Here is the Mensuration Hacks (Part II). We have already provided you the
Mensuration Hacks (Part I). So we suggest you to read the Mensuration Hacks (Part I) before reading this part. This way you can get the whole concepts of Mensuration.
Mensuration is purely formula based topic. So here, we will try to solve
different types of mensuration questions that have appeared in SSC Exams and in this process we will share the
important mensuration concepts / formulas / tricks / hacks.
Q1. A piece of wire when bent to form a circle will have a radius of 84cm. If the wire is bent to form a square the length of a side of the square is:
A. 132 cm
B. 225 cm
C. 152 cm
D. 168 cm
Solution:
Note:- This a famous question. Just remember whenever you are forming a circle and then a square, the side of that square is given by, a = 1.6*r (approx.) where r = radius of the circle.
I have written approx. because the actual formula is 1.57*r but it will make the calculation a bit lengthy. So just find 1.6*r and the answer will be little less than that.
Now,
1.6*84=134.4, so the answer be little less than 134.4. Check the options given above. Answer will be 132cm.
Answer: (A)
Q2: At each corner of a triangular field of sides 26m, 28m, and 30m, a cow is tethered by a rope of length 7m. The area in (m square) ungrazed by the cows is:
A. 336
B. 259
C. 154
D. 77
Solution:
Note:- In such question the grazed area is given by
(Pi*r^2)/2
Now,
(22/7)*7*7/2= 77 sq m
So the un-grazed area = area of the triangle - grazed area.
Area of the triangle= apply heron's formula.
Semi perimeter = (a+b+c)/2 = 42 m
Area = 336 sq m
Un-grazed area = 336 - 77= 259 sq m
Answer: (B)
Q3: A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges just to touch each other. Then the volume of the cylinder is:
A. 100/pi cu. cm
B. 60/pi cu. cm
C. 180/pi cu. cm
D. 120/pi cu. cm
Solution:
Note: If the given rectangular sheet of paper [length = l, breadth = b] is rolled across its length to form a cylinder, having a height b, then volume of cylinder = (l*l*b)/4*pi.
Whenever we used the term pi, it's mean (22/7).
If rolled across its breadth then volume of cylinder = (b*b*l)/4 )*pi.
In the above question the sheet is rolled along its length, so volume = (l*l*b)/4 pi = 12*12*5/(4 pi)
Volume = 180/pi
Answer: (C)
Q4: If the surface areas of two spheres are in the ratio 4:9, then the ratio of their volumes will be:
A. 4:9
B.16:27
C. 8:27
D. 16:9
Solution:
Note: In this question the ratio of surface areas is given and they are asking the ratio of volumes. The word sphere is useless here. In such questions just imagine area as A^2 and volume as A^3.
Now A^3 is given and you have to find A^3. How will do it? Simple, first take the square root of A^2 to convert it into A, and then take the cube of A to find A^3.
Answer is 8:27.
Answer: (C)
Q5: If the ratio of the areas of two similar triangles is 4:9 then the ratio of their corresponding sides is:
A. 2:3
B. 3:2
C. 4:9
D. 9:4
Solution:
Here again ratio of areas is given, that means A2 is given, and we have to find A. So 4:9 will become 2:3.
Answer: (A)
Q6: On increasing the diameter of a circle by 75% the percentage increases in the perimeter is:
A. 80%
B. 65%
C. 70%
D. 75%
Solution:
Note: The diameter and the perimeter are directly proportional, P=D*pi, where P is the perimeter and D is the diameter.
Hence a 75% increase in diameter means a 75% increase in perimeter.
Answer: (D)
Q7. If the area of the base of a cone is increased by 100% then the volume is increased by:
A. 141%
B. 100%
C. 200%
D. 182%
Solution:
Note:- the area of base and the volume of a cone are directly proportional V = A*h/3 where V is volume and h is the height of the cone.
Hence, a 100% increase in the area of the base would mean a 100% increase in the volume.
