AREA & VOLUME (OTHER)AREA AND VOLUME (MENSURATION)MATHS

AREA AND VOLUME (MENSURATION) PART-2

AREA AND VOLUME (MENSURATION) PART-2

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Area and Volume (Mensuration) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

4. Grade-II DASS Exam conducted by Delhi Staff Subordinate Services (DSSSB)

QUERY 11

The radius of a cylinder is 10 cm and height is 4 cm. The number of centimeters that may be added either to the radius or to the height to get the same increase in the volume of the cylinder is?

A) 4 cm
B) 5 cm
C) 10 cm
D) 6 cm

RONNIE BANSAL
Let the number to be added either to the radius or to the height of the cylinder be x

Now the volume of the cylinder after increase in the radius
= π[(10+x)²](4)
And volume after increase in the height
= π(10²)(4+x)
Must you remember that volume of a cylinder = π r²h

As the volume of the cylinder has to show the same increase after that addition, we can say,
π[(10+x)²](4) = π(10²)(4+x)
=> x² – 5x = 0
=> x(x – 5) = 0
=> x=0 or x=5
But 0 is not possible as it does not show any increase in something
Hence 5 is the required number (option ‘B’)

NOTE: In an objective question there are answer options; it will become much easier if solved by them.


QUERY 12

A solid right circular cylinder and a solid hemisphere stand on equal bases and have the same height. The ratio of their whole surface area is?

A) 4:3
B) 3:4
C) 2:3
D) 3:2

MAHA GUPTA
Whole surface area of the cylinder = 2πr(h+r)
Whole surface are of the hemisphere = 3πr²

Therefore, whole surface area of the cylinder : Whole surface are of the hemisphere = 2πr(h+r) : 3πr²

But their height is the same and their bases being equal, so height of the cylinder and radius of the hemisphere will be equal.

According to above, the ratio can be re-written as 2πr(r+r) : 3πr²
=> 4πr² : 3πr²
=> 4 : 3 (option ‘A’)


QUERY 13

A cylindrical can whose base is horizontal and is of internal radius 3.5 cm, contains sufficient water so that when a solid sphere is placed inside, water just covers the sphere. The sphere fits in the can exactly. The depth of water in the can before the sphere was put is?

A) 7/3 cm
B) 14/3 cm
C) 35/3 cm
D) 17/3 cm

MAHA GUPTA
According to the question
Volume of the can up to the length sphere is placed – Volume of the sphere = Volume of the can up to the depth of water that was in the can before the sphere was placed inside

πr²h – (4/3)πr³ =πr²H, where ‘h’ is the height of the can up to sphere is placed i.e. 2r = 2*3.5 = 7; r = 3.5 and H = depth of the water in the can before the sphere was put in (to find)

Now dividing both sides of the above equation by πr²
h – (4/3)r = H
=> 7 – (4/3)*3.5 = H
=> 7/3 = H

So 7/3 cm (option ‘A’) is the required answer.


QUERY 14

What is the ratio of the volume of a cube to that of a sphere, which will exactly fit inside the cube?

A) 5 : π
B) 6 : π
C) 5 : 3
D) 6 : 3

MAHA GUPTA
In such a question one should be careful enough to see which is fitted in what; then the solution doesn’t remain to be much.

Well, here it can easily be thought that the diagonal of the sphere is of the same length as is the length of the side of the cube.

Therefore, we can say if the radius of the sphere (r); side of the cube is 2r

Hence the required ratio = (2r)³ = (4/3)πr³

=> 6 : π (option ‘B’)

QUERY 15

The height of a circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface of the cylinder increases is?

A) 2/3
B) 3/2
C) 2
D) 1/2

MAHA GUPTA
Curved/lateral surface area of the cylinder = 2πrh; where r = radius and h = height of the cylinder.

Let R be the radius and H be the height of the new cylinder

As the area of the base is now decreased to 1/9 of the previous; therefore πR² = (πr²)/9
=> R = r/3

And ‘h’ is now increased by 6 six times; so H = 6h

Now lateral surface area of the new cylinder = 2πRH = 2π(r/3)*(6h) = 4πrh = 2(2πrh)
Means increase by 2 times (option ‘C’)


QUERY 16

If the diameter of a sphere is decreased by 25%, its curved surface area will decrease by?

