Question
Sheetal rides a bike at an average speed of 40 kmph and
reaches her destination in 5 hours. Hemlata covers the same distance in 4 hrs. If Sheetal increases her average speed by 10 kmph and Hemlata increases her average speed by 10 kmph, what would be the difference in their time taken to reach destination?Solution
Sheetal’s speed= 40 km/hr. Time taken by Sheetal = 5 hrs. Distance = Speed ×  Time = 40 × 5 = 200 km. ∴ Distance covered by Hemlata  = 200  km. Time taken by Hemalata = 4 hrs. ∴ Hemlata’s Speed = (Distance )/Time = (200 )/4 = 50 km/hr. Sheetal’s increased speed= 50 km/hr. ∴ Time taken by Sheetal = (200 )/50 = 4 hrs.  Hemlata’s increased speed = 60 km/hr. ∴ Time taken by Hemlata = (200 )/60 = 31/3 hrs. = 3hrs 20 minutes. ∴ Required difference = 4 hrs – 3hrs 20 minutes = 40 minutes.
I. 2x² - 7x + 3 = 0
II. 8y² - 14y + 5 = 0
I. x2 - 20x + 96 = 0
II. y2 - 23y + 22 = 0
I. x2 – 36 = 0
II. y2 - 7y + 6 = 0
I: 2x² - 8x + 6 = 0
II: 3y² - 12y + 9 = 0
I. 5x² - 24 x + 28 = 0  Â
II. 4y² - 8 y - 12= 0  Â
I. 96x² + 52x - 63 = 0
II. 77y² + 155y + 72 = 0
I. 35x² - 46x – 16 = 0
II. 35y² - 116y + 96 = 0
I. 3y² - 20y + 25 = 0
II. 3x² - 8x + 5 = 0
Find the roots of the equation 6p² – 5p – 6 = 0.
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 20x + 96 = 0
Equation 2: y² - 18y + 72 = 0
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