# AREA AND VOLUME (MENSURATION) PART-1

#### AREA AND VOLUME (MENSURATION) PART-1

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 1

**If the length of a rectangle is increased in the ratio 6:7 and breadth is diminished in the ratio 5:4 then its area will be diminished in the ratio?**

A) 15:14

B) 15:13

C) 14:15

D) 10:15

**RONNIE BANSAL**

Let the length of the rectangle before being altered be 6; therefore it will be 7 after it being increased

and the breadth before being altered be 5; therefore it will be 4 after it being diminished

Now the area before = 6*5 = 30

and the area after = 7*4 = 28

Hence the area of the rectangle will be diminished in the ratio 30 : 28 = 15 : 14 (option ‘A’)

QUERY 2

**A copper wire is bent in the form of square of area 81 sq cm. If the same wire is bent in the form of a semi circle the radius of semicircle is?**

A) 7

B) √7

C) 49

D) 14

**JAYANTCHARAN CHARAN
**Side of the square = √81 = 9

So the perimeter = 9 x 4 = 36

This should be equal to the circumference of the semi-circle i.e. the diameter + the semi circular arc

means 2r + πr (where r is the radius of the semi-circle)

= r(2 + π)

=> r(2 + π) = 36

=> r = 36/(2 + π)

=> r = 36/[(14 + 22)/7]

=> r = 7 (option ‘A’)

QUERY 3

**Three horses are tethered at 3 corners of a triangular plot of land having sides 20 m, 30 m and 40 m each with a rope of length 7 m. The area of region of this plot which can be grazed by the horses is**

A) 77/3

B) 75

C) 77

D) 80

**RONNIE BANSAL**

Think carefully, you’ll see that each horse will be covering the area of a sector of radius 7 m each. As the sum of the angles of a triangle is 180°, the total of the angles of all the sectors also will be 180°. means a semi-circular area.

So area of the region of the plot which can be gazed by the horses = πr²/2 = (22/7 x 7 x 7)/2 = 77.

So option ‘C’ is correct.

QUERY 4

**A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68 m/min. The area of the field is?**

A) 30 m²

B) 40 m²

C) 50 m²

D) 60 m²

**KANW@LJEET**

Walking at a rate of 52 m/min for 15 seconds means total distance covered by A = 52*15/60 = 13 m

Similarly distance covered by B to cross the same field along its sides walking at the rate of

68 m/min = 68*15/60 = 17 m. (Obviously he will have to walk along the length and breadth once each to cross the field)

Now what we have is diagonal of that rectangular field as 13 m and sum of its sides as 17 m

Letting ‘a’ and ‘b’ the length and breadth respectively

x+y = 17 ……………………eq (1)

x² + y² = 13² (using pythagoras theorem)……….eq (2)

Solving this xy = 60, which is the value of the area of the field (option ‘D’)

**JAYANTCHARAN CHARAN**

15 sec is 1/4 of a minute and 1/4 of 52 is 13, so it can be said that 13 is the diagonal of that rectangular field; in other words the hypotenuse of the right triangular area formed by half of it We know that with 13 as hypotenuse there is only one pythagoras triplet 5, 12, 13. So we can say that 5 m and 12 m are the two sides of that rectangular field, so the area of that field = 5 x 12 = 60 (option ‘D’)

QUERY 5

**The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, the cost of leveling it at the rate of Rs 5 per m² is?**

A) 4750

B) 5000

C) 6250

D) 7920

**AMIT JHA**

Area covered in one revolution = Curved surface area of the roller

The shape of a roller is always cylindrical

So the curved surface area= 2πrh where r is the radius and h the height

Now the radius = diameter/2 = 84/2 = 42 cm;

and the height is obviously the length of the roller i.e. 120 cm

So the curved surface (area covered by it in one revolution) = 2 x (22/7) x 42 x 120 cm²

Thus area covered by it in 500 revolutions = 2 x (22/7) x 42 x 120 x 500 cm²

And the cost of leveling @ Rs 5 per square meter

= [2 x (22/7) x 42 x 120 x 500 x 5]/10000 (It is divided by 10000 because 1 square meter

= 10,000 cm² = Rs 7920 (option ‘D’)

QUERY 6

**A right circular cylinder is circumscribing a hemisphere such that their bases are common. The ratio of their volume will be?**

A) 1:3

B) 1:2

C) 2:3

D) 3:4

**KANW@LJEET**

just put the formula of volumes but remember h = r here

Thus u will get ratio as 2:3 (option ‘C’)

QUERY 7

**Anil makes a Plasticine cuboid whose sides are 5 cm, 2 cm and 5 cm. How many such cuboids will be needed to make a cube?**

A) 30

B) 15

C) 20

D) 25

**VAIBHAV VATS**

Dimensions of the two sides of the given cuboid is same i.e. 5 cm each. Now we’ll have to try thinking only how many rectangles of side 5cm & 2cm will be needed to make a square. Obviously equal to the LCM of these two sides i.e. 10

Now the number of such cuboids needed to make the cube = Volume of cube/volume of cuboids

=> 10*10*10/5*2*5 = 20 (option ‘C’)

QUERY 8

**Three spherical balls of radii 1 cm, 2 cm, 3 cm are melted to form a single spherical ball. In this process loss of material is 25%. The radius of new ball is?**

A) 6 cm

B) 5 cm

C) 3 cm

D) 2 cm

**HEMANT SINGH
**Volume of the 3 balls after being melted = 4/3π(1³ + 2³ + 3³) = 4/3π36 (Volume of a sphere = 4/3πr³

but 25% (i.e. 1/4) loss of material is there to form a single ball, so the remaining material must be 3/4 of the same

Thus the remaining material = {4/3π36}*(3/4) = 36π, which is equal to the volume of the new ball.

Letting the radius of the new ball as R, the equation will be

4/3πR³ = 36π

=> R³ = 36*3/4 = 27

=> R = 3

So radius of the new ball is 3 cm (option ‘C’)

QUERY 9

**The volume (in cubic meters) of rain water that can be collected from 1.5 hectares of ground in a rainfall of 5 cm is**

A) 75

B) 750

C) 7500

D) 75000

**AMIT JHA
**Hectare (symbol ‘ha’) is a metric unit of area and 1 Hectare = 10,000 m²

We have to find the volume in cubic meters, so convert every unit in meters.

Now 1.5 hectares = 15,000 m²

and 5 cm = 0.05 meter

We know volume of a thing is the product of area of the base and its height

and the rainfall is the height of water that has gathered on a particular surface

Hence the volume of rain water to be collected = rain = 15,000 x 0.05 = 750 m³ (option ‘B’)

QUERY 10

**66 cm³ of silver is drawn into a wire 1 mm in diameter. The length of the wire in meters will be?**

A) 36

B) 84

C) 87

D) 174

**RONNIE BANSAL
**66 cm³ (66000000 meter) is the volume of that cubic thing. And a wire is, if not stated otherwise, is always cylindrical in shape.

We know the volume of cylinder = πr²h, where r = diameter/2 = 1/2 mm = 1/2000 m, and ‘h’ is the height of the cylinder means length of the wire

Also we can say that volume of the cubic object and the wire is same

Hence 66000000 = (22/7) (1/2000)²h

=> h = (66000000 x 7 x 2000 x 2000)/22 = 84

So the length of the wire is 84 m (option ‘B’)