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Two small circular parks of diameters 6 m and 8 m are to be replaced by a bigger circular park. What would be the radius of this new park, in meter, if the new park occupies the same space as the two small parks (in meter)?

A.

5

B.

10

C.

15

D.

20

Answer with explanation

Answer: Option AExplanation

Area of the new circular park = sum of the areas of the 2 smaller parks

⇒ π (6/2)^{2} + π (8/2)^{2} = π(9+ 16) = 25π ⇒ 25 π = π R^{2}. ∴ R^{2} = 25 ⇒ R = 5 m

Workspace

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

A.

20

B.

30

C.

50

D.

40

Answer with explanation

Answer: Option CExplanation

Let x and y be the length and breadth of the rectangle respectively.

Then, x – 4 = y + 3 or x – y = 7 —-(i)

Area of the rectangle =xy; Area of the square = (x – 4) (y + 3)

(x – 4) (y + 3) =xy <=> 3x – 4y = 12 —-(ii)

Solving (i) and (ii), we get x = 16 and y = 9.

Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

Workspace

The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field ?

A.

18 meter

B.

20 meter

C.

22 meter

D.

25 meter

Answer with explanation

Answer: Option BExplanation

Area of rectangle = 460 m^{2}

Length = 115% of Breadth

Formula used:

Area of a rectangle = Length × breadth

Calculation:

Let the breadth be x m

Then the breadth will be 1.15x

The area of the rectangle will be

x × 1.5x = 460

⇒ 1.15x^{2} = 460

⇒ x^{2} = 460/1.15

⇒ x^{2} = 400

⇒ x = 20 m

∴ The breadth of the rectangle is 20 m.

Workspace

If the diameters of a circle is increased by 100% . Its area is increased by ?

A.

100%

B.

200%

C.

300%

D.

400%

Answer with explanation

Answer: Option CExplanation

Area of circle = πd^{2}/4

diameters increased by 100%, then new area = π(2d)^{2} /4 => πd^{2}

Then, New area = (πd^{2} – πd^{2}/4) => 3πd^{2}/4

increase percent =[(3πd^{2}/4) / ( πd^{2}/4) ]*100 %

= [(3πd^{2}/4) x (4/πd^{2})]*100 %

= 300%

Workspace

The area of a rectangle is equal to the area of a square whose diagonal is 12√2 metre. The difference between the length and the breadth of the rectangle is 7 metre. What is the perimeter of rectangle ? (in metre).

A.

68 metre

B.

50 metre

C.

62 metre

D.

64 metre

Answer with explanation

Answer: Option BExplanation

d = a√2

12√2 = a√2

a = 12

l * b = a² = (12²) = 144

l – b = 7 ; l = b + 7

(b + 7)*(b) = 144

b² + 7b – 144 = 0

b = 9; l = 16

2(l + b) = 2(16 + 9) = 50m

Workspace

If the length is increased by 25%, by what percent the width of a rectangle should be decreased so as to keep the area same.

A.

25%

B.

10%

C.

20%

D.

30%

Answer with explanation

Answer: Option CExplanation

Let the original length be l and the width be b

Therefore, the area = l*b

Now, as the length is increased by 25%, the new length is (1.25*l) and let the new width be x.

As the area is same, 1.25*l*x = l*b

x = b/1.25 = 0.8b

Therefore, the width is to be decreased by 20%.

Workspace

A man walked 20m to cross a rectangular field diagonally if the length of the field is 16cm then the breadth of the field is

A.

11m

B.

12m

C.

13m

D.

14m

Answer with explanation

Answer: Option BExplanation

Let a rectangular field ABCD

A/Q a man cross rectangular field diagonally and he had to walk 20m

So,Diagonal of Rectangle =20m

Length=16m (given)

We know that Each angle of a rectangle is 90°

ADC forms a right angled triangle

AC=16m(perpendicular)

AD=20m(Hypotenus)

CD=?(base)

According to Pythagorean Theorem

h2=b2+p2

(20)2=b2+(16)2

400=b2+256

b2=144

b=√144

b=12m

Workspace

If the area of the trapezium, whose parallel sides are 6 cm and 10 cm is 32 sq. cm, what will be the distance between the parallel sides?

