MCQ: Area - 2 - SSC CGL MCQ

# MCQ: Area - 2 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Area - 2

MCQ: Area - 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Area - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Area - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Area - 2 below.
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MCQ: Area - 2 - Question 1

### If the length of a rectangle is increased by 25% and the breadth is reduced by 33.33% then what will be the effect on its diagonal(approximately)?

Detailed Solution for MCQ: Area - 2 - Question 1

Let the length of the rectangle = 4 units
And breadth of the rectangle = 3 units
Then diagonal of the rectangle = √(42 + 32) = 5 units
According to the question, the length of a rectangle is increased by 25% and the breadth is reduced by
33.33%
New length = 125% of 4 units = 5 units
New breadth = 66.66% of 3 units = 2 units
In the new rectangle, New diagonal= √(52 + 22) = √29 = approximately 5.38 units

Hence, option A is correct.

MCQ: Area - 2 - Question 2

### The perimeter of rectangle of length 2 (x + 3) cm and breadth 2 (x + 1) cm is double the perimeter of a square of area 225 cm2. Find the area of rectangle.

Detailed Solution for MCQ: Area - 2 - Question 2

Area of square = 225 cm2
Side of square = √225 = 15 cm
Perimeter of square = 15 × 4 = 60 cm
So, perimeter of rectangle = 2 × 60 = 120 cm
2 (2 (x + 3) + 2 (x + 1)) = 120
2x + 6 + 2x + 2 = 60
4x + 8 = 60
4x = 52
x = 13
So, length of rectangle = 2 (13 + 3) = 32 cm
And, breadth of rectangle = 2 (13 + 1) = 28 cm
Therefore, area of rectangle = 32 × 28 = 896 cm2
Hence, option B is correct.

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MCQ: Area - 2 - Question 3

### The difference between the area of a circle and the area of the rectangle is 394.94 cm2. If the length of rectangle is 25% more and breadth is 10% less than the radius of the circle, then find the perimeter of the rectangle? (Take π = 3.14)

Detailed Solution for MCQ: Area - 2 - Question 3

According to the question,
Let the radius of the circle be r cm.
So, the length of the rectangle = 1.25 r cm and the breadth of the rectangle = 0.90r cm
Given that,
πr2 – 1.25 r x 0.90 r = 394.94
3.14r2 – 1.125r2 = 394.94

r2 = 196
r = 14
length = 1.25 × 14 = 17.5 cm
breadth = 0.90 × 14 = 12.6 cm
Perimeter of rectangle = 2(17.5 + 12.6) = 60.2 cm
Hence, option C is correct.

MCQ: Area - 2 - Question 4

A track of width 1.5 m is made along the inside edge of a rectangular park of dimensions 50 m and 35 m. What is the total cost of laying bricks on the track at the rate of Rs 125 per 10m2?

Detailed Solution for MCQ: Area - 2 - Question 4

Dimensions of rectangle inside the track = 47m (50 – 2 x 1.5) and 32m (35 – 2 x 1.5)
Area of track = area of park – area of rectangle inside the track
= (50 × 35 – 47 × 32) = (1750 – 1504) = 246m2
Cost of laying bricks at the rate of Rs 125 per 10m

Hence, option B is correct

MCQ: Area - 2 - Question 5

The sum of perimeter of a rectangle (R) and a square (S) is 210 cm. If the length and breadth of rectangle (R) is 40% more and 20% less than the side of square(S), what is the length of diagonal of rectangle (R)?

Detailed Solution for MCQ: Area - 2 - Question 5

Let the side of square be ‘s’ cm
So, length of rectangle = 1.4s cm
And, breadth of rectangle = 0.8s cm
According to the question,
2(1.4s + 0.8s) + 4s = 210
4.4s + 4s = 210
8.4s = 210

So, length of rectangle = 1.4 × 25 = 35 cm
And, breadth of rectangle = 0.8 × 25 = 20 cm
So, diagonal of rectangle

So option B is the correct answer.

MCQ: Area - 2 - Question 6

A jeweler wants to build a 1250 square meters rectangular jewellery shop. Since she has only 100 m spiky wire, she barbed only three sides of the jewellery shop letting her home wall acting as the remaining side of the shop. Find the dimension of the
jewellery shop where the spiky wires are being used?

Detailed Solution for MCQ: Area - 2 - Question 6

Length = L; Width = B; Area = 1250;
L = 1250/B

Assume that the side CD in the figure is fenced by his home wall.
So, the total 100 meter wire is used to fence sides AC, AB & BD
AC + AB + BD = 100
AC = BD = B
AB = L
(2 x B) + L = 100

On solving, we get, B = 25 and L = 50 respectively
Therefore, L = 50 and B = 25 meters
Hence, option B is correct.

MCQ: Area - 2 - Question 7

Find the height of equilateral triangle if its area is 36√3 m2?

Detailed Solution for MCQ: Area - 2 - Question 7

We know that,
Area of an equilateral triangle of side

We know that,
Height of an equilateral triangle = √3a/2

⇒ Height of the equilateral triangle
∴ Height of the equilateral triangle = 6√3 m
Hence, option D is correct.

