Competitive Exams : Quantitative Aptitude - Mensuration (2D/3D) (Set 1) | Concept of Maths

Question 1

1. A rectangular field is $32 \text{ m}$ long and $24 \text{ m}$ wide. A path of uniform width $3 \text{ m}$ is constructed inside the field along its boundary. Find the cost of paving this path with bricks at the rate of ₹$25$ per square meter.

A) ₹$8,250$

B) ₹$7,500$

C) ₹$6,900$

D) ₹$9,100$

Answer

Correct Answer: B

Solution: Area of the outer rectangle = $32 \times 24 = 768 \text{ m}^2$. Width of path = $3 \text{ m}$. Dimensions of inner rectangle = $(32 - 2 \times 3) \times (24 - 2 \times 3) = 26 \times 18 = 468 \text{ m}^2$. Area of the path = Outer Area - Inner Area = $768 - 468 = 300 \text{ m}^2$. Total cost = $300 \times 25 = ₹7,500$.

Question 2

2. A solid metallic cylinder of base radius $12 \text{ cm}$ and height $15 \text{ cm}$ is melted and recast into small solid cones, each of radius $3 \text{ cm}$ and height $5 \text{ cm}$. How many such cones can be formed?

A) $124$

B) $136$

C) $144$

D) $152$

Answer

Correct Answer: C

Solution: Volume of Cylinder = $\pi R^2 H = \pi \times 12^2 \times 15$. Volume of one Cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 3^2 \times 5$. Number of cones ($n$) = $\frac{\text{Volume of Cylinder}}{\text{Volume of Cone}}$ $n = \frac{\pi \times 144 \times 15}{\frac{1}{3} \pi \times 9 \times 5} = \frac{144 \times 15}{3 \times 5} = 144$.

Question 3

3. A wire is in the form of a circle of radius $28 \text{ cm}$. If it is bent into the shape of a square, what is the difference between the area of the circle and the area of the square? (Use $\pi = \frac{22}{7}$)

A) $528 \text{ cm}^2$

Question 4

4. If the total surface area of a sphere is equal to the total surface area of a hemisphere, find the ratio of their volumes.

A) $1 : \sqrt{2}$

B) $3\sqrt{3} : 8$

C) $\sqrt{3} : 2$

D) $3\sqrt{3} : 4$

Answer

Correct Answer: D

Solution: Let radius of sphere be $R$ and hemisphere be $r$. TSA of Sphere = $4\pi R^2$; TSA of Hemisphere = $3\pi r^2$. Given: $4\pi R^2 = 3\pi r^2 \Rightarrow \frac{R^2}{r^2} = \frac{3}{4} \Rightarrow \frac{R}{r} = \frac{\sqrt{3}}{2}$. Ratio of Volumes = $\frac{\frac{4}{3}\pi R^3}{\frac{2}{3}\pi r^3} = \frac{2 R^3}{r^3} = 2 \times \left(\frac{\sqrt{3}}{2}\right)^3 = 2 \times \frac{3\sqrt{3}}{8} = \frac{3\sqrt{3}}{4}$.

Question 5

5. The perimeter of a rhombus is $100 \text{ cm}$ and one of its diagonals is $30 \text{ cm}$. Find the area of the rhombus.

A) $500 \text{ cm}^2$

B) $750 \text{ cm}^2$

C) $600 \text{ cm}^2$

D) $450 \text{ cm}^2$

Answer

Correct Answer: C

Solution: Perimeter = $4 \times \text{side} = 100 \Rightarrow \text{side} (s) = 25 \text{ cm}$. Diagonals of a rhombus bisect each other at $90^\circ$. Let diagonals be $d_1 = 30$ and $d_2$. Half-diagonals are $15$ and $\frac{d_2}{2}$. Using Pythagoras: $15^2 + (\frac{d_2}{2})^2 = 25^2 \Rightarrow 225 + (\frac{d_2}{2})^2 = 625$. $(\frac{d_2}{2})^2 = 400 \Rightarrow \frac{d_2}{2} = 20 \Rightarrow d_2 = 40 \text{ cm}$. Area = $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 30 \times 40 = 600 \text{ cm}^2$.

Question 6

6. A rectangular water tank of base dimensions $8 \text{ m} \times 6 \text{ m}$ and height $2.5 \text{ m}$ is completely full of water. If $3000$ liters of water are drawn out, find the drop in the water level.

