SAT Mathematics Practice Test 6

Let n be the number of friends originally in the group. Since the cost of the trip was 800 dollars, the share, in dollars, for each friend was originally $\frac{800}{\mathit{n}}$.When two friends decided not to go on the trip, the number of friends who split the 800 dollars cost became $\mathit{n}-2$, and each friend’s cost became $\frac{800}{\mathit{n}-2}$ . Since this share represented a 20 dollars increase over the original share, the equation $\frac{800}{\mathit{n}}+20=\frac{800}{\mathit{n}-2}$ must be true. Multiplying each side of $\frac{800}{\mathit{n}}+20=\frac{800}{\mathit{n}-2}$ by$\mathit{n}\left(\mathit{n}-2\right)$ to clear all the denominators gives.$800\left(\mathit{n}-2\right)+20\mathit{n}\left(\mathit{n}-2\right)=80\mathit{n}$This is a quadratic equation and can be rewritten in the standard form by expanding,simplifying, and then collecting like terms on one side, as shown below:$800\mathit{n}+1600+20{\mathit{n}}^2-40\mathit{n}=800\mathit{n}$
$40\mathit{n}-80+{\mathit{n}}^2-2\mathit{n}=40\mathit{n}$
${\mathit{n}}^2-2\mathit{n}-80=0$After factoring, this becomes $\left(\mathit{n}+8\right)\left(\mathit{n}-10\right)=0.$
The solutions of this equation are −8 and 10. Since a negative solution makes no sense for thenumber of people in a group, the number of friends originally in the group was 10.