Circumference = No.of revolutions * Distance covered Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.Number of revolutions per min. =(1100/4.4) = 250.
Area of equilateral triangle = ?3a2/4 = x ......(i)And perimeter = 3a = y ? a = y/3 ....(ii)Now, Putting the value of a from Eq. (ii) in Eq. (i). we get?3 (y/3)2/4 = x ? x = ?3 x y2/36? x = y2/3?3x = y2/12?312?3 x = y2On squaring both sides, we get y4 = 432x2
Let the breadth of floor be 'b' m. Then, length of the floor is 'l = (b + 25)' Area of the rectangular floor = l x b = (b + 25) × b According to the question, (b + 15) (b + 8) = (b + 25) × b b 2 + 8 b + 15 b + 120 = b 2 + 25 b 2b = 120 b = 60 m. l = b + 25 = 60 + 25 = 85 m. Area of the floor = 85 × 60 = 5100 sq.m.
Let original length = x metres and original breadth = y metres. Original area = (xy) m2. New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 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 ∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%
Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12Let length of the sides be 20k, 15k and 12k.Then, according to the question,20k + 15k + 12k = 94? 47k = 94? k = 94/47 = 2Smallest side = 12k = 12 x 2 = 24 cm
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