# 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 = ^{1⁄3πr²h}⁄_{4⁄3πR³ }= ^{1⁄3(30²)45}⁄_{4⁄3(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 = ^{1000}⁄_{70} = ^{100}⁄_{7}

Decrease in width = 20 – ^{100}⁄_{7} = ^{40}⁄_{7}

Decrease on 20 = ^{40}⁄_{7}

Hence, decrease on 100 = [(^{40}⁄_{7})/20]*100 = ^{200}⁄_{7} = 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 = (^{40}⁄_{100+40})*100 = ^{200}⁄_{7} = 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) ^{102}⁄_{7}

B) ^{128}⁄_{7}

C) ^{162}⁄_{7}

D) ^{182}⁄_{7}

**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 = (^{22}⁄_{7})*4² = ^{352}⁄_{7}

Area of the square = (^{1}⁄_{2})*diagonal² = (^{1}⁄_{2})*8a = 32

Therefore, The area of the portion between circle and square = ^{352}⁄_{7} – 32 = ^{128}⁄_{7} (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 = ^{15}⁄_{300} = 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}⁄_{2π} = ^{42}⁄_{11}

Hence the its volume = πr²h = (^{22}⁄_{7})*(^{42}⁄_{11})*(^{42}⁄_{11})*22 = 1008 cc (option ‘C’)