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

AREA AND VOLUME (MENSURATION) PART-3

AREA AND VOLUME (MENSURATION) PART-3

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 21

If each side of a rectangle is increased by 50%, then its area will be increased by?

A) 50%
B) 100%
C) 125%
D) 250%

MAHA GUPTA
Remember this:
CASE-I: If both the quantities/amounts increase, the net effect: (a + b) + ab/100

CASE-II: If both the quantities/amounts decrease, the net effect = (a + b) – ab/100

CASE-III: If one of the quantities/amounts increase and other decrease, the net effect = (a –b) – ab/100 OR (-a + b) – ab/100 (as the case may be)

Here it’s CASE-I; so area of the triangle will increase by = (a + b) + ab/100 = (50 + 50) + (50*50)/100 = 125% (option ‘C’)


QUERY 22

Length of each equal side of an isosceles triangle is 10 cm and the included angle between those two sides is 45°. Find the area of the triangle.

A) 25√2 cm²
B) 35√2 cm²
C) 5√2 cm²
D) 15√2 cm²

MAHA GUPTA
Area of any triangle with two sides and included angle = (product of the two sides/2)*sinA; where A is the measurement of the angle.

Therefore area of the above triangle = [(10*10)/2]*sin45
= 50*(1/√2) = 25√2 (option ‘A’)


QUERY 23

If a metallic cone of radius 30 cm and height 45 cm is melted and recast into metallic spheres of radius 5 cm; find the number of spheres.

A) 81
B) 41
C) 80
D) 40

MAHA GUPTA
Number of spheres = Volume of cone/volume of a sphere

So, number of spheres = 1⁄3πr²h4⁄3πR³ = 1⁄3(30²)454⁄3(5³) ——– Cancelling π from both

Solving, number of spheres = 81 (option ‘A’)


QUERY 24

A rectangle has 20 cm as its length and 200 cm² as its area. If area is increased by 6/5 times the original area by increasing its length only; then the perimeter of the rectangle so formed is?

A) 100 cm
B) 104 cm
C) 108 cm
D) 110 cm

MAHA GUPTA
Breadth of the original rectangle = 200/20 = 10 cm

New area of the rectangle = Original area + increased area = 200 + 200*(6/5) 440 cm²

But the breadth has to be unchanged; so length of the new rectangle = 440/10 = 44 cm

Hence, the new perimeter = 2*(44+10) = 108 cm (option ‘C’)


QUERY 25

A cuboid piece of brass having a base 11 dm² and height as 0.2 dm is drawn into a cylindrical wire of 0.25 cm in diameter. Find the length of the wire.

A) 400 meter
B) 425 meter
C) 448 meter
D) 500 meter

 

MAHA GUPTA
Radius of the wire = 0.25 cm = (25/100)/2 = 1/8 cm = 1/80 dm

Here, volume of the cuboid = volume of the cylinder
=> area of the base of the cuboid*its height = πr²h

Therefore 11*0.2 = (22/7)*[(1/80)²]*h
=> h = 4480 dm = 448 meter
Obviously h is the length of the wire; so 448 meter is the answer (option ‘C’)


QUERY 26

A man takes 2 minutes to cross the diagonal of a square field at 3 km/h. Find the area of the field.

A) 4000 m²
B) 4500 m²
C) 5000 m²
D) 7000 m²

MAHA GUPTA
Distance covered by the man in 1 hour = 3 km = 3000 meters
So, distance covered by that man in 2 minutes = 100 meters

So, we can say the diagonal of the field = 100 meters

Therefore area of the field = diagonal²/2 = 100²/2 = 5000 m² (option ‘C’)


QUERY 27

The length of a rectangle is increased by 40%. By what percentage the width of the rectangle have to be decreased to maintain the same area?

A) 28.5%
B) 28%
C) 27.75%
D) 35%

MAHA GUPTA
Let the length = 50 and the width = 20
Then the area = 50*20 = 1000

Length when increased = 50 + 40% of 40 = 70
But the area has to be the same

Therefore, New length*new breadth = 1000
=> New breadth = 100070 = 1007

Decrease in width = 20 – 1007 = 407

Decrease on 20 = 407
Hence, decrease on 100 = [(407)/20]*100 = 2007 = 28.5% (option ‘A’)

TRICK
When increase in length or breadth is there
Percentage decrease in Breadth or length = (percentage given/100+Percentage)*100

Therefore here decrease in breadth = (40100+40)*100 = 2007 = 28.50% (option ‘A’)


QUERY 28

Length of the diagonal of a square is 8 cm. A circle has been drawn circumscribing the square. The area of the portion between circle and square is?

A) 1027
B) 1287
C) 1627
D) 1827

MAHA GUPTA
Area of the portion between circle and square = area of the circle – area of the square

As the circle is circumscribing the square, the radius of the circle will be half of the diagonal i.e. 8/2 = 4 cm
Thus, the area of the circle = (227)*4² = 3527

Area of the square = (12)*diagonal² = (12)*8a = 32

Therefore, The area of the portion between circle and square = 3527 – 32 = 1287 (option ‘B’)


QUERY 29

There is a rectangular field with dimensions 31 m × 10 m. We have to dig a rectangular hole in the field with dimensions 4 m × 2.5 m and 1.5 m deep and the volume of that hole is to be spread on the round equally, tell how much the level of the ground will raise.

A) 0.5 m
B) 0.05 m
C) 0.75 m
D) 1.5 m

SUBHASH CHANDRA
Volume of the hole = 4 × 2.5 × 1.5 = 15 m³

Area of the field on which earth of the hole is to be spread = Area of the field – Area of the surface of the hole
= 31 × 10 – 4 × 2.5
= 300 m²

Rise in the level of the ground = volume of the earth/remaining area of the field = 15300 = 0.05 m (option ‘B’)


QUERY 30

A right circular cylinder is formed by rolling a rectangular sheet of metal of length 24 cm and breadth 22 cm along its length. The volume of the cylinder is?

A) 924 cc
B) 462 cc
C) 1008 cc
D) 528 cc

MAHA GUPTA
The cylinder is made by rolling the sheet along its length, means the height of the cylinder will be the breadth of the sheet and length of the sheet as the circumference of the base of the cylinder.

Therefore height (h) of the cylinder = 22 cm
And circumference of its base (2πr) = 24 cm

So radius (r) of the base = 24 = 4211

Hence the its volume = πr²h = (227)*(4211)*(4211)*22 = 1008 cc (option ‘C’)

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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:

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