Samsung – Aptitude Questions & Answers for Placement Tests

Ans: C

The incenter is the point where the angle bisectors intersect. The incenter is equidistant from the sides of the triangle. The area of △PQR = Area of △POQ + Area of △POR + Area of △QOR = 15 + 20 + 25 = 60 cm2

Ans: D

Power of a point theorem. The product of the lengths of the two segments from a point to the intersections of a circle with a line through the point is constant. Applying this to point P and R, the products are the same. Hence, PA x PB = RA x RS (but the question says SB, there is a mistake here). Considering QA x RB. Then PA x PB = QA x RB, is not correct. Considering PS x PA. Consider the second circle. Applying the Power of point theorem to the second circle, QA x RB = PS x PA

Ans: B

Since sin α = 12/13, we can consider a right triangle with opposite side 12 and hypotenuse 13. Using the Pythagorean theorem, the adjacent side is √(13^2 – 12^2) = √(169 – 144) = √25 = 5. Since α is obtuse, cos α must be negative. Therefore, cos α = -5/13.

Ans: B

sin²15° + sin²75° = sin²15° + sin²(90°-15°) = sin²15° + cos²15° = 1

Ans: A

In triangle YIZ, ∠IYZ + ∠IZY + ∠YIZ = 180°. Therefore, ∠IYZ + ∠IZY = 180° – 130° = 50°. Since I is the incenter, ∠IYZ = ∠XYZ/2 and ∠IZY = ∠YZX/2. Also, ∠IXY = ∠XYZ/2 = 30° Thus ∠XYZ = 60°. Therefore ∠YZX = 2 * (50° – 60°/2) = 2 * 20° = 40°

Ans: B

Using similar triangles, height of tree/shadow of tree = height of pole/shadow of pole. Let h be the height of the tree. h/15 = 5/3. h = (5/3)*15 = 25 meters.

Ans: A

By the Angle Bisector Theorem, PQ/PR = QS/SR. Also, QS + SR = QR = 12. Let QS = x. Then 10/14 = x/(12-x). Solving for x: 10(12-x) = 14x => 120 – 10x = 14x => 24x = 120 => x = 5.

Ans: B

Let ‘s’ be the side of the square. The diagonal is given by s√2. Thus, s√2 = 10√2. So, s = 10 cm. The area of the square is s² = 10² = 100 cm².

Ans: B

Let the length of the ladder be L. The distance from the foot of the ladder to the wall is adjacent to the angle 30 degrees formed with the wall. Therefore, sin(30°) = opposite/L, and cos(30°) = 5/L. cos(30°) = √3/2. Hence, 5/L = √3/2. So, L = 10/√3. The question states the foot is 5 meters from the wall, this refers to the base of a right triangle. Then sin(60) = 5/L or cos(60) = x/L. The actual relationship uses sin(60) = opposite/hypotenuse. 5 is the base so we use Cos(60) = 5/L. As Cos(60) = 1/2, then 1/2=5/L therefore L=10.

Ans: D

We know that 1 + tan²Θ = sec²Θ. Therefore, (1+tan²Θ) * cos²Θ = sec²Θ * cos²Θ. Since secΘ = 1/cosΘ, we have sec²Θ = 1/cos²Θ. Hence, sec²Θ * cos²Θ = (1/cos²Θ) * cos²Θ = 1.