Question
A 164-liter mixture consists of milk and water in a
ratio of 17:24. The milkman sells 82 liters of this mixture and then adds 123 liters of water to the remaining portion. What percentage of milk in the original mixture is compared to the percentage of milk in the final mixture?Solution
Quantity of milk in 164 litre mixture = 17/(17 + 24) x 164= 17/41 x 164= 17 x 4= 68 litresQuantity of water in 164 litre mixture = 164 - 68 = 96 litresPercentage of milk in the initial mixture = 68/164 x 100 %Quantity of milk in 82 litre mixture = 17/41 x 82 = 34 litresQuantity of water in 82 litre mixture = 82 - 34 = 48 litresQuantity of milk in the final mixture = 68 - 34 = 34 litresQuantity of water in the final mixture = 96 - 48 + 123 = 171 litresPercentage of milk in the final mixture = 34/(34 + 171) x 100= 34/205 x 100 %Required percentage = [68/164 x 100]/[ 34/205 x 100] = 250 %
I. 15y2 + 4y – 4 = 0
II. 15x2 + x – 6 = 0
I. 27(p + 2) = 2p(24 – p)
II. 2q2 – 25q + 78 = 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 - 32x + 252 = 0
Equation 2: y² - 30y + 221 = 0
If α, β are the roots of the equation x² – px + q = 0, then the value of α2+β2+2αβ isÂ
...I. 7x² + 27x + 18 = 0
II. 19y² - 27y + 8 = 0
I). 4p + 3q = 25
II). 5p + 2q = 26Â
I. 8x – 3y = 85
II. 4x – 5y = 67
I. 2y2 + 13y + 15 = 0
II. 2x2 + 11 x + 12 = 0
In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer Â
I. x2 – ...