# Mensuration Volumes & Area Question and Answers : Aptitude

1.

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%

Answer: Option 'D'

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.
Ratio of volumes of the two cones v1 : v2 is 3 : 4.
If v1 is 3x, v2 will be 4x
Hence, volume of cylinder V= 7x
The volumes of the cylinder, cone 1 and cone 2 are πR2h, 1/3 πr12h and 1/3 πr22h
We know the ratio of the volumes V : v1, v2 is 7 : 3 : 4
So, πR2h : 1/3 πr12h : 1/3 πr22h is 7 : 3 : 4
Cancelling π and h, which are common to all terms, we get R2 : 1/3 r12 : 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.
Flat surface area of cylinder (sum of the areas of the two circles at the top and bottom of the cylinder) = 2 * π * R2
Flat surface area of cone 1 & 2 are: π * r12 & π * r22 respectively (areas of the circle at the bottom of each of the cones).
Ratio of the flat surface area of cylinder to that of the two cones is 2 * π * R2 : (π * r12 + π * r22)
Cancelling π on both sides of the ratio we get 2R2 : (r12 + r22 Or 14k : 21k

2.

What is the are of an equilateral triangle of side 16 cm?

A.) 48√3 cm2
B.) 128√3 cm2
C.) 9.6√3 cm2
D.) 64√3 cm2

Answer: Option 'D'

64√3 2
Area of an equilateral triangle = √3/4 S2
If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm2;

3.

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

A.) 544 cm2
B.) 484 cm2
C.) 250 cm2
D.) 384 cm2

Answer: Option 'D'

Area of a parallelogram = base × height = 24 × 16 = 384 cm2

4.

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

A.) 220 cm2
B.) 110 cm2
C.) 120 cm2
D.) 320 cm2

Answer: Option 'C'

The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle = 1/2 × 24 × 10 = 120 cm2

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

Answer: Option 'A'

What is to be found?
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.
Perimeter of the square = 200 ft.
Therefore, side of the square field = 50 ft.
The diagonal of the square field is the diameter of the circle that circumscribes it.
Measure of the diagonal of the square of side 50 ft = 50√2 ft.
Therefore, inner diameter of the circular road = 50√2.
Hence, inner radius of the circular road = 25√2 ft.
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 = 250800/2 = Rs.1,25,400

6.

If x units are added to the length of the radius of a circle, what is the number of units by which the circumference of the circle is increased?

A.) 2 π
B.) 2 π x
C.) x
D.) x2

Answer: Option 'B'

Let the radius of the circle be 'r' units.
The circumference of the circle will therefore be 2 π r units.
If the radius is increased by 'x' units, the new radius will be (r + x) units.
The new circumference will be 2 π(r + x) = 2 πr + 2 πx
Or the circumference increases by 2πx units.

7.

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

Answer: Option 'B'

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
= 6x2 : 21y2
= 6 × (2y)2 : 21y2
= 6 × 4y2 : 21y2
= 8 : 7

8.

A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of the base radius to the side of the square?

A.) 1/2π
B.) √2/π
C.) 1/(√2π)
D.) 1/π

Answer: Option 'D'

Surface area of the cylinder = Surface area of the square 2πh = a2
Here height 'h' of the cylinder = side of the square since it is rolled along its length.2πa = a2
Base radius r = a/2π
Ratio of base radius to side of square = a/2π : a = 1/2π

9.

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 cm2
B.) 35 cm2
C.) 25 cm2
D.) 55 cm2

Answer: Option 'B'

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 × 28/2 = 35 cm2

10.

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 cm2
B.) 355 cm2
C.) 385 cm2
D.) 585 cm2

Answer: Option 'A'

Area of a trapezium = 1/2 (sum of parallel sides) × (perpendicular distance between them) = 1/2 (20 + 18) × (15) = 285 cm2