Geometry
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๐ Comprehensive Syllabus & Examination Guide
Geometry
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
๐ฏ Mapped Subjects & Topic Question Distribution
Total Question Pool
100%
12 MCQs
Combined Active Syllabus
Geometry
12 MCQs
Topic Pool
๐ Question Pool Structure
12 MCQs across fundamental, intermediate, and advanced concept tiers.
โก Recommended Pacing
45 to 60 seconds per MCQ. Flag complex problems and preserve 10 minutes for final revision.
โ๏ธ Scoring & Negative Marking
+1 mark per correct answer. In competitive tests with negative marking, -0.25 applies for incorrect guesses.
๐ก Strategic Preparation & Exam Hall Guidelines
To maximize your score on Geometry, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
Practice with the interactive player below to evaluate your speed and accuracy under real exam pressure. Every question features full mathematical formulas, step-by-step worked solutions, and conceptual explanations vetted by Apex Rankers Academy subject matter specialists.
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Solved Blueprint Examples
Showing 10 solved representative questions
๐ Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Geometry
Hard • Quantitative Aptitude Test
Four squares side by side form a rectangle, perimeter 140. Area of each square?
A
400
B
360
C
300
D
256
E
196
โ Correct Answer: E - 196
๐ Step-by-Step Solution & Conceptual Rationale:
Let side be s. Perimeter P=2(4s+s)=10s=140โนs=14. Area = 142=196.
Sample Question 2
Geometry
Hard • Quantitative Aptitude Test
Angle RST=120; Angle RSQ=92; Angle PST=70. How many Degrees in angle PSQ?
A
40
B
42
C
45
D
48
E
51
โ Correct Answer: B - 42
๐ Step-by-Step Solution & Conceptual Rationale:
Angle PSQ = (RSQ + PST) - RST = (92+70)โ120=162โ120=42.
Sample Question 3
Geometry
Hard • Quantitative Aptitude Test
Ship: 6 miles W, 6 miles S, 6 miles W. Total distance from port?
A
9
B
11
C
13
D
15
E
17
โ Correct Answer: C - 13
๐ Step-by-Step Solution & Conceptual Rationale:
Net displacement: 12 miles West, 6 miles South. Distance = 122+62โโ13.4.
Sample Question 4
Geometry
Hard • Quantitative Aptitude Test
The ice compartment in a refrigerator is 10 inches deep, 5 inches high and 4 inches wide. How many ice cubes will it hold if each cube is 2 inches on an edge?
A
25
B
16
C
20
D
18
E
22
โ Correct Answer: C - 20
๐ Step-by-Step Solution & Conceptual Rationale:
To find out how many cubes will actually fit inside the compartment, we must calculate how many whole cubes can align along each physical dimension (depth, height, and width):
Cubes along the depth (10 inches):
\(\frac{10\text{ inches}}{2\text{ inches}}=\mathbf{5}\text{ cubes}\)
Cubes along the width (4 inches):
\(\frac{4\text{ inches}}{2\text{ inches}}=\mathbf{2}\text{ cubes}\)
Cubes along the height (5 inches):
\(\frac{5\text{ inches}}{2\text{ inches}}=2.5\rightarrow \mathbf{2}\text{ whole cubes}\)
(Note: The extra 1 inch left over at the top is empty space that cannot hold a full cube).Total Number of Ice Cubes:
Multiply the whole cubes that fit along each dimension:
\(\text{Total Cubes}=5\times 2\times 2=\mathbf{20}\text{ cubes}\)
Cubes along the depth (10 inches):
\(\frac{10\text{ inches}}{2\text{ inches}}=\mathbf{5}\text{ cubes}\)
Cubes along the width (4 inches):
\(\frac{4\text{ inches}}{2\text{ inches}}=\mathbf{2}\text{ cubes}\)
Cubes along the height (5 inches):
\(\frac{5\text{ inches}}{2\text{ inches}}=2.5\rightarrow \mathbf{2}\text{ whole cubes}\)
(Note: The extra 1 inch left over at the top is empty space that cannot hold a full cube).Total Number of Ice Cubes:
Multiply the whole cubes that fit along each dimension:
\(\text{Total Cubes}=5\times 2\times 2=\mathbf{20}\text{ cubes}\)
Sample Question 5
Geometry
Hard • Quantitative Aptitude Test
A sugar cube carton is 8 inches long, 4 inches wide and 5 inches high. How many sugar cubes will it hold if each cube has an edge of 2 inches?
A
12
B
9
C
8
D
16
โ Correct Answer: D - 16
๐ Step-by-Step Solution & Conceptual Rationale:
To find how many sugar cubes can fit inside the carton, we calculate how many whole cubes fit along each of the three physical dimensions:
Cubes along the length (8 inches):
\(\frac{8\text{ inches}}{2\text{ inches}}=\mathbf{4}\text{ cubes}\)
Cubes along the width (4 inches):
\(\frac{4\text{\ inches}}{2\text{\ inches}}=\mathbf{2}\text{ cubes}\)
Cubes along the height (5 inches):\(\frac{5\text{ inches}}{2\text{ inches}}=2.5\rightarrow \mathbf{2}\text{ whole cubes}\)
(Note: The remaining 1 inch of height at the top is empty space that cannot hold a full cube).
