Competitive Exams : Quantitative Aptitude - Mensuration (2D/3D) (Set 2) | Concept of Maths

Question 1

1. A wire in the form of a square with a side of $22\text{ cm}$ is reshaped into the form of a circle. What is the area of the circle formed? (Take $\pi = \frac{22}{7}$)

A) $484\text{ cm}^2$

B) $616\text{ cm}^2$

C) $154\text{ cm}^2$

D) $308\text{ cm}^2$

Answer

Correct Answer: B

Solution: 1. Perimeter of the square = $4 \times \text{side} = 4 \times 22 = 88\text{ cm}$. 2. When reshaped into a circle, the circumference of the circle is equal to the perimeter of the square. $$2\pi r = 88$$ $$2 \times \frac{22}{7} \times r = 88 \implies r = \frac{88 \times 7}{44} = 14\text{ cm}$$ 3. Area of the circle = $\pi r^2 = \frac{22}{7} \times 14 \times 14 = 22 \times 28 = 616\text{ cm}^2$.

Question 2

2. A solid metallic cylinder of radius $6\text{ cm}$ and height $15\text{ cm}$ is melted to form a number of small solid cones, each of radius $3\text{ cm}$ and height $9\text{ cm}$. How many such cones can be formed?

A) 15

B) 18

C) 20

D) 24

Answer

Correct Answer: C

Solution: 1. Volume of Cylinder = $\pi R^2 H = \pi \times 6^2 \times 15 = 540\pi \text{ cm}^3$. 2. Volume of one Cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 3^2 \times 9 = 27\pi \text{ cm}^3$. 3. Number of cones = $\frac{\text{Volume of Cylinder}}{\text{Volume of Cone}} = \frac{540\pi}{27\pi} = 20$.

Question 3

3. The length of a rectangular park exceeds its breadth by $8\text{ m}$. If the area of the park is $240\text{ m}^2$, find the perimeter of the park.

A) $60\text{ m}$

B) $64\text{ m}$

C) $68\text{ m}$

Question 4

4. A sphere is exactly inscribed inside a cube. What is the ratio of the volume of the cube to the volume of the sphere?

A) $6 : \pi$

B) $\pi : 6$

C) $3 : 4\pi$

D) $4 : 3\pi$

Answer

Correct Answer: A

Solution: 1. Let the side of the cube be $a$. 2. Since the sphere is inscribed in the cube, its diameter is equal to the side of the cube, so radius $r = \frac{a}{2}$. 3. Volume of Cube $= a^3$. 4. Volume of Sphere $= \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (\frac{a}{2})^3 = \frac{4}{3} \pi \frac{a^3}{8} = \frac{\pi a^3}{6}$. 5. Ratio $= a^3 : \frac{\pi a^3}{6} = 1 : \frac{\pi}{6} = 6 : \pi$.

Question 5

5. The diagonals of a rhombus are $16\text{ cm}$ and $12\text{ cm}$. Find the height of the rhombus if its area is considered with one of its sides as the base.

A) $8.4\text{ cm}$

B) $9.2\text{ cm}$

C) $9.6\text{ cm}$

D) $10.4\text{ cm}$

Answer

Correct Answer: C

Solution: 1. Area of rhombus $= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 16 \times 12 = 96\text{ cm}^2$. 2. Side of rhombus $s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{ cm}$. 3. Area can also be expressed as Base $\times$ Height: $s \times h = 96$. $$10 \times h = 96 \implies h = 9.6\text{ cm}$$.

Question 6

6. A right prism has a triangular base with sides $3\text{ cm}$, $4\text{ cm}$, and $5\text{ cm}$. If the height of the prism is $10\text{ cm}$, what is the total surface area?

