Mensuration in Geometry is a fairly intuitive subject. Structural and spatial understanding of the forms, other than knowing simple formulae, makes one nail Mensuration questions in the test. A candidate can easily expect 3-5 Mensuration questions in CAT based on logarithm and also in other MBA exams such as NMAT, XAT, SNAP, etc.
1. A solid right circular cone of height 27 cm is cut into pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is
A) 243 B) 232 C) 256 D) 264 [CAT 2020]
Answer
If height becomes 1/3rd then radius also becomes 1/3rd
If both height and radius become 1313rd, Volume will be 1/27th
So, Top part’s volume = V
Remaining part = 26V
Totally = 27V
Given 26 V – V = 225
V = 9
27 V = 243
2. The length, breadth, and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will (CAT 19 – Slot 1)
1. Remain the same
2. Dec. by 13.64%
3. Dec. by 15%
4. Dec. by 18.75%
5. Dec. by 30%
Answer
in the present case, let length = l = 3x, breadth = b = 2x, height = h = x
then, area of four walls = 2 (l + b) h = 2(3x + 2x) x = 10×2.
now as length gets doubled = 6x, breadth halved = x, height halved = x/2.
new area of four walls = 2 (6x + x) x/2 = 7×2.
thus there is a decrease of 30%. hence, the fifth option is the answer.
3. Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is
A) 3:4 B) 2:3 C) 5:6 D) 4:5 [CAT 2019]
Answer
In given case , figure can be drawn as below
Let side of equilateral triangle ABC = 3a
So side of hexagon = a
Area of triangle = (root3 /4 )*(3a)^2 = 9a^2 *(root3/4)
Area of hexagon with side a = 6*(root3 /4 )*(a)^2
Ratio of areas of hexagon to that of triangle = { 6*(root3 /4 )*(a)^2 } / { 9a^2 *(root3/4)} = 6/9 = 2/3
Thus required ratio = 2 : 3
4. A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is [CAT 2019]
A) 8464π B) 928π C) 1026(1 + π) D) 1044(4 + π)
Answer
As cylinders have the same volume and each of these has radius 3 cm. So volume of each cylinder will be equal to HCF of (405, 783 and 351) which is 27.
So volume of each cylinder = 27
No of cylinders = [ 405/27 ] + [783/27] + [351/27] = 15 + 29 + 13 =57
Using V = πr^2 h
27 = 22/7 *9 *h
So h = 3/ π cm
• Impact of the cavity on a solid figure’s surface area
• Length of longest rod in cube shape room
• Impact of the solids melting to create new solids
• Slant surface and Entire Pyramid surface
• Dimensions of Pyramid etc.
Learn, study and remember the Curved Surface Area, Gross Surface Area, and Volume for the following solid figures or mensuration topics
SOLID | Total Surface Area | Lateral/ Curved Surface area | Volume | Length of Leading Diagonal/ Slant Height |
Cube | 2(LB+ BH+ HL) | 2H (L + B) | LBH | √ (L^{2} + H^{2} + B^{2}) |
Cuboid | 6a^{2} | 4a^{2} | a^{3} | √3a |
Cylinder | 2Πr (r + h) | 2Πrh | Πr^{2}h | No Slant height or |
Cone | Πr (r + l) | Πrl | ⅓Πr^{2}h | √(h^{2} + r^{2}) |
Sphere | 4Πr^{2} | 4Πr^{2} | ^{4}/_{3}Πr^{3} | No Slant height or |
Hollow Cylinder | 2Π(r₁+r₂) (r₂-r₁+h) | 2Πh(r₁+r₂) | Πh(r₂²-r₁²) | No Slant height or |
Frustum | Π(R_{1} + R_{2})s + (R_{1}^{2} + R_{2}^{2}) | Π(R_{1} + R_{2})s | ⅓Πh(R_{1}^{2} + R_{2}^{2} + R_{1}R_{2}) | √(h^{2} + (R_{1} – R_{2})^{2}) |
Hemisphere
| 3Πr^{2} | 2Πr^{2} | ^{2}/_{3}Πr^{3} | No Slant height or |
Level 1
Level 2
Write down the formulas on a sheet of paper and put them on the wall so you can see them regularly. It’ll help memorize them for you.
Level 3
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