100+ Problems on Area and Perimeter Aptitude Questions Pdf - 1

Question: 1

A lawn is in the shape of a rectangle of 80 m length and 50 m width. Outside the lawn there is a footpath of uniform 1 m width bordering the lawn. The area of the footpath is

(A) 164 m2

(B) 264 m2

(C) 284 m2

(D) 4264 m2

Ans: B

Area of the footpath = [{80 + 2} × (50 + 2) – (80 × 50)]m2

= [(82 × 52) – (80 × 50]m2 = (4264 - 4000) m2 = 264 m2.

Question: 2

A rectangular lawn 80 metres by 60 metres has two roads each 10 m wide running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling them at Rs. 30 per square metre.

(A) Rs. 3600

(B) Rs. 36000

(C) Rs. 39000

(D) Rs. 49000

Ans: C

Area of the roads = (80 × 10 + 60 × 10 – 10 × 10) m2 = 1300 m2.

∴ Cost of gravelling = Rs. (1300 × 30) = Rs. 39000.

Question: 3

The total cost of flooring a room at Rs. 8.50 per square metre is Rs. 510. If the length of the room is 8 m, its breadth is

(A) 7.5 m

(B) 8.5 m

(C) 9.5 m

(D) 10.5 m

Ans: B

Area of the floor = $$({510} / {8.50})$$m2 = 60 m2

∴ Breadth of the room = $$({60} / {8})$$ = 7.5 m.

Question: 4

A rectangular courtyard, 3.78 m long and 5.25 m broad, is to paved exactly with square tiles, all of the same size. Find the least number of square tiles covered.

(A) 250

(B) 350

(C) 450

(D) 550

Ans: C

Area of the room = (378 × 525) cm2.

Size of largest square tile = H.C.F. of 378 cm and 525 cm = 21 cm.

Area of 1 tile = (21 x 21) cm2.

∴ Number of tiles required = $$({378 × 525} / {21 × 21})$$ = 450.

Question: 5

The area of a rectangular field is 52000 m2. This rectangular area has been drawn on a map to the scale 1 cm to 100 m. The length is shown as 3.25 cm on the map. The breadth of the rectangular field is

(A) 150 m

(B) 160 m

(C) 200.5 m

(D) 300.5 m

Ans: B

Length of the field = (3.25 × 100) m = 325 m.

∴ Breadth of the field = $$({52000} / {325})$$m = 160 m.

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