Algebra(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

ALGEBRA MCQs

Total Questions : 240 | Page 1 of 24 pages

Question 1. If α and β are the roots of the equation x² + px + q = 0 and α⁴ and β⁴ are the roots of the equation x²– rx + s = 0, then the equation x² – 4qx + 2q² – r = 0 has always:

2 real roots
2 positive roots
2 negative roots
1 positive and 1 negative root

Discuss Question

Answer: Option D. -> 1 positive and 1 negative root
:
D
Given question isflexible. So, we can assume the roots and can make an other quardatic equation.
Let α = 1 and β = 2, then the expression becomes x² – 3x + 2 = 0. So p = -3, q = 2, r = 17, s = 16. The new expression is x² – 8x – 9 = 0. 1 root is positive and other is negative.

Question 2. If

\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = r

then, r cannot take any value except :(CAT 2004)

12
-1
12 or -1
−12 or -1

Discuss Question

Answer: Option C. -> 12 or -1
:
C
Option (c)
As

\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = \frac{a + b + c}{b + c + c + a + a + b}

= \frac{a + b + c}{2 (a + b + c)} = r = \frac{1}{2}

(Assuming a + b + c ≠ 0)
If a + b + c = 0

\frac{a}{b + c}

\frac{a}{a + b + c - a}

(by adding and subtracting a in the denominator)

= \frac{a}{0 - a} = \frac{a}{- a} = r = - 1

(similarly

\frac{b}{c + a} = \frac{c}{a + b}

= -1)
Hence r can take only

\frac{1}{2}

or -1 as values. Choice (c)

Question 3.

t3+t2+t+1
t3+t2-5
t3+ √5t2 +√5t -2
t3-1
t3

Discuss Question

Answer: Option D. -> t3-1
:
D
Option (d)Assume t= 2The question changes to[(5-√3i)(5+√3i)]/4 [25-(√3i)²]/4, since (a-b)(a+b)=a²-b²(25+3)/4, since i²=-1=7Only answer option (d) gives 7 on substitution of t=2

Question 4. Given that p and q are the roots of the equation x² – ax +b =0 and D_n= pⁿ+qⁿ. Find the value of D_n+1

aDn-bDn-1
aDn
bDn
aDn + bDn-1
Dn

Discuss Question

Answer: Option A. -> aDn-bDn-1
:
A
Option (a)Best way to proceed is by assumption of values.Assume a quadratic equation. Let’s take x²+5x-6=0. Here the roots are p=1 and q=-6Thus, a = sum of roots = -5 and b= product of roots = -6D₁= p¹+q¹ = -5D₀ = 2D₂ = 37Assume n=1We need to find D_n+1 = D₂ = 37Look in the answer options for 37Option a is the only one which gives (-5)(-5)-(-6)(2) = 37

Question 5. What is the value of X ?

3
2
1
0
none of these

Discuss Question

Answer: Option C. -> 1
:
C
WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.
Comparing we get => X = 1 and Z = 0

Question 6. Given that b³+mb² +nb + c is divisible by (b-s) if s³ + ms² + ns + c = 0Also given that d³+d²+ed+1 is divisible by (d-1) and d³-4d²+fd-3 is divisible by (d-3)What is the value of e+f ?

0
-1
1
2
none of these

Discuss Question

Answer: Option C. -> 1
:
C
Option (c)d³+d²+ed+1 is divisible by (d-1). As 1 is a root, Put d=1 in the equation and equate it to 0, to get the value of e1+1+e+1=0 e=-3Similarly put d=3 in d³-4d²+fd-3 to get the value of f 27-36+3f-3=0 3f=12 f=4e+f= -3+4=1

Question 7. The roots of the equation (x+1)(x+8) +10 = 0 are given as p and q. Find the roots of (x+p)(x+q)-4=0

-1, 3
7, 8
2,7
–2, 7
–2, 3

Discuss Question

Answer: Option C. -> 2,7
:
C
Option c (x+1)(x+8)+10 can be expanded as x²+9x+8+10=0 x²+9x+18=0 Sum of roots p+q= -9 Product of roots pq = 18(x+p)(x+q)-4=0 can be written as x²+ (p+q)x+pq-4=0 x²-9x+18-4=0x²-9x+14=0Roots are 7 and 2

Question 8. If a and b are the roots of the equation x²+px+1=0 and ; c and d are the roots of the equation x²+qx+1=0, then (a-c)(b-c)(a+d)(b+d)=?

p2-q2
a2+b2
q2-p2
q2+p2
p2

Discuss Question

Answer: Option C. -> q2-p2
:
C
Shortcut: AssumptionOption (c)Product of roots=1 is the only constraint.Assume values that satisfy the constraint and substitute, take a= -1 and b= -1, then p= 2Assume c= - 1/2 and d= - 2; q= 5/2 Question is = (a-c)(b-c)(a+d)(b+d)=9/4Only option (c) satisfies this => q² –p² = 25/4 – 4 =9/4

Question 9. The largest interval for which x¹² – x⁹ + x⁴ – x + 1 > 0 is:

-4 < x < 0
0 < x < 1
-100 < x < 100

Discuss Question

Answer: Option D. -> q2-p2
:
D
Use answer options to arrive at the answer at the earliest
Since the highest power of x is even (12) , it will always be greater than its negative counterpart (9), hence, irrespective of the value of x, the net value will be positive. Answer is option (d)

Question 10.