Quantitative Aptitude

Mensuration

2,507 Questions
All Direct Formula Application 1067 Shape Transformations 343 Mensuration Ratio Equations 277 Combined & Hollow Solids 247 Successive Percentage Changes 178 Inscribed & Circumscribed Figures 165 Cost Estimation Problems 100 Liquid Pour & Submergence 67 Unit Conversions 35 Path & Border Calculations 17 Time & Work Integration 7 Scale & Model Proportions 2

Mensuration Questions

Multiple choice

In each of the following questions, read the given statement and compare the Quantity I and Quantity II on its basis. निम्नलिखित प्रत्येक प्रश्न में दिए गए कथन को पढ़िए और उसके आधार पर मात्रा I और मात्रा II की तुलना कीजिए। Quantity : I if the radius of a circle is increased by 20 % then change in Area of circle. Quantity : II if length of a rectangle is increased by 50 % and width decreased by 20 % then change in area of rectangle. मात्रा : I यदि किसी वृत्त की त्रिज्या में 20% की वृद्धि की जाती है तो वृत्त के क्षेत्रफल में परिवर्तन होता है। मात्रा : II यदि एक आयत की लंबाई में 50% की वृद्धि की जाती है और चौड़ाई में 20% की कमी की जाती है तो आयत के क्षेत्रफल में परिवर्तन होता है।

  1. Quantity : I > Quantity : II मात्रा : I > मात्रा : II

  2. Quantity : I ≥ Quantity : II मात्रा : I ≥ मात्रा : II

  3. Quantity : I < Quantity : II मात्रा : I < मात्रा : II

  4. Quantity : II ≥ Quantity : I मात्रा : II ≥ मात्रा : I

  5. Quantity I = Quantity II or relation can't be established मात्रा I = मात्रा II या संबंध स्थापित नहीं किया जा सकता
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quantity I: New radius = 1.2r. New area = π(1.2r)² = 1.44πr², a 44% increase. Quantity II: New dimensions = 1.5l × 0.8w. New area = 1.5 × 0.8 × lw = 1.2lw, a 20% increase. Since 44% > 20%, Quantity I > Quantity II.

Multiple choice

In each of the following questions, a question & then two statements – I & II are given. You have to determine that the data provided in these statements is enough to answer the question or not. Study both the statement: \ What is the capacity of cylindrical tank? \ Statement I –Radius of the base is half of its height. Area of base is 616 sqm. height of the cylinder is 28m. \ Statement II – Radius of the cylinder equal to the height of a cone whose volume is 1224 cbm, height of the cone equal to the side of square whose perimeter is 96m

  1. Only Statement I alone.

  2. Only Statement II alone.

  3. Both Statements I and II together.

  4. Neither Statement I nor II is sufficient.

  5. Either Statement I or II.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Statement I provides base area (616 sqm) and height (28m). From base area = πr², we can find r = √(616/π). Then volume = πr²h can be calculated. Statement II only gives the cylinder's radius (via cone volume and square perimeter relations) but doesn't provide the cylinder's height, so volume cannot be determined. Therefore, only Statement I is sufficient.

Multiple choice

In each of the following questions, read the given statement and compare the Quantity I and Quantity II on its basis. निम्नलिखित प्रत्येक प्रश्न में दिए गए कथन को पढ़िए और उसके आधार पर मात्रा I और मात्रा II की तुलना कीजिए। Quantity I : A sphere of diameter 14 cm is melted and cast into a right circular cone of height 7 cm. The radius of the base of the cone is ? Quantity II: A circle having area 154cm2 radius is मात्रा I : 14 सेमी व्यास वाले एक गोले को पिघलाकर 7 सेमी ऊंचाई वाले एक लंब वृत्तीय शंकु में ढाला जाता है। शंकु के आधार की त्रिज्या है ? मात्रा II: 154cm2 क्षेत्रफल वाले एक वृत्त त्रिज्या है

  1. Quantity : I > Quantity : II मात्रा : I > मात्रा : II

  2. Quantity : I ≥ Quantity : II मात्रा : I ≥ मात्रा : II

  3. Quantity : I < Quantity : II मात्रा: I < ) मात्रा: II

  4. Quantity : II ≥ Quantity : I मात्रा: II ≥ मात्रा: I

  5. Quantity I = Quantity II or relation can't be established मात्रा I = मात्रा II या संबंध स्थापित नहीं किया जा सकता
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sphere: diameter 14cm, radius = 7cm, volume = (4/3)π(7)³ = 1436.76 cm³. Cone: same volume, height = 7cm, so πr²(7) = 1436.76, giving r = 14cm. Circle: area 154 cm² = πr², so r = 7cm. Quantity I (14cm) > Quantity II (7cm), so option A is correct.

Multiple choice

A spherical drop of water carrying a charge of 0.032 nC has a potential of 512 V at its surface. If two such drops with the same radius and charge were to combine to form a single drop, what would be the potential at the surface of the new drop?

  1. 136.2 V

  2. 113 V

  3. 999.08 v

  4. 812.756 V
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Potential V = kQ/R. When two drops combine, volume doubles, so R_new = R * 2^(1/3). Charge Q_new = 2Q. V_new = k(2Q) / (R * 2^(1/3)) = V * 2^(2/3) = 512 * 1.5874 = 812.756 V.

Multiple choice

A TV tower has a height of $100$m. The area covered by the TV broadcast, if radius of the earth is $6400$ km, will be?
  1. $380\times 10^7m^2$
  2. $402\times 10^7m^2$
  3. $595\times 10^7m^2$
  4. $440\times 10^7m^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The area covered by a TV tower of height h is A = 2 * pi * R * h. Given h = 100 m = 0.1 km and R = 6400 km, A = 2 * 3.14 * 6400 * 0.1 = 4019.2 sq km. Converting to square meters: 4019.2 * (1000 m)^2 = 4019.2 * 10^6 m^2 = 401.92 * 10^7 m^2, which is approximately 402 * 10^7 m^2.

