Area Problems with Solutions for SSC, IBPS Bank Exams - 1

Question: 1

A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?

(A) 54

(B) 55

(C) 56

(D) 58

Ans: C

Perimeter of the plot = 2(90 + 50) = 280 m.

∴ Number of poles = $$({280} / {5})$$ = 56.

Question: 2

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 complete one round in 8 minutes, then the area of the park (in sq.m) is

(A) 15360 sq.m

(B) 151600 sq.m

(C) 153600 sq.m

(D) 30720 sq.m

Ans: C

Perimeter = Distance covered in 8 min = $$({12000} / {60} × 8)$$m = 1600 m.

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

Length = 480 m and Breadth = 320 m.

∴ Area = (480 × 320) m2 = 153600 m2.

Question: 3

How many metres of carpet 63 cm wide will be required to cover the floor of a room 14 m by 9 m?

(A) 185 m

(B) 190 m

(C) 195 m

(D) 200 m

Ans: D

Area of the floor = (14 × 9) m2 = 126 m2.

∴ Length of the carpet = $$({126} / {63} × 100)$$m = 200 m.

Question: 4

A room 5 m and 8 m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is Rs. 18 per sq.metre, the cost of carpeting the room will be

(A) Rs. 673.92

(B) Rs. 682.46

(C) Rs. 691.80

(D) Rs. 702.60

Ans: A

Area of the carpet = [(5 – 0.20) × (8.20)]m2 = (4.8 × 7.8) m2 = 37.44 m2.

∴ Cost of carpeting = Rs. (37.44 × 18) = Rs. 673.92.

Question: 5

A garden is 24 m long and 14 m wide. There is a path 1 m wide outside the garden along its sides. If the path is to be constructed with square marble tiles 20 cm x 20 cm, the number of tiles required to cover the path is

(A) 1800

(B) 2000

(C) 2150

(D) 2250

Ans: B

Area of the path

= [(26 × 16) – (24 × 14)] m2 = (416 – 336) m2 = 80m2.

∴ Number of tiles required to cover the path

= $${Area of path} / {Area of each tile}$$ = $$({80 × 100 × 100} / {20 × 20})$$ = 2000.

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