Examination fee is partly constant and varies directly to the number of subject entered. When the examination fee is 550 three subjects are entered. When the examination fee is 700 five subjects are entered. Write the law of variation and find the number of subject entered if the examination fee is 1000
Let x be the constant part of the examination fee and y be the number of subjects entered.
From the problem, we know that:
When y = 3, x + 3k = 550 (where k is the constant of proportionality)
When y = 5, x + 5k = 700
We can solve for x and k by subtracting the first equation from the second:
2k = 150
k = 75
Substituting k back into the first equation:
x + 3(75) = 550
x = 325
So the law of variation is:
examination fee = 325 + 75(number of subjects entered)
To find the number of subjects entered if the examination fee is 1000, we can solve for y:
1000 = 325 + 75y
675 = 75y
y = 9
Therefore, if the examination fee is 1000, the number of subjects entered is 9.
To find the law of variation and the number of subjects entered if the examination fee is 1000, we can analyze the given information:
From the information provided, we know that the examination fee is partly constant and varies directly with the number of subjects entered. This indicates a direct variation relationship.
Let's assign variables to the given information:
- Let "x" represent the number of subjects entered.
- Let "y" represent the examination fee.
Using the given information, we can set up two equations to represent the relationship:
Equation 1: When the examination fee is 550 and three subjects are entered.
y = kx
Substituting the values into the equation:
550 = k * 3
Equation 2: When the examination fee is 700 and five subjects are entered.
700 = k * 5
Now, we have a system of equations to solve.
To find the value of "k," we can solve the first equation for "k":
550 = k * 3
Divide both sides by 3:
550/3 = k
k = 550/3 ≈ 183.33
Now, we can use this value of "k" in the second equation to find the number of subjects entered when the examination fee is 1000:
1000 = k * x
Substituting the value of "k" we found:
1000 = 183.33 * x
To isolate "x," divide both sides by 183.33:
x = 1000/183.33 ≈ 5.46
Rounding to the nearest whole number, the number of subjects entered, if the examination fee is 1000, would be 5 subjects.
Therefore, the law of variation is y = 183.33x, and approximately 5 subjects would be entered if the examination fee is 1000.
Let's denote the examination fee as F and the number of subjects entered as N.
From the given information:
When F = 550, N = 3, and when F = 700, N = 5.
We can use these values to write the law of variation.
Since the examination fee varies directly with the number of subjects entered, we can write this relationship as:
F = kN
where k is the constant of variation.
To find the value of k, we can plug in the values from either of the given scenarios:
550 = k * 3
Solving this equation, we can find k:
k = 550 / 3
k = 183.33 (rounded to two decimal places)
Now that we have the value of k, we can use it to find the number of subjects entered when the examination fee is 1000.
F = 1000
k = 183.33
Substituting these values into the formula:
1000 = 183.33 * N
Rearranging the equation to solve for N:
N = 1000 / 183.33
N ≈ 5.46 (rounded to two decimal places)
Therefore, if the examination fee is 1000, the number of subjects entered would be approximately 5.46. Since the number of subjects should be a whole number, we can round it to the nearest whole number.
Therefore, if the examination fee is 1000, the number of subjects entered would be approximately 5.