# Math Help - World Problem with Rational Expressions

1. ## World Problem with Rational Expressions

I know what the answer is (1400) but I can't figure out how to express the steps. Here is the question:

1. Last year, the total bill for yearbooks at the school was $109 200. When the bill arrived, 100 students had graduated and the remaining students each had to pay an additional$6.00 to cover the cost. How many students ordered yearbooks last year?

Thanks.

I know what the answer is (1400) but I can't figure out how to express the steps. Here is the question:

1. Last year, the total bill for yearbooks at the school was $109 200. When the bill arrived, 100 students had graduated and the remaining students each had to pay an additional$6.00 to cover the cost. How many students ordered yearbooks last year?

Thanks.
Are we to assume that the 100 students who graduated were among those who ordered yearbooks? But that, because they graduated before the yearbooks arrived they did not pay for them? If not, I see no way to do this problem.

Assuming those things, let the number of students who ordered year books be "x". If they all paid for year books, they would have to pay $\frac{109200}{x}$ dollars each. Because 100 of then are not paying for the yearbooks, there are x- 100 who are and each will have to pay $\frac{109200}{x- 100}$. We are told that that second cost is "an additional $6". That is, we have $\frac{109200}{x- 100}= \frac{109200}{x}+ 6$. That is the equation you need to solve. I recommend multiplying both sides by x(x-100). 3. Hello, needhelpplease1! Another set-up . . . 1. Last year, the total bill for yearbooks at the school was$109,200.
When the bill arrived, 100 students had graduated and the remaining students
had to pay an additional $6.00 each to cover the cost. How many students ordered yearbooks last year? Let: . $\begin{Bmatrix}N &=& \text{no. of students last year} \\ x &=& \text{cost per student last year} \end{Bmatrix}$ Originally, $N$ students would pay $x$ dollars each ... totalling$109,200.

. . $Nx \:=\:109,200$ .[1]

Now there are only $N-100$ students who will pay $x+6$ dollars each.

. . $(N-100)(x+6) \;=\;109,\!200$ .[2]

From [2], we have: . $Nx + 6N - 100x - 600 \:=\:109,\!200 \quad\Rightarrow\quad Nx + 6N - 100x \:=\:109,\!800$

Substitute [1]: . $109,200 + 6N - 100x \:=\:109,\!800 \quad\Rightarrow\quad 6N - 100x \:=\:600$

Solve for $x\!:\;\;x \:=\:\frac{6N - 600}{100} \:=\:\frac{3N-300}{50}$

Substitute into [1]: . $N\left(\frac{3N-300}{50}\right) \:=\:109,\!200 \quad\Rightarrow\quad 3N^2 - 300N \:=\:5,\!460,\!000$

We have: . $N^2 - 100N - 1,\!820,\!000 \:=\:0 \quad\Rightarrow\quad (N - 1400)(N + 1300) \:=\:0$

. . Hence: . $N \;=\;1400,\:-1300$

Therefore, 1400 students ordered yearbooks last year.