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

Exams > Cat > Quantitaitve Aptitude

ALGEBRA MCQs

Total Questions : 240 | Page 2 of 24 pages

Question 11. The number of values of k for which the system of equations have infinite solution-
(k+1)x + 8y = 4k
kx + (k+3)y = 3k – 1

0
1
2
Infinite

Discuss Question

Answer: Option B. -> 1
:
B
option (b)
Sol:- k+1/k = 8/k+3 = 4k/3k-1
Only k = 1 satisfies this.
2nd method: - Forinfinite solution

The Number Of Values Of K For Which The System Of Equations ...

Question 12. The number of real solutions of the equation |x|² – 3|x| + 2 = 0 is:

Discuss Question

Answer: Option A. -> 4
:
A
The expression can be rewritten as
(|x| - 1)(|x| - 2) = 0. x = {-1, +1, -2, +2}
Or, we can say, If x is greater than 0.
x²– 3x + 2 = 0
(x - 2) (x - 1) = 0
x = {1, 2}
If x is less than 0.
x²+ 3x + 2 = 0
(x + 2) (x + 1) = 0
x = { - 1, - 2}
Option (a)

Question 13. If f (0, y) = y + 1, and f(x +1, y) = f (x, f(x, y)). Then, what is the value of f(1,2) ? (CAT 2000)

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Answer: Option D. -> 4
:
D
SOLN:(d) f (x + 1, y) = f(x, f(x, y))put x = 0, f (1, y) = f(0, f(0, y)) = f(0, y + 1)= y + 1 + 1 = y + 2Put y = 2, f (1, 2) = 4.

Question 14. For what value of x; f(x) = g(x – 3)

-3
– 4
None of these
22

Discuss Question

Answer: Option C. -> – 4
:
C
Option (c) f(x) = g(x – 3) 2x + 3 = (x – 3 –3 )/2 = (x – 6)/24x + 6 = x – 63x = -12x = -4

Question 15. The function f(x) = |x - 2| + |2.5 - x| + |3.6 - x|, where x is a real number, attains minimum at

x = 2.3
x = 2.5
x = 2.7
None of these
22

Discuss Question

Answer: Option B. -> x = 2.5
:
B
Option (b)
f(x) = | x – 2| + | 2.5 – x | + | 3.6 – x | can attain minimum value when either of the terms = 0.
Case I :
When | x – 2 | = 0 => x = 2, value of f(x) = 0.5 + 1.6 = 2.1.
Case II.
When | 2.5 – x | = 0 => x = 2.5
value of f(x)
= 0.5 + 0 + 1.1 = 1.6.
Case III.
When | 3.6 – x | = 0 => x = 3.6
f(x) = 1.6 + 1.1 + 0 = 2.7. Hence the minimum value of f(x) is 1.6 at x = 2.5.

Question 16. fog (x) =

1
gof (x)
0
1/x
22

Discuss Question

Answer: Option B. -> gof (x)
:
B
Option (b)
fog (x) = f{g(x)} = f {(x-3)/2}= 2 {(x-3)/2} + 3 = x
gof (x) = g{f(x)} = g (2x + 3) = (2x + 3 – 3)/2= x
therefore fog (x) = gof (x)

Question 17. Let u _n+1 = 2u_n +1, (n = 0, 1, 2,…) and u₀ = 0. Then u₁₀ would be nearest to: (CAT 1993)

1023
2047
4095
8195
22

Discuss Question

Answer: Option A. -> 1023
:
A
Ans: (a) u _n+1 = 2u_n + 1 (n = 0, 1, 2,…..)Put n = 0, u₁ = 1n = 1, u₂ = 3n = 2, u₃ = 7n = 3, u₄ = 15n = 4, u₅ = 31Seeing the pattern it is clear that u_n = 2ⁿ – 1Hence u₁₀ = (2)¹⁰ – 1 = 1,023.

Question 18. What is the value of (gofofogogof) (x) (fogofog) (x)

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Answer: Option B. -> x2
:
B
Option (b) {go fo fo go go f(x)} { fo go fo g (x)} from Q. 3, We have fog (x) = gof (x) = xTherefore above expression becomes (x) . (x) = x²

Question 19. What is the value of S[M(D(A(a, b), 2), D(A(a, b), 2)), M(D(S(a, b), 2), D(S(a, b), 2))]?

a2 + b2
ab
a2 – b2
a / b

Discuss Question

Answer: Option B. -> ab
:
B
Soln:(b)
S [M (D (A (a, b),2), D (A (a, b), 2)), M (D (S (a, b), 2), D (S (a, b), 2))]
S [M (D (a + b, 2), D (a + b, 2)), M (D (a – b, 2),D (a – b, 2))]
S [M(a + b)/2), (a + b) /2), M(

What Is The Value Of S[M(D(A(a, B), 2), D(A(a, B), 2)), M(D(...

)]
S [ (

] = [(a +b)² – (a – b)² ]/ 2²

Question 20. An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …., 9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can however potentially create confusion, when read upside down-for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?(2003)

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Answer: Option D. -> 69
:
D
Option (d)
The available digits are 0,1,2,3……………..9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digit repetition are not allowed). Thus the code can be made in 9