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A Learning Portal from Recruitment India

The area of the path 1 m wide surrounding a playground 60 m long and 40 m broad is

A.

2604 sq. m.

B.

240 sq. m.

C.

200 sq. m.

D.

204 sq. m.

Answer with explanation

Answer: Option DExplanation

Length of the playground (L1) = 60 m

Breadth of the playground (B1) = 40 m

∴ area of the playground = length * breadth = 60 * 40 = 2400 sq. m

Also,

Length of the playground including length of the path (L2) = 60 + 2 = 62 m

Breadth of the playground including length of the path (B2) = 40 + 2 = 42 m

∴ area of the playground including area of the path = 62 * 42 = 2604 sq. m

Thus, area of the path = 2604 – 2400 = 204 sq. m

Workspace

A rectangular lawn 55 m by 35 m has two roads each 4 m wide running in the middle of it, one parallel to length and the other parallel to breadth. The cost of gravelling the roads at 75 paise per sq. metre is:

A.

Rs. 257.50

B.

Rs. 275

C.

Rs. 262.50

D.

Rs. 258

Answer with explanation

Answer: Option DExplanation

area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m

cost of graveling = 344 x (75/100) = Rs. 258

Workspace

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required ?

A.

86 feet

B.

84 feet

C.

88 feet

D.

82 feet

Answer with explanation

Answer: Option CExplanation

We are given with length and area, so we can find the breadth.

as Length * Breadth = Area

=> 20 * Breadth = 680

=> Breadth = 34 feet

Area to be fenced = 2B + L = 2*34 + 20

= 88 feet

Workspace

A wire is in the form of a circle. The radius of the circle is 28 cm. The wire is then moulded to form a square. Find the side of the square formed?

A.

11 cm

B.

44 cm

C.

66 cm

D.

22 cm

Answer with explanation

Answer: Option BExplanation

Radius of the circle (r) = 28 cm.

Length of the wire (Circumference) = 2πr=2π28=176 cm

Let side of the square be ‘a’ cm.

Perimeter of square (4a) = Circumference of the circle = 176 cm

or, 4a = 176 cm

or, a = 44 cm

Thus, side of the square is 44 cm.

Workspace

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is

A.

B.

C.

D.

Answer with explanation

Answer: Option BExplanation

Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.

We are having speed and time so we can calculate the distance or perimeter in this question.

Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:

Perimeter = Distance travelled in 8 minutes,

=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]

As per question length is 3x and width is 2x

We know perimeter of rectangle is 2(L+B)

So, 2(3x+2x) = 1600

=> x = 160

So Length = 160*3 = 480 meter

and Width = 160*2 = 320 meter

Finally, Area = length * breadth

= 480 * 320 = 153600

Workspace

Shyam’s room has dimensions 6m x 5m x 4m. The room also has a door of dimension 2.5m x 1.2m and a window of dimensions 1m x 1m. What will be cost of painting the four walls of his room if rate of painting is Rs. 2 per sq.m.?

A.

Rs. 840

B.

Rs. 410

C.

Rs. 168

D.

Rs. 84

Answer with explanation

Answer: Option CExplanation

Workspace

the cross-section of a canal is a trapezium in shape. if the canal is a 12m wide at the top ,8mwide a the bottom and the area of cross-section is 84m,find its depth

A.

8.4 meters

B.

8 meters

C.

4.6 meters

D.

5 meters

Answer with explanation

Answer: Option AExplanation

area of a trapezium = 1/2 (b + a) h

where a and b are parallel sides and h is the altitude or distance between a and b.

In a canal, the floor of canal and the top water surface are parallel. Height of trapezium is the depth of canal.

84 meters = 1/2 ( 12 + 8) Height

Height = 168/20 = 8.4 meters

Workspace

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

A.

50m

B.

120m

C.

40m

D.

None of these

Answer with explanation

Answer: Option DExplanation

Workspace

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A.

2.91 m

B.

5.82 m

C.

3 m

D.

None of these

Answer with explanation

Answer: Option CExplanation

Workspace

The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m.What is the area of the field?

A.

77500

B.

57500

C.

37500

D.

47500

Answer with explanation

Answer: Option CExplanation

Let length = x meters, then breadth = 0.6x

Given that perimeter = 800 meters

=> 2[ x + 0.6x] = 800

=> x = 250 m

Length = 250m and breadth = 0.6 x 250 = 150m

Area = 250 x 150 = 37500 sq.m

Workspace

Water flows a tank 200 x 150m through a rectangular pipe 1.5 x 1.25m at the rate at 20kmph. In what time will be water rises by 2 meters?

A.

96min

B.

90min

C.

80min

D.

76 min

Answer with explanation

Answer: Option AExplanation

Volume of water flown in the tank

=(200 x 150 x 2) Cu.m = 60000 Cu.m

Volume flown per hour

=(3/2 x 125/100 x 20 x 1000) Cu.m =37500Cu.m

Time taken = 60000/37500 hrs = 8/5 hrs

=(8/5 x 60)min = 96 min

Workspace

The length of a rectangle is 3/5th of the side of a square. The radius of a circle is equal to side of the square. The circumference of the circle is 132 cm. What is the area of the rectangle, if the breadth of the rectangle is 15 cm?

