Question
One year ago from now, the ages of 'A' and 'B' were in
the ratio 3:2, respectively. Six years ago from now, the ages of 'A' and 'B' were in the ratio 2:1. Present age of 'A' is how much percent more than that of 'B'?Solution
Let the ages of 'A' and 'B', 6 years ago from now was, '2x' years and 'x' years, respectively. ATQ; {(2x + 5)/(x + 5)} = (3/2) Or, 2 X (2x + 5) = 3 X (x + 5) Or, 4x + 10 = 3x + 15 Or, x = 5 So, present age of 'A' = 2 X 5 + 6 = 16 years And present age of 'B' = 5 + 6 = 11 years So, required percentage = {(16 - 11)/11} X 100 = 45.45%
I. y² - 7 y – 18 = 0
II. x² + 10 x + 16 = 0
I. 2x² - 11x + 12 = 0
II. 12y² + 29y + 15 = 0
I. 40x² + 81x + 35 = 0
II. 63y² + 103y + 42 = 0
I. 3p² - 11p + 10 = 0
II. 2q² + 13q + 21 = 0
I. 2x2 – 19x + 45 = 0
II. y2 – 14y + 48 = 0
If the roots of the quadratic equation 6m² + 7m + 8 = 0 are α and β, then what is the value of [(1/α) + (1/β)]?
 If x satisfies x² – 14x + 40 = 0, find x.
I. 81x - 117√x + 40 = 0
II. 81y - 225√y + 136 = 0
If the roots of the quadratic equation 5x² + 4x + 6 = 0 are α and β, then what is the value of [(1/α) + (1/β)]?
...I. 4x² - 21 x + 20 = 0
II. 8y² - 22 y + 15 = 0
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