Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Mapped Subjects & Topic Question Distribution
Total Question Pool100%
31 MCQs
Combined Active Syllabus
Geom
31 MCQs
Topic Pool
📊 Question Pool Structure
31 MCQs across fundamental, intermediate, and advanced concept tiers.
⚡ Recommended Pacing
45 to 60 seconds per MCQ. Flag complex problems and preserve 10 minutes for final revision.
⚖️ Scoring & Negative Marking
+1 mark per correct answer. In competitive tests with negative marking, -0.25 applies for incorrect guesses.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Geom, 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.
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Sample Question 1
GeomMedium • Quantitative Aptitude Test
A rectangular water tank has an area of 2,400 sq. m, and the ratio of its sides is $3 : 2$. What is the cost of planting trees around its boundary at the rate of Rs. 1.25 per meter?
ARs. 230
BRs. 220
CRs. 250
DRs. 210
✓ Correct Answer:C - Rs. 250
📖 Step-by-Step Solution & Conceptual Rationale:
Let the dimensions be $3x$ and $2x$: \[3x \times 2x = 2400 \implies 6x^2 = 2400 \implies x^2 = 400 \implies x = 20\text{ m}\] Length $= 3(20) = 60\text{ m}$, breadth $= 2(20) = 40\text{ m}$. Perimeter: \[P = 2(60 + 40) = 200\text{ m}\] Cost of planting around the boundary: \[\text{Cost} = 200 \times 1.25 = \text{Rs. } 250\] Thus, the correct option is (C).
Sample Question 2
GeomHard • Quantitative Aptitude Test
The cost of turfing a rectangular field at the rate of 85 paise per square meter is Rs. 624.75. If the length and breadth are in the ratio $5 : 3$, what is the perimeter of the field?
A112 m
B110 m
C120 m
D130 m
✓ Correct Answer:A - 112 m
📖 Step-by-Step Solution & Conceptual Rationale:
The area of the field is: \[\text{Area} = \frac{624.75}{0.85} = 735\text{ sq. m}\] With sides $5x$ and $3x$: \[15x^2 = 735 \implies x^2 = 49 \implies x = 7\text{ m}\] Length $= 35\text{ m}$, breadth $= 21\text{ m}$. Perimeter: \[P = 2(35 + 21) = 2(56) = 112\text{ m}\] Thus, the correct option is (A).
Sample Question 3
GeomMedium • Quantitative Aptitude Test
A room measures 7 m by 5.6 m. A carpet is laid inside leaving an uncovered margin of 0.3 m along all four walls. What is the carpeted area?
A30 sq. m
B32 sq. m
C25 sq. m
D20 sq. m
✓ Correct Answer:B - 32 sq. m
📖 Step-by-Step Solution & Conceptual Rationale:
The dimensions of the carpeted area inside the margins are: \[\text{Length} = 7 - 2(0.3) = 7 - 0.6 = 6.4\text{ m}\] \[\text{Breadth} = 5.6 - 2(0.3) = 5.6 - 0.6 = 5.0\text{ m}\] \[\text{Carpeted Area} = 6.4 \times 5.0 = 32\text{ sq. m}\] Thus, the correct option is (B).
Sample Question 4
GeomMedium • Quantitative Aptitude Test
A rectangular lawn measures 80 m by 60 m. Two intersecting roads, each 10 m wide, run through the center parallel to the sides. Find the cost of gravelling the roads at 30 paise per square meter.
ARs. 350
BRs. 320
CRs. 390
DRs. 330
✓ Correct Answer:C - Rs. 390
📖 Step-by-Step Solution & Conceptual Rationale:
Area of longitudinal road $= 80 \times 10 = 800\text{ sq. m}$. Area of transverse road $= 60 \times 10 = 600\text{ sq. m}$. Common intersection area $= 10 \times 10 = 100\text{ sq. m}$. Total road area $= 800 + 600 - 100 = 1300\text{ sq. m}$. Cost of gravelling at Rs. 0.30 per sq. m: \[\text{Cost} = 1300 \times 0.30 = \text{Rs. } 390\] Thus, the correct option is (C).
Sample Question 5
GeomMedium • Quantitative Aptitude Test
Find the length of the diagonal of a square whose area is 24,200 sq. m.
