Question
A milkman initially prepared a mixture of milk and water
in the ratio of 11:3. Later, he replaced half of the mixture with 30 liters of water, resulting in a new mixture where the milk content became 55%. Determine the original total quantity of the mixture.Solution
Let the original quantity of milk and water in the mixture be '11x' litres and '3x' litres, respectively After replacing half the mixture with 30 litres of water, Quantity of milk remaining in the mixture = 11x ÷ 2 = '5.5x' litres Quantity of water remaining in the mixture = 3x ÷ 2 + 30 = (1.5x + 30) litres According to the question, 5.5x/(1.5x + 30 + 5.5x) = 55/100 Or, 550x = 385x + 1650 So, 165x = 1650 So, x = 1650 ÷ 165 = 10 So, original quantity of the mixture = 11x + 3x = 14x = 14 X 10 = 140 litres
I. 4p² + 17p + 15 = 0
II. 3q² + 19q + 28 = 0
I:Â x2Â - 33x + 242 = 0
II:Â y2Â - 4y - 77 = 0
I. x² - 208 = 233
II. y² + 47 - 371 = 0
I. 2b2 - 37b + 143 = 0
II. 2a2 + 15a - 143 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 45x + 450 = 0
Equation 2: y² - 48y + 540 = 0�...
I). p2 + 22p + 72 = 0,
II). q2 - 24q + 128 = 0
I. √(74x-250 )– x=15
II. √(3y²-37y+18)+ 2y=18
I. 3y² - 20y + 25 = 0
II. 3x² - 8x + 5 = 0
I. 2y2 - 15y + 18 = 0
II. 2x2 + 9x - 18 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 103x² - 470x + 367 = 0
Equation 2: 107y² - 504y + 397 = 0
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