Maths Olympiad Prep

Library / /174 of 520

Algebra Difficulty 2.6 Junior Find the answer

Angelina drove at an average rate of 8080 kmh and then stopped 2020 minutes for gas. After the stop, she drove at an average rate of 100100 kmh. Altogether she drove 250250 km in a total trip time of 33 hours including the stop. Which equation could be used to solve for the time tt in hours that she drove before her stop?
(A) 80t+100(83t)=250\textbf{(A)}\ 80t+100\left(\frac{8}{3}-t\right)=250(B) 80t=250\textbf{(B)}\ 80t=250(C) 100t=250\textbf{(C)}\ 100t=250(D) 90t=250\textbf{(D)}\ 90t=250(E) 80(83t)+100t=250\textbf{(E)}\ 80\left(\frac{8}{3}-t\right)+100t=250

Multiple choice: answer with the letter of the option you want.

Solution

The answer is AA because she drove at 8080 kmh for tt hours (the amount of time before the stop), and 100100 kmh for 83t\frac{8}{3}-t because she wasn't driving for 2020 minutes, or 13\frac{1}{3} hours. Multiplying by tt gives the total distance, which is 250250 km. Therefore, the answer is 80t+100(83t)=25080t+100\left(\frac{8}{3}-t\right)=250 \Rightarrow (A)\boxed{(A)}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.