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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
Answer is: BLet the two parallel sides of the trapezium be a cm and b cm.
Then, a â€“ b = 4 â€¦.(i)
And, Â½ x (a + b) x 19 = 475
⇒ a + b = 50 â€¦..(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
Answer is: CLet breath = X cm. Then, length = (160/100)x cm = (8/5)X cm.
So, [(8/5)X â€“ X] = 24
(3/5)X = 24
X = (24 x 5)/3 = 40
âˆ´ Length = 64 cm, Breadth = 40 cm.
Area = (64 x 40) cmÂ² = 2560 cmÂ².

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
Answer is: BLet breadth = X meters. Then, length = (X + 20) meters.
Perimeter = (5300/26.50) m = 200 m.
âˆ´ 2 [(X + 20) + X] = 200
â‡’ 2X + 20 = 100
â‡’ 2X = 80
â‡’ 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
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.
âˆ´ Length = 480 m and Breadth = 320 m.
âˆ´ Area = (480 x 320) mÂ² = 153600 mÂ².

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
Answer is: DLet breadth = X meters. Then, length = (115X/100) meters.
âˆ´ X x (115X/100) = 460
â‡’ XÂ² = [(460 x 100)/115] = 400
â‡’ 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
Answer is: DLength of diagonal = [52 x (15/60) m = 13 m.
Sum of length and breadth = [68 x (15/60)] m = 17 m.
âˆ´ âˆšlÂ² + bÂ² = 13
â‡’ lÂ² + bÂ² = 169
â‡’ l + b = 17
Area = lb = Â½(2 lb) = Â½ [(l + b)Â² - (lÂ² + bÂ²)]
Area = Â½ [(17)Â² - (169)] = Â½ (289 â€“ 169) = 60 mÂ².

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%
Answer is: CLet original length = X meters and original breadth = Y meters.
Original area = (XY) mÂ²
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 mÂ².
âˆ´ 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%
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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