Q. 1
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
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?
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Step-by-Step Explanation & Concept 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).
Q. 2
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
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?
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Step-by-Step Explanation & Concept 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).
Q. 3
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
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?
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Step-by-Step Explanation & Concept 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).
Q. 4
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
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.
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Step-by-Step Explanation & Concept 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).
Q. 5
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Find the length of the diagonal of a square whose area is 24,200 sq. m.
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Step-by-Step Explanation & Concept 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).
Q. 6
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
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?
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Step-by-Step Explanation & Concept 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).
Q. 7
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
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.
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Step-by-Step Explanation & Concept 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).
Q. 8
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
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.
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Step-by-Step Explanation & Concept 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).
Q. 9
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The area of a triangle is 48 sq. cm and its base is 12 cm. Find its corresponding altitude.
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Step-by-Step Explanation & Concept 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).
Q. 10
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The area of an isosceles right-angled triangle is 200 sq. cm. Find the length of its hypotenuse.
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Step-by-Step Explanation & Concept 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).
Q. 11
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The perimeter of an equilateral triangle is 18 cm. Find its area.
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Step-by-Step Explanation & Concept Rationale
Side of the equilateral triangle: \[s = \frac{18}{3} = 6\text{ cm}\] \[\text{Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (6)^2 = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\text{ cm}^2\] Thus, the correct option is (B).
Q. 12
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Find the area of a triangle with sides measuring 150 cm, 120 cm, and 200 cm.
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Step-by-Step Explanation & Concept Rationale
Semi-perimeter $s = \frac{150 + 120 + 200}{2} = 235\text{ cm}$. Using Heron's formula: \[\text{Area} = \sqrt{235(235 - 150)(235 - 120)(235 - 200)} = \sqrt{235 \times 85 \times 115 \times 35} = \sqrt{80,396,875} \approx 8966.43\text{ sq. cm}\] Thus, the correct option is (B).
Q. 13
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The sides of a triangular field are 120 m, 160 m, and 200 m. Find the cost of ploughing the field at the rate of 25 paise per square meter.
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Step-by-Step Explanation & Concept Rationale
Check if right-angled: $120^2 + 160^2 = 14400 + 25600 = 40000 = 200^2$. The area is: \[\text{Area} = \frac{1}{2} \times 120 \times 160 = 9600\text{ sq. m}\] Cost of ploughing at Rs. 0.25 per sq. m: \[\text{Cost} = 9600 \times 0.25 = \text{Rs. } 2400\] Thus, the correct option is (A).
Q. 14
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
The perimeter of a triangular plot is 540 m, and its sides are in the ratio $25 : 17 : 12$. Find the area of the plot.
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Step-by-Step Explanation & Concept Rationale
Sum of ratio parts = $25 + 17 + 12 = 54$. One ratio unit = $\frac{540}{54} = 10\text{ m}$. The sides are $a = 250\text{ m}$, $b = 170\text{ m}$, and $c = 120\text{ m}$. Semi-perimeter $s = \frac{540}{2} = 270\text{ m}$. Using Heron's formula: \[\text{Area} = \sqrt{270(270 - 250)(270 - 170)(270 - 120)} = \sqrt{270 \times 20 \times 100 \times 150} = \sqrt{81,000,000} = 9000\text{ sq. m}\] Thus, the correct option is (C).
Q. 15
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Find the area of a triangle whose sides measure 9 cm, 12 cm, and 15 cm.
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Step-by-Step Explanation & Concept Rationale
Since $9^2 + 12^2 = 81 + 144 = 225 = 15^2$, this is a right-angled triangle with base $12\text{ cm}$ and height $9\text{ cm}$. \[\text{Area} = \frac{1}{2} \times 9 \times 12 = 54\text{ sq. cm}\] Thus, the correct option is (B).
Q. 16
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
A triangle has sides of length 5 cm, 12 cm, and 13 cm. What is the length of the perpendicular drawn from the opposite vertex to the longest side (13 cm)?
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Step-by-Step Explanation & Concept Rationale
Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$, the triangle is a right-angled triangle. Its area is: \[\text{Area} = \frac{1}{2} \times 5 \times 12 = 30\text{ sq. cm}\] Expressing the area using the hypotenuse ($13\text{ cm}$) as the base: \[\frac{1}{2} \times 13 \times h = 30 \implies h = \frac{60}{13} \approx 4.615\text{ cm}\] Thus, the correct option is (A).
Q. 17
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The perimeter of a triangular garden is 240 dm, and two of its sides are 50 dm and 78 dm. Find the cost of watering the garden at the rate of Rs. 2.75 per 100 sq. m.
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Step-by-Step Explanation & Concept Rationale
The third side is: \[c = 240 - (50 + 78) = 112\text{ dm}\] Semi-perimeter $s = \frac{240}{2} = 120\text{ dm}$. Using Heron's formula: \[\text{Area} = \sqrt{120(120 - 50)(120 - 78)(120 - 112)} = \sqrt{120 \times 70 \times 42 \times 8} = \sqrt{2822400} = 1680\text{ sq. dm} = 16.8\text{ sq. m}\] Cost of watering at Rs. 2.75 per 100 sq. m (2.75 paise/sq. m): \[\text{Cost} = 16.8 \times \frac{2.75}{100}\text{ Rs.} = 0.462\text{ Rs.} = 46.2\text{ paise}\] Thus, the correct option is (B).
