- 1
- 2
- 3
- …
- 5
- Next Page »

A Learning Portal from Recruitment India

How many bricks each measuring 25cm x 11.25cm x 6cm, will be needed to build a wall 8m x 6m x 22.5m

A.

5600

B.

600

C.

6400

D.

7200

Answer with explanation

Answer: Option CExplanation

No. of bricks= volume of the wall/volume of 1 brick

= (800 x 600 x 22.5) / (25 x 11.25 x 6) = 6400

Workspace

50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 cu.m , then the rise in the water level in the tank will be:

A.

20 cm

B.

25 cm

C.

35 cm

D.

50 cm

Answer with explanation

Answer: Option BExplanation

Workspace

The length, width, and height of a room are 7 m, 8 m, and 9 m respectively. What is the total area of the walls?

A.

270 sq. m.

B.

260 sq. m.

C.

265 sq. m.

D.

255 sq. m.

Answer with explanation

Answer: Option AExplanation

Area of the 4 walls of a room = 2 (Length + Breadth) * Height

Given that, Length = 7 meters, Width or Breadth = 8 meters and Height = 9 meters

So, the required area, A = 2 (7 + 8) * 9

= 2 * 15 * 9

= 270 sq. m.

Workspace

What is the area of the largest circle that can be drawn inside a square of side 20 cm?

A.

90 π cm^{2}

B.

100 π cm^{2}

C.

110 π cm^{2}

D.

95 π cm^{2}

Answer with explanation

Answer: Option BExplanation

The diameter of the largest circle that can be drawn inside a square = length of the side of the square

As per the question: The side of the square = 20 cm.

So, radius of the circle = 20/2 = 10 cm

Area of a circle = π × radius^{2}

The area of the largest circle = π × 102 = 100 π cm^{2}

Workspace

The difference between the length and breadth of a rectangle is 25 meters. If the perimeter of this rectangle is 210 meters, what is its area?

A.

3600 m^{2}

B.

2600 m^{2}

C.

3000 m^{2}

D.

2000 m^{2}

Answer with explanation

Answer: Option BExplanation

The perimeter: 2 (L + B) = 210 or L+ B = 105 …… (i)

As per the question: (L – B) = 25 ……. (ii)

Solving the above two equations:

L + B = 105

B = 105 – L

Put the value of B in (ii).

L – (105 – L) = 25

L – 105 + L = 25

2L = 25 + 105

2L = 130

L = 65

So, – B = 105 – 65 = 40

Now, area of a rectangle: L *B= 65 * 40 = 2600 m^{2}

Workspace

The height of a room to its semi-perimeter is 2: 5. It costs Rs. 260 to paper the walls of the room with paper 50 cm wide at Rs. 2 per meter allowing an area of 15 cm^{2} for doors and windows. The height of the room is

A.

2.6 m

B.

3.9 m

C.

4 m

D.

4.2 m

Answer with explanation

Answer: Option CExplanation

Workspace

The sides of a rhombus are 10 cm in length, and one diagonal is 16 cm. The area of the rhombus is

A.

96 cm^{2}

B.

95 cm^{2}

C.

94 cm^{2}

D.

93 cm

Answer with explanation

Answer: Option AExplanation

When d_{1} and d_{2} are the diagonals of rhombus then,

The Side of rhombus =

Squaring both sides

20^{2} = 16^{2} + d_{2}^{2}

400 = 256 + d_{2}^{2}

d_{2}^{2} = 400 – 256

d_{2} = √144

d_{2} = 12

Area of rhombus =1/2( d_{1} × d_{2} )

=1/2 ( 16 × 12 )=96 cm^{2}

Workspace

A rectangular grass field has a length of 85 meters and breadth of 65 meters. It is surrounded by a 5-meter wide path. Find the area of the Path and the cost of constructing it at Rs. 5 per square meter?

A.

4000

B.

6000

C.

8000

D.

7000

Answer with explanation

Answer: Option CExplanation

Area of the entire grass field including the path = Length (85 + 5 +5) * Breadth (65 + 5+5)

= 95 * 75 = 7125 sq. m.

Area of the grass field excluding the path = 85 * 65 = 5525 sq. m.

So, area of the path = 7125 – 5525 = 1600 sq. m.

The cost of constructing is Rs. 5 per square meter.

So, the cost of constructing the entire path = 1600 * 5 = Rs. 8000

Workspace

The perimeter of the rectangular field is 480 meters and the ratio between the length and breadth is 5:3. Find the area of the field.

A.

7200 m^{2}

B.

15000 m^{2}

C.

13500 m^{2}

D.

54000 m^{2}

Answer with explanation

Answer: Option CExplanation

Let the length of the rectangle be 5x, and breadth be 3x,

Perimeter of a rectangle = 2(l +b) = 480

2(5x + 3x) = 480

2 x 8x = 480

16x = 480

x = 480/16 = 30

∴ Length = 5 x 30 = 150 m

Breadth = 3 x 30 = 90 m

And, Area of rectangle = L x B

= 150 x 90 = 13500 m^{2}

Workspace

Find the perimeter , if the area of a rectangular box is 360 m ^{2 }and length is 12m.

A.

84m

B.

30m

C.

42m

D.

21m

Answer with explanation

Answer: Option AExplanation

Length = 12m

Breadth = Xm

Area = 360m = L×B = 12×X

= 360m = 12x

X = 360/12

X = 30m

so, perimeter = 2(L+B)

= 2(12+30)

= 2(42)

= 84m.

Workspace

A ground is in the shape of rectangle of length 30m and width 20 m inside the lawn there is a lane pf uniform width 2m bordering the ground .the area of lane in m^2 is

A.

184m ^{2 }

B.

166m ^{2 }

C.

225m ^{2 }

D.

196m ^{2 }

Answer with explanation

Answer: Option AExplanation

Length of the rectangular ground = 30 m

Breadth of the rectangular ground = 20 m

Area of the rectangular ground = L*B

⇒ 30*20

= 600 sq m

Given that the lane of uniform width 2 m bordering the ground.

Then,

Length of the inner rectangular portion = 30 – (2 + 2)

= 30 – 4

= 26 m

Breadth of the inner rectangular portion = 20 – (2 + 2)

= 20 – 4

= 16 m

Area of the inner rectangular portion = 26*16

= 416 sq m

Area of the lane of uniform width of 2 m = Area of the rectangular ground – Area of the inner rectangular portion

⇒ 600 – 416

= 184 sq m

Workspace

- 1
- 2
- 3
- …
- 5
- Next Page »

Correct Answer 👍

Wrong Answer 👎

RecruitmentIndia.in is Blog where we will update the information by exploring various online and offline sources of information. Our aim is to provide the latest Education related news as fast as possible to the students for free of cost.

Exams.Recruitmentindia.in is a Preparation portal where you can prepare all the competitive related questions like Aptitude, Reasoning, English Questions and Current affairs for free of cost

You can Contact us at :