 # SSC CPO Mensuration Quiz 2 5 Steps - 3 Clicks

# SSC CPO Mensuration Quiz 2

### Introduction

Mensuration  is a topic in Geometry which is a branch of mathematics. Mensuration deals with length, area, and volume of different kinds of shape- both 2D and 3D. The article SSC CPO Mensuration Quiz 2 provides information about menuration related questions and answers. SSC CPO Mensuration Quiz 2 is very useful to crack different competitive examinations like IBPS RRB, RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC CGL, SSC CHSL, IBPS, SBI PO, SBI Clerks, CAT and etc.

### Quiz

1. A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7√2 ft. and cost of construction is Rs.100 per sq. ft., find the lowest possible cost to construct 50% of the total road.

A. Rs.125,400
B. Rs.140,800
C. Rs.235,400
D. None of these

Explanation:
We have to find the lowest cost to construct 50% of the total road.
Data Given
A circular road is constructed outside a square field. So, the road is in the shape of a circular ring.
If we have to determine the lowest cost of constructing the road, we have to select the smallest circle that can be constructed outside the square.
Therefore, the inner circle of the ring should circumscribe the square.
The perimeter of the square = 200 ft.
Therefore, the side of the square field = 50 ft.
The diagonal of the square field is the diameter of the circle that circumscribes it.
The measure of the diagonal of the square of side 50 ft = 50√2 ft.
Therefore, the inner diameter of the circular road = 50√2.
Then, outer radius = 25√2 + 722 = 32√2
The area of the circular road
= π ro2 – π ri2, where ro is the outer radius and ri is the inner radius.
= 22/7 × {(32 √2)2 – (25√2)2}
= 22/7 × 2 ×(32 + 25)×(32 – 25)
= 2508 sq. ft.
If per sq. ft. cost is Rs. 100, then cost of constructing the road = 2508 × 100 = Rs.2,50,800.
Cost of constructing 50% of the road = 50% of the total cost = $$\frac{250800}{2}$$ = Rs.1,25,400

2. A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?

A. 9%
B. 15%
C. 25%
D. 50%

Explanation:
Let radius of cylinder, cone 1, cone 2 = R, r1, r2 respectively.
Let their volumes be V, v1, v2 respectively.
Cylinder is melted and recast into cone 1 & cone 2
So V = v1 + v2.
The ratio of volumes of the two cones v1: v2 is 3: 4.
If v1 is 3x, v2 will be 4x
Hence, the volume of cylinder V = 7x
The volumes of the cylinder, cone 1 and cone 2 are πR2h, $$\frac{1}{3}$$ πr12h, and $$\frac{1}{3}$$ πr22h
We know the ratio of the volumes V: v1, v2 is 7 : 3: 4
So, πR2h : $$\frac{1}{3}$$ πr12h : $$\frac{1}{3}$$ πr22h is 7 : 3 : 4
Canceling π and h, which are common to all terms, we get R2: $$\frac{1}{3}$$ r12: $$\frac{1}{3}$$ r22 = 7 : 3: 4
Or R2 : r12 : r22 = 7 : 3 : 4.
So, if R2 is 7k, r12 will be 9k and r22 will be 12k.
The flat surface area of the cylinder (sum of the areas of the two circles at the top and bottom of the cylinder) = 2 * π * R2
The flat surface area of cones 1 & 2 is π * r12 & π * r22 respectively (areas of the circle at the bottom of each of the cones).
The ratio of the flat surface area of cylinder to that of the two cones is 2 * π * R2 : (π * r12 + π * r22)
Canceling π on both sides of the ratio we get 2R2 : (r12 + r22 Or 14k: 21k

3. The perimeter of a triangle is 28 cm and the inradius of the triangle is 2.5 cm. What is the area of the triangle?

A. 45 c $${m}^{2}$$
B. 35 c $${m}^{2}$$
C. 25 c $${m}^{2}$$
D. 55 c $${m}^{2}$$

Explanation:

Area of a triangle = r × s
Where r is the in radius and s is the semi perimeter of the triangle.
Area of triangle = 2.5 × $$\frac{28}{2}$$ = 35 c $${m}^{2}$$

4. Find the area of a parallelogram with base 24 cm and height 16 cm.

A. 544 c$${m}^{2}$$
B. 484c$${m}^{2}$$
C. 250 c$${m}^{2}$$
D. 384 c$${m}^{2}$$

Explanation:
Area of a parallelogram = base × height = 24 × 16 = 384 c$${m}^{2}$$

5. Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.

A. 285 c$${m}^{2}$$
B. 355 c$${m}^{2}$$
C. 385 c$${m}^{2}$$
D. 585 c$${m}^{2}$$

Explanation:
Area of a trapezium = 1/2 (sum of parallel sides) × (perpendicular distance between them) = 1/2 (20 + 18) × (15) = 285 c$${m}^{2}$$

1. If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area?

A. 220 c$${m}^{2}$$
B. 110 c$${m}^{2}$$
C. 120 c$${m}^{2}$$
D. 320 c$${m}^{2}$$

Explanation:
The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle = $$\frac{1}{2}$$ × 24 × 10 = 120 c$${m}^{2}$$

2. An order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2. Subsequently, the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter. Find the ratio of the areas of the carpets in both the cases.

