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

A dish in the shape of a frustum of a cone has a height of $6$cm. Its top and its bottom have radii of $24$cm and $16$cm respectively. Find its curved surface area. ( in $\displaystyle cm^{2}$)
  1. $200 \displaystyle \pi $
  2. $300 \displaystyle \pi $
  3. $400 \displaystyle \pi $
  4. $500 \displaystyle \pi $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The slant height l = sqrt(h^2 + (R-r)^2) = sqrt(6^2 + (24-16)^2) = sqrt(36 + 64) = 10. The curved surface area of a frustum is pi * (R+r) * l = pi * (24+16) * 10 = 400 * pi.

Multiple choice

Consider a pyramid with a square base side of $6$ inches and with a height of $12$ inches, as shown below. If we cut off the top of the pyramid parallel to the base $3$ inches from the tip, what is the volume of the remaining solid?
  1. $141.75$
  2. $140$
  3. $135.48$
  4. $144$
  5. $130$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The original pyramid has height 12 and base side 6. The cut is made 3 inches from the tip, creating a smaller pyramid with height 3. By similar triangles, the base side of the smaller pyramid is (3/12) * 6 = 1.5. Volume = (1/3) * base_area * height. Original volume = (1/3) * 36 * 12 = 144. Smaller volume = (1/3) * (1.5^2) * 3 = 2.25. Remaining volume = 144 - 2.25 = 141.75.

Multiple choice

Metal spheres ,each of radius $2$ cm, are packed into a rectangular box of internal dimensions $16$ cm $\times $ $8$ cm $\times $ $8$ cm. When $16$ spheres are packed the box filled with preservative liquid. Find the volume of this liquid. (Give the answer to the nearest integer value)
  1. $482 cm^3$
  2. $478 cm^3$
  3. $488 cm^3$
  4. $490 cm^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume of box = 16 * 8 * 8 = 1024 cm^3. Volume of 16 spheres = 16 * (4/3) * pi * (2^3) = 16 * (4/3) * 3.14159 * 8 = 536.16 cm^3. Liquid volume = 1024 - 536.16 = 487.84, which rounds to 488.

Multiple choice

A circle touches the sides $\overline { \mathrm { BC } } , \overline { \mathrm { CA } } , \overline { \mathrm { AB } }$ of $\Delta \mathrm { ABC }$ at points D, E, F respectively. BD = x CE =y , AF = z then area of $\Delta A B C = \sqrt { x y z ( x + y + z ) }$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The area of a triangle with an incircle touching sides at D, E, F is indeed given by the formula sqrt(r_a * r_b * r_c * s) where r_i are the segments from vertices to tangency points. This is a standard geometric property.

Multiple choice

The radii ${ r }{ 1 },{ r }{ 2 },{ r }_{ 3 }$ escribed circle of the traingle $ABC$ are in H.P. If its area is $24$ square cm and its perimeter is $24$ cm, then lengths of its sides are
  1. $4,6,8$
  2. $3,9,11$
  3. $6,8,10$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Area = 24, perimeter = 24, so semi-perimeter s = 12. r = Area/s = 24/12 = 2. The exradii are r1 = Area/(s-a), r2 = Area/(s-b), r3 = Area/(s-c). Since they are in HP, (s-a), (s-b), (s-c) are in AP. This implies a, b, c are in AP. For sides 6, 8, 10, s = 12, area = sqrt(12*6*4*2) = 24. This matches.