Area Problems with Solutions for SSC, IBPS Bank Exams - 2

Ans: B

Maximum possible size of a flower bed

= (H.C.F of 110, 130, 190) sq.m. = 10 sq.m

∴ Maximum possible length = $$({10}/ {2})$$m = 5 m.

Ans: C

Side of the square = 12 cm.

Area of rectangle = [(12 × 12) – 4]cm² = 140 cm².

∴ Breadth = $${Area} / {Length}$$ = $${140}/ {14}$$ = 10 cm.

Hence, perimeter = 2(l + b) = 2(14 + 10) cm = 48 cm.

Ans: D

Let the breadth of the rectangle be x metres. Then, length of the rectangle = (2x) metres.

2(2x + x) = 60 ⇒ 6x = 60 ⇒ x = 10.

So, length = 20 m, breadth = 10 m.

∴ Area = (20 × 10)m² = 200 m².

Ans: C

Original breadth = 3m,

Original length = (1.44 × 3) m = 4.32 m.

New breadth = (125% of 3) m = $$({125} / {100} × 3)$$ m = 3.75 m.

New length = (140% of 4.32) m = $$({140} / {100} × 4.32)$$ m = 6.048 m.

Original area = (4.32 × 3) m² = 12.96 ²

New area = (6.048 × 3.75) m² = 22.68 m²

Increase in area = (22.68 – 12.96) m² = 9.72 m²

∴ Increase in cost = Rs. (9.72 × 45) = Rs. 437.40.

Ans: B

Area of the plot = (110 × 65)m² = 7150 m².

Area of the plot exchanging the path

= [(110 - 5) × (65 – 5)] m² = 6300 m².

∴ Area of the path = (7150 – 6300) m² = 850 m².

Cost of gravelling the path = Rs. $$(850 × {80}/ {100})$$ = Rs. 680.