3D Figures, Heights and Distances Previous Year Questions (PYQs) for XAT: 4+ Solved Questions with Step-by-Step Solutions

Solve 4+ 3D Figures, Heights and Distances previous year questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: 3D Figures, Heights and Distances

Chapter Journey

1. 3D Solids & Surface Area

Mastering volumes, surface areas, frustums, combinations of solids, and shortest paths on 3D surfaces.

2. Heights, Distances & Angles of Elevation

Applying trigonometric ratios to solve for unknown heights and horizontal distances.

The Shift to Three Dimensions

The Core Distinction

When moving from 2D to 3D geometry, we shift our focus from flat boundaries to solid spaces.

Volume

The 3D space enclosed. Measured in cubic units ( $c m^{3}$ , $m^{3}$ ).

Surface Area

The 2D area covering the exterior. Measured in square units ( $c m^{2}$ , $m^{2}$ ).

Two Types of Surface Area

Curved / Lateral (CSA / LSA): Only the side walls.
Total (TSA): Side walls PLUS all top and bottom bases.

3D Figures, Heights and Distances: Solved Questions with Step-by-Step Explanations (4 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

An industrial robot manufacturing company is tasked to design humanoid robots to be used in warehouses where the robots need to pick items from a stack of shelves. The height of the topmost shelf from the ground is 7 feet. To operate, the robot has to move on a track, running parallel to the stack of shelves. The track is fixed 1 foot away from the base of the stack of shelves. Further, the robot cannot raise its arms by more than 60° from the horizontal plane.

If the robot’s arms are attached to its shoulder, what should be the minimum height of the robot from the ground to the shoulder for its arms to reach the topmost shelf?

A.
$7 - 3$ feet
B.
7 feet
C.
$7 - \frac{1}{3}$ feet
D.
None of the other options is correct
E.
$6 + 3$ feet

Correct Answer:

Step-by-Step Solution

Key idea: This is a max-reach angle of elevation problem. The robot's arm forms the hypotenuse of a right triangle, with the horizontal distance and the vertical reach forming the other two sides.

Step 1: Identify the given values. The topmost shelf is at a height of 7 feet. The horizontal distance from the track to the shelves is 1 foot. The maximum angle of elevation of the arm is 60°.

Step 2: Let the height of the robot's shoulder be $h$ . The vertical distance the arm needs to reach from the shoulder to the topmost shelf is $7 - h$ .

Step 3: Use the tangent ratio for the right triangle formed by the arm, the horizontal distance, and the vertical reach.

$tan (6 0^{\circ}) = \frac{Opposite}{Adjacent} = \frac{7 - h}{1}$

Step 4: Solve for $h$ .

$3 = 7 - h ⟹ h = 7 - 3$ feet.

Answer: $7 - 3$ feet.

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

An archer, from the point A, is aiming at a target, placed on the top of a 56 feet tall tree. The base of the tree is at the point C. From the archer’s position, the angle of elevation to the target is 45° from his eye level. The archer, facing the tree, moves backwards on the straight line joining the points A and C, to a new position at the point B. From the point B, the angle of elevation from his eye level to the target becomes 30°.

How far did the archer move from A to B(in feet) if his eye level is at a height of 6 feet from the ground?

A.
$503$
B.
$563$
C.
$50 (3 - 1)$
D.
$50 (3 + 1)$
E.
$\frac{50}{3}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a moving observer problem with an eye-level trap. The angles of elevation are measured from the observer's eye level, not from the ground. We must adjust the effective height of the target.

Step 1: Identify the effective height of the target. The tree is 56 feet tall, and the archer's eye level is 6 feet. The vertical distance from the eye level to the target is $56 - 6 = 50$ feet.

Step 2: Let the initial horizontal distance from the archer at A to the tree base C be $x$ . The angle of elevation is 45°.

$tan (4 5^{\circ}) = \frac{50}{x} ⟹ 1 = \frac{50}{x} ⟹ x = 50$ feet.

