Area Aptitude Questions and Answers for Bank Exams - 1

Question: 1

The area of a square is 1024 sq.cm. What is the ratio of the length to the breadth of a rectangle whose length is twice the side of the square and breadth is 12 cm less than the side of this square?

(A) 5 : 18

(B) 14 : 5

(C) 16 : 5

(D) 32 : 5

Ans: C

Area of square = $$√1024$$cm = 32 cm.

Length of rectangle = (2 × 32) cm = 64 cm.

Breadth of rectangle = (32 – 12) cm = 20 cm.

∴ Required ratio = 64 : 20 = 16 : 5.

Question: 2

The area of a rectangle is 252 cm2 and its length and breadth are in the ratio of 9 : 7 respectively. What is its perimeter?

(A) 62 cm

(B) 64 cm

(C) 68 cm

(D) 96 cm

Ans: B

Let the length and breadth of the rectangle be (9x) cm and (7x) cm respectively.

Then, 9x + 7x = 252 ⇒ 63 x2 ⇒252 ⇒ x2 = 4 ⇒ x = 2.

So, length = 18 cm, breadth = 14 cm.

∴ Perimeter = 2(18 + 14) cm = 64 cm.

Question: 3

A rectangular garden (60 m x 40 m) is surrounded by a road of width 2m, the road is covered by tiles and the garden is fenced. If the total expenditure is Rs.51600 and rate of fencing is Rs.50 per metre, then the cost of covering 1 sq.m of road by tiles is

(A) Rs. 10

(B) Rs. 100

(C) Rs. 150

(D) Rs. 250

Ans: B

Length of the fence = 2(60 + 40) m = 200 m.

Cost of fencing = Rs. (200 × 50) = Rs. 10000.

Area of the road = [(64 × 44) – (60 × 40)] m2.

= (2816 – 2400) m2 = 416 m2.

Let the cost of tiling the rood be Rs. x per sq.m.

∴ 416x + 10000 = 51600 ⇒ 416x = 41600 ⇒ x = Rs. 100.

Question: 4

Find the largest size of a bamboo that can be placed in a square of area 100 sq.m.

(A) 14.11 m

(B) 14.13 m

(C) 14.14 m

(D) 14.17 m

Ans: C

Side of the square = $$√100$$m = 10 m.

Largest size of bamboo = Length of diagonal of the square

= 10$$√2$$ m. = (10 × 1.414)m = 14.14 m.

Question: 5

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

(A) 1350 sq.m

(B) 1550 sq.m

(C) 15500 sq.m

(D) 13500 sq.m

Ans: D

Let the length and breadth of the field be (5x) metres and (3x) metres respectively.

Then, 2(5x + 3x) = 480 ⇒ 8x = 240 ⇒ x = 30.

So, length = 150 m, breadth = 90 m.

∴ Area of the field = (150 × 90) sq.m = 13500 sq.m.

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