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Jerry has a large car which holds 22 gallons of fuel and gets 20 miles per gallon. Kate has a smaller car which holds 16.5 gallons of fuel and gets 30 miles per gallon. If both cars have a full tank of fuel now and drive the same distance, in how many miles will the remaining fuel in each tank be the same?
Choose 1 answer:
(A) 320
(B) 325
(c) 330
(D) 335

Jerry has a large car which holds 2222 gallons of fuel and gets 2020 miles per gallon. Kate has a smaller car which holds 1616.55 gallons of fuel and gets 3030 miles per gallon. If both cars have a full tank of fuel now and drive the same distance, in how many miles will the remaining fuel in each tank be the same?\newlineChoose 11 answer:\newline(A) 320320\newline(B) 325325\newline(C) 330330\newline(D) 335335

Full solution

Q. Jerry has a large car which holds 2222 gallons of fuel and gets 2020 miles per gallon. Kate has a smaller car which holds 1616.55 gallons of fuel and gets 3030 miles per gallon. If both cars have a full tank of fuel now and drive the same distance, in how many miles will the remaining fuel in each tank be the same?\newlineChoose 11 answer:\newline(A) 320320\newline(B) 325325\newline(C) 330330\newline(D) 335335
  1. Set up an equation based on the fuel consumption: Let xx be the number of miles driven.\newlineFor Jerry's car: \newlineFuel consumption rate: 2020 miles per gallon \newline Initial fuel: 2222 gallons \newlineFor Kate's car: \newlineFuel consumption rate: 3030 miles per gallon \newline Initial fuel: 16.516.5 gallons \newlineSetting up the equation: \newline 22x20=16.5x3022 - \frac{x}{20} = 16.5 - \frac{x}{30}
  2. Isolating the term xx: 22x20=16.5x3022 - \frac{x}{20} = 16.5 - \frac{x}{30} \newlineAdd x20\frac{x}{20} to both sides of the equation: \newline 22x20+x20=16.5x30+x2022 - \frac{x}{20} + \frac{x}{20} = 16.5 - \frac{x}{30} + \frac{x}{20} \newline 22=16.5x30+x2022 = 16.5 - \frac{x}{30} + \frac{x}{20} \newlineSubtract 16.516.5 from both sides: \newline 2216.5=16.5x30+x2016.522 - 16.5 = 16.5 - \frac{x}{30} + \frac{x}{20} - 16.5 \newline 5.5=x20x305.5 = \frac{x}{20} - \frac{x}{30}
  3. Simplify the equation: To simplify, let's find a common denominator, which is 6060: \newline5.5=3x602x605.5 = \frac{3x}{60} - \frac{2x}{60} \newline5.5=x605.5 = \frac{x}{60}
  4. Solving for the value of xx: To solve for xx multiply both sides of the equation by 6060:\newline5.5×60=x60×605.5 \times 60 = \frac{x}{60} \times 60\newline x=330x = 330 \newlineThe remaining fuel in each tank will be the same in 330330 miles.

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