Step 3: The archer moves backwards by a distance $d$ to point B. The new horizontal distance is $x + d = 50 + d$ . The new angle of elevation is 30°.

$tan (3 0^{\circ}) = \frac{50}{50 + d} ⟹ \frac{1}{3} = \frac{50}{50 + d}$

Step 4: Solve for $d$ .

$50 + d = 503 ⟹ d = 503 - 50 = 50 (3 - 1)$ feet.

Answer: $50 (3 - 1)$ feet.

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A solid trophy, consisting of two parts, has been designed in the following manner: the bottom part is a frustum of a cone with the bottom radius 30 cm, the top radius 20 cm, and height 40 cm, while the top part is a hemisphere with radius 20 cm. Moreover, the flat surface of the hemisphere is the same as the top surface of the frustum.

If the entire trophy is to be gold-plated at the cost of Rs. 40 per square cm, what would the cost for gold-plating be closest to?

A.
Rs. 1,12,000
B.
Rs. 3,60,000
C.
Rs. 4,73,000
D.
Rs. 5,23,000
E.
Rs. 3,72,000

Correct Answer:

Step-by-Step Solution

Key idea: This is a surface area of joined solids problem. When two solids are joined, the surfaces that touch each other become hidden and are not included in the total exposed surface area.

Step 1: Identify the components and their dimensions.

Bottom part: Frustum of a cone with bottom radius $R = 30$ cm, top radius $r = 20$ cm, height $h = 40$ cm.

Top part: Hemisphere with radius $r = 20$ cm.

Step 2: Calculate the slant height of the frustum ( $l$ ).

$l = h^{2} + (R - r)^{2} = 4 0^{2} + (30 - 20)^{2} = 1600 + 100 = 1700 = 1017$ cm.

Step 3: Calculate the exposed surface areas.

Curved Surface Area (CSA) of frustum = $π (R + r) l = π (30 + 20) (1017) = 500 π 17$ cm².
Curved Surface Area of hemisphere = $2 π r^{2} = 2 π (20)^{2} = 800 π$ cm².
Bottom base of the frustum = $π R^{2} = π (30)^{2} = 900 π$ cm².

(The top base of the frustum is covered by the hemisphere, so it is not exposed).

Step 4: Total exposed surface area = $500 π 17 + 800 π + 900 π = π (50017 + 1700)$ .

Using $π \approx 3.1416$ and $17 \approx 4.123$ :

Area $\approx 3.1416 \times (500 \times 4.123 + 1700) = 3.1416 \times (2061.5 + 1700) = 3.1416 \times 3761.5 \approx 11817$ cm².

Step 5: Calculate the cost of gold plating.

Cost = $11817 \times 40 = 472, 680$ Rs.

The closest option is Rs. 4,73,000.

Answer: Rs. 4,73,000.

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

There are three rectangular tanks in a building. The length, width and height of the first tank are m meters each, and the length, width and height of the second tank are n meters each. However, the length, width and height of the third tank are m meters, n meters and 1 meter, respectively.Initially, the first tank is full of water, while the second and the third are empty. When the second and the third tanks are completely filled with water transferred from the first tank, 85,000 liters of water is still left in the first tank.

If both m and n are positive integers, what is the value of m?(1 meter=1000 liters)

A.
None of the other options is correct
B.
7
C.
5
D.
6
E.
10

Correct Answer:

Step-by-Step Solution

Key idea: This is a volume conservation problem involving fluid transfer, recognizable because water is moved from one container to fill others, leaving a specific remainder.

Step 1: Convert the remaining water from liters to cubic meters to match the dimensions of the tanks. Since $1 m^{3} = 1000 liters$ , the remaining water is $85, 000/1000 = 85 m^{3}$ .

Step 2: Calculate the volume of each tank using the formula $Volume = length \times width \times height$ .

