P is partly constant and partly varies as Q, when Q is 5, P is 20 and when Q is 8, P is 26. Find P when Q is 4.
p = mq+b
Using the two data points, you have
5m+b = 20
8m+b = 26
clearly, 3m=6, so now just finish it off
Why did the chicken cross the road? To find out what P is when Q is 4, of course!
Let's solve this riddle with a bit of math. We know that P is partly constant and partly varies as Q. So, we can assume there's a constant term and a variable term in the equation for P.
Let's call the constant term "C" and the variable term "V." Then, we can say that P = C + V.
Given that P is 20 when Q is 5, we can substitute these values into our equation:
20 = C + 5V
Similarly, when P is 26 and Q is 8:
26 = C + 8V
Now, let's subtract the first equation from the second equation to eliminate C:
26 - 20 = C + 8V - (C + 5V)
Simplifying this, we get:
6 = 3V
Dividing both sides by 3, we find:
2 = V
Now, substitute the value of V back into the first equation to find C:
20 = C + 5(2)
20 = C + 10
C = 10
Now that we have found C and V, we can find P when Q is 4 by substituting these values into our original equation:
P = C + V
P = 10 + 2
P = 12
So, when Q is 4, P is 12. Ta-da!
To find the value of P when Q is 4, we can first determine the constant part, and then use that to find the variable part.
Let's find the constant part (C) first. We can use the formula:
C = P - (Q * V) (where P is the value of P, Q is the value of Q, and V is the variable part)
From the given information, when Q is 5, P is 20. So we can substitute these values into the formula:
20 = C - (5 * V) (equation 1)
Next, when Q is 8, P is 26. Using the same formula:
26 = C - (8 * V) (equation 2)
Now we have a system of equations (equation 1 and 2) with two variables (C and V). We can solve for C using these equations.
Subtract equation 2 from equation 1:
20 - 26 = C - (5 * V) - (C - (8 * V))
-6 = C - 5V + (8V - C)
-6 = 3V
Divide both sides by 3:
-6/3 = 3V/3
-2 = V
Now substitute the value of V (which is -2) into one of the equations to solve for C. Let's use equation 1:
20 = C - (5 * (-2))
20 = C + 10
Subtract 10 from both sides:
20 - 10 = C + 10 - 10
10 = C
So, we have found that the value of C is 10 and the value of V is -2.
Now, to find P when Q is 4, we can use the formula:
P = C + (Q * V)
Substituting the values we found:
P = 10 + (4 * -2)
P = 10 - 8
P = 2
Hence, when Q is 4, P is 2.
To solve this problem, we can use the given information to set up a linear equation and then solve it for the unknown value of P when Q is 4.
Let's first find the constant component of P. We can assume that P consists of two parts: a constant term, let's call it 'a,' and a variable term, which is 'Q multiplied by some coefficient,' let's call it 'b.'
So, the equation representing the relationship between P and Q can be written as:
P = a + bQ
Now, let's substitute the given values into the equation to form two equations:
When Q is 5: P = 20
20 = a + 5b ---(Equation 1)
When Q is 8: P = 26
26 = a + 8b ---(Equation 2)
We have two equations with two unknowns (a and b). We can solve them using any method, such as substitution or elimination. Let's use the substitution method.
First, solve Equation 1 for 'a':
a = 20 - 5b
Now, substitute this value of 'a' into Equation 2:
26 = (20 - 5b) + 8b
Simplifying the equation:
26 = 20 + 3b
Move 20 to the right side:
6 = 3b
Divide both sides by 3:
b = 2
Next, substitute the value of 'b' back into Equation 1 to solve for 'a':
a = 20 - 5(2)
a = 20 - 10
a = 10
So, we have found that the constant term, a, is 10, and the coefficient of Q, b, is 2.
Now, we can substitute Q = 4 into the equation we formed initially to find P:
P = 10 + 2(4)
P = 10 + 8
P = 18
Therefore, when Q is 4, P is 18.