A. 7 : 8
B. 8 : 7
C. 6 : 7
D. 5 : 6

Explanation:
Let the length and breadth of the carpet in the first case be 3x units and 2x units respectively.
Let the dimensions of the carpet in the second case be 7y, 3y units respectively.
From the data,.
2(3x + 2x) = 2(7y + 3y)
=> 5x = 10y
=> x = 2y
Required ratio of the areas of the carpet in both the cases
= 3x × 2x : 7y : 3y
= 6×2 : 21y2
= 6 × $${2Y}^{2}$$ : 21$${Y}^{2}$$
= 6 × 4$${Y}^{2}$$ : 21$${Y}^{2}$$
= 8 : 7

3. A 5 cubic centimeter cube is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted?

A. 9
B. 61
C. 98
D. 54

Explanation:
When a 5 cc cube is sliced into 1 cc cubes, we will get 5*5*5 = 125 1 cc cubes.

In each side of the larger cube, the smaller cubes on the edges will have more than one of their sides painted.
Therefore, the cubes which are not on the edge of the larger cube and that lie on the facing sides of the larger cube will have exactly one side painted.

In each face of the larger cube, there will be 5*5 = 25 cubes. Of these, there will be 16 cubes on the edge and 3*3 = 9 cubes which are not on the edge.
Therefore, there will be 9 1-cc cubes per face that will have exactly one of their sides painted.
In total, there will be 9*6 = 54 such cubes.

4. The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel?

A. 20 ft
B. 25 ft
C. 750 ft
D. 900 ft

Explanation:
The circumference of the front wheel is 30 ft and that of the rear wheel is 36 feet.

Let the rear wheel make n revolutions. At this time, the front wheel should have made n+5 revolutions.

As both the wheels would have covered the same distance, n*36 = (n+5)*30

36n = 30n + 150

6n = 150

n = 25.

Distance covered = 25*36 = 900 ft

5. If the sides of a triangle measure 72, 75 and 21, what is the measure of it’s in radius?

A. 37.5
B. 24
C. 9
D. 15

Explanation:
The sides of the triangle happen to be a Pythagorean triplet. Hence, the triangle is a right angled triangle.

For a right-angled triangle,

The measure of in radius = (product of perpendicular sides) / (perimeter of the triangle)

= $$\frac{(72 * 21)}{(72 + 21 +75)}$$ = $$\frac{1512}{168}$$ = 9

1. If each interior angle of a regular polygon is 150 degrees, then it is

A. Octagon
B. Decagon
C. Dodecagon
D. Tetrahedron

Explanation:
The sum of the exterior angle and interior angle of a regular polygon is 3600.

As the measure of each interior angle is 1500, the measure of each exterior angle will be 300.

The number of sides of a regular polygon is given by the following relationship n = $$\frac{360}{exterior angle}$$

As the value of each exterior angle is 300, the number of sides = 12

2. Four horses are tethered at 4 corners of a square field of side 70 metres so that they just cannot reach one another. The area left ungrazed by the horses is:

A. 1050 sq.m
B. 3850 sq.m
C. 950 sq.m
D. 1075 sq.m

Explanation:
The length of the rope in which the horses tied should be equal to half of the side of the square plot so that they just cannot reach one another.

Therefore, the length of the rope is 35m (70/2).

The area covered by each horse should be equal to the area of the sector with a radius of $$\frac{70}{2}$$ = 35m(length of the rope).

The total area covered by the four horses = 4* area of the sector of radius 35 meters = Area of the circle of radius 35m.

Area left ungrazed by the horses = Area of the square field – Area covered by four horses.

= 702 – (22/7)*35*35 = 4900 – 3850 = 1050 sq.m

3. A 4 cm cube is cut into 1 cm cubes. What is the percentage increase in the surface area after such cutting?

A. 4%
B. 300%
C. 75%
D. 400%

Explanation:
Volume of 4 cm cube = 64 cc. When it is cut into 1 cm cube, the volume of each of the cubes = 1cc
Hence, there will be 64 such cubes. Surface area of small cubes = 6 (12) = 6 sqcm.
Therefore, the surface area of 64 such cubes = 64 * 6 = 384 sqcm.
The surface area of the large cube = 6(42) = 6*16 = 96.
% increase = $$\frac{384-96}{96}$$ = 300%

4. A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?

A. 9%
B. 16%
C. 25%
D. 50%

Explanation:
Let radius of cylinder, cone 1, cone 2 = R, r1, r2 respectively.

Let their volumes be V, v1, v2 respectively.

The cylinder is melted and recast into cone 1 & cone 2

So V = v1 + v2.

The ratio of volumes of the two cones v1: v2 is 3: 4.
If v1 is 3x, v2 will be 4x

Hence, the volume of cylinder V= 7x

5. A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7√2 ft. and cost of construction is Rs.100 per sq. ft., find the lowest possible cost to construct 50% of the total road.

A. Rs.70,400
B. Rs.125,400
C. Rs.140,800
D. Rs.235,400

Explanation:
A circular road is constructed outside a square field. So, the road is in the shape of a circular ring.

If we have to determine the lowest cost of constructing the road, we have to select the smallest circle that can be constructed outside the square.

Therefore, the inner circle of the ring should circumscribe the square.

The perimeter of the square = 200 ft.

Therefore, the side of the square field = 50 ft.