Multiple choice

The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius $= \sqrt{3}$ is :
  1. $\displaystyle \frac{4}{3} \sqrt{3} \pi$
  2. $2 \pi$
  3. $\displaystyle \frac{8}{3} \sqrt{3} \pi$
  4. $4 \pi$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a cylinder of radius r and height h inscribed in a sphere of radius R, r^2 + (h/2)^2 = R^2. Volume V = pi * r^2 * h = pi * (R^2 - h^2/4) * h = pi * (R^2*h - h^3/4). Maximize V by setting dV/dh = 0: R^2 - 3h^2/4 = 0, so h^2 = 4R^2/3, h = 2R/sqrt(3). Then r^2 = R^2 - R^2/3 = 2R^2/3. Max V = pi * (2R^2/3) * (2R/sqrt(3)) = 4*pi*R^3 / (3*sqrt(3)). With R=sqrt(3), V = 4*pi*(3*sqrt(3)) / (3*sqrt(3)) = 4*pi.

Multiple choice

The radius of a right circular cylinder increases at the rate of $0.1 cm/min$, and the height decreases at the rate of $0.2 cm/min$. The rate of change of the volume of the cylinder, in $cm^3/min.$, when the radius is $2 cm$ and the height is $3 cm$ is
  1. $-2p$
  2. $-\displaystyle \frac{8\pi}{5}$
  3. $-\displaystyle \frac{3\pi}{5}$
  4. $\displaystyle \frac{2\pi}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

V = pi * r^2 * h. dV/dt = pi * (2rh * dr/dt + r^2 * dh/dt). Given r=2, h=3, dr/dt=0.1, dh/dt=-0.2. dV/dt = pi * (2*2*3*0.1 + 2^2*-0.2) = pi * (1.2 - 0.8) = 0.4pi = 2pi/5.

Multiple choice

A circular metal plate is heated so that its radius increases at a rate of $0.1$ $mm/ minute$. Then the rate at which the plate's area is increasing when the radius is $50$ $cm$ is
  1. $10 \pi$ $mm^2 /minute$
  2. $100 \pi$ $mm^2 /minute$
  3. $ \pi$ $mm^2 /minute$
  4. $- \pi$ $mm^2 /minute$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area A = pi * r^2. Rate of change dA/dt = 2 * pi * r * (dr/dt). Given r = 50 cm = 500 mm and dr/dt = 0.1 mm/min. dA/dt = 2 * pi * 500 * 0.1 = 100 * pi mm^2/min.

Multiple choice

The radius of a right circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as radius. When the radius is $1cm$ the altitude is $6 cm$. When the radius is $6cm$, then volume is increasing at the rate of $1$ Cu $cm/sec$. When the radius is $36cm$, the volume is increasing at a rate of $n$ cu. $cm/sec$. The value of '$n$' is equal to:
  1. $12$
  2. $22$
  3. $30$
  4. $33$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

r = r(t), h = 3r + c. At r=1, h=6, so 6 = 3(1) + c => c=3. h = 3r + 3. V = pi*r^2*h = pi*r^2(3r+3) = 3pi(r^3 + r^2). dV/dt = 3pi(3r^2 + 2r) * dr/dt. At r=6, dV/dt = 1: 1 = 3pi(3*36 + 12) * dr/dt => 1 = 3pi(120) * dr/dt => dr/dt = 1/(360pi). At r=36, dV/dt = 3pi(3*36^2 + 2*36) * (1/(360pi)) = 3pi(3888 + 72) / 360pi = 3960 / 120 = 33.

Multiple choice

Water is flowing out at the rate of $\displaystyle 6::m^{3}/min$ from a reservoir shaped like a hemispherical bowl of radius$ R = 13$ m The volume of water in the hemispherical bowl is given by $\displaystyle v=\frac{\pi }{3}.y^{2}\left ( 3R-y \right )$ when the water is $y$ meter deep Find a t what rate is the water level changing when the water is $8$ m deep.
  1. $\displaystyle -\frac{1}{12\pi }m/min$
  2. $\displaystyle -\frac{1}{18\pi }m/min$
  3. $\displaystyle -\frac{1}{24\pi }m/min$
  4. $\displaystyle -\frac{1}{30\pi }m/min$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given V = (pi/3) * y^2 * (3R - y). Differentiate with respect to t: dV/dt = (pi/3) * [2y * (3R - y) * dy/dt + y^2 * (-dy/dt)]. Given dV/dt = -6, R = 13, y = 8. -6 = (pi/3) * [16 * (39 - 8) * dy/dt - 64 * dy/dt]. -6 = (pi/3) * [496 - 64] * dy/dt = (pi/3) * 432 * dy/dt = 144 * pi * dy/dt. dy/dt = -6 / (144 * pi) = -1 / (24 * pi).

Multiple choice

The height of the cone of maximum volume inscribed in a sphere of radius R is
  1. $\displaystyle \frac{\mathrm{R}}{3}$
  2. $\displaystyle \frac{2\mathrm{R}}{3}$
  3. $\displaystyle \frac{4\mathrm{R}}{3}$
  4. $\displaystyle \frac{4\mathrm{R}}{\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume V = 1/3 * pi * r^2 * h. In a sphere of radius R, r^2 = R^2 - (h-R)^2 = 2Rh - h^2. V = 1/3 * pi * (2Rh^2 - h^3). dV/dh = 1/3 * pi * (4Rh - 3h^2) = 0. h = 4R/3.