Suppose a tap mixes hot water and cold water in a ratio that depends linearly on the proportion of opening. Water out of the tap has temperature 40°C when the tap is half-open, and 30°C when it is three-fourths open. To get water at 50°C, the tap should be_______________.
Tap Temperature Linear Relationship Calculation
The problem states that the water temperature ($T$) from the tap depends linearly on the proportion of the tap opening ($x$). This can be represented by the equation:
$T = mx + c$
where $m$ is the slope and $c$ is the y-intercept.
Determining the Linear Equation
We are given two points:
- When the tap is half-open ($x = \frac{1}{2}$), the temperature is $T = 40$°C. $40 = m \left( \frac{1}{2} \right) + c \quad (1)$
- When the tap is three-fourths open ($x = \frac{3}{4}$), the temperature is $T = 30$°C. $30 = m \left( \frac{3}{4} \right) + c \quad (2)$
Subtract equation (1) from equation (2) to find $m$:
$ (30 - 40) = \left( \frac{3}{4}m + c \right) - \left( \frac{1}{2}m + c \right) $
$ -10 = \frac{3}{4}m - \frac{1}{2}m $
$ -10 = \frac{1}{4}m $
$ m = -40 $
Substitute $m = -40$ into equation (1) to find $c$:
$ 40 = (-40) \left( \frac{1}{2} \right) + c $
$ 40 = -20 + c $
$ c = 60 $
The linear equation is:
$ T = -40x + 60 $
Calculating Tap Opening for 50°C
We need to find the tap opening ($x$) when the temperature ($T$) is 50°C:
$ 50 = -40x + 60 $
$ 40x = 60 - 50 $
$ 40x = 10 $
$ x = \frac{10}{40} = \frac{1}{4} $
Therefore, the tap should be $\frac{1}{4}$ (one-fourth) open to get water at 50°C.
Important Questions from Mixture Problems (Notes)
- In 80 litres mixture of milk and water, the ratio of amount of milk to that of amount of water is 7 : 3. In order to make this ratio 2 : 1, how many litres of water should be added ?
- A $595$ litre of mixture contains milk and water in the ratio $17:18$. How much milk must be added to the mixture so that it contains milk and water in the proportion of $3:2$?
- How many liters of water should be added to a 30-liter mixture containing milk and water in the ratio 7:3 such that the resultant mixture has 40% water in it?
Two solutions $X$ and $Y$ containing ingredients $A, B$ and $C$, in proportions $a:b:c$ and $c:b:a$, respectively, are mixed. For the resultant mixture to have $A, B$ and $C$ in $1:1:1$ proportion, it is necessary that $a:b:c$ is
- A 360 ml aqueous solution contains 40% alcohol. How much will be the approximate percentage of alcohol if 3600 ml of water is added to the solution?
