Aptitude AreaPage 3
17. | | The difference between two parallel sides of a trapezium is 4 cm. The perpendicular distance between them is 19 cm. If the area of the trapezium is 475 cm², find the lengths of the parallel sides ? | | (a)27 cm, 33 cm | | (b)27 cm, 23 cm | | (c)29 cm, 33 cm | | (d)29 cm, 23 cm |
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Answer is: BLet the two parallel sides of the trapezium be a cm and b cm.
Then, a – b = 4 ….(i)
And, ½ x (a + b) x 19 = 475
⇒ a + b = 50 …..(ii)
solving (i) and (ii), we get
a = 27, b = 23.
So, the two parallel sides are 27 cm and 23 cm.
18. | | The length of a rectangular plot is 60% more than its breadth. If the difference between the length and the breadth of that rectangle is 24 cm, what is the area of that rectangle ? | | (a)2400 sq. cm | | (b)2480 sq. cm | | (c)2560 sq. cm | | (d)Data inadequate |
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Answer is: CLet breath = X cm. Then, length = (160/100)x cm = (8/5)X cm.
So, [(8/5)X – X] = 24
(3/5)X = 24
X = (24 x 5)/3 = 40
∴ Length = 64 cm, Breadth = 40 cm.
Area = (64 x 40) cm² = 2560 cm².
19. | | The length of a rectangular plot is 20 m more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per meter is Rs. 5300, what is the length of the plot in meters ? | | (a)40 m | | (b)60 m | | (c)120 m | | (d)Data inadequate |
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Answer is: BLet breadth = X meters. Then, length = (X + 20) meters.
Perimeter = (5300/26.50) m = 200 m.
∴ 2 [(X + 20) + X] = 200
⇒ 2X + 20 = 100
⇒ 2X = 80
⇒ X = 40
Hence, length = X + 20 = 60 m.
20. | | 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 kmph completes one round in 8 minutes, then the area of the park (in sq. m) is: | | (a)15360 sq. m | | (b)153600 sq. m | | (c)30720 sq. m | | (d)307200 sq. m |
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Answer is: BPerimeter = Distance covered in 8 min. = [(12000/60) x 8] m = 1600 m.
Let length = 3X meters and breadth = 2X meters.
Then, 2(3X + 2X) = 1600 or X = 160 m.
∴ Length = 480 m and Breadth = 320 m.
∴ Area = (480 x 320) m² = 153600 m².
21. | | The area of a rectangle is 460 sq. m. If the length is 15% more than the breadth, what is the breadth of the rectangular field ? | | (a)15 m | | (b)26 m | | (c)34.5 m | | (d)20 m |
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Answer is: DLet breadth = X meters. Then, length = (115X/100) meters.
∴ X x (115X/100) = 460
⇒ X² = [(460 x 100)/115] = 400
⇒ X = 20 m.
22. | | A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68 m/min. The area of the field is ? | | (a)30 sq. m | | (b)40 sq. m | | (c)50 sq. m | | (d)60 sq. m |
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Answer is: DLength of diagonal = [52 x (15/60) m = 13 m.
Sum of length and breadth = [68 x (15/60)] m = 17 m.
∴ √l² + b² = 13
⇒ l² + b² = 169
⇒ l + b = 17
Area = lb = ½(2 lb) = ½ [(l + b)² - (l² + b²)]
Area = ½ [(17)² - (169)] = ½ (289 – 169) = 60 m².
23. | | The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is: | | (a)40% | | (b)42% | | (c)44% | | (d)46% |
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Answer is: CLet original length = X meters and original breadth = Y meters.
Original area = (XY) m²
New length = (120X/100) m = 6X/5 m,
New breadth = (120Y/100) m = 6Y/5 m.
New area = (6X/5) x (6Y/5) = (36/25)XY m².
∴ Increase % = [(11/25)XY x (1/XY) x 100]% = 44%.
24. | A rectangle has width a and length b. If the width is decreased by 20% and the length is increased by 10%, then what is the area of the new rectangle in percentage compared to ab ? | | (a)80% | | (b)88% | | (c)110% | | (d)120% |
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Answer is: BNew area = [(80/100)a x (110/100)b] = (4/5 x 11/10 x ab) = (22/25)ab.
∴ Required percentage = [(22/25)ab x (1/ab) x 100] = 88%
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