Volume of Tank 1 = $m \times m \times m = m^{3}$
Volume of Tank 2 = $n \times n \times n = n^{3}$
Volume of Tank 3 = $m \times n \times 1 = mn$

Step 3: Set up the conservation of volume equation. The initial volume of water in Tank 1 equals the sum of the volumes used to fill Tanks 2 and 3, plus the water left in Tank 1.

$m^{3} = n^{3} + mn + 85$

Step 4: Rearrange the equation to group the variables.

$m^{3} - n^{3} - mn = 85$

Step 5: Use the given condition that $m$ and $n$ are positive integers. Since $m^{3}$ must be strictly greater than 85, $m$ must be at least 5 (because $4^{3} = 64 < 85$ ). We can test integer values for $m$ :

If $m = 5$ : $125 - n^{3} - 5 n = 85 ⟹ n^{3} + 5 n = 40$ . Testing integers, $n = 3$ gives $27 + 15 = 42 \neq = 40$ . No integer solution.
If $m = 6$ : $216 - n^{3} - 6 n = 85 ⟹ n^{3} + 6 n = 131$ . Testing integers, $n = 4$ gives $64 + 24 = 88$ , and $n = 5$ gives $125 + 30 = 155$ . No integer solution.
If $m = 7$ : $343 - n^{3} - 7 n = 85 ⟹ n^{3} + 7 n = 258$ . Testing integers, $n = 6$ gives $216 + 42 = 258$ . This is a perfect match!

Step 6: The value of $m$ that satisfies the equation with a positive integer $n$ is 7.

Answer: 7

A question from this chapter

Question 1

An industrial robot manufacturing company is tasked to design humanoid robots to be used in warehouses where the robots need to pick items from a stack of shelves. The height of the topmost shelf from the ground is 7 feet. To operate, the robot has to move on a track, running parallel to the stack of shelves. The track is fixed 1 foot away from the base of the stack of shelves. Further, the robot cannot raise its arms by more than 60° from the horizontal plane. If the robot’s arms are attached to its shoulder, what should be the minimum height of the robot from the ground to the shoulder for its arms to reach the topmost shelf?

Question 2

An archer, from the point A, is aiming at a target, placed on the top of a 56 feet tall tree. The base of the tree is at the point C. From the archer’s position, the angle of elevation to the target is 45° from his eye level. The archer, facing the tree, moves backwards on the straight line joining the points A and C, to a new position at the point B. From the point B, the angle of elevation from his eye level to the target becomes 30°. How far did the archer move from A to B(in feet) if his eye level is at a height of 6 feet from the ground?

Question 3

A solid trophy, consisting of two parts, has been designed in the following manner: the bottom part is a frustum of a cone with the bottom radius 30 cm, the top radius 20 cm, and height 40 cm, while the top part is a hemisphere with radius 20 cm. Moreover, the flat surface of the hemisphere is the same as the top surface of the frustum. If the entire trophy is to be gold-plated at the cost of Rs. 40 per square cm, what would the cost for gold-plating be closest to?

Question 4

There are three rectangular tanks in a building. The length, width and height of the first tank are m meters each, and the length, width and height of the second tank are n meters each. However, the length, width and height of the third tank are m meters, n meters and 1 meter, respectively.Initially, the first tank is full of water, while the second and the third are empty. When the second and the third tanks are completely filled with water transferred from the first tank, 85,000 liters of water is still left in the first tank. If both m and n are positive integers, what is the value of m?(1 meter=1000 liters)

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3D Figures, Heights and Distances Previous Year Questions (PYQs) for XAT: 4+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: 3D Figures, Heights and Distances

Chapter Journey

1. 3D Solids & Surface Area

2. Heights, Distances & Angles of Elevation

The Shift to Three Dimensions

The Core Distinction

Volume

Surface Area

Two Types of Surface Area

3D Figures, Heights and Distances: Solved Questions with Step-by-Step Explanations (4 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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Built around you, not around a syllabus PDF

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Questions calibrated to the real exam

Notes written for recall, not for volume

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Honest progress

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