Question:
A 480 L mixture of milk and water in the ratio 5: 3 If 'X' L of the mixture is replaced by water and again '(X - 24)' L of the mixture is replaced by water. If the amount of water in the final mixture is 300 L, then find the value of 'X + 15'.
A 480 L mixture of milk and water in the ratio 5: 3 If 'X' L of the mixture is replaced by water and again '(X - 24)' L of the mixture is replaced by water. If the amount of water in the final mixture is 300 L, then find the value of 'X + 15'.
Updated On: Sep 10, 2024
- 125
- 120
- 135
- 110
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The Correct Option is C
Solution and Explanation
The correct option is (C): 135.
Milk = 480*\(\frac{5}{8}\) = 300 L
Water = 480 - 300 = 180 L
Amount of milk in the mixture when 'X' L is replaced by water = 300 - X *\(\frac{5}{8}\) = (300 - \(\frac{5x}{8}\))
Amount of water in the mixture when 'X' L is replaced by water = 180 - X* \(\frac{3}{8}\) + X = (180 + \(\frac{5x}{8}\))
Amount of water in the mixture when 'X - 24' L is replaced by water =(180 + \(\frac{5x}{8}\)) - (X - 24) * \(\frac{(180 + \frac{5X}{8})}{480} \)+ (X - 24) = 300
If X = 125 - 15 = 110
The amount of water in the final mixture = (180 + 5 * \(\frac{110}{8}\)) - (110 -24) * \(\frac{(180 + 5 * \frac{110}{8})}{480}\) + (110 - 24)
= \(\frac{248.75 - (21392.5)}{480} \)+ 86 [not satisfied]
If X = 120 - 15 = 105
The amount of water in the final mixture = (180 + 5 * \(\frac{105}{8}\)) - (105 -24) *\(\frac{(180 + 5 * \frac{105}{8})}{480}\) + (105 - 24)
= 245.625 - 41.44 + 81 [Not satisfied]
If X = 135 - 15 = 120
The amount of water in the final mixture = (180 + 5 * \(\frac{120}{8}\)) - (120 -24) *\(\frac{(180 + 5 * \frac{120}{8})}{480}\)+ (120 - 24)
= 255 - 51 + 96
= 300 [satisfied].
Milk = 480*\(\frac{5}{8}\) = 300 L
Water = 480 - 300 = 180 L
Amount of milk in the mixture when 'X' L is replaced by water = 300 - X *\(\frac{5}{8}\) = (300 - \(\frac{5x}{8}\))
Amount of water in the mixture when 'X' L is replaced by water = 180 - X* \(\frac{3}{8}\) + X = (180 + \(\frac{5x}{8}\))
Amount of water in the mixture when 'X - 24' L is replaced by water =(180 + \(\frac{5x}{8}\)) - (X - 24) * \(\frac{(180 + \frac{5X}{8})}{480} \)+ (X - 24) = 300
If X = 125 - 15 = 110
The amount of water in the final mixture = (180 + 5 * \(\frac{110}{8}\)) - (110 -24) * \(\frac{(180 + 5 * \frac{110}{8})}{480}\) + (110 - 24)
= \(\frac{248.75 - (21392.5)}{480} \)+ 86 [not satisfied]
If X = 120 - 15 = 105
The amount of water in the final mixture = (180 + 5 * \(\frac{105}{8}\)) - (105 -24) *\(\frac{(180 + 5 * \frac{105}{8})}{480}\) + (105 - 24)
= 245.625 - 41.44 + 81 [Not satisfied]
If X = 135 - 15 = 120
The amount of water in the final mixture = (180 + 5 * \(\frac{120}{8}\)) - (120 -24) *\(\frac{(180 + 5 * \frac{120}{8})}{480}\)+ (120 - 24)
= 255 - 51 + 96
= 300 [satisfied].
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