Algebra (Test 4)

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Algebra

Algebra
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Q.1
If x2 + y2 = 29 and xy = 10, where x>0, y>0, x>y then the value of c is
A. -7/3
B. 7/3
C. 3/7
D. -3/7
Answer : Option B
Explaination / Solution:

Given, x2 + y2 = 29 and xy = 10, where x > 0, y > 0, x > y 



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Q.2
What will be the value of if a+b+c = 0 and a2+b2+c2 = 2abc ?
A. abc
B. a2b2c2
C. a3b3c3
D. 4abc
Answer : Option B
Explaination / Solution:

Given a+b+c = 0 
⇒b+c = -a 
⇒(b+c)2 = a2 
⇒b2 + c2 + 2bc = a2 
⇒a2 + b2 + c2 = 2(a2 - bc) 
⇒(a2 - bc) = (1/2)(a2 + b2 + c2
Similarly, 
(b2 - ac) = (1/2)(a2 + b2 + c2
And, (c2 - ab) = (1/2)(a2 + b2 + c2
Now,
 

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Q.3
If p3 – q3 = (p – q) {(p + q )2 – x p q} then the value of x is
A. 1
B. -1
C. 2
D. -2
Answer : Option A
Explaination / Solution:


On comparing the coefficient, we get 


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Q.4
LCM of two polynomials is x3 + x2 - 4x - 4 is and their HCF is (x-2) and one of the polynomials is (x2-4) then which of the following is the 2nd polynomial?
A. x2-x-2
B. x2-3x+2
C. x2+x-6
D. x2-5x+6
Answer : Option A
Explaination / Solution:

LCM = x3 + x2 - 4x - 4 = x(x2-4)+1(x2-4) = (x+1)(x-2)(x+2) 
HCF = (x-2) 
Som 2nd polynomials = (x+1)(x-2)(x+2)(x-2)/(x2-4) = (x+1)(x-2) = x2-x-2

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Q.5
If  then the value of (a4 – a) is:
A. 0
B. 1
C. 2
D. -1
Answer : Option A
Explaination / Solution:



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Q.6
What is the value of m in the quadratic equation x2 + mx +24=0 if one of its roots is 
A. -45/2
B. 16
C. -21/2
D. -35/2
Answer : Option D
Explaination / Solution:

X =  3/2 in x2 + mx + 24 = 0
(3/2) + m*(3/2) + 24 = 0
(9/4) + (3m/2) + 24 = 0


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Q.7
a = 1000, b = 1001, c = 1002. then find the value of .
A. 0
B. 1
C. 2
D. 3
Answer : Option D
Explaination / Solution:



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Q.8
Sum of three consecutive even integers is 54. Find the least among them
A. 18
B. 15
C. 14
D. 16
Answer : Option D
Explaination / Solution:

Let three consecutive even integers by 2x, 2x + 2 and 2x + 4. respectively
2x + 2x + 2 + 2x + 4 = 54
6x + 6 = 54 


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Q.9
 +  = ?
A. 1
B. 0
C. -1
D. 2
Answer : Option A
Explaination / Solution:

 + 
 +  = 1

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Q.10
If  +  = 2 and  +  = 1 than find (ZQ – PX)2 + (ZX + PQ)2
A. -1
B. 0
C. 4/9
D. 2
Answer : Option D
Explaination / Solution:

Z2 + P2 = 2 ; X2 + Q2 = 1 
Let Z = 1 and X =  1/√2
P = 1 Q =  1/√2
(ZQ – PX )2 + (ZX + PQ)2 =  ((1/√2) - (1/√2)2 + ((1/√2) + (1/√2))2
0 + (4/2) = 2

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