2. | An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is: | ||||||||||||||
Answer: Option D Explanation:
100 cm is read as 102 cm. A1 = (100 x 100) cm2 and A2 (102 x 102) cm2. (A2 – A1) = [(102)2 – (100)2] = (102 + 100) x (102 – 100) = 404 cm2.
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3. | The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? | |||||||||||||||||||||||
Answer: Option B Explanation:
2l + 2b = 5b 3b = 2l
Then, Area = 216 cm2 l x b = 216
l2 = 324 l = 18 cm. |
4. | The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is: | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Answer: Option C Explanation:
Let original length = x metres and original breadth = y metres. Original area = (xy) m2.
The difference between the original area = xy and new-area 36/25 xy is = (36/25)xy – xy = xy(36/25 – 1) = xy(11/25) or (11/25)xy
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5. | A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road? | |||||||
Answer: Option B Explanation:
Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. Area of the crossroads = (2400 – 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x – x2 = 291 x2 – 100x + 291 = 0 (x – 97)(x – 3) = 0 x = 3. |