MCQ: Area - 2 - Question 8

The ratio of the breadth to the length of a rectangular field is 1 : 3. The total cost of cutting the grass of the field at the rate of Rs 6 per m2 is Rs. 8712. Find the cost of fencing the boundary of the field at the rate of Rs. 11 per m.

Detailed Solution for MCQ: Area - 2 - Question 8

Let the length and breadth of the field is 3x and x respectively

Therefore, the length of the field = 3x = 66 m
And the breadth of the field = x = 22 m
Required cost = 2 x (66 + 22) x 11 = Rs. 1936
Hence, option D is correct

MCQ: Area - 2 - Question 9

The ratio of area of a rectangle to that of a square is 3 : 5. If the perimeter of thesquare is 100 cm then what can be the perimeter of the rectangle if the breadth of therectangle is 66.67% more than that of length?

Detailed Solution for MCQ: Area - 2 - Question 9

The side of the square = 1000/4 = 25 cm

The area of the square = 25 × 25 = 625 sq. m
The area of the rectangle = 3x625 / 5 = 375 sq. cm
Let the length of the rectangle = 3x and breadth of the rectangle = 166.67% of 3x = 5x then perimeter =
2(l + b)
And area = 375 = 3x x 5x
x = 5 cm
Perimeter = 2(3x + 5x) = 16x = 80 cm
Hence, option B is correct.

MCQ: Area - 2 - Question 10

A cuboid of dimensions 15 m x 20 m x 18 m was painted from inside as well as outside at the rate of 5 paisa per sq. m. If the cuboid was made of metal of negligible thickness then how much money will be required (In Rupees) to paint the cuboid for inside as well as outside?

Detailed Solution for MCQ: Area - 2 - Question 10

The total surface area of the cuboid = 2(l x b + b × h + h x l) = 2(15 × 20 + 20 × 18 + 18 × 15) = 2 × (300 +
360 + 270) = 2 x 930 = 1860 sq. m
The cuboid is made of metal of negligible thickness then outer surface area = inner surface area
The required answer = 2 x 1860 x 5 = 18600 paisa = Rs. 186
Hence, option C is correct.

MCQ: Area - 2 - Question 11

In a parallelogram shaped field, one of its side is 15 meters and the length of the perpendicular distance between the opposite sides is 16 m. In the field, if wheat seeds were to sow at the rate of Rs. 15 per sq. meter, then how much money will be needed
to sow wheat seeds in the field?

Detailed Solution for MCQ: Area - 2 - Question 11

The boundary meets perpendicularly the opposite side it means it the perpendicular distance between the opposite sides.
The area of a parallelogram = base × perpendicular distance between the opposite sides = 15 × 16 = 240 sq. m
The total money required = 240 × 15 = Rs. 3600
Hence, option B is correct.

MCQ: Area - 2 - Question 12

The height of a cylindrical shaped wood is 15 cm less than its circumference of the base and the curved surface area is 154 cm2 , then what is the volume (in cm3) of the cylinder shaped wood?

Detailed Solution for MCQ: Area - 2 - Question 12

Let Radius = r then Circumference = 2πr cm and height = (2πr – 15) cm
Curved surface area = 2πrh = 2πr x (2πr – 15) = 154
By solving, r = 3.5 cm
Circumference of the base = 2πr cm
Height = 22 – 15 = 7 cm

Hence, option B is correct.

MCQ: Area - 2 - Question 13

Sum of the volume of a cylinder and a cone is 2190π cm2. The radius of both cylinder and cone is same i.e., 10 cm. If the height of cone is 15 cm, then find the ratio of height of cylinder to height of cone?

Detailed Solution for MCQ: Area - 2 - Question 13

Let Height of cylinder = h'
Volume of cylinder = πr2h

Hence, option B is correct.

MCQ: Area - 2 - Question 14

A triangular field has to be fenced with iron wire. The cost of fencing is Rs.15 per meter. If the sum of lengths of two of the three sides of the triangular field is 25 meters, then which of the following cannot be the cost of fencing the field?

Detailed Solution for MCQ: Area - 2 - Question 14

For a triangle, Sum of two sides is always greater than third side.
Let the third side of the triangle be C
If sum of two side is 25
∴ 0 < C < 25 holds true for range of C.
C cannot be zero and cannot be more than 25
Perimeter of triangle
= 25 < (perimeter of triangle) < 50
Cost of fencing at Rs 15 per meter will lie between 375 – 750
Hence, option D Rs 800 cannot be the cost of fencing

MCQ: Area - 2 - Question 15

Area of a circle is 616 cm2 breadth of Rectangle is half than the radius of the Circle. The length of rectangle is 37.5% of the perimeter of Rectangle. If the side of a square is double the length of rectangle, then find the area of square.

Detailed Solution for MCQ: Area - 2 - Question 15

Area of Circle = A = πr2

Breadth of Rectangle = 14/2 = 7
Perimeter of Rectangle = 2 (L + B)

4L = 3L + 21
L = 21
Side of Square = 2 x 21 = 42
Area of Square = 42 x 42 = 1764
Hence, option A is correct

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## Quantitative Aptitude for SSC CGL

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