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C = 1.4S
H = 1.15C => H = 1.15 * 1.4S

H = 1.61S => H is 61% greater than S

Answer D.
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1. The problem asks us to find what percent is Harry’s gas mileage greater than that of Stan’s.

2. We can mark Chris's, Harry's, and Stan's gas mileages as C, H, and S, respectively.

3. Remember that "x% higher than y" is the same as (1 + \(\frac{x}{100}\))y.

4. From the problem two equations can be derived: C = 1.4S and H = 1.15C. Since we are comparing H to S, both equations can be turned into H = 1.15 \(*\) 1.4S = 1.61S = (1 + \(\frac{61}{100}\))S. That means our answer is 61%.
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Let :
c = Chris's car's gas mileage
s = Stan'scar's gas mileage
h = Harry's car's gas mileage
we have c=1.4s and h=1.15c so h=1.15*1.4s=1.61s
=> +61%
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