Aptitude AreaPage 4
25. | | If the length and breadth of a rectangular plot be increased by 50% and 20% respectively, then how many times will its area be increased ? | | (a)4/3 | | (b)2 | | (c)17/5 | | (d)None of these |
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Answer is: DLet original length = X meteres and original breadth = Y meteres.
Original area = XY.
New length = (150X/100) m = (3X/2) m,
New breadth = (120Y/100) m = (6Y/5) m.
New area = [(3X/2) x (6Y/5)] m² = (9XY/5) m².
∴ Increase = [(4XY/5)/XY] = 4/5 times.
26. | | 2 metres broad pathway is to be constructed around a rectangular plot on the inside. The area of the plot is 96 sq. m. The rate of construction is Rs. 50 per square metre. Find the total cost of the construction ? | | (a)Rs 2400 | | (b)Rs 4000 | | (c)Rs 4800 | | (d)Data inadequate |
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Answer is: Dlb = 96 (given)
Area of pathway = [(l – 4)(b – 4) – lb] = 16 – 4(l + b)
Which cannot be determined
So, data is inadequate
27. | | A park square in shape has a 3 meters wide road inside it running along its sides. The area occupied by the road is 1764 square meters. What is the perimeter along the outer edge of the road ? | | (a)576 m | | (b)600 m | | (c)640 m | | (d)Data inadequate |
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Answer is: BLet the length of the outer edge be X meters, Then, length of the inner edge (X – 6) m.
∴ X² - (X – 6)² = 1764
⇒ 12X – 36 = 1764
⇒ X = 1800/12 = 150.
⇒ Required perimeter = 4 x 150 = 600 m.
28. | | The area of a triangle is 216 cm² and its sides are in the ratio 3 : 4 : 5. The perimeter of the triangle ? | | (a)6 cm | | (b)12 cm | | (c)36 cm | | (d)72 cm |
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Answer is: DLet a = 3X cm, b = 4X and c = 5X. Then,
s = 6X cm.( Half Perimeter)
A = √s(s – a) (s – b) (s – c) = √6X x 3X x 2X x X = 6X² cm²
∴ 6X² = 216
∴ X = 6.
∴ a = 18, b = 24 and c = 30
Perimeter = (18 + 24 + 30) = 72 cm.
29. | | If the height of a triangle is decreased by 40% and its base is increased by 40%, what will be the effect on its area ? | | (a)No change | | (b)8% decrease | | (c)16% decrease | | (d)16% increase |
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Answer is: CLet initial base = b cm and initial height = h cm. Then,
Initial area = ½ bh cm²
New base = 140b/100 = 7b/5 cm.
New height = 60h/100 = 3h/5 cm.
New area = (1/2 x 7b/5 x 3h/5) = 21bh/50 cm².
Area decreased = (bh/2 – 21bh/50) = 4bh/50 cm².
Percentage decreased = (4bh/50 x 2/bh x 100)% = 16%.
30. | | The area of a rhombus is 150 cm². The length of one of its diagonals is 10 cm. The length of the other diagonal is: | | (a)25 cm | | (b)30 cm | | (c)35 cm | | (d)40 cm |
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Answer is: B½ d1 x d2 = 150.
⇒ ½ x 10 x d2 = 150.
⇒ d2 = 30 cm.
31. | | A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What is the area of the circle ? | | (a)88 cm² | | (b)154 cm² | | (c)1250 cm² | | (d)616 cm² |
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Answer is: D2πR = 2(l + b)
2πR = 2(26 + 18)
∴ R = 14 cm.
∴ Area of the circle = πR² = (22/7 x 14 x 14) = 616 cm².
32. | | A circular ground whose diameter is 35 meters, has a 1.4 m broad garden around it. What is the area of the garden in square meters ? | | (a)160.16 | | (b)147.84 | | (c)196.16 | | (d)Data inadequate |
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Answer is: BRadius of the ground = 17.5 m.
Radius of the inner circle = (17.5 – 1.4) = 16.1 m.
Area of the garden = π x [(17.5)² - (16.1)²] m².
Area of the garden = [ 22/7 x (17.5 + 16.1) (17.5 – 16.1)]
Area of the garden = 22/7 x 33.6 x 1.4 = 147.84 m².
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