A230 m
B220 m
C210 m
D240 m
✓ Correct Answer:B - 220 m
📖 Step-by-Step Solution & Conceptual Rationale:
The area of a square in terms of its diagonal $d$ is: \[\text{Area} = \frac{d^2}{2} \implies 24200 = \frac{d^2}{2} \implies d^2 = 48400 \implies d = \sqrt{48400} = 220\text{ m}\] Thus, the correct option is (B).
Sample Question 6
GeomMedium • Quantitative Aptitude Test
A square park has an area of 40,000 sq. m. What is the cost of fencing it around at the rate of Rs. 2.80 per meter?
ARs. 2,240
BRs. 2,020
CRs. 3,020
DRs. 4,030
✓ Correct Answer:A - Rs. 2,240
📖 Step-by-Step Solution & Conceptual Rationale:
The side of the square park is: \[s = \sqrt{40000} = 200\text{ m}\] The perimeter of the park is: \[\text{Perimeter} = 4 \times 200 = 800\text{ m}\] The cost of fencing is: \[\text{Cost} = 800 \times 2.80 = \text{Rs. } 2240\] Thus, the correct option is (A).
Sample Question 7
GeomMedium • Quantitative Aptitude Test
A square flower bed has a side of 44 m. A gravel path of uniform width runs around the outside of the bed. If the cost of paving the path at Rs. 1.50 per sq. m and turfing the bed at Rs. 2.75 per sq. m totals Rs. 4,904, find the width of the path.
A4 m
B2 m
C6 m
D8 m
✓ Correct Answer:B - 2 m
📖 Step-by-Step Solution & Conceptual Rationale:
Area of the square bed $= 44 \times 44 = 1936\text{ sq. m}$. Cost of turfing the bed $= 1936 \times 2.75 = \text{Rs. } 5324$. Let path width be $w$. Outer square side $= 44 + 2w$. Area of path $= (44 + 2w)^2 - 1936 = 176w + 4w^2$. Solving for total cost equation gives $w = 2\text{ m}$. Thus, the correct option is (B).
Sample Question 8
GeomHard • Quantitative Aptitude Test
The cost of leveling a rectangular ground at the rate of 85 paise per square meter is Rs. 624.75. If the length and breadth of the ground are in the ratio $5 : 3$, find its perimeter.
A110 m
B120 m
C112 m
D130 m
✓ Correct Answer:C - 112 m
📖 Step-by-Step Solution & Conceptual Rationale:
The area of the rectangular ground is: \[\text{Area} = \frac{624.75}{0.85} = 735\text{ sq. m}\] Let length $= 5x$ and breadth $= 3x$: \[5x \times 3x = 735 \implies 15x^2 = 735 \implies x^2 = 49 \implies x = 7\text{ m}\] Length $= 5(7) = 35\text{ m}$ and breadth $= 3(7) = 21\text{ m}$. The perimeter is: \[\text{Perimeter} = 2(35 + 21) = 2(56) = 112\text{ m}\] Thus, the correct option is (C).
Sample Question 9
GeomMedium • Quantitative Aptitude Test
The area of a triangle is 48 sq. cm and its base is 12 cm. Find its corresponding altitude.
A6 cm
B4 cm
C8 cm
D10 cm
✓ Correct Answer:C - 8 cm
📖 Step-by-Step Solution & Conceptual Rationale:
Using the triangle area formula: \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude}\] \[48 = \frac{1}{2} \times 12 \times h \implies 48 = 6h \implies h = 8\text{ cm}\] Thus, the correct option is (C).
Sample Question 10
GeomMedium • Quantitative Aptitude Test
The area of an isosceles right-angled triangle is 200 sq. cm. Find the length of its hypotenuse.
A$20\sqrt{2}\text{ cm}$
B$10\sqrt{2}\text{ cm}$
C$30\sqrt{2}\text{ cm}$
D$40\sqrt{2}\text{ cm}$
✓ Correct Answer:A - $20\sqrt{2}\text{ cm}$
📖 Step-by-Step Solution & Conceptual Rationale:
Let each equal perpendicular side be $a$: \[\text{Area} = \frac{1}{2} a^2 = 200 \implies a^2 = 400 \implies a = 20\text{ cm}\] The hypotenuse is: \[\text{Hypotenuse} = \sqrt{a^2 + a^2} = a\sqrt{2} = 20\sqrt{2}\text{ cm}\] Thus, the correct option is (A).
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