A) $120\text{ cm}^2$

B) $126\text{ cm}^2$

C) $132\text{ cm}^2$

D) $144\text{ cm}^2$

Answer

Correct Answer: C

Solution: 1. Base is a right-angled triangle (since $3^2 + 4^2 = 5^2$). 2. Base Area $= \frac{1}{2} \times 3 \times 4 = 6\text{ cm}^2$. 3. Perimeter of Base $= 3 + 4 + 5 = 12\text{ cm}$. 4. Lateral Surface Area $= \text{Perimeter} \times \text{Height} = 12 \times 10 = 120\text{ cm}^2$. 5. Total Surface Area $= \text{LSA} + 2 \times \text{Base Area} = 120 + 2(6) = 132\text{ cm}^2$.

Question 7

7. Find the area of an isosceles triangle whose base is $12\text{ cm}$ and each of its equal sides is $10\text{ cm}$.

A) $48\text{ cm}^2$

B) $60\text{ cm}^2$

C) $36\text{ cm}^2$

D) $54\text{ cm}^2$

Answer

Correct Answer: A

Solution: 1. Base $b = 12$, equal sides $a = 10$. 2. Height $h = \sqrt{a^2 - (\frac{b}{2})^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}$. 3. Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 8 = 48\text{ cm}^2$.

Question 8

8. A hollow cylindrical pipe is $7\text{ cm}$ long. The external radius of the pipe is $5\text{ cm}$ and the volume of the metal used in making the pipe is $352\text{ cm}^3$. Find the thickness of the pipe. (Use $\pi = \frac{22}{7}$)

A) $1\text{ cm}$

B) $1.5\text{ cm}$

C) $2\text{ cm}$

D) $2.5\text{ cm}$

Answer

Correct Answer: C

Solution: 1. Volume of metal $= \pi(R^2 - r^2)h$, where $R=5, h=7, V=352$. $$352 = \frac{22}{7} \times (5^2 - r^2) \times 7$$ $$352 = 22 \times (25 - r^2) \implies 16 = 25 - r^2$$ $$r^2 = 9 \implies r = 3\text{ cm}$$. 2. Thickness $= R - r = 5 - 3 = 2\text{ cm}$.

Question 9

9. The length of the arc of a sector of a circle is $44\text{ cm}$ and the radius of the circle is $14\text{ cm}$. What is the area of the sector?

A) $154\text{ cm}^2$

B) $308\text{ cm}^2$

C) $616\text{ cm}^2$

D) $440\text{ cm}^2$

Answer

Correct Answer: B

Solution: 1. Area of a sector can be calculated using the arc length ($L$) and radius ($r$): $$\text{Area} = \frac{1}{2} \times L \times r$$ $$\text{Area} = \frac{1}{2} \times 44 \times 14 = 22 \times 14 = 308\text{ cm}^2$$.

Question 10

10. The radii of the top and bottom of a bucket (frustum of a cone) are $20\text{ cm}$ and $8\text{ cm}$ respectively. If the height of the bucket is $16\text{ cm}$, find its capacity in terms of $\pi$.

A) $3120\pi \text{ cm}^3$

B) $3224\pi \text{ cm}^3$

C) $3328\pi \text{ cm}^3$

D) $3416\pi \text{ cm}^3$

Answer

Correct Answer: C

Solution: 1. Volume of Frustum $= \frac{1}{3} \pi h (R^2 + r^2 + Rr)$. 2. Here $R = 20, r = 8, h = 16$. $$V = \frac{1}{3} \pi \times 16 \times (20^2 + 8^2 + 20 \times 8)$$ $$V = \frac{1}{3} \pi \times 16 \times (400 + 64 + 160) = \frac{1}{3} \pi \times 16 \times 624$$ $$V = \pi \times 16 \times 208 = 3328\pi \text{ cm}^3$$.

Accepted Answer

Correct Answer: B

Solution: Let breadth $= B$. Then length $L = B + 8$. Area $= L \times B = (B + 8)B = 240$. $$B^2 + 8B - 240 = 0$$ $$(B + 20)(B - 12) = 0 \implies B = 12\text{ m}$$ (Breadth cannot be negative). Length $L = 12 + 8 = 20\text{ m}$. Perimeter $= 2(L + B) = 2(20 + 12) = 2(32) = 64\text{ m}$.

Competitive Exams : Quantitative Aptitude - Mensuration (2D/3D) (Set 2) | Concept of Maths