4, then the value of a3+b3 is:

a) 7760

b) 8000

c) 8240

d) 240

Question 13: If x + y = 15 and xy = 14, then the value of x — y is:

a) 13

b) 12

c) 11

d) 14

Question 14: If a = 355, b = 356, c = 357,then find the value of a3+b3+c33abc.

a) 3204

b) 3206

c) 3202

d) 3208

Question 15: If x+y=14;x3+y3=1064, then the value of (xy)2 is:

a) 36

b) 64

c) 81

d) 100

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Question 16: If a+b=8 and a+a2b+b+ab2=128 then the positive value of a3+b3 is:

a) 344

b) 96

c) 224

d) 152

Question 17: The coefficient of y in the expansion of (2y5)3, is:

a) 150

b) 50

c) -30

d) -150

Question 18: The given table represents the revenue (in ₹ 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) 44.5

b) 31.2

c) 43.6

d) 45.4

Question 19: If a+b+c=19,ab+bc+ca=120, then what is the value of a3+b3+c33abc?

a) 18

b) 23

c) 31

d) 19

Question 20: Solve the following:
(a + b + c)(ab + bc + ca) – abc = ?

a) (a + b)(b + c)(c – a)

b) (a + b)(b – c)(c + a)

c) (a + b)(b + c)(c + a)

d) (a – b)(b – c)(c – a)

Question 21: If x6512y6=(x2+Ay2)(x4Bx2y2+Cy4), then what is the value of (A+BC)?

a) -80

b) -72

c) 72

d) 48

Question 22: If x,y,z are three integers such that x+y=8,y+z=13 and z+x=17, then the value of x2yz is:

a) 75

b) 1811

c) 1

d) 0

Question 23: On simplification, x3y3x[(x+y)23xy]÷y[(xy)2+3xy]x3+y3×(x+y)2(xy)2x2y2 is equal to:

a) 4

b) 1

c) 12

d) 14

Question 24: If (x3+1x3k)2+(x+1xp)2=0 where k and p are real numbers and x ≠ 0, then kp is equal to:

a) P2+1

b) P2+3

c) P21

d) P23

Question 25: For real a, b, c if a2+b2+c2=ab+bc+ca, the value of (a+b)c

a) 3

b) 1

c) 2

d) 0

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Answers & Solutions:

1) Answer (C)

Given,  x=35

 x=(35)2

 x=3+5215

 x=8215 ……………(1)

 x2=(8215)2

 x2=64+603215

 x2=1243215 ………..(2)

  x216x+6=124321516(8215)+6

=1243215128+3215+6

=130128

=2

Hence, the correct answer is Option C

2) Answer (B)

Given, x=32

1+x+1x1+x1x=1+x+1x1+x1x×1+x+1x1+x+1x

=1+x+1x+2(1+x)(1x)1+x(1x)

=2+2(1x2)2x

=1+1x2x

=1+1(32)232

=1+13432

=1+1232

=32×23

=3

Hence, the correct answer is Option B

3) Answer (D)

Given,  a+b=2

a3+b3=62

 (a+b)(a2ab+b2)=62

 (2)(a2+2ab+b23ab)=62

 (a+b)23ab=31

 (2)23ab=31

 43ab=31

 3ab=27

 ab=9

Hence, the correct answer is Option D

4) Answer (B)

Given,  ab=18 and

a3b3=324

 (ab)(a2+ab+b2)=324

  (18)(a22ab+b2+3ab)=324

 (ab)2+3ab=18

 (18)2+3ab=18

 324+3ab=18

 3ab=306

 ab=102

Hence, the correct answer is Option B

5) Answer (D)

Given,  a2+2a2=16

72a2a4+2+8a2=72a2a2(a2+2a2+8)

=72a2+2a2+8

=7216+8

=7224

=3

Hence, the correct answer is Option D

6) Answer (C)

Given, a+3b=13

 (a+3b)2=122

 a2+9b2+6ab=144

 a2+9b2+6(9)=144

 a2+9b2+546ab+6ab=144

 a2+9b26ab+6ab=90

 (a3b)2+6ab=90

 (a3b)2+6(9)=90

 (a3b)2+54=90

 (a3b)2=36

 a3b=6

Hence, the correct answer is Option C

7) Answer (C)