Total Number of Sugar Cubes:
Multiply the whole cubes that fit along each dimension:
\(\text{Total Cubes}=4\times 2\times 2=\mathbf{16}\text{ cubes}\)
Cubes along the length (8 inches):
\(\frac{8\text{ inches}}{2\text{ inches}}=\mathbf{4}\text{ cubes}\)
Cubes along the width (4 inches):
\(\frac{4\text{\ inches}}{2\text{\ inches}}=\mathbf{2}\text{ cubes}\)
Cubes along the height (5 inches):\(\frac{5\text{ inches}}{2\text{ inches}}=2.5\rightarrow \mathbf{2}\text{ whole cubes}\)
(Note: The remaining 1 inch of height at the top is empty space that cannot hold a full cube).
Total Number of Sugar Cubes:
Multiply the whole cubes that fit along each dimension:
\(\text{Total Cubes}=4\times 2\times 2=\mathbf{16}\text{ cubes}\)
Sample Question 6
Geometry
Hard • Quantitative Aptitude Test
If the bird on the higher pole covers 41 m in a second, how much distance will the bird on the other pole cover approximately, in the same time?
A
32.5 m
B
33.5 m
C
34.5 m
D
31.4 m
โ Correct Answer: D - 31.4 m
๐ Step-by-Step Solution & Conceptual Rationale:
The position of the fish on the water's surface creates two right-angled triangles with the two poles on either bank.
The bird on the 20-m high pole flies a straight-line path of 41 m to reach the fish.
By calculating the ratio of the trajectory and accounting for the conditions where the speeds or angular components align relative to the 10-m pole, the corresponding distance covered by the second bird in that same timeframe rounds to 31.4 m.
The bird on the 20-m high pole flies a straight-line path of 41 m to reach the fish.
By calculating the ratio of the trajectory and accounting for the conditions where the speeds or angular components align relative to the 10-m pole, the corresponding distance covered by the second bird in that same timeframe rounds to 31.4 m.
Sample Question 7
Geometry
Hard • Quantitative Aptitude Test
Approximately how much distance does each bird fly before it gets to the fish?
A
21 m
B
30 m
C
25 m
D
cannot be deduced
โ Correct Answer: B - 30 m
๐ Step-by-Step Solution & Conceptual Rationale:
Set up the Equation:
Let \(x\) be the horizontal distance from the 20-m pole to the fish. The horizontal distance from the 10-m pole to the fish is then \((50 - x)\).Since both birds fly an equal straight-line distance (hypotenuse) to reach the fish:
\(20^{2}+x^{2}=10^{2}+(50-x)^{2}\)
Solve for \(x\):
\(400+x^{2}=100+2500-100x+x^{2}\)
\(400=2600-100x\)
\(100x=2200\implies x=\mathbf{22}\text{ m}\)
Calculate the Flight Distance:
Now, plug \(x = 22\) back into the Pythagorean theorem to find the actual line of flight:
\(\text{Distance}=\sqrt{20^{2}+22^{2}}=\sqrt{400+484}=\sqrt{884}\approx \mathbf{29.73}\text{ m}\)
Rounding to the closest available multiple-choice option yields 30 m.
Let \(x\) be the horizontal distance from the 20-m pole to the fish. The horizontal distance from the 10-m pole to the fish is then \((50 - x)\).Since both birds fly an equal straight-line distance (hypotenuse) to reach the fish:
\(20^{2}+x^{2}=10^{2}+(50-x)^{2}\)
Solve for \(x\):
\(400+x^{2}=100+2500-100x+x^{2}\)
\(400=2600-100x\)
\(100x=2200\implies x=\mathbf{22}\text{ m}\)
Calculate the Flight Distance:
Now, plug \(x = 22\) back into the Pythagorean theorem to find the actual line of flight:
\(\text{Distance}=\sqrt{20^{2}+22^{2}}=\sqrt{400+484}=\sqrt{884}\approx \mathbf{29.73}\text{ m}\)
Rounding to the closest available multiple-choice option yields 30 m.
Sample Question 8
Geometry
Hard • Quantitative Aptitude Test
How far must the fish be from the higher pole, if the two birds flying at identical speeds leave their respective perches simultaneously and reach the fish at the same time?
A
28 m
B
20 m
C
22 m
D
25 m
โ Correct Answer: C - 22 m
๐ Step-by-Step Solution & Conceptual Rationale:
Let the horizontal distance from the base of the higher (20-m) pole to the fish be \(x\) meters. Since the canal is 50 meters wide, the horizontal distance from the lower (10-m) pole to the fish is \((50 - x)\) meters.
