An examination fee is partly constant and partly varies directly as the number of subject entered. When the examination fee is 55000 three subjects are entered when that examination fee is 700, five subjects are entered. Write a law of variation and find the number of subjects if the examination fee is 1000
Write a law connecting fee and subject entered
White a law connecting fee and subject entered
To solve this problem, we need to find a law of variation for the examination fee.
Let's denote:
- x as the number of subjects entered
- y as the examination fee
We are given two data points:
1) When the examination fee is 55000, three subjects are entered. This can be represented as (55000, 3).
2) When the examination fee is 700, five subjects are entered. This can be represented as (700, 5).
We can use these two points to find the law of variation.
The law of variation can be written as:
y = kx
Where k is the constant of variation.
Using the first data point (55000, 3), we can substitute the values into the equation to get:
55000 = k * 3
Solving for k:
k = 55000/3
k ≈ 18333.33
Now we can write the equation for the law of variation:
y = 18333.33 * x
To find the number of subjects if the examination fee is 1000, we can substitute y = 1000 into the equation and solve for x:
1000 = 18333.33 * x
Dividing both sides by 18333.33:
x = 1000/18333.33
x ≈ 0.0546
Therefore, if the examination fee is 1000, approximately 0.0546 subjects are entered.
To write the law of variation, let's break down the information given. We know that the examination fee consists of a constant part and a variable part that is directly proportional to the number of subjects entered.
Let:
x = number of subjects entered
C = constant part of the examination fee
k = constant of proportionality (variable part of the examination fee)
As per the given information, when three subjects are entered, the examination fee is 55000, and when five subjects are entered, the fee is 700. Using this, we can set up two equations:
1) When three subjects are entered:
C + 3k = 55000
2) When five subjects are entered:
C + 5k = 700
From these equations, we can solve for C and k.
Multiplying both sides of equation 1 by 5 gives:
5C + 15k = 275000
Multiplying both sides of equation 2 by 3 gives:
3C + 15k = 2100
Subtracting equation 2 from equation 1 eliminates k:
5C - 3C + 15k - 15k = 275000 - 2100
2C = 273900
Dividing both sides by 2, we find:
C = 136950
Now we can substitute the value of C back into either equation to solve for k.
Using equation 1:
136950 + 3k = 55000
Rearranging the equation:
3k = 55000 - 136950 = -81950
Dividing both sides by 3, we find:
k = -27316.67 (approx.)
Now that we have the value of k, we can use the law of variation to find the number of subjects (x) when the examination fee is 1000.
C + kx = examination fee
Substituting the known values:
136950 - 27316.67x = 1000
Rearranging the equation:
27316.67x = 135950
Dividing both sides by 27316.67, we find:
x ≈ 4.98
Therefore, when the examination fee is 1000, the number of subjects entered would be approximately 4.98. Since the number of subjects must be a whole number, we can round up to the nearest whole number. Hence, the approximate number of subjects would be 5.