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On a 130130 mile trip, a car traveled an average speed of \newline5555mph and then reduced its speed to 4040 mph for the remainder of the trip. The trip took 2.52.5 hours. For how long did the car travel at \newline4040mph ?

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Q. On a 130130 mile trip, a car traveled an average speed of \newline5555mph and then reduced its speed to 4040 mph for the remainder of the trip. The trip took 2.52.5 hours. For how long did the car travel at \newline4040mph ?
  1. Set Up Total Time Equation: Let t1 t_1 be the time the car traveled at 5555 mph and t2 t_2 be the time at 4040 mph. The total time is given as 22.55 hours. Set up the equation for total time:\newlinet1+t2=2.5 t_1 + t_2 = 2.5
  2. Set Up Distance Equations: Use the distance formula D=R×T D = R \times T for each part of the trip. The total distance is 130130 miles. Set up the equations for distances:\newline55t1+40t2=130 55t_1 + 40t_2 = 130
  3. Solve System of Equations: Solve the system of equations:\newline11. t1+t2=2.5 t_1 + t_2 = 2.5 \newline22. 55t1+40t2=130 55t_1 + 40t_2 = 130 \newlineMultiply the first equation by 4040:\newline40t1+40t2=100 40t_1 + 40t_2 = 100 \newlineSubtract this from the second equation:\newline55t1+40t2(40t1+40t2)=130100 55t_1 + 40t_2 - (40t_1 + 40t_2) = 130 - 100 \newline15t1=30 15t_1 = 30 \newlinet1=2 t_1 = 2
  4. Substitute and Solve: Substitute t1=2 t_1 = 2 back into the total time equation:\newline2+t2=2.5 2 + t_2 = 2.5 \newlinet2=0.5 t_2 = 0.5

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