The charge c of a telephone company is partly constant and partly varies as the number of units of call u.The cost of 50 units is #2500 and the cost of 120 units is #3000.Find:
a.the formula connecting c and u.
b.u when c =#4000
c = mu + b
so, using the two given data points, you just need to solve
50m+b = 2500
120m+b = 3000
Now, having m and b, you can find u when c=4000
Or, even without knowing m and b, note that m grew by 70 when c grew by 500. So, if c grows another 1000, m will grow another 140.
So, u=260 when c=4000.
To find the formula connecting c and u, we need to determine the constant part of the charge and the variable part.
Let's assume the constant part of the charge is x and the variable part is y.
Based on the given information:
1. When 50 units are used, the cost is #2500:
c = x + 50y
2500 = x + 50y
2. When 120 units are used, the cost is #3000:
c = x + 120y
3000 = x + 120y
Now, we have a system of equations. We can solve these equations to find the values of x and y.
Let's multiply the first equation by 24 and the second equation by 5 to eliminate the variable x:
(24)(2500) = (24)(x + 50y)
60000 = 24x + 1200y
(5)(3000) = (5)(x + 120y)
15000 = 5x + 600y
Now, subtract the second equation from the first:
60000 - 15000 = 24x + 1200y - (5x + 600y)
45000 = 19x + 600y
To simplify the equation further, let's divide everything by 50:
900 = 19x + 12y
We now have a new equation after eliminating x.
From this new equation, we can see that the constant part of the charge (x) = 900, and the variable part (y) is 12.
a. The formula connecting c and u is: c = 900 + 12u.
b. To find u when c = #4000, we substitute c = 4000 into the formula:
4000 = 900 + 12u
Now, solve for u:
12u = 4000 - 900
12u = 3100
u = 3100/12
u ≈ 258.33 (rounded to two decimal places)
Therefore, when c = #4000, the number of units of call (u) is approximately 258.33.
To find the formula connecting the charge (c) and the number of units of call (u), we need to determine the constant part of the charge and the rate at which the charge varies per unit.
Let's assume the constant part of the charge is "a" and the rate of variation is "b".
According to the given information:
For 50 units, the cost (c) is #2500. Therefore, we can write the equation:
c = a + 50b .......(Equation 1)
And for 120 units, the cost (c) is #3000. We can write the equation:
c = a + 120b .......(Equation 2)
Now, we can solve these two equations simultaneously to find the values of "a" and "b".
Subtracting Equation 1 from Equation 2, we get:
(a + 120b) - (a + 50b) = #3000 - #2500
70b = #500
b = #500 / 70 = #7.14 (approx.)
Substituting the value of "b" back into Equation 1, we can find the value of "a":
c = a + 50b
#2500 = a + 50 * #7.14
a + 357 = #2500
a = #2500 - 357 = #2143
Therefore, the formula connecting c and u is:
c = #2143 + 7.14u
To find the value of u when c = #4000, substitute c = #4000 into the formula:
#4000 = #2143 + 7.14u
Rearranging the equation to solve for u, we have:
7.14u = #4000 - #2143
7.14u = #1857
u = #1857 / 7.14
Calculating the value of u, we get:
u ≈ 260.07
So, when the charge (c) is #4000, the number of units of call (u) is approximately 260.