Question
A seller marked the price of an item at Rs. 6,000. The
seller gave successive discounts of (c + 6)% and (c - 6)% to a customer. If the customer paid Rs. 4,848 for the item, then find the value of 'c'.Solution
ATQ; SP of the item = 6000 × (1 - (c + 6)/100) × (1 - (c - 6)/100) = 4848 or, {(100 - c - 6)/100} × {(100 - c + 6)/100} = 4848/6000 or, (94 - c) × (106 - c) = (4848/6000) × 100 × 100 or, 94 × 106 - 94c - 106c + c² = 8092 or, c² - 200c + 9964 - 8092 = 0 or, c² - 200c + 1872 = 0 or, c² - 10c - 190c + 1872 = 0 or, c × (c - 10) - 187 × (c - 10) = 32 or, (c - 10) × (c - 187) = 32 So, c = 10 or c = 187 But, c = 187 is not possible because the discount cannot be more than 100%. Therefore, c = 10
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 3x2 + 6√7x - 315 = 0    Â
E...
I). p2 + 8p + 15 = 0
II). q2 + 9q + 20 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 30x + 221 = 0
Equation 2: y² - 28y + 189 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...
I. p2 - 19p + 88 = 0  Â
II. q2Â - 48q + 576 = 0
I. 2x2 + 13x + 21 = 0
II. 3y2 + 34y + 63 = 0
I. 2y² - 11 y + 15 = 0
II. 2x² + 3x – 14 = 0
What will be the product of smaller roots of both equations.Â
(i) x² – 3x – 40 = 0
(ii) y² + 11y + 30 = 0
I. 99 x² + 31 x – 110 = 0
II. 6y² - 31y + 35 = 0
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