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Kajal invested Rs. (y+2000) on compound interest at the rate of 18% per annum compounded annually in scheme P. Interest after 2 years from scheme P = (y+2000) of (100+18)% of (100+18)% - (y+2000) = (y+2000) of 118% of 118% - (y+2000) = (y+2000) x 1.18 x 1.18 - (y+2000) = 1.3924(y+2000) - (y+2000) = (y+2000)x[1.3924 - 1] = 0.3924(y+2000) She invested Rs. ‘y’ on 20% per annum on simple interest in scheme Q. Interest after 2 years from scheme Q = y x 20% x 2 = 0.4y After two years, if the interest obtained from scheme P is Rs. 739.2 more than the interest obtained from scheme Q. 0.3924(y+2000) = 0.4y+739.2 0.3924y+784.8 = 0.4y+739.2 784.8-739.2 = 0.4y-0.3924y 0.0076y = 45.6 Value of ‘y’ = 6000
I. 4x2 + 9x - 9 = 0
II. 4y2 - 19y + 12 = 0
I. x − √2401 = 0
II. y2 − 2401 = 0
I. p² - 10p +21 = 0
II. q² + q -12 = 0
Equation 1: x² - 220x + 12100 = 0
Equation 2: y² - 210y + 11025 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the...
What will be the product of smaller roots of both equations.
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 33x² - 186x + 240 = 0
Equation 2: 35y² - 200y + ...
How many values of x and y satisfy the equation 2x + 4y = 8 & 3x + 6y = 10.
I. 2x² - 9x + 10 = 0
II. 3y² + 11y + 6 = 0
I. x2 - 17x + 70 = 0
II. y2 - 11y + 28 = 0