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- 16. The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 10 cm. If the area of the trapezium is 250cm2 find the lengths of the parallel side
A.27, 23
B.25, 25
C.24, 26
D.23, 29
Answer & Explanation
Answer: Option A 
Let the two parallel sides of the trapezium be a cm and b cm.
Then, a-b = 4 ------ (1)
And, (1/2) x (a+b) x 10 = 475 =>(a+b) =((250 x 2)/10) => a + b = 50------- (2)
Solving 1 and 2, we get: a=27,b=23
So, the two parallel sides are 27 cm and 23 cm.
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- 17. Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 2244 sq. metres.
A.25.4
B.26.7
C.24.3
D.22.5
Answer & Explanation
Answer: Option B
Clearly, the cow will graze a circular field of area 2244 sq. metres and radius
equal to the length of the rope.
Let the length of the rope be R metres.
Then, Π(R)2= 2244 => R2 = (2244 X (7/22)) = 714=> R = 26.72
Length of the rope = 26.72 m.
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- 18. The area of a circular field is 17.56 hectares. Find the cost of fencing it at the rate of Rs. 3 per metre approximately
A.4457
B.4567
C.4235
D.4547
Answer & Explanation
Answer: Option A
Area = (17.56 x 10000) m2= 175600 m2.
ΠR2 = 175600 ⇔ (R)2 = (175600 x (7/22)) ⇔ R = 236.37 m.
Circumference = 2ΠR = (2 x (22/7) x 236.37) m =1485.78 m.
Cost of fencing = Rs. (1485.78 x 3) = Rs. 4457.
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- 19. The diameter of the driving wheel of a bus is 120 m. How many revolution, per minute must the wheel make in order to keep a speed of 60 kmph approximately
A.245
B.257
C.262
D.265
Answer & Explanation
Answer: Option D
Distance to be covered in 1 min. = (60 X 1000)/(60) m = 1000 m.
Circumference of the wheel = (2 x (22/7) x 0.60) m = 3.77 m.
Number of revolutions per min. =(1000/3.77) = 265
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- 20. A wheel makes 1200 revolutions in covering a distance of 60 km. Find the radius of the wheel.
A.10.5m
B.11m
C.11.25m
D.11.45m
Answer & Explanation
Answer: Option D
2πR = 72 => 2 x (22/7) x R = 72 => R = 72 x (7/44) = 11.45 m.
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