How do I solve this rate problem?

1 Answer

Answer :

answer:The trick to this problem is realizing that you really have TWO simultaneous rate / time problems. What we know is that Alex drove 220 miles at one rate and 80 miles at another rate (15 mph slower) for two different lengths of time that together added up to 6 hours. We can express this with our two rates R and r and times T and t, where R and T represent the 220 mile leg, and r and t represent the 80 mile leg: R * T + r * t = 300 R * T = 220 r * t = 80 Also: Since T + t = 6, then t = 6 – T and R = r + 15 Substituting for t: r * (6 – T) = 80 6r – Tr = 80 Substituting for r: 6 * (R – 15) – T * (R – 15) = 80 And completing the operations: 6R – 80 – TR – 15T = 80 Then substituting for R (since R = 220 / T): 6 * (220 / T) – 80 – T * (220 / T) – 15T = 80 And completing the operations: 1320 / T – 80 – 220 – 15T = 80 1320 / T – 380 – 15T = 0 Then multiplying both sides by T: 1320 – 380T – 15T^2 = 0 Multiplying both sides by -1 and rearranging terms: 15T^2 + 380T – 1320 = 0 And simplifying: 3T^2 + 76T – 264 = 0 T^2 + 26T – 88 = 0 Solving for the roots: T + 22 = 0 (irrational: no one could drive for -22 hours) or T – 4 = 0 The time at the higher speed was 4 hours. The rest is easy to solve. 220 miles in 4 hours = 55 mph 80 miles in 2 hours = 40 mph

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