Important Questions on Algebra for MAH MBA CET

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Important Questions On Algebra For MAH MBA CET Exam
Important Questions On Algebra For MAH MBA CET Exam

Algebra Questions for MAH MBA CET Exam

Download MAH MBA CET Algebra Questions and Answers PDF covering the important questions. Most expected Algebra questions with explanations for MAH MBA CET / MMS CET 2021 exam.

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Question 1: 8+57+38+108+169

a) 4

b) 6

c) 8

d) 10

Question 2: If a * b = 2a + 3b – ab, then the value of (3 * 5 + 5 * 3) is

a) 10

b) 6

c) 4

d) 2

Question 3: If a * b = ab, then the value of 5 * 3 is

a) 125

b) 243

c) 53

d) 15

Question 4: If x=1+2+3 , then the value of (2x48x35x2+26x28) is __?

a) 66

b) 0

c) 36

d) 26

Question 5: If a2+b2+c2=2(a2bc3) then the value of a+b+c is

a) 3

b) 0

c) 2

d) 4

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Question 6: Find the simplest value of 250+1872 is __? (2=1.414).

a) 9.898

b) 10.312

c) 8.484

d) 4.242

Question 7: If a3b3c3=0 then the value of a9b9c93a3b3c3 is

a) 1

b) 2

c) 0

d) -1

Question 8: If p2q2+q2p2=1 then the value of (p6+q6) is

a) 0

b) 1

c) 2

d) 3

Question 9: If (m+1)=n+3 the value of 12(m36m2+12m8nn)

a) 0

b) 1

c) 2

d) 3

Question 10: If x=aba+b,y=bcb+c,z=cac+a then (1x)(1y)(1z)(1+x)(1+y)(1+z) is equal to

a) 1

b) 0

c) 2

d) 12

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Question 11: If 717+17+171=a+7b the values of a and b are respectively

a) 7,1

b) 7,1

c) 0,23

d) 23,0

Question 12: If m = – 4, n = – 2, then the value of m33m2+3m+3n+3n2+n3 is

a) – 126

b) 124

c) – 124

d) 126

Question 13: If x+1x=1 then the value of x2+3x+1x2+7x+1

a) 1

b) 37

c) 12

d) 2

Question 14: If x=a3ba3b then the value of x is

a) ab35

b) a3b3

c) a5b3

d)  a2ba4

Question 15: If the cube root of 79507 is 43, then the value of 79.5073+0.0795073+0.0000795073
is

a) 0.4773

b) 477.3

c) 47.73

d) 4.773

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Question 16: If xy=34 the ratio of (2x+3y) and (3y2x) is

a) 2 : 1

b) 3 : 2

c) 1 : 1

d) 3 : 1

Question 17: If m – 5n = 2, then the vlaue of (m3125n3 – 30 mn) is

a) 6

b) 7

c) 8

d) 9

Question 18: If x+1x=2 then the value of x12+1x12 is

a) 2

b) -4

c) 0

d) 4

Question 19: If 5x + 9y = 5 and 125x3 + 729y3 = 120 then the value of the product of x and y is

a) 19

b) 1135

c) 45

d) 135

Question 20: What is the value of (941+149)2+(941149)2(941×941+149×149)?

a) 10

b) 2

c) 1

d) 100

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Answers & Solutions:

1) Answer (A)

Start from the root of 169 then second root will reduce to 11, thrid root will reduce to 7, fourth root will reduce to 8, and finally it reduce to value 4

2) Answer (A)

For 3*5 put a=3 and b=5 in given equation
and for 5*3 put a=5 and b=3 in equation
now add both values

3) Answer (A)

Put a=5 and b=3 in given equation
hence it will be 53 = 125

4) Answer (A)

x = 1+ 2+3
(x1)2 = (2+3)2
x2+12x=5+26
x22x=4+26 ( eq. (1) )
(x22x)2=x4+4x24x3=40+166 eq (2)
Now in 2x48x35x2+26x28
or 2(x44x3)5x2+26x28 ( putting values from eq (1) and eq (2) )
After solving we will get it reduced to 66

5) Answer (B)

Given a2+b2+c2=2(a2bc3),

So, (a1)2+(b+2)2+(c1)2=0

Hence, a=1, b=-2 and c=1

So, the sum of the equation is

6) Answer (A)

Given equation can be reduced in the form of 102+3262=72
Hence  72 will be around 9.898

7) Answer (C)

shortcut :

put c = 0 in  a3b3c3=0  a3=b3

a9b9(0)93a3b3(0)3a9b9(a3)3(b3)3 =  (a)3(a)3 = 0  ( a3=b3 )

so the answer is option C.

normal method :

a3b3c3=0

a3=b3+c3

cubing on both sides,

(a3)3=(b3+c3)3

a9=b9+c9+3b3c3(b3+c3)

a9=b9+c9+3b3c3(a3)

a9b9c93a3b3c3=0

so the answer is option C.