Answer: (B)
Q8: If each edge of a circle is increased by 50%, the percentage increase in surface area is:
A. 125%
B. 50%
C. 100%
D. 75%
Solution:
A is increased by 50% hence A^2 or surface area will increase by (1.5*1.5 - 1)*100% =125%
Answer: (A)
Note:- similarly A^3 or volume will increase by (1*5*1.5*1.5 -1)*100%= 237.5%
Q9: A rectangular block of metal has dimensions 21cm, 77cm, and 24cm. The block has been melted into a sphere. The radius of the sphere is (take pi as 22/7):
A. 21cm
B. 7cm
C. 14 cm
D. 28cm
Solution:
Note:- Whenever the word melting is used in mensuration, it means only one thing => equate the volume.
Volume of the rectangular block = l*b*h= 21*77*24 cu. cm.
Now this volume will be equal to the volume of the sphere formed after melting the block.
Volume of the sphere = (4/3)*pi*r3= 21*77*24
Hence r= 21cm
Answer: (A)
Q10: Marbles of a diameter 1.4cm are dropped into a cylinder beaker containing some water and are fully submerged. The diameter of the beaker is 7cm. Find how many marbles have been dropped in it if the water rises by 5.6cm.
A. 50
B. 150
C. 250
D. 350
Solution:
The water rises by 5.6cm. Take this 5.6cm as the height of the cylinder beaker and find its volume.
Volume of the marbles(spherical in shape)= (4/3)*pi*r3
= (4/3)*pi*0.7*0.7*0.7
No. of marbles dropped = volume of beaker / volume of a marble
= 150
Answer: (B)
Q11: A large solid sphere is melted and moulded to form identical right circular cones with base radius and height same as the radius of the sphere. One of these cones is melted and moulded to form a smaller solid sphere. Then the ratio of the surface area of the smaller to the surface area of the larger sphere is:
A. 1:4^(4/3)
B. 1:2^(3/2)
C. 1:3^(2/3)
D. 1:3^(4/3)
Solution: Let the radius of the big sphere be R.
Volume of a cone= (1/3)*pi*R^3 (since radius and volume are same as the radius of the sphere).
Let the radius of the smaller sphere =r
Then volume of cone = volume of smaller sphere
(1/3)*pi*R^3= (4/3)*pi*r^3.
r:R = 1:2^(2/3).
Surface area of smaller sphere(s) = 4*pi*r^2.
Surface area of larger sphere(S) = 4*pi*R^2.
S/s = (r/R)^2 = 1:2^(4/3)
Answer: (D)
Q12: From a solid right circular cylinder of length 4cm and diameter 6cm, a conical cavity of the same height and base is hollowed out. The whole surface of the remaining solid ( in sq cm) is:
A. 25pi
B. 24pi
C. 48pi
D. 63pi
Solution: When a cone is hollowed out from a cylinder, we get the below fig;
The whole surface area of the remaining solid = area of A + area of B + area of C.
A = curved surface area of the cone
B= curved surface area of the cylinder.
C = area of the cylindrical base.
A = pi*r*l
Where l= slant height of the cone which is square root of (r^2+h^2) or square root of (3^2+4^2) = 5cm.
Hence, A = pi*3*5 = 15 pi
B = 2pi*r*h = 24pi
C= pi*r^2 = 9pi
The whole surface of the remaining solid = 48pi
Answer: (C)
Q13: A spherical ball of radius 1cm is dropped into a conical vessel of radius 3cm and slant height 6cm. The volume of water (in cu.cm) that can just immerse the ball is:
A. Pi/3
B. 4pi/3
C. 5pi/3
D. 3pi
Solution:
Given, AB = 3 cm, BC = 6 cm and OF = 1 cm
Height of the cone (AC) = √(6^2 - 3^2 ) = 3√3 cm
Triangles ABC and CFO are similar (RHS similarity)
So, OC/BC = OF/AB
OC = 2 cm, therefore CG = 3 cm (OG = 1 cm)
Now, ABC and CEG are similar
GE/AB = CG/AC
So, GE =
Required volume = Volume of cone (CDE) - Volume of Sphere
= 3π – (4/3)π
= (5/3)π
Answer: (C)
If you have any doubt in this article, please drop a comment.
Keep reading :)
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