Question
The average of the present ages of R, S and T is 28 years.
Triple the present age of S is 45 more than twice the present age of R. If T’s age after 6 years will be 25% higher than his current age, determine the present age of R.Solution
ATQ,
Present age of T = 6 / 0.25 = 24 years
Let present ages of R and S be r and s, respectively.
So, r + s = 28 × 3 – 24 = 60 ………………(1)
And, 3s – 2r = 45 ………………(2)
Equation (1) × 3 – Equation (2), we get
3r + 3s – 3s + 2r = 60 × 3 – 45
Or, 5r = 135
Or, r = 27
So, present age of R = 27 years
I. 24x² - 58x + 23 = 0
II. 20y² + 24y – 65 = 0
I. 6x2 + 23x + 10 = 0
II. 2y2 - 3y - 5 = 0
I. p2 – 2p – 15 = 0
II. q2 + 4q – 12 = 0
Equation 1: x² + 16x + 63 = 0
Equation 2: y² + 10y + 21 = 0
Solve: x² − 7x + 12 = 0
I. 15y2 + 4y – 4 = 0
II. 15x2 + x – 6 = 0
I. x² + 4x + 4 = 0
II. y² - 8y + 16 = 0
I. 18p²- 21p + 6 = 0   Â
II. 16q² - 24q +9 = 0
I. 5x² - 24 x + 28 = 0  Â
II. 4y² - 8 y - 12= 0  Â
If x + 1/x = 3, find x² + 1/x².