1. The milk and water in two vessels P and Q are in the ratio 5:2 and 2:3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel R... 1. The milk and water in two vessels P and Q are in the ratio 5:2 and 2:3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel R containing milk and half water? 2. How many liters of water should be added to a 30 liters mixture of milk and water in the ratio of 7:3, such that the resultant mixture has 60% milk in it?
Understand the Problem
This question has two parts. The first part asks about mixing liquids from two vessels (P and Q) with ratios 5:2 and 2:3, respectively, to create a new mixture in vessel R with equal parts milk and water. The goal is to find the ratio in which the liquids from P and Q should be mixed. The second part asks to find how many liters of water need to be added to a 30-liter mixture (milk and water in a 7:3 ratio) to make the resultant mixture contain 60% milk.
Answer
$7:15$, $5$
Answer for screen readers
The ratio in which the liquids in both vessels be mixed to obtain a new mixture in vessel R containing milk and half water $7:15$. The amount of water should be added to a 30 liters mixture of milk and water in the ratio of 7:3, such that the resultant mixture has 60% milk in it is 5 liters.
Steps to Solve
- Define variables for the first part
Let the ratio in which liquids from vessels P and Q are mixed be $x:y$. Milk to water ratio in P is $5:2$, so the fraction of milk in P is $\frac{5}{7}$ and the fraction of water is $\frac{2}{7}$. Milk to water ratio in Q is $2:3$, so the fraction of milk in Q is $\frac{2}{5}$ and the fraction of water is $\frac{3}{5}$. In vessel R, the milk and water are in equal proportions, so milk fraction is $\frac{1}{2}$ and water fraction is $\frac{1}{2}$.
- Set up an equation for the milk fraction in Vessel R
The milk fraction in the mixture is given by: $$ \frac{x \cdot (\frac{5}{7}) + y \cdot (\frac{2}{5})}{x + y} = \frac{1}{2} $$
- Solve the equation
Multiply both sides by $2(x+y)$ and by $35$ to get rid of fractions, we will have:
$2(\frac{5}{7}x + \frac{2}{5}y) = x+y$ $2(\frac{25x + 14y}{35}) = x+y$ $2(25x + 14y) = 35(x+y)$ $50x + 28y = 35x + 35y$ $15x = 7y$ $\frac{x}{y} = \frac{7}{15}$
So, the liquids from P and Q should be mixed in the ratio $7:15$.
- Solve the second part
Let $w$ be the liters of water added to the mixture. The original mixture is 30 liters with a milk to water ratio of $7:3$. The amount of milk in the original mixture is $\frac{7}{10} \times 30 = 21$ liters. The amount of water in the original mixture is $\frac{3}{10} \times 30 = 9$ liters.
- Set up an equation for the second part
After adding $w$ liters of water, the total volume of the mixture is $30 + w$ liters. The amount of milk remains at 21 liters. We want the new mixture to be 60% milk, so: $\frac{21}{30 + w} = 0.60$
- Solve for $w$ (the liters of water)
$21 = 0.6(30 + w)$ $21 = 18 + 0.6w$ $3 = 0.6w$ $w = \frac{3}{0.6} = 5$
Therefore, 5 liters of water should be added.
The ratio in which the liquids in both vessels be mixed to obtain a new mixture in vessel R containing milk and half water $7:15$. The amount of water should be added to a 30 liters mixture of milk and water in the ratio of 7:3, such that the resultant mixture has 60% milk in it is 5 liters.
More Information
Finding the ratio in which two mixtures must be mixed to get a desired ratio involves using the concept of alligation or mixture problems. For the second part of the question calculating the quantity to be added to get a mixture of desired concentration is again an application of ratios and mixtures.
Tips
- For part 1, a common mistake is to add the ratios directly without considering the fractions of milk and water in each vessel.
- For part 2, a common mistake is to calculate the milk and water amounts incorrectly from the given ratios.
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