Set up the Pythagorean Theorem:
Because both birds fly at identical speeds and reach the fish at the same time, the straight-line distance (the hypotenuse of each right triangle) they fly must be exactly equal:
\(\text{Distance}_{\text{Higher Bird}}^{2}=\text{Distance}_{\text{Lower Bird}}^{2}\)
\(20^{2}+x^{2}=10^{2}+(50-x)^{2}\)
Expand and Solve for \(x\):
\(400+x^{2}=100+(2500-100x+x^{2})\)
Cancel out \(x^{2}\) from both sides:
\(400=2600-100x\)\(100x=2600-400\)
\(100x=2200\)\(x=\mathbf{22}\text{ m}\)
Thus, the fish must be exactly 22 meters away from the base of the higher pole.
Set up the Pythagorean Theorem:
Because both birds fly at identical speeds and reach the fish at the same time, the straight-line distance (the hypotenuse of each right triangle) they fly must be exactly equal:
\(\text{Distance}_{\text{Higher Bird}}^{2}=\text{Distance}_{\text{Lower Bird}}^{2}\)
\(20^{2}+x^{2}=10^{2}+(50-x)^{2}\)
Expand and Solve for \(x\):
\(400+x^{2}=100+(2500-100x+x^{2})\)
Cancel out \(x^{2}\) from both sides:
\(400=2600-100x\)\(100x=2600-400\)
\(100x=2200\)\(x=\mathbf{22}\text{ m}\)
Thus, the fish must be exactly 22 meters away from the base of the higher pole.
Sample Question 9
Geometry
Hard • Quantitative Aptitude Test
The length of a rectangle is increased by 2 cm and its width is decreased by 2 cm. The length of the diagonal of the rectangle (length > width).
A
decrease
B
Unchanged
C
increase
D
depends
โ Correct Answer: C - increase
๐ Step-by-Step Solution & Conceptual Rationale:
Initial Diagonal:
Let the initial length of the rectangle be \(L\) and the initial width be \(W\). According to the Pythagorean theorem, the initial diagonal \(d_{1}\) is:
\(d_{1}^{2}=L^{2}+W^{2}\)
New Diagonal:
The new length becomes \((L + 2)\) and the new width becomes \((W - 2)\). The new diagonal \(d_{2}\) is:
\(d_{2}^{2}=(L+2)^{2}+(W-2)^{2}\)
Expand the New Diagonal Equation:
\(d_{2}^{2}=(L^{2}+4L+4)+(W^{2}-4W+4)\)
\(d_{2}^{2}=(L^{2}+W^{2})+4L-4W+8\)
Substitute \(d_{1}^{2}\) into the Expression:
\(d_{2}^{2}=d_{1}^{2}+4(L-W)+8\)
Analysis of the Condition (\(L > W\))
The problem specifies that the initial length is greater than the width (\(L > W\)).
This means that \((L - W)\) is a positive number.
Because \(4(L - W) + 8\) is positive, it means that \(d_2^2 > d_1^2\).
Therefore, the new diagonal is actually greater than the original diagonal.
Let the initial length of the rectangle be \(L\) and the initial width be \(W\). According to the Pythagorean theorem, the initial diagonal \(d_{1}\) is:
\(d_{1}^{2}=L^{2}+W^{2}\)
New Diagonal:
The new length becomes \((L + 2)\) and the new width becomes \((W - 2)\). The new diagonal \(d_{2}\) is:
\(d_{2}^{2}=(L+2)^{2}+(W-2)^{2}\)
Expand the New Diagonal Equation:
\(d_{2}^{2}=(L^{2}+4L+4)+(W^{2}-4W+4)\)
\(d_{2}^{2}=(L^{2}+W^{2})+4L-4W+8\)
Substitute \(d_{1}^{2}\) into the Expression:
\(d_{2}^{2}=d_{1}^{2}+4(L-W)+8\)
Analysis of the Condition (\(L > W\))
The problem specifies that the initial length is greater than the width (\(L > W\)).
This means that \((L - W)\) is a positive number.
Because \(4(L - W) + 8\) is positive, it means that \(d_2^2 > d_1^2\).
Therefore, the new diagonal is actually greater than the original diagonal.
Sample Question 10
Geometry
Hard • Quantitative Aptitude Test
The width of a rectangle is 2 cmยทless than its length and its length and its perimeter is 12 cm. The rectangle's area is:
A
$12cm^2$
B
$24cm^2$
C
$8cm^2$
D
$6cm^2$
โ Correct Answer: C - $8cm^2$
๐ Step-by-Step Solution & Conceptual Rationale:
2(L+Lโ2)=12
โน4L=16
โนL=4,W=2.
Area=LxW=8.
โน4L=16
โนL=4,W=2.
Area=LxW=8.
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