Texas Previous Years Solved Sample Placement Papers
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Two dice are thrown at the same time. Find the probability of getting different values on both.
A. 5/6
B. 1/2
C. 1/6 Answer: C
D. 1/36
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If matrices A and B are inverse of each other then
A. AB = BA
B. AB = BA = I Answer: B
C. AB = BA = 0
D. AB = 0, BA = I
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The point diametrically opposite to the point P (1, 0) on the circle x² + y² + 2x + 4y - 3 = 0 is
A. (-3, -4) Answer: A
B. (-3, 4)
C. (3, 4)
D. (-4, -1)
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The conjugate of a complex number is 1/(i - 1). Then the complex number is
A. -1/(i - 1)
B. 1/(i + 1) Answer: B
C. 1/(i - 1)
D. -1/(i + 1)
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What is the probability that the minute and hour hands of a clock will form an acute angle at any given time?
A. P > 0.5
B. P = 0.5
C. P < 0.5 Answer: C
D. P = 0.25
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A single die is rolled. The probability of getting 1 or an even number is
A. 1/6
B. 4/6
C. 5/6
D. 3/6 Answer: D
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The perpendicular bisector of the line segment joining P (1, 4) and Q (k, 3) has y-intercept -4. Then a possible value of k is
A. 1
B. -4 Answer: B
C. 3
D. 2
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The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?
A. a = 0, b = 7
B. a = 5, b = 2
C. a = 3, b = 4 Answer: C
D. a = 2, b = 4
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The line passing through the points (5, 1, a) and (3, b, 1) crosses the yz-plane at the point (0, 17/2, -13/2). Then
A. a = 2, b = 8
B. a = 4, b = 6
C. a = 6, b = 4 Answer: C
D. a = 8, b = 2
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The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is
A. -2
B. -4
C. -12 Answer: C
D. 8
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How many real solutions does the equation x⁷ + 14x⁵ + 16x³ + 30x – 560 = 0 have?
A. 1 Answer: A
B. 4
C. 7
D. 5
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Each of the letters of the word PILOTS is on separate cards, face down on the table. If you pick a card at random, what is the probability that the letter will be a T or an L?
A. 1/6
B. 1/3
C. 1/2
D. 2/3 Answer: D
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If the system of linear equations
x + 2ay + az = 0
x + 3by + bz = 0
x + 4cy + cz = 0
has a non-zero solution, then a, b, c
A. are in A. P.
B. are in G. P.
C. are in H. P. Answer: C
D. satisfy a + 2b + 3c = 0