Mensuration Questions and Answers

Answer: D

Let the length of the brick be 3X, breadth of the brick be 2X and height of the brick be 4X.
Given: Volume of a brick is 3000 cm³
WKT, Volume of brick = l x b x h
Volume of brick = 3X*2X*4X
3000 = 24X³
X = 5
Length of the brick = 3X = 3(5) = 15 cm
Breadth of the brick = 2X = 2(5) = 10 cm
Height of the brick = 4X = 4(5) = 20 cm

WKT, Total surface area of brick = 2(lb + bh + hl)
Total surface area of brick = 2[(15 x 10) + (10 x 20) + (20 x 15)]
= 2[150 + 200 + 300]
= 1300²
The cost = 1300 x (50/100) = Rs. 650.

Answer: B

Angles in opposite segments of a circle - equivalently, the opposite angles of a cyclic quadrilateral - add to \( 180^{\circ} \), so they are supplementary.

\[ \angle A + \angle C = 180^{\circ} \]

This follows from the inscribed-angle theorem: each angle is half the arc it faces, and the two arcs together make the whole \( 360^{\circ} \).

Complementary would mean adding to \( 90^{\circ} \), a different relation altogether.

Answer: C

Circular measure is the system that measures an angle by the arc it subtends on a circle, divided by the radius - which is exactly the radian.

\[ \theta = \frac{\text{arc}}{\text{radius}} \]

It stands beside sexagesimal measure (degrees, minutes, seconds) as the other standard way of quantifying angles.

Answer: D

A rectangular water tank is 80 m x 40 m. Water flows into it through a pipe of 40 sq.cm at the opening at a speed of 10 km/hr. The water level will rise in the tank in half an hour is

Volume of water filled by pipe in 30 minutes

-----> (40 *1000000/2) cu.cm
-----> = 2000000 cu.cm

Therefore, Height of water level

-----> (2000000/8000 *4000)
-----> = (5/8) cm.

Hence the answer is: (5/8) cm

Answer: C

Drop a horizontal from the shorter fort's top to the taller one. That gives a right triangle whose legs are the ground gap and the height difference.

\[ \text{height difference} = 15 - 8 = 7 \text{ cm}, \qquad \text{ground gap} = 24 \text{ cm} \]

\[ d = \sqrt{24^{2} + 7^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm} \]

7-24-25 is a Pythagorean triple, which is why the answer comes out exact. Answering 24 cm ignores the height difference; 31 cm adds the numbers instead of squaring them.

Answer: B

The metal is what lies between the outer box and the hollow inside it. The box is open, so only the depth loses one thickness, while the length and breadth lose two.

\[ V_{\text{outer}} = 50 \times 40 \times 23 = 46{,}000 \text{ cm}^{3} \]

\[ V_{\text{inner}} = (50-6)(40-6)(23-3) = 44 \times 34 \times 20 = 29{,}920 \text{ cm}^{3} \]

\[ V_{\text{metal}} = 46{,}000 - 29{,}920 = 16{,}080 \text{ cm}^{3} \]

\[ \text{weight} = 16{,}080 \times 0.5 = 8040 \text{ g} = 8.04 \text{ kg} \]

Answer: C

The liquid's volume is conserved, so divide the bowl's by one bottle's. The bottle's radius is \( 3 \div 2 = 1.5 \) cm.

\[ V_{\text{bowl}} = \frac{2}{3}\pi (9)^{3} = \frac{2}{3}\pi (729) = 486\pi \text{ cm}^{3} \]

\[ V_{\text{bottle}} = \pi (1.5)^{2}(4) = 9\pi \text{ cm}^{3} \]

\[ n = \frac{486\pi}{9\pi} = 54 \]

A hemisphere is two thirds of \( \pi r^{3} \), not one half of the sphere formula misapplied - the \( \tfrac{2}{3} \) is what makes the count come out whole.

Answer: A

Correct answer: 1540 cu. cm

\[V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 10 = \frac{22}{7} \times 490 = 1540 \text{ cu. cm}\]

The radius is squared but the height is not, so doubling the radius would quadruple the volume while doubling the height merely doubles it.

Answer: A

A chord splits the circle into two segments. The smaller one - the piece not containing the centre - is the minor segment.

The larger is the major segment, and a chord through the centre (a diameter) makes the two equal, each then being a semicircle.

Answer: D

Arc length and central angle are related by \( s = r\theta \) with \( \theta \) in radians.

\[ \theta = \frac{s}{r} = \frac{19}{30} = 0.6333 \text{ rad} \]

Converting to degrees:

\[ 0.6333 \times \frac{180}{\pi} = 36.28^{\circ} \]

The formula \( s = r\theta \) works only in radians; in degrees it needs the extra factor \( \tfrac{\theta}{360} \times 2\pi r \).