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

The total surface area of a solid cylinder is $616$ $cm^2$. If the ratio between its curved surface area and total surface area is $1 : 2$; find the volume of the cylinder.
  1. $1078$ cu. cm
  2. $1070$ cu.cm
  3. $1700$$ cu.cm
  4. $1087$ cu. cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total Surface Area (TSA) = 2*pi*r*(r+h) = 616. Curved Surface Area (CSA) = 2*pi*r*h. Given CSA/TSA = 1/2, then CSA = 308. Thus 2*pi*r*h = 308 and 2*pi*r^2 = 308. From 2*pi*r^2 = 308, r^2 = 49, so r = 7. Then 2*pi*7*h = 308 gives h = 7. Volume = pi*r^2*h = (22/7)*49*7 = 1078.

Multiple choice

The sum of the height and the radius of a solid cylinder is $35\ cm$ and its total surface area is $3080\ cm^2$; find the volume of the cylinder.
  1. $12,936\ cu. cm$
  2. $12,933\ cu.cm$
  3. $12,930\ cu.cm$
  4. $12,900\ cu. cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given h + r = 35 and 2*pi*r(h + r) = 3080. 2 * (22/7) * r * 35 = 3080. 44 * 5 * r = 3080. 220 * r = 3080, so r = 14. Then h = 35 - 14 = 21. Volume = pi * r^2 * h = (22/7) * 14 * 14 * 21 = 22 * 2 * 14 * 21 = 12,936.

Multiple choice

Two solid cylinders, one with diameter $60$ cm and height $30$ cm and the other with radius $30$ cm and height $60$ cm are melted and recasted into a third solid cylinder of height $10$ cm. Find the diameter of the cylinder formed.
  1. $150$ cm
  2. $180$ cm
  3. $140$ cm
  4. $160$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The volume of the first cylinder is pi * 30^2 * 30 = 27000 * pi. The volume of the second is pi * 30^2 * 60 = 54000 * pi. The total volume is 81000 * pi, which must equal pi * r^2 * 10, so r^2 = 8100, r = 90. The diameter is 2 * r = 180 cm.

Multiple choice

The total surface area of a solid hemisphere of radius 3.5 m is covered with canvas at the rate of Rs. 20 per $m^{2}$. The total cost to cover the hemisphere is $\displaystyle \left(Take\, \pi\, =\, \frac{22}{7}\right)$

  1. Rs. 2210

  2. Rs. 2310

  3. Rs. 2220

  4. Rs. 2420
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total surface area of a solid hemisphere is 3 * pi * r^2. With r = 3.5 m, the area is 3 * (22/7) * 3.5 * 3.5 = 115.5 m^2. Multiplying this area by the rate of Rs. 20 per m^2 gives a total cost of 115.5 * 20 = Rs. 2310.

Multiple choice

A right circular metallic cone of height $20$ cm and radius of base $5$ cm is melted and recast into a sphere. Find the radius of the sphere.
  1. $4$ cm
  2. $3$ cm
  3. $5$ cm
  4. $10$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the cone is melted and recast into a sphere, the volume remains constant. The volume of the cone is (1/3) * pi * r^2 * h = (1/3) * pi * 5^2 * 20 = 500/3 * pi, which equals the volume of the sphere (4/3) * pi * R^3, solving to R = 5 cm.

Multiple choice

How many spherical lead shots each $4.2 cm $ in diameter can be obtained from a rectangular solid (cuboid) of lead with dimensions $66 cm, 42 cm, 21 cm$. (Take $\pi\, =\, \dfrac {22}{7}$)
  1. $1500$
  2. $1200$
  3. $1300$
  4. $1600$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of cuboid = 66 * 42 * 21 = 58212 cm^3. Volume of one spherical shot = (4/3) * pi * r^3 = (4/3) * (22/7) * (2.1)^3 = (4/3) * (22/7) * 9.261 = 38.808 cm^3. Number of shots = 58212 / 38.808 = 1500.