Q. 18
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The cost of turfing a triangular field at Rs. 45 per 100 sq. m is Rs. 900. If the base of the triangle is 2.5 times its height, find its height.
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Step-by-Step Explanation & Concept Rationale
The total area of the field is: \[\text{Area} = \frac{900}{45} \times 100 = 2000\text{ sq. m}\] Given base $b = 2.5 h$: \[\text{Area} = \frac{1}{2} \times 2.5h \times h = 1.25 h^2\] \[1.25 h^2 = 2000 \implies h^2 = 1600 \implies h = 40\text{ m}\] Thus, the correct option is (C).
Q. 19
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The perimeter of a right-angled triangle is 60 cm, and its hypotenuse is 26 cm. Find the area of the triangle.
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Step-by-Step Explanation & Concept Rationale
Let the perpendicular legs be $a$ and $b$. \[\text{Perimeter} = a + b + 26 = 60 \implies a + b = 34\] By the Pythagorean theorem: \[a^2 + b^2 = 26^2 = 676\] Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$: \[34^2 = 676 + 2ab \implies 1156 = 676 + 2ab \implies 2ab = 480 \implies ab = 240\] The area is: \[\text{Area} = \frac{1}{2} ab = \frac{1}{2} \times 240 = 120\text{ sq. cm}\] Thus, the correct option is (B).
Q. 20
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The area of an isosceles triangle is 60 sq. cm and the length of each of its two equal sides is 13 cm. What are the possible lengths of the base?
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Step-by-Step Explanation & Concept Rationale
Let the base be $b$ and altitude to base be $h$. For an isosceles triangle with equal sides $13\text{ cm}$: \[h = \sqrt{13^2 - (b/2)^2} = \sqrt{169 - \frac{b^2}{4}}\] \[\text{Area} = \frac{1}{2} b h = 60 \implies b \sqrt{169 - \frac{b^2}{4}} = 120\] Squaring both sides: \[b^2 \left(169 - \frac{b^2}{4}\right) = 14400 \implies 169 b^2 - \frac{b^4}{4} = 14400 \implies b^4 - 676 b^2 + 57600 = 0\] Factoring the quadratic in $b^2$: \[(b^2 - 100)(b^2 - 576) = 0 \implies b = 10\text{ cm or } b = 24\text{ cm}\] Thus, the base can be either $24\text{ cm}$ or $10\text{ cm}$. Thus, the correct option is (A).
Q. 21
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Find the area of an equilateral triangle whose side is 9 cm.
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Step-by-Step Explanation & Concept Rationale
The area of an equilateral triangle with side $s$ is: \[\text{Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (9)^2 = \frac{\sqrt{3}}{4} \times 81\text{ cm}^2 \approx 35.07\text{ cm}^2\] Thus, the correct option is (D).
Q. 22
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The base of a triangular field is 2.5 times its height. The cost of turfing the field at the rate of Rs. 35 per 100 sq. m is Rs. 700. Find the length of the base.
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Step-by-Step Explanation & Concept Rationale
The total area of the field is: \[\text{Area} = \frac{700}{35} \times 100 = 2000\text{ sq. m}\] Given base $b = 2.5h$: \[\text{Area} = \frac{1}{2} \times b \times h = \frac{1}{2} \times (2.5h) \times h = 1.25 h^2\] \[1.25 h^2 = 2000 \implies h^2 = \frac{2000}{1.25} = 1600 \implies h = 40\text{ m}\] Therefore, the base is: \[b = 2.5 \times 40 = 100\text{ m}\] Thus, the correct option is (B).
Q. 23
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find the perimeter of the rhombus.
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Step-by-Step Explanation & Concept Rationale
The diagonals of a rhombus bisect each other at right angles: \[\text{Half-diagonals} = \frac{24}{2} = 12\text{ cm},\quad \frac{10}{2} = 5\text{ cm}\] Using the Pythagorean theorem, the side of the rhombus is: \[\text{side} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}\] The perimeter is: \[\text{Perimeter} = 4 \times 13 = 52\text{ cm}\] Thus, the correct option is (B).
Q. 24
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The area of a rhombus is 60 sq. cm. If one of its diagonals is 12 cm, find the length of the other diagonal.
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Step-by-Step Explanation & Concept Rationale
The area of a rhombus in terms of its diagonals $d_1$ and $d_2$ is: \[\text{Area} = \frac{1}{2} d_1 d_2 \implies 60 = \frac{1}{2} \times 12 \times d_2 \implies 60 = 6 d_2 \implies d_2 = 10\text{ cm}\] Thus, the correct option is (A).
Q. 25
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
The area of a parallelogram is 72 sq. cm. If its altitude is twice its corresponding base, determine the length of the base.
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Step-by-Step Explanation & Concept Rationale
Let the base be $b\text{ cm}$. The altitude is $h = 2b\text{ cm}$. \[\text{Area} = b \times h = b \times 2b = 2b^2\] \[2b^2 = 72 \implies b^2 = 36 \implies b = 6\text{ cm}\] Thus, the base is $6\text{ cm}$. Thus, the correct option is (D).
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