The charge C of a telephone company is partly constant and partly varies as the number of unit of call u.the cost of 50 unit is #2500 and the cost of 120 units is #3000. Find : the formular connecting C and u
C = mu + b
when u = 50, C = 2500
50u + b = 2500
when u = 120, C = 3000
120u + b = 3000
subtract the two equations:
70u = 500
u = 50/7
in 50u + b = 2500
2500/7 + b = 2500
b = 15000/7
Cost = (50/7)u + 15000/7
I didn't get it
To find the formula connecting C and u, let's first identify the constant part and the variable part of the charge.
Let's assume that the constant part of the charge is represented by "a" and the variable part is represented by "b."
Given that the cost of 50 units is #2500, we can form the equation:
50u = a + b ✦ Equation 1
Similarly, with the cost of 120 units being #3000, we can form another equation:
120u = a + b ✦ Equation 2
Now, let's solve these equations to find the values of "a" and "b."
Subtracting Equation 1 from Equation 2, we get:
(120u - 50u) = (a + b) - (a + b)
70u = 0
This implies that u = 0, which is not possible. So, there might be a mistake in the given information.
Please recheck the cost values for 50 units and 120 units to proceed further with finding the formula connecting C and u.
To find the formula connecting C (the charge) and u (the number of call units), we can start by setting up a system of equations using the given information.
Let's denote the constant part of the charge as a (independent of u), and the variable part of the charge as b (dependent on u).
From the given information, we have two data points:
1) When the number of call units u is 50, the cost C is #2500.
C = a + b * u
#2500 = a + b * 50 --- (Equation 1)
2) When the number of call units u is 120, the cost C is #3000.
C = a + b * u
#3000 = a + b * 120 --- (Equation 2)
Now, we have a system of two linear equations with two variables (a and b). We can solve this system of equations to find the values of a and b.
To do this, we can use a method called substitution or elimination. Let's use substitution:
Solve Equation 1 for a:
a = #2500 - b * 50
Substitute this value into Equation 2:
#3000 = (#2500 - b * 50) + b * 120
#3000 = #2500 - 50b + 120b
#3000 = #2500 + 70b
70b = #3000 - #2500
70b = #500
b = #500 / 70
b = #7.14 (rounded to two decimal places)
Now, substitute the value of b back into Equation 1 to find a:
#2500 = a + #7.14 * 50
#2500 = a + #357.14
a = #2500 - #357.14
a = #2142.86 (rounded to two decimal places)
Therefore, the formula connecting C and u is:
C = #2142.86 + (#7.14 * u)
So, the charge C for a given number of call units u can be calculated using this formula.