A quantity P is partly constant and partly varies as the inverse of Q².When Q=2,P=49 1/4 and when Q=3 P=49 2/3 .Determine the value of P when Q=5

1 answer

Let's assume that the constant part of P is represented by the variable "a".
Thus, P can be written as:

P = a + (k/Q²)

where "k" is the variable part of P that varies inversely with Q².

Using the given information, when Q = 2, P = 49 1/4, which can be written as P = 49.25.

Plugging these values into the equation, we get:

49.25 = a + (k/2²)
49.25 = a + k/4
49.25 = 4a + k

Similarly, when Q = 3, P = 49 2/3, which can be written as P = 49.67.

Again, plugging these values into the equation:

49.67 = a + (k/3²)
49.67 = a + k/9
49.67 = 9a + k

Now, we have a system of two equations:

49.25 = 4a + k (equation 1)
49.67 = 9a + k (equation 2)

We can solve this system of equations to find the values of "a" and "k".

Subtracting equation 1 from equation 2, we get:

49.67 - 49.25 = 9a - 4a + k - k
0.42 = 5a
a = 0.42/5
a = 0.084

Substituting the value of "a" into equation 1, we get:

49.25 = 4(0.084) + k
49.25 = 0.336 + k
49.25 - 0.336 = k
48.914 = k

Now, we have the values of "a" and "k":

a = 0.084
k = 48.914

Finally, let's substitute these values back into the equation to find the value of P when Q = 5:

P = a + (k/Q²)
P = 0.084 + (48.914/5²)
P = 0.084 + 48.914/25
P = 0.084 + 1.95656
P = 2.04056

Therefore, when Q = 5, P ≈ 2.04056.