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ATQ, Usual speed of the cyclist = (240/6) = 40 km/h New speed of the cyclist = 0.25 × 40 = 10 km/h New time taken = (240/10) = 24 hours Alternate Solution: Let the usual speed of the cyclist be 10x km/hr. Therefore, decreased speed = 2.5x km/hr Since the distance covered is the same and time taken is inversely proportional to the speed: Ratio of time taken with usual speed to time taken with reduced speed = 6:24 = 1:4 Required time taken = 6 × (4/1) = 24 hours
I. 104x² + 9x - 35 = 0
II. 72y² - 85y + 25 = 0
l). p² - 26p + 153 = 0
ll). q² - 17q + 72 = 0
Equation 1: x² - 144x + 5184 = 0
Equation 2: y² - 130y + 4225 = 0
I. 18p²- 21p + 6 = 0
II. 16q² - 24q +9 = 0
Equation 1: x² - 180x + 8100 = 0
Equation 2: y² - 170y + 7225 = 0
I. 2x2- 15x + 25 = 0
II. 3y2- 10y + 8 = 0
I. 2x2 – 19x + 45 = 0
II. y2 – 14y + 48 = 0
I. 17x² - 26x – 16 = 0
II. 17y²- 26y + 9 = 0
I. x2 + 91 = 20x
II. 10y2 - 29y + 21 = 0
I. x2 – 18x + 81 = 0
II. y2 – 3y - 28 = 0
