Sector at center of circle having an angle of 180° is called
Answer: A
A sector whose central angle is \( 180^{\circ} \) covers half the circle, bounded by a diameter and the arc - a semicircle.
Its area is \( \tfrac{1}{2}\pi r^{2} \) and its perimeter \( \pi r + 2r \), the arc plus the diameter.
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The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square?
Answer: B
\[ \text{rectangle} = 125 \times 64 = 8000 \text{ cm}^{2} \]
\[ \text{square} = 5 \times 8000 = 40{,}000 \text{ cm}^{2} \]
\[ \text{side} = \sqrt{40000} = 200 \text{ cm} \]
\[ \text{perimeter} = 4 \times 200 = 800 \text{ cm} \]
The question asks for the perimeter, so the square root must be multiplied by four - stopping at the side gives 200.
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By converting angle from degrees to radians, 40° is equal to
Answer: B
\[ \pi \text{ radians} = 180^{\circ} \]
\[ 40^{\circ} = 40 \times \frac{\pi}{180} = \frac{2\pi}{9} \]
\[ = 0.698 \text{ rad} \]
A radian is about \( 57.3^{\circ} \), so \( 40^{\circ} \) must come to a little under 1 radian - a quick check on the size of the answer.
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Two identical cubes each of total surface area of 6 cm² are joined end to end. Which of the following is the total surface area of the cuboid so formed?
Answer: C
Each cube has total surface area 6 cm², so a single face is 1 cm² and the edge is 1 cm.
Joined end to end they form a cuboid of \( 2 \times 1 \times 1 \):
\[ \text{TSA} = 2(lb + bh + hl) = 2(2 + 1 + 2) = 10 \text{ cm}^{2} \]
The two faces that meet are sealed inside and no longer count. Adding the cubes' areas would give 12 cm², which is exactly 2 cm² too many - the pair of hidden faces.
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The perimeter of a triangle is 28 cm and the inradius of the triangle is 2.5 cm. What is the area of the triangle?
Answer: D
The inradius relates the area to the semi-perimeter:
\[ A = r \times s, \qquad s = \frac{28}{2} = 14 \]
\[ A = 2.5 \times 14 = 35 \text{ cm}^{2} \]
35 cm² is not among the choices, so the answer is none of these. The 70 cm² on offer comes from using the full perimeter instead of the semi-perimeter.
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Two circular cylinders of equal volume have their heights in the ratio 1: 2. Ratio of their radii is
Answer: B
Equal volumes with heights in the ratio \( 1 : 2 \):
\[ \pi r_{1}^{2} h = \pi r_{2}^{2} (2h) \]
\[ r_{1}^{2} = 2 r_{2}^{2} \quad\Longrightarrow\quad \frac{r_{1}}{r_{2}} = \sqrt{2} \]
So the ratio is \( \sqrt{2} : 1 \). The shorter cylinder must be wider, and because volume depends on \( r^{2} \), halving the height needs only a \( \sqrt{2} \) increase in radius rather than a doubling.
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A rectangular field measures 60 m by 40 m. What is the cost of fencing it at Rs. 25 per metre ?
Answer: B
Correct answer: Rs. 5000
Fencing follows the perimeter, not the area.
\[\text{Perimeter} = 2(60 + 40) = 200 \text{ m}\]
\[\text{Cost} = 200 \times 25 = \text{Rs. } 5000\]
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Line segment which joins distinct points on circumference is called
Answer: B
A chord is a line segment whose two endpoints both lie on the circle.
A radius joins the centre to a point on the circle, so only one of its ends is on the circumference. The longest chord of a circle is its diameter, which passes through the centre.
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A conical cavity is drilled in a circular cylinder of 15 cm height and 16 cm base diameter. The height and the base diameter of the cone are the same as those of the cylinder. Determine the total surface area of the remaining solid.
Answer: A
The cone is drilled out of the cylinder, so three surfaces remain exposed: the cylinder's curved side, its base, and the cone's curved inner surface. The cylinder's top face is gone, replaced by the cavity's opening.
With \( r = 8 \) cm and \( h = 15 \) cm, the cone's slant height is
\[ l = \sqrt{r^{2} + h^{2}} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ cm} \]
\[ \text{TSA} = 2\pi r h + \pi r^{2} + \pi r l \]
\[ = 2\pi(8)(15) + \pi(64) + \pi(8)(17) \]
\[ = 240\pi + 64\pi + 136\pi = 440\pi \text{ cm}^{2} \]
The cavity's wall is exposed and must be counted, while the flat top is not - the drilling removes it entirely.
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The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 cu. meters. Find the breadth of the wall.
Answer: D
Let the breadth be \( b \). Then the height is \( 5b \) and the length \( 8 \times 5b = 40b \).
\[ V = b \times 5b \times 40b = 200b^{3} = 12.8 \]
\[ b^{3} = 0.064 \quad\Longrightarrow\quad b = 0.4 \text{ m} = 40 \text{ cm} \]
"Eight times as long as it is high" refers to the height \( 5b \), not the breadth - reading it as \( 8b \) would give a different wall.
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