Numerical Ability Test

Result

Numerical Ability Test - 28

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Weekly Quiz Competition

Question 1
5 / -1

The number of revolutions made by a bicycle wheel 56 cm in diameter. in covering a distance of 1.1 km is

A
56.25

B
31.25

C
62.5

D
625

Solution

GIVEN:

A bicycle wheel of 56 cm diameter in covering a distance of 1.1 km

FORMULA USED:

Circumference of wheel in bicycle = 2πr

Diameter = 2 × radius

Where r = radius of circumference

CALCULATION:

Here, diameter = 56 cm

⇒ diameter = 2 × radius = 56 cm

⇒ radius = 28 cm = 28/100 m = 0.28m

Circumference of wheel in bicycle = 1 revolution

⇒ 2πr = 1 revolution

⇒ 2 × 22/7 × 0.28 = 1 revolution

As, 1.76 distance is covered in 1 revolution

So,

Number of revolutions in 1.1km

⇒ 1.1km = 1.1 × 1000m = 1100 m

[As, 1 km = 1000 m]

⇒ Number of revolutions = 1100 /1.76 = 625

∴ Number of revolutions is 625.
Question 2
5 / -1

The area of a triangle with sides 21 cm, 24 cm and 27 cm is:

A
216√5 cm²

B
72 cm²

C
108√5 cm²

D
72√5 cm²

Solution

Given:

The three sides of the triangle are 21 cm, 24 cm, and 27 cm, respectively.

Calculation:

Let us assume that a, b, and c are the sides of the triangle.

⇒ Area of triangle = √{s(s - a) (s - b) (s - c)}

⇒ s = (a + b + c)/2 = (21 + 24 + 27)/2 = 36 cm

⇒ √{36 × (36 - 21) × (36 - 24) × (36 - 27)}

⇒ √(36 × 15 × 12 × 9)

⇒ √(6² × 3 × 5 × 3 × 4 × 3²)

⇒ (6 × 3 × 3 × 2)√5

⇒ The area of triangle = 108√5 cm²

∴ The required result will be 108√5.
Question 3
5 / -1

The length of one diagonal of a rhombus is 12 cm and the area is 108 cm². The length of the second diagonal of the rhombus is:

A
18 cm

B
22 cm

C
36 cm

D
20 cm

Solution

Given:

Length of one diagonal (d₁) = 12 cm

Area of the rhombus (A) = 108 cm²

Formula used:
Question 4
5 / -1

The area of square is equal to four times the area of a rectangle of length 117 cm and breadth 13 cm. What is the perimeter of the square?

A
310 cm

B
308 cm

C
312 cm

D
300 cm

Solution

Given:

The area of a square is equal to four times the area of a rectangle of length 117 cm and breadth 13 cm.

Formula used:

Area of rectangle = length × breadth

Area of square = 4 × Area of rectangle

Perimeter of square = 4 × side

Calculation:

Area of rectangle = 117 × 13 = 1521 cm²

Area of square = 4 × 1521

⇒ Area of square = 6084 cm²

Side of square = √6084

⇒ Side of square = 78 cm

Perimeter of square = 4 × 78 = 312 cm

∴ The correct answer is option (3).
Question 5
5 / -1

In a parallelogram two adjacent sides are in the ratio 2 : 3 and the perimeter is 60 cm. The length of each of the two shorter sides of this parallelogram is:

A
18 cm

B
13 cm

C
19 cm

D
12 cm

Solution

Given:

Two adjacent sides of a parallelogram are in the ratio 2:3

Perimeter = 60 cm

Formula used:

Perimeter of a parallelogram = 2(a + b)

Where, a and b are the lengths of the adjacent sides

Calculation:

Let the adjacent sides be 2x and 3x

Perimeter = 2(2x + 3x)

60 = 2(5x)

⇒ 60 = 10x

⇒ x = 6

Length of the shorter side = 2x = 2 × 6 = 12 cm

∴ The correct answer is option 4.
Question 6
5 / -1

The area (in cm²) of an equilateral triangle whose altitude from vertex is 12 cm, is:

A
27√3

B
48√3

C

D

Solution

Given:

Altitude from vertex is 12 cm.

Formula used:

For an equilateral triangle, the altitude (h) is given by:

⇒ a = 8√3 cm

Now, using the area formula:
Question 7
5 / -1

If the length of a rectangle is increased by 1/3rd and the width is decreased by 1/3rd, then the area of the rectangle is decreased by the fraction

A
2/3

B
1/6

C
1/9

D
1/8

Solution

Given:

Length of the rectangle = L

Width of the rectangle = W

Formula used:

Area of the rectangle, A = L × W

Calculation:

New length = L + (1/3)L = (4/3)L

New width = W - (1/3)W = (2/3)W

New area, A_new = (4/3)L × (2/3)W

⇒ A_new = (8/9)LW

Decrease in area = Original area - New area

⇒ Decrease in area = LW - (8/9)LW

⇒ Decrease in area = (1/9)LW

Fractional decrease in area = (Decrease in area) / (Original area)

⇒ Fractional decrease = ((1/9)LW) / (LW)

⇒ Fractional decrease = 1/9

∴ The correct answer is option (3).
Question 8
5 / -1

Tarika has a decorative wall-hanging in the shape of a regular hexagon having side 10 cm. What is the area (in cm²) of the decorative wall (Take √3 = 1.7)

A
250

B
260

C
255

D
265

Solution

Given:

Side of the regular hexagon (s) = 10 cm

Formula Used:

⇒ Area = 150 × 1.7

⇒ Area = 255 cm²

The area of the decorative wall-hanging is 255 cm².
Question 9
5 / -1

The area of a trapezium is 480 cm², the distance between two parallel sides is 15 cm and one of the parallel side is 20 cm. The other parallel side is:

A
20 cm

B
34 cm

C
44 cm

D
50 cm

Solution

Given:

Area of trapezium (A) = 480 cm²

Distance between parallel sides (h) = 15 cm

One parallel side (a) = 20 cm

Formula used:

A = 1/2 × (a + b) × h

Where, b = other parallel side

Calculation:

480 = 1/2 × (20 + b) × 15

⇒ 480 = 150 + 7.5b

⇒ 480 - 150 = 7.5b

⇒ 330 = 7.5b

⇒ b = 44 cm

∴ The correct answer is option (3).
Question 10
5 / -1

The area of the largest triangle that can be inscribed in a semi-circle of radius 6 cm is:

A
36 cm²

B
72 cm²

C
38 cm²

D
76 cm²

Solution

Given:

The area of the largest triangle that can be inscribed in a semi-circle of radius 6 cm

Formula used:

Area of the largest triangle in a semi-circle = (1/2) x base x height

For the largest triangle, the base = diameter of the semi-circle, height = radius of the semi-circle

Calculation:

Radius (r) = 6 cm

Diameter (base) = 2 x radius = 2 x 6 = 12 cm

Height = radius = 6 cm

Area = (1/2) x base x height

⇒ Area = (1/2) x 12 x 6

⇒ Area = 36 cm²

∴ The correct answer is option (1).