8) Answer (A)

Expression : p2q2+q2p2 = 1

=> p4+q4p2q2 = 1

=> p4+q4=p2q2 ————–Eqn(1)

Now, to find : (p6+q6)

=> (p2)3+(q2)3

Using the formula, a3+b3=(a+b)(a2+b2ab)

=> (p2+q2)(p4+q4p2q2)

From eqn (1), we get :

=> (p2+q2)(p2q2p2q2)

=> (p2+q2)0

= 0

9) Answer (A)

If (m+1)=n+3

=> m2=n ————–Eqn(1)

to find : 12(m36m2+12m8nn)

(m2)3=m36m2+12m8

=> 12((m2)3nn)

Using eqn(1), we get :

=> 12((n)3nn)

=> 12(nn)

= 0

10) Answer (A)

If x=aba+b

=> (1x)=1(aba+b)

=> (1x)=2ba+b

Similarly, (1+x)=2aa+b

Applying the same method, we get :

=> (1y)=2cb+c and => (1+y)=2bb+c

=> (1z)=2ac+a and => (1+z)=2cc+a

Putting above values in the equation : (1x)(1y)(1z)(1+x)(1+y)(1+z)

=> (2ba+b)(2cb+c)(2ac+a)(2aa+b)(2bb+c)(2cc+a)

=> 2a2b2c2a2b2c

= 1

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11) Answer (C)

717+17+171=a+7b

L.H.S. = 717+17+171

= (71)2(7+1)2(71)(7+1)

= (7+127)(7+1+27)71

= 476

= 273

Now, comparing with R.H.S. a+7b

we get,

a=0 and b=23

12) Answer (A)

We are given that m = -4 and n = -2

Expression : m33m2+3m+3n+3n2+n3

= (m33m2+3m1)+(n3+3n2+3n+1)

= (m1)3+(n+1)3

= (41)3+(2+1)3

= (5)3+(1)3

= 1251=126

13) Answer (C)

Expression : x+1x=1

=> x2+1=x ——Eqn(1)

To find : x2+3x+1x2+7x+1

= (x2+1)+3x(x2+1)+7x

Using eqn(1),we get :

= x+3xx+7x=48

= 12

14) Answer (D)

x=a3ba3b.

here we know that b×b = b

and a3=aa

hence,x=a3ba3b = a2ba4

15) Answer (D)

Since 795073 = 43

=> 79.5073 = 4.3

=> 0.0795073 = 0.43

=> 0.0000795073 = 0.043

=> 4.3+0.43+0.043 = 4.773

16) Answer (D)

Let x=3k and y=4k

=> 2x+3y3y2x

= 6k+12k12k6k

= 186

= 31 = 3 : 1

17) Answer (C)

Using the formula, (xy)3=x3y33xy(xy)

=> (m5n)3=m3125n315mn(m5n)

=> 23=m3125n315mn2

=> m3125n330mn=8

18) Answer (A)

Expression : x+1x=2

Squaring both sides

=> x2+1x2+2=4

=> x2+1x2=2

Cubing both sides

=> x6+1x6+3.x.1x(x+1x)=8

=> x6+1x6=86=2

Again, squaring both sides, we get :

=> x12+1x12+2=4

=> x12+1x12=2

19) Answer (B)

Expression : 5x+9y=5

Cubing both sides, we get :

=> (5x+9y)3=125

=> 125x3+729y3+135xy(5x+9y)=125

=> 125x3+729y3+135xy5=125

Since, 125x3 + 729y3=120

=> xy=55135=1135

20) Answer (B)

Expression : (941+149)2+(941149)2(941×941+149×149)

= (9412+1492+2.941.149)+(9412+14922.941.149)9412+1492

= 2(9412+1492)9412+1492

= 2

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