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Let, rate of interest be ‘r’% per annum So, 645 = 2000 × {(1 + r/100)2 – 1} 129/400 = {(1 + r/100)2 – 1} 529/400 = (1 + r/100)2 (23/20)2 = (1 + r/100)2 23/20 = 1 + r/100 3/20 = r/100, r = 15 So, simple interest earned = 2300 × 0.15 × 2 = Rs. 690 Required amount = 690 – 645 = Rs. 45
The equation x2 – px – 60 = 0, has two roots ‘a’ and ‘b’ such that (a – b) = 17 and p > 0. If a series starts with ‘p’ such...
I. 4 x ² - 4 x + 1 = 0
II. 4 y ² + 4 y + 1 = 0
...I. 2y² - 11 y + 15 = 0
II. 2x² + 3x – 14 = 0
I. 104x² + 9x - 35 = 0
II. 72y² - 85y + 25 = 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. 66x² - 49x + 9 = 0
II. 46y² - 37y - 30 = 0
I. 3x2 - 14x + 15 = 0
II. 15y2- 34 y + 15 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 21x² - 82x + 80 = 0
Equation 2: 23y² - 132y + 85 = 0
In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer
I. x² - 8x + 15 = 0 ...
I. x2 – 18x + 81 = 0
II. y2 – 3y - 28 = 0