Description

ABInBev Aptitude Test Past Questions Study Pack

ABInBev Aptitude Test Past Questions Study pack contains solved test covering the new odyssey assessments.

• Problem solving
• numerical literacy
• sentence correction
• Vocabulary
• Spelling
• Clerical Accuracy
• Executive code instructions.

We also included questions from previous years:  Numerical Analysis, Verbal Analysis and Diagrammatic Analysis. (Online test)

With this pack, you will practice likely repeated questions from the online odyssey assessments and guarantee your success.

ABInBev Aptitude Test Past Questions study pack, Samples

VERBAL REASONING

1. Which of the following word is nearly most opposite in meaning to the word “BRUSQUE”?
(A) Gold
(B) Opulent
(C) Gracious
(D) Suspect
Answer: (C)

2. Which of the following is the Antonym of the word “ORNATE”?
(A) Spartan
(B) Blemished
(C) Sturdy
(D) Counterfeit
Answer: (B)

3. Which of the following is the Antonym of the word “EXTENUATE”?
(A) Alleviate
(B) Preclude
(C) Worsen
(D) Subdue
Answer: (C)

4. Which of the following is the synonym of the word “NUANCE”?
(A) New word
(B) Sense
(C) Interpretation
(D) Subtle meaning
Answer: (D)

5. Which of the following is the synonym of the word “PLUTOCRAT”?
(A) Magnate
(B) Priest
(C) Judge
(D) Astronaut
Answer: (A)

6. Which of the following is the synonym of the word “BULWARK”?
(A) Conundrum
(B) Festival
(C) Rampart
(D) Confuse
Answer: (C)

NUMERICAL REASONING

1. An equilateral triangle of side √3cm is inscribed in a circle. Find the radius of the circle.
A. 2/3 cm
B. 2 cm
C. 1 cm
D. 3 cm
Answer: C
Explanation
Since the inscribed triangle is equilateral, therefore the angles at all the points = 60°
Using the formula for inscribed circle,
2R = asinA=bsinB=csinC
where R = radius of the circle; a, b and c are the sides of the triangle.
⇒ 2R = 3√sin60
2R = 3√3√2
2R = 2
R = 1cm

2. 3y = 4x – 1 and Ky = x + 3 are equations of two straight lines. If the two lines are perpendicular to each other, find K.
A. -4/3
B. -3/4
C. 3/4
D. 4/3
Answer: A
Explanation
Grad of 3y = 4x – 1
y = 4x/3 – 1/3
Grad = 4/3
Grad of Ky = x + 3
y = x/k + 3/4
Grad = 1/k
Since two lines are perpendicular,
1/k = -3/4
-3k = 4
k = -4/3

3. if P and Q are fixed points and X is a point which moves so that XP = XQ, the locus of X is
A. A straight line
B. a circle
C. the bisector of angle PXQ
D. the perpendicular bisector of PQ
Answer: D

4. In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon.
A. 8
B. 6
C. 4
D. 3
Answer:  B
Explanation
2x + x = 180°, => 3x = 180°, and thus x = 60°
Each exterior angle = 60° but size of ext. angle = 360°/n
Therefore 60° = 360°/n
n = 360°/60° = 6 sides

5. A predator moves in a circle of radius √2 centre (0,0), while a prey moves along the line y = x. If 0 ≤ x ≤ 2, at which point(s) will they meet?
A. (1,1) only
B. (1,1) and (1,2)
C. (0,0) and (1,1)
D. (√2,√2) only
Answer: Option A
Explanation
x2 + y2 = (√2)2
x2 + y2 = 2
but y = x
Thus; x2 + x2 = 2
2×2 = 2
x2 = + or – 1
But x2 + y2 = 2
12 + y2 = 2
1 + y2 = 2
y2 = 2 – 1
y2 = 1
y = + or – 1
Thus point (x,y) = (1,1) only.

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