Given, x36x2+ax+b is divisible by (x23x+2)

Let the quotient when x36x2+ax+b is divisible by (x23x+2) be xp

(x23x+2)(xp)=x36x2+ax+b

  x33x2+2xpx2+3px2p=x36x2+ax+b

 x3(3+p)x2+(2+3p)x2p=x36x2+ax+b

Comparing both sides,

(3+p)=6

 p=3

2+3p=a

 2+3(3)=a

 a=11

2p=b

 2(3)=b

 b=6

  a = 11 and b = -6

Hence, the correct answer is Option C

8) Answer (D)

Given, a + b + c + d = 2

We know that, AM   GM

Arithmetic mean of (1 + a),(1 + b),(1 + c),(1 + d)   Geometric mean of (1 + a),(1 + b),(1 + c),(1 + d)

 (1+a)+(1+b)+(1+c)+(1+d)4[ (1+a)(1+b)(1+c)(1+d)]14

 4+a+b+c+d4[(1+a)(1+b)(1+c)(1+d)]14

 4+24[(1+a)(1+b)(1+c)(1+d)]14

 64[ (1+a)(1+b)(1+c)(1+d)]14

 [(1+a)(1+b)(1+c)(1+d)]14 32

 (1+a)(1+b)(1+c)(1+d) (32)4

(1+a)(1+b)(1+c)(1+d) 8116

 Maximum value of (1 + a)(1 + b)(1 + c)(1 + d) = 8116

Hence, the correct answer is Option D

9) Answer (C)

Given,  x+1x=4

=>  (x+1x)2=42

=>  x2+1x2+2.x. 1x=16

=>  x2+1x2+2=16

=>  x2+1x2=14

=>  (x2+1x2)2=142

=>  x4+1x4+2.x2. 1x2=196

=>  x4+1x4+2=196

=>  x4+1x4=1962

=>  x4+(1x)4=194

Hence, the correct answer is Option C

10) Answer (C)

Given,

xy=13 and  xy=25

=>  (xy)2=132

=>  x2+y22xy=169

=>  x2+y2+2xy4xy=169

=>  (x+y)24xy=169

=>  (x+y)24(25)=169

=>  (x+y)2100=169

=>  (x+y)2=269

=>  x+y=269

  x2y2=(x+y)(xy)=(269)(13)=13269

Hence, the correct answer is Option C

11) Answer (A)

4x2+2x3y+9y212xy=4x2+9y212xy+2x3y

=(2x3y)2+2x3y

=(2x3y)(2x3y+1)

  2x+3y is factor of 4x2+2x3y+9y212xy

Hence, the correct answer is Option A

12) Answer (A)

Given,  a+b=20  and  ab=4

=>  (a+b)3=203

=>  a3+b3+3ab(a+b)=8000

=>  a3+b3+3(4)(20)=8000

=>  a3+b3+240=8000

=>  a3+b3=8000240

=>  a3+b3=7760

Hence, the correct answer is Option A

13) Answer (A)

Given, x+y =15  and xy=14

=(x+y)2=152

=x2+y2+2xy=225

=x2+y2+2xy2xy+2xy=225

=x2+y22xy+4xy=225

=(xy)2+4(14)=225

=(xy)2+=22556

=(xy)2+=169

=xy=13

Hence, the correct answer is Option A

14) Answer (A)

Given, a = 355, b = 356, c = 357

a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcca)

=12(a+b+c)(2a2+2b2+2c22ab2bc2ca)

=(a+b+c)2[(ab)2+(bc)2+(ca)2]

=(355+356+357)2[(355356)2+(356357)2+(357355)2]

=10682[1+1+4]

=10682[6]

=3204

Hence, the correct answer is Option A

15) Answer (A)

Given, x+y=14

x3+y3=1064

=>  (x+y)(x2+y2xy)=1064

=>  14(x2+y2+2xy3xy)=1064

=>  (x+y)23xy=76

=>  (14)23xy=76

=>  1963xy=76

=>  3xy=120

=>  xy=40

  (xy)2=x2+y22xy

=x2+y2+2xy4xy

=(x+y)24xy

=(14)24(40)

