Directions (58-60): Read the information carefully and answer the following question given below. A right circular cylindrical vessel was filled with milk and water in the ratio 4: 5. If x ml of mixture was sold and replaced with x ml of pure milk , then the ratio of milk to water in the final mixture becomes 57:60.The quantity of mixture sold is 108 ml less than the initial quantity of mixture in the vessel.

A right circular cylindrical vessel was filled with milk and water in the ratio 4: 5. If x ml of mixture was sold and replaced with x ml of pure milk , then the ratio of milk to water in the final mixture becomes 57:

If the radius of vessel is 0.7 cm and 37 cm³ of the vessel is empty, then find the height of the vessel.

To find the height of the cylindrical vessel, we can break down the problem into two parts: determining the volume of the mixture in the vessel, and then using the cylinder's volume formula to find the height.

Step 1: Determine the initial quantity of the mixture
Let the initial quantity of the mixture in the vessel be $V$ ml.
The initial ratio of milk to water is $4:5$.
Let the initial quantity of milk be $4y$ and water be $5y$.
So, the total initial volume of the mixture is:

$V=4y+5y=9y$

When $x$ ml of the mixture is sold (removed):
Milk removed = $\frac{4}{9}x$
Water removed = $\frac{5}{9}x$

After replacing the sold mixture with $x$ ml of pure milk, the new quantities are:
New quantity of milk = $4y-\frac{4}{9}x+x=4y+\frac{5}{9}x$
New quantity of water = $5y-\frac{5}{9}x$

The final ratio of milk to water is given as $57:60$, which simplifies to $19:20$ by dividing both terms by 3.
Therefore, we can set up the ratio equation:

$\frac{4y+\frac{5}{9}x}{5y-\frac{5}{9}x}$

Cross-multiplying to solve for $x$ and $y$:

$20\left(4y+\frac{5}{9}x\right)=19\left(5y-\frac{5}{9}x\right)$

$80y+\frac{100}{9}x=95y-\frac{95}{9}x$

$\frac{100}{9}x+\frac{95}{9}x=95y-80y$

$\frac{195}{9}x=15y$

Simplifying the fraction $\frac{195}{9}$ by dividing the numerator and denominator by 3 gives $\frac{65}{3}$:

$\frac{65}{3}x=15y$

$65x=45y$

Dividing both sides by 5:

$13x=9y$

Since the initial volume of the mixture is $V=9y$, we have:

$V=13x$

We are given that the quantity of mixture sold ($x$) is 108 ml less than the initial quantity of mixture ($V$):

$V-x=108$

Substituting $V=13x$ into this equation:

$13x-x=108$

$12x=108$

$x=9\text{ml}$

Thus, the volume of the mixture in the vessel is:

$V=13\times 9=117{\text{cm}}^3$

Step 2: Find the height of the cylindrical vessel
The total volume capacity of the vessel is equal to the volume of the mixture plus the empty space:

$V_{\text{total}}=117{\text{cm}}^3+37{\text{cm}}^3=154{\text{cm}}^3$

The formula for the volume of a right circular cylinder is:

$V_{\text{total}}=\pi r^{2}h$

Given that the radius $r=0.7\text{cm}$ and taking $\pi \approx \frac{22}{7}$:

$154=\frac{22}{7}\times {\left(0.7\right)}^2\times h$

$154=\frac{22}{7}\times 0.49\times h$

$154=22\times 0.07\times h$

$154=1.54\times h$

$h=\frac{154}{1.54}=100\text{cm}$

Thus, the height of the vessel is 100 cm.