Members of a college chemistry department agree to contribute equal amounts of money to make up a scholarship fund of $280. Then the department hires three new members, resulting in each members share being reduced by $30. What is the new size of the department?

7 members

Step-by-step explanation:

Lets say that the department has n members. If each member contributes with an equal amount x and make a total of $280, we know that:

n*x = 280

Where each one colaborated with:

n*x/n = 280/n

Or what is the same that: (as the n on the left eliminate themselves)

x = 280/n (*)

Now the department hires 3 more members, so we now have n+3 members. And we know that the pass from contributing x to contribute $30 less, it is, x-30 (I omit the $ symbol for simplicity). So, know we have:

x-30 = 280 / (n+3)

We can replace x here by the formulation of x we did in the (*) equation:

(280/n) - 30 = 280 / (n+3)

using common denominator:

(280/n) - 30 n/n = 280 / (n+3)

(280 - 30 n) / n = 280 / (n+3)

Cross multiplying the denominators:

(280 - 30 n) * (n+3) = 280 * (n)

280n - 30n^2 + 840 - 90n = 280n

Subtracting 280 n in both sides:

30n^2 + 840 - 90n = 0

Dividing both sides by 30:

-n^2 - 3n + 28 = 0

Factorizing this term (you can notice the factor immediately or solving with baskara resolution, I omit the baskara for simplicity):

- (n + 7) (n - 4) = 0

So, n is equal to 4 and - 7. As we are dealing with people we can not have - 7 people, so we keep only n=4.

So, previous the new hiring there were 4 members, now there are 7.

The used to collaborate with 280/4 = $70, now the collaborate with 280/7=$40, and we see the collaboration decreased in $30.