=196160

=36

Hence, the correct answer is Option A

16) Answer (D)

Given, a+b=8

a+a2b+b+ab2=128

=>  a(1+ab)+b(1+ab)=128

=>  (1+ab)(a+b)=128

=>  (1+ab)8=128

=>  1+ab=1288

=>  1+ab=16

=>  ab=15

  a3+b3=(a+b)(a2+b2ab)

=8(a2+b2+2ab3ab)

=8((a+b)23ab)

=8(823(15))

=8(6445)

=152

Hence, the correct answer is Option D

17) Answer (A)

(2y5)3

= 8y3125+150y60y2

((ab)3=a3b33a2b+3ab2)

The coefficient of y = 150

18) Answer (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 = 538370370×100 = 45.4%

19) Answer (D)

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

192=a2+b2+c2+2(120)

a2+b2+c2=361240=121

a3+b3+c33abc

=(a+b+c)(a2+b2+c2abbcca)

= 19(121×120)=19

20) Answer (C)

(a + b + c)(ab + bc + ca) – abc

= a2b+a2c+ab2+cb2+bc2+ac2+2abc

= a2(b+c)+bc(b+c)+a(b2+c2+2bc)

= a2(b+c)+bc(b+c)+a(b+c)2

=(b+c)(a2+bc+ab+ac)

= (b + c)[a(a + b) + c(a + b)]

= (a + b)(b + c)(c + a)

21) Answer (A)

x6512y6=(x2+Ay2)(x4Bx2y2+Cy4)

(x2)3+(8y2)3=(x2+Ay2)(x4Bx2y2+Cy4)

(a3+b3=(a+b)(a2ab+b2))

by comparison –

8y2=Ay2

A = -8

Bx2y2=x2×8y2

B = -8

(8y2)2=Cy4

C = 64

The value of (A+BC) = -8 – 8 – 64 = -80

22) Answer (B)

x + y = 8 —-(1)

y + z = 13 —-(2)

z + x = 17 —-(3)

Eq (1) + (2) + (3),

2(x + y + z) = 8 + 13 + 17

x + y + z = 38/2 = 19 —-(4)

From eq (3) and (4),

x + 13 = 19

x = 6

From eq(3),

z + x = 17

z + 6 = 17

z = 11

From eq(2),

y + z = 13

y + 11 = 13

y = 2

x2yz

622×11

= 36/22 = 18/11

23) Answer (A)

x3y3x[(x+y)23xy]÷y[(xy)2+3xy]x3+y3×(x+y)2(xy)2x2y2

= x3y3x[(x+y)23xy]×x3+y3y[(xy)2+3xy]×(x+y)2(xy)2x2y2

= (xy)(x2+xy+y2)x[(x+y)23xy]×(x+y)(x2xy+y2)y[(xy)2+3xy]×(x2+y2+2xy)(x2+y22xy)(x+y)(xy)

= x2+xy+y2x[x2xy+y2]×x2xy+y2y[x2+xy+y2]×4xy = 4

24) Answer (D)

(x3+1x3k)2+(x+1xp)2=0

It will be zero, if the individual terms will be zero.

So, (x3+1x3k)2=0 and (x+1xp)2

So, k=x3+1x3 and (x+1xp)2=0

k=x3+1x3 and (x+1x)=p

Now, (x+1x)=p taking cube of both side,

(x+1x)3=p3

x3+1x3+3(x+1x)=p3

Now substituting the values in the above,

k+3p=p3

k=p33p

kp=p23

25) Answer (C)

Given,    a2+b2+c2=ab+bc+ca

Multiplying both sides by “2”, it becomes

2(a2+b2+c2)=2(ab+bc+ca)

2a2+2b2+2c2=2ab+2bc+2ca

a2+a2+b2+b2+c2+c22ab2bc2ca=0

(a2+b22ab)+(b2+c22ca)+(c2+a22ca)=0

(ab)2+(bc)2+(ca)2=0

Sum of squares is zero so each term should be zero

=>    (ab)2=0, (bc)2=0, (ca)2=0

=>      ab=0,       bc=0,       ca=0,

=>         a=b,               b=c,                c=a

=>         a=b=c

Therefore       a+bc=  a+aa=  2aa=2

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