A) 43.25%
B) 43.50%
C) 43.75%
D) 44.25%

MAHA GUPTA
You have to remember here that decrease in percentage will be the same in the radius also

Now let the radius of the sphere = 1
Therefore radius after decrease = 3/4 (decrease is 25%)

Surface area of the sphere = (4/3)πr²
So, surface area before = 4π/3
And the surface area after = (4/3)π(3/4)² = 3π/4

Hence the decrease in the surface area = 4π/3 – 3π/4 = 7π/12

Decrease on 4π/3 = 7π/12
Decrease on 100 = (7π/12)*(3/4π)*100 = 43.75 (OPTION ‘C’)


QUERY 17

How many tiles, each 40 dm², will be required to cover the floor of a room 8 m long and 6 m broad?

A) 200
B) 260
C) 280
D) 120

MAHA GUPTA
The number of tiles = Area of the floor of the room/area of 1 tile

But the units of measurement must be the same
Now area of the floor of the room = 8*6 = 48 m²
Area of 1 tile is given already i.e. 40 dm² = 40/100 m² = 2/5 m²

So the number of tiles required = 48/(2/5) = 120 (option ‘D’)


QUERY 18

The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km/h?

A) 250
B) 400
C) 300
D) 350

MAHA GUPTA
As the speed to keep is 66 km/h, distance to be covered in 1 minute by the wheel = 66/60 km = (66/60)*1,000 meters = 1,100 meters = 1,10,000 cm

Now we need to find the circumference of the wheel to know the distance to be covered by it 1 revolution

Radius of the wheel = Diameter/2 = 140/2 = 70 cm
Therefore the circumference = 2πr = 2*(22/7)*70 = 440 cm

Hence the number of revolutions to be made by the wheel in 1 minute = 1,10,000/440 = 250 (option ‘A’)


QUERY 19

Two triangular plots each having two sides equal to 100 m. Find maximum perimeter difference of two if area is greater than 4800 for each.

A) 97 m
B) 79 m
C) 80 m
D) 90 m

MAHA GUPTA
Both the plots are isosceles triangular plots in which two of each sides 100 m each; means 100m + 100m + b meters in each

Now the area of an isosceles triangle = (b/4)*√(4a² – b²), where a = equal sides and b = the third side

Now let the area of the first plot = 4800 m²
Therefore (b/4)*√(4a² – b²) = 4800
Solving b = 120
Thus the sides are 100 m, 100 m, 120 m
Now we have to see what can be the maximum possible value of ‘b’ above

We know the sum of two sides of a triangle is always greater than the third side, hence ‘b’ can’t be greater than 199 m as if it’s 200 or more than the above rule will fail.

So the minimum perimeter possible is greater than 100 + 100 +120 = 320

Maximum perimeter possible = 100 + 100 + 199 = 399

Hence maximum perimeter difference = 399 – 320 = 79 (option ‘B’)


QUERY 20

Forty two cubes each of side 1 cm are glued together to form a cuboid. If the perimeter of the base of the cuboid is 18 cm then its height in cm?

A) 2
B) 3
C) 4
D) 1

MAHA GUPTA
The perimeter of the base of the cuboid is 18 cm; means the sum of its length and breadth = 18/2 = 9

But the cuboid is made of cubes being glued together; means the length and breadth must be in natural numbers. So possible pairs are (8, 1), (7, 2), (6, 3) and (4, 4); so its area being either of (8*1 = 8), (7*2 =14), (6*3 = 18) and (4*4 = 16)

Now the total area = Area of the 42 cubes of 1 cm each; means 42*1 = 42 cm²

As the whole game is to played in natural numbers; so it must be divisible by the area of the base of the cuboid. You see 42 is divisible only by 14 among above.

Hence the height of the cuboid = 42/14 = 3 cm (option ‘B’)

Previous post

AREA AND VOLUME (MENSURATION) PART-1

Next post

AREA AND VOLUME (MENSURATION) PART-3

Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)