A.

112 cm²

B.

149 cm²

C.

189 cm²

D.

Cannot be determined

Answer with explanation

Answer: Option CExplanation

Circumference of the circle = 132

2πR = 132; R = 21 cm

Side of square = 21 cm

Length of the rectangle = 3/5 * 21 = 63/5

Area of the rectangle = 63/5 * 15 = 189 cm²

Workspace

A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is

A.

rs.58

B.

rs.158

C.

rs.258

D.

rs.358

Answer with explanation

Answer: Option CExplanation

area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m

cost of graveling = 344 x (75/100) = Rs. 258

Workspace

A hall 20 m long and 15 m broad is surrounded by a verandah of uniform width of 2.5 m. The cast of flooring the verandah at the rate of Rs 3.50 per sq. Meter is:

A.

Rs.500

B.

Rs. 700

C.

Rs. 600

D.

Rs. 800

Answer with explanation

Answer: Option BExplanation

Area of verandah = [(25 x 20) – (20 x 15)]Sq.m = 20 Sq.m

Cost of flooring = Rs. (200 x 3.50) = Rs. 700

Workspace

The perimeter of a square is equal to twice the perimeter of a rectangle of length 8 cm and breadth 7 cm. What is the circumference of a semicircle whose diameter is equal to the side of the square ?

A.

47.47 cm

B.

38.57 cm

C.

23.57 cm

D.

42.46 cm

Answer with explanation

Answer: Option BExplanation

Perimeter of square = 2 x Perimeter of rectangle

= 2 * 2 (8+7) = 60 cm.

Side of square = 60/4 = 15 cm = Diameter of semi-circle

Circumference of semi-circle = πd/2 + d

= (22/7) * 2 * 15 + 15 = 38.57 cm

Workspace

A.

Rs. 480

B.

Rs. 384

C.

Rs. 352

D.

Rs. 176

Answer with explanation

Answer: Option DExplanation

Workspace

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

A.

27 and 23

B.

24 and 23

C.

25 and 23

D.

22 and 23

Answer with explanation

Answer: Option AExplanation

Let the two parallel sides of the trapezium be a cm and b cm.

Then, a – b = 4

Workspace

The perimeter of one square is 48 cm and that of another is 20 cm. Find the perimeter and the diagonal of a square which is equal in area to these two combined?

A.

15√2 cm

B.

17√2 cm

C.

13√2 cm

D.

16√2 cm

Answer with explanation

Answer: Option CExplanation

4a = 48 4a = 20

a = 12 a = 5

a^{2} = 144 a^{2} = 25

Combined area = a^{2} = 169 => a = 13

d = 13√2

Workspace

A.

15m, 9m

B.

17m, 9m

C.

14m, 7m

D.

16m, 7m

Answer with explanation

Answer: Option BExplanation

Length = x; Breadth =y

xy – (x-5)(y+3) = 9

3x – 5y – 6 = 0 —(i)

(x+3)(y+2) – xy = 67

2x + 3y -61 = 0 —(ii)

solving (i) and (ii)

x = 17m ; y = 9m

Workspace

The rainwater from a flat roof 16 m long and 10 m wide is collected in a tank whose internal measurements are 2 m long, 2.4 m wide and 3 m deep. Before it started raining it was half-full and after the rain it was full. Find the height of rainfall.

A.

0.9 cm

B.

1.8 cm

C.

4.5 cm

D.

9 cm

Answer with explanation

Answer: Option CExplanation

Let h be the height of the rainfall = h

1.6 × 10 × h = 1/2(2 × 2.4 × 3)

h = .045m = 4.5cm

Workspace

A circular swimming pool is surrounded by a concrete wall 4ft wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius of the pool is?

A.

10ft

B.

20ft

C.

30ft

D.

40ft

Answer with explanation

Answer: Option BExplanation

let the radius of the pool be Rft

Radius of the pool including the wall = (R+4)ft

Workspace

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

A.

140 m

B.

148 m

C.

150 m

D.

200 m

Answer with explanation

Answer: Option DExplanation

Let the triangle and parallelogram have common base b,

let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then

Workspace

A copper wire when bent in the form of a square encloses an area of 121 cm^{2}. If the same wire is bent in the form of a circle, it encloses an area equal to?

A.

121 cm^{2}

B.

144 cm^{2}

C.

154 cm^{2}

D.

168 cm^{2}

Answer with explanation

Answer: Option CExplanation

As is given, area of square = a^{2} = 121 cm^{2}

So, side of square, a = 11 cm

length of the wire = Perimeter of square of side a = 4a = 4*11 = 44cm

Let the radius of circle be r cm. Then,

Perimeter of circle = Length of wire

=> 2πr = 44 cm

=> r = 44/2π = 7 (π = 22/7)

Now, Area of circle = πr^{2} = (22/7) *7*7 = 154 cm^{2}

Workspace

If each side pair of opposite sides of a square is increased by 20 m, the ratio of the length and breadth of the rectangular so formed becomes 5:3. The area of the old square is?

A.

990m²

B.

900m²

C.

930m²

D.

945m²

Answer with explanation

Answer: Option BExplanation

(x+20) / x = 5 / 3

3x + 60 = 5x

x = 30m; Area = 900m²

Workspace

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