P is partly constant and varies partly as Q. If P = 32 when Q = 16 and P = 20 when Q = 12, find P when Q = 28

Answer Details

The statement "P is partly constant and varies partly as Q" means that P can be written as the sum of a constant part and a part that is directly proportional to Q. In algebra, this is expressed as:

\[ P = a + bQ \]

where \( a \) is the constant part and \( bQ \) represents the part that varies directly as \( Q \).

We know that:

• When \( Q = 16 \), \( P = 32 \)
• When \( Q = 12 \), \( P = 20 \)

Let's use these values to form two equations:

• \( 32 = a + b \times 16 \)
• \( 20 = a + b \times 12 \)

Subtract the second equation from the first to eliminate \( a \):

\[ 32 - 20 = (a + 16b) - (a + 12b) \] \[ 12 = 4b \] \[ b = 3 \]

Now, substitute \( b = 3 \) back into one of the original equations, such as:

\[ 20 = a + 12 \times 3 \] \[ 20 = a + 36 \] \[ a = 20 - 36 = -16 \]

So the equation relating \( P \) and \( Q \) is:

\[ P = -16 + 3Q \]

Now, to find \( P \) when \( Q = 28 \):

\[ P = -16 + 3 \times 28 \] \[ P = -16 + 84 \] \[ P = 68 \]

The correct answer is 68.

Why this is correct:
This method works because when a variable is "partly constant and partly varies as" another, you always represent it as a sum of a constant and a variable part. By using two given values for \( P \) and \( Q \), you can solve for both the constant and variable components, then apply these to find unknown values of \( P \) for any value of \( Q \).