#dailyquizadda #mensuration #18Question #sscAdvancemaths
Mensuration is a pure formula-based topic and tricks/shortcuts are seldom applied here. So in this series I will try to solve all the mensuration problems that have appeared in CGL lately and in the process I will share the important concepts/formulas.
For Prism and Calendar (figures with uniform girth) -
Lateral Surface Area = Height * Perimeter of the Base
Volume = Height * Area of the Base
In this question the Total surface area is being asked
Total Surface Area of a Prism = Lateral Surface Area + Area of the two bases
Height of the prism = 10 cm
Perimeter of the base = 5 + 12 + 13 (Calculate the hypotenuse with Pythagoras Theorem) = 30 cm
So Lateral Surface Area = 10 * 30 = 300 cm
Area of the base = 1/2 * base * height = 1/2 * 5 * 12 = 30 cm
So Total Surface Area = 300 + 2*30 = 360 cm
Answer : (A)
Put the value of r/R from equation (1)
Area of base/quadrilateral = 84 + 54 = 138 cm
Volume = Height * Area of the Base
2070 = Height * 138
So, Height of the prism = 15 cm
Lateral Surface Area = Height * Perimeter of the Base
Perimeter of the base = AB + BC + CD + DA = 48 cm
Lateral Surface Area = 48 * 15 = 720 cm^2
Answer : (A)
Volume of the prism = Area of the base * Height =√3/4 * a^2 * h ... (1)
Lateral surface Area of the prism = Perimeter of the base * Height = 3a * h ...(2)
Divide equation (1) by (2)
Volume/Area = (1/4√3) * a
40√3/120 = a/4√3 [Since Volume = 40√3 and Lateral surface Area = 120]
a = 160 * 3/120
Answer: (C)
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 A2 and volume as A3. Now A3 is given
and you have to find A3. How will you do it? Simple, first take the square-root of A2 to convert it into A,
and then take the cube of A to find A3.
So for solving this question, we just have to take the square-root of 4:9. The ratio will become 2:3. Then
take the cube of 2:3. Hence the answer is 8:27
Answer: (C)
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)
Diameter and perimeter are directly proportional, P = D*π, where P is the perimeter and D is the
diameter.
Hence a 75% increase in diameter means a 75% increase in perimeter
The area of base and the volume of a cone are directly proportional V = A * h/3, where V = volume and h
= height of the cone
Hence a 100% increase in the area of the base would mean a 100% increase in the volume
Answer: (B)
Answer: (A)
Where ever the word “melting” is used in mensuration, it means only one thing – equate the volume
The volume of the rectangular block = l*b*h = 21*77*24 cm3
Now this volume will be equal to the volume of the sphere formed after melting the block
Volume of sphere = (4/3) * π * r3 = 21*77*24
Hence, r = 21 cm
Answer: (A)
The water rises by 5.6 cm. Take this 5.6 cm as the height of the cylindrical beaker and find its volume.
Volume of a cylinder = π*r*r*h = π * (7/2) * (7/2) * 5.6
Volume of the marbles (spherical in shape) = (4/3) * π * r3 = (4/3) * π * 0.7 * 0.7 * 0.7
No. of marbles dropped = Volume of beaker/Volume of a marble = 150
Answer: (B)
Answer: (C)
Mensuration is a pure formula-based topic and tricks/shortcuts are seldom applied here. So in this series I will try to solve all the mensuration problems that have appeared in CGL lately and in the process I will share the important concepts/formulas.
Q. 1)
For Prism and Calendar (figures with uniform girth) -
Lateral Surface Area = Height * Perimeter of the Base
Volume = Height * Area of the Base
In this question the Total surface area is being asked
Total Surface Area of a Prism = Lateral Surface Area + Area of the two bases
Height of the prism = 10 cm
Perimeter of the base = 5 + 12 + 13 (Calculate the hypotenuse with Pythagoras Theorem) = 30 cm
So Lateral Surface Area = 10 * 30 = 300 cm
Area of the base = 1/2 * base * height = 1/2 * 5 * 12 = 30 cm
So Total Surface Area = 300 + 2*30 = 360 cm
Answer : (A)
Q. 2)
In such questions, remember one thing SIMILARITY
r/R = h/H ... (1)
where r = radius of small cone
R = radius of Big cone
h = height of small cone
H = height of big cone
Volume of cone = 1/3 * π* r2 * h
Given, Volume of big cone = 27 * Volume of small cone
1/3 * π * R2
* H = 27 * 1/3 * π * r2 * h
27 * (r/R)^2=H/hPut the value of r/R from equation (1)
27 * (h/H)^2 = H/h
27 * h3
= H3
Put H = 30 cm
So h = 10 cm
The question asks us the height above the base, which is (30 - h) = 30 - 10 = 20 cm
Answer : (B)
Q. 3)
The base of the prism looks like the figure above.
