Question
1600 metres long train crosses a man who is moving in
the same direction with a certain speed in 25 seconds. If the same train can cross a pole in 20 seconds with the same speed, then find the speed of the man.Solution
Let the speed of the man be ‘x’ m/sec. Speed of train = 1600/20 = 80 m/sec Relative speed of the train = (80 – x) m/sec According to the question, => (80 – x) = 1600/25 => x = 80 – 64 => x = 16 Therefore the speed of the man = 16 m/sec
I. 3x2 - 14x + 15 = 0
II. 15y2 - 34 y + 15 = 0
I. 2x2 + 5x + 2 = 0
II. 4y2 = 1
If x2Â - 3x - 18 = 0 and y2Â + 9y + 18 = 0, which of the following is true?
I. 12x2 - 55x + 63 = 0
II. 10y2 - 47y + 55 = 0
I. Â 3y2Â + 13y - 16 = 0
II. 3x2 – 13x + 14 = 0
I. 27x² + 120x + 77 = 0
II. 56y² + 117y + 36 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 38x + 352 = 0
Equation 2: y² - 38y + 312 = 0
I. x2 + x – 42 = 0
II. y2 + 6y – 27 = 0
I. 84x² - 167x - 55 = 0
II. 247y² + 210y + 27 = 0
I. 6x2 + 19x + 10 = 0
II. y2 + 10y + 25 = 0
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