A.

2 cm

B.

4 cm

C.

5 cm

D.

8 cm

Answer with explanation

Answer: Option BExplanation

Parallel sides of a trapezium = 6 cm, and 10 cm

Area of trapezium =1/2 ( sum of the parallel sides ) × distance between the parallel sides

32=1/2 ( 6+10 ) × distance

32=8 × distance

Distance = 32/8 = 4 cm

So, the distance between the parallel lines of trapezium = 4 cm.

**Height = 4 cm**

**Step-by-step explanation:**

**Parallel sides = 6 cm and 10 cm**

**Area of trapezium = 32 cm²**

**⇒ *(sum of || sides)* h = 32**

**⇒ * (6 + 10) * h = 32**

**⇒ 16 / 2 * h = 32**

**⇒ 8 * h = 32**

**⇒ h = 32 / 8**

**⇒ h = 4**

Workspace

The ratio of the areas of a square and rhombus whose base is same is:

A.

1:2

B.

2:1

C.

1:1

D.

3:1

Answer with explanation

Answer: Option CExplanation

When parallelogram is on same and located between the same parallels then its area will be equal. It is given rhombus and square are two parallelograms on same base and located between the same parallels.

∴ Area of rhombus = Area of square

∴ Ratio = 1:1

Workspace

The sides of a rectangle are in the ratio 4:3 and its area is 972sq.m find the perimeter of the rectangle?

A.

120m

B.

122m

C.

124m

D.

126m

Answer with explanation

Answer: Option DExplanation

let 4x and 3x be sides of the rectangle

We know that area of rectangle = l×b

4x × 3x = 972

12×2 = 972

X2 = 81

X = 9

Therefore length = 4x = 4×9 = 36m

Breadth = 3x = 3×9 = 27m

Therefore perimeter = 2 (l+b) = 2(36+27) = 126m

Workspace

The ratio between the breadth and perimeter of a rectangle is 2:10. If the area of the rectangle is 428 sq. cm, what is the length of the rectangle?

A.

45.4 cm

B.

25.4 cm

C.

30.4 cm

D.

40.4 cm

Answer with explanation

Answer: Option BExplanation

per the question = 2(L+B)/B = 10/2

2(L+B)/B = 5/1

2L + 2B = 5B

2B – 5B = – 2L

– 3B = – 2L

B = 2/3 L

Given Area = 430 sq. cm.

Area = Length * Breadth

430 = L * 2/3 L

430 = 2/3 L^{2}

Or L^{2} = 1290 / 2

L^{2} = 645

L = 25.4 cm

Workspace

How many cubes of 10cm edge can be put in a cubical box of 1m edge

A.

10

B.

100

C.

1000

D.

10000

Answer with explanation

Answer: Option CExplanation

Number of cubes= (100x100x100) / (10x10x10)= 1000

(OR)

The volume of cubical box =100^3 = 1000000cubic cm

Volume of cubes =10^3 =1000cubic cm

Now, no. Of cubes that can be fit in the cubical box = [(The volume of cubical box)/(volume of 1 cube)]

= 1000000/1000

= 1000 cubes

Workspace

The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 cu. meters. Find the breadth of the wall.

A.

40cm

B.

30cm

C.

20cm

D.

10cm

Answer with explanation

Answer: Option AExplanation

(or)

Given: The volume of a wall is 12.8 m^{3}.

Height is 5 times the breadth and length is 8 times the height.

Let breadth = x

∴ Height = 5x

⇒ Length = 8 × (5x) = 40x

The volume of a Cuboid is given by l × b × h (here l, b, h are, length, breadth, height respectively)

∴ l × b × h = 12.8

⇒ x × 5x × 40x = 12.8

⇒ 200x^{3} = 12.8

⇒ x^{3} = 12.8/200 m^{3} = 8/125

⇒ x= ∛ (8/125)m = 2/5 m = 0.4 m

∴ x = 0.4m = 0.4 × 100 = 40cm

That is Breadth is: 40cm

Workspace

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

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