AD = 12 cm, AB = 9 cm
Hence BD = 15 cm (Pythagoras Theorem)
Area of base = Area of triangle ABD + Area of triangle BDC
Area of triangle ABD = 1/2 * 9 * 12 = 54 cm
Area of triangle BDC = 84 cm (Apply Heron's formula)Area of base/quadrilateral = 84 + 54 = 138 cm
Volume = Height * Area of the Base
2070 = Height * 138
So, Height of the prism = 15 cm
Lateral Surface Area = Height * Perimeter of the Base
Perimeter of the base = AB + BC + CD + DA = 48 cm
Lateral Surface Area = 48 * 15 = 720 cm^2
Answer : (A)
Q. 4)
Area of the base = √3/4 * a^2, where a is the side of the equilateral triangle
Perimeter of the base = 3aVolume of the prism = Area of the base * Height =√3/4 * a^2 * h ... (1)
Lateral surface Area of the prism = Perimeter of the base * Height = 3a * h ...(2)
Divide equation (1) by (2)
Volume/Area = (1/4√3) * a
40√3/120 = a/4√3 [Since Volume = 40√3 and Lateral surface Area = 120]
a = 160 * 3/120
a = 4 cm
Answer : (A)
Q. 5)
This question is about 'Pyramid'. So let me just give you some basic
understanding of Pyramids. CGL can ask questions about two types of
Pyramids - Pyramid with a Triangular Base and Pyramid with a square
base. Both these pyramids have different formulas. Look at the below
figure and understand the labellings, i.e., Slant edge and Slant Height
In the below image, I have written formulas for both types of Pyramids. The formula for Volume is same for both the Pyramids-
V = 1/3 * A * h
where A = Area of the base (calculation of A will be different for both)
h = Height of the Pyramid
When lateral surface area is asked, you will first calculate the 'Slant
Height'. Then with the help of slant height you will find the area of
one lateral face (let's call this area M). If the pyramid is having a
triangular base then multiply M with 3, to get the lateral surface area
of the pyramid. And if the pyramid is having a square base, then
multiply M with 4.
The area of the square is 324, hence its side is 18 cm
Volume of the pyramid = 1/3 * Area of the base * Height
1296 = 1/3 * 324 * Height
So Height = 12 cm
Slant Height of the pyramid with square base = √ (h^2 + a^2/4) = √(12^2 + 18^2/4)
Slant Height = 15cm
Area of the lateral face = 1/2 * Base * Height = 1/2 * 18 * 15 = 135 cm^2
Pyramid with a square base has 4 lateral faces, so lateral surface area of the pyramid = 4 * 135 = 540 cm^2
Answer : (D)
Q.6)
This is 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 calculations a bit lengthy. So, just find 1.6*r and the answer
will be little less than that. Like here
Side = 1.6*84 = 134.4, so the answer is 132 cm
Answer: (A)
Method:
When the wire is bent in the form of a circle of radius
84cm, that means the circumference (or the length of the wire) of the circle is
2*π*84 = 44*12 cm
Now this wire forms a square of (let’s say) side ‘a’
Then, 4a (perimeter of the square) = 44*12
Hence a = 132 cm
Q. 7)
Q. 8)
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)/4π
If rolled across its breadth, then = (b*b*l)/4π
In this question the sheet is rolled
along its length, so volume = (l*l*b)/4π = 12*12*5/(4* π)
Volume = 180/ π
cm3
Answer: (C)
Q. 9)
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 A2 and volume as A3. Now A3 is given
and you have to find A3. How will you do it? Simple, first take the square-root of A2 to convert it into A,
and then take the cube of A to find A3.
So for solving this question, we just have to take the square-root of 4:9. The ratio will become 2:3. Then
take the cube of 2:3. Hence the answer is 8:27
Answer: (C)
Q. 10)
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)
Q. 11)
Diameter and perimeter are directly proportional, P = D*π, where P is the perimeter and D is the
diameter.
Hence a 75% increase in diameter means a 75% increase in perimeter
Answer: (D)
Q. 12)
The area of base and the volume of a cone are directly proportional V = A * h/3, where V = volume and h
= height of the cone
Hence a 100% increase in the area of the base would mean a 100% increase in the volume
Answer: (B)
Q. 13)
A is increased by 50% hence A2 (or
surface area) will increase by (1.5*1.5 – 1)*100 % = 125%
Note: Similarly A3 (or volume) will
increase by (1.5*1.5*1.5 – 1)*100 % = 237.5%
Q. 14)
Where ever the word “melting” is used in mensuration, it means only one thing – equate the volume
The volume of the rectangular block = l*b*h = 21*77*24 cm3
Now this volume will be equal to the volume of the sphere formed after melting the block
Volume of sphere = (4/3) * π * r3 = 21*77*24
Hence, r = 21 cm
Answer: (A)
Q. 15)
Volume of a cylinder = π*r*r*h = π * (7/2) * (7/2) * 5.6
Volume of the marbles (spherical in shape) = (4/3) * π * r3 = (4/3) * π * 0.7 * 0.7 * 0.7
No. of marbles dropped = Volume of beaker/Volume of a marble = 150
Answer: (B)
Q. 16)
Let
the radius of the big sphere be R.
Volume
of a cone = (1/3) * π * R3 (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)
* π * R3 = (4/3) * π * r3
r
: R = 1 : 22/3
Surface
area of smaller sphere(s) = 4 * π * r2
Surface
area of larger sphere(S) = 4 * π * R2
S/s
= (r/R)2 = 1 : 24/3
Answer: (D)
Q. 17)
When
a cone is hollowed out from a cylinder, we get the above figure
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
= π * r * l, where l = slant height of the cone, which is
or
Hence
A = π*3*5 = 15π
B
= 2πrh = 2π*3*4 = 24π
C
= πr2 = π*32 = 9π
The
whole surface area of the remaining solid = 15π + 24π + 9π = 48π
Answer: (C)
Q. 18)
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
dailyquizadda #SSC #AdvanceMaths
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