A) $5.5 \text{ cm}$

B) $6.25 \text{ cm}$

C) $7.5 \text{ cm}$

D) $4.75 \text{ cm}$

Answer

Correct Answer: B

Solution: Volume drawn out = $3000$ liters. Since $1000$ liters = $1 \text{ m}^3$, Volume = $3 \text{ m}^3$. Let the drop in water level be $h$ meters. Volume of water removed = Area of base $\times$ Drop in height. $3 = (8 \times 6) \times h \Rightarrow 3 = 48h \Rightarrow h = \frac{3}{48} = \frac{1}{16} \text{ m}$. In cm: $h = \frac{1}{16} \times 100 = 6.25 \text{ cm}$.

Question 7

7. The area of a trapezium is $480 \text{ cm}^2$. The ratio of its parallel sides is $3:5$ and the distance between them is $12 \text{ cm}$. Find the length of the shorter parallel side.

A) $30 \text{ cm}$

B) $50 \text{ cm}$

C) $45 \text{ cm}$

D) $24 \text{ cm}$

Answer

Correct Answer: A

Solution: Let parallel sides be $3x$ and $5x$. Height $h = 12$. Area = $\frac{1}{2} \times (\text{sum of parallel sides}) \times h$. $480 = \frac{1}{2} \times (3x + 5x) \times 12 \Rightarrow 480 = 6 \times 8x$. $480 = 48x \Rightarrow x = 10$. Shorter side = $3x = 3 \times 10 = 30 \text{ cm}$.

Question 8

8. A solid right circular cone is cut into two parts at one-third of its height from the top by a plane parallel to its base. Find the ratio of the volume of the small cone to the volume of the resulting frustum.

A) $1 : 27$

B) $1 : 8$

C) $1 : 26$

D) $1 : 9$

Answer

Correct Answer: C

Solution: Let the height of the original cone be $H$ and the small cone be $h = \frac{H}{3}$. By similarity, the ratio of radii is also $1:3$. Ratio of Volumes = $(\text{ratio of heights})^3 = (\frac{1}{3})^3 = \frac{1}{27}$. If Volume of small cone = $1$ unit, then Volume of original cone = $27$ units. Volume of frustum = Original Volume - Small Cone Volume = $27 - 1 = 26$ units. Ratio = $1 : 26$.

Question 9

9. Find the length of the arc of a sector of a circle with radius $21 \text{ cm}$ and a central angle of $120^\circ$.

A) $22 \text{ cm}$

B) $33 \text{ cm}$

C) $66 \text{ cm}$

D) $44 \text{ cm}$

Answer

Correct Answer: D

Solution: Length of arc = $\frac{\theta}{360} \times 2\pi r$. Length = $\frac{120}{360} \times 2 \times \frac{22}{7} \times 21$. Length = $\frac{1}{3} \times 2 \times 22 \times 3 = 44 \text{ cm}$.

Question 10

10. The lateral surface area of a hollow cylinder is $4224 \text{ cm}^2$. It is cut along its height and formed into a rectangular sheet of width $33 \text{ cm}$. Find the perimeter of the rectangular sheet.

A) $256 \text{ cm}$

B) $322 \text{ cm}$

C) $294 \text{ cm}$

D) $310 \text{ cm}$

Answer

Correct Answer: B

Solution: LSA of cylinder = $2\pi rh = 4224$. When cut along height, the width of the sheet equals the height of the cylinder ($h = 33 \text{ cm}$). The length of the sheet ($L$) equals the circumference of the base ($2\pi r$). $L \times 33 = 4224 \Rightarrow L = \frac{4224}{33} = 128 \text{ cm}$. Perimeter of sheet = $2(L + W) = 2(128 + 33) = 2(161) = 322 \text{ cm}$.

Accepted Answer

Correct Answer: A

Solution: Length of wire = Circumference of circle = $2\pi r = 2 \times \frac{22}{7} \times 28 = 176 \text{ cm}$. Area of circle = $\pi r^2 = \frac{22}{7} \times 28 \times 28 = 2464 \text{ cm}^2$. Perimeter of square = $176 \text{ cm} \Rightarrow$ Side of square ($a$) = $\frac{176}{4} = 44 \text{ cm}$. Area of square = $a^2 = 44^2 = 1936 \text{ cm}^2$. Difference = $2464 - 1936 = 528 \text{ cm}^2$.

Competitive Exams : Quantitative Aptitude - Mensuration (2D/3D) (Set 1) | Concept of Maths