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In a certain parking lot, the number of blue cars is \(\frac{2}{3}\) the number of red cars, and the number of yellow cars is \(\frac{1}{4}\) greater than the number of blue cars. If there are 14 more red cars than yellow cars, how many blue cars are in the parking lot?

A) 12

B) 28

C) 42

D) 56

E) 70

Blue = (2/3)Red

Yellow = Blue + (1/4)Blue = (5/4)Blue = (5/4)(2/3)Red = (5/6)Red

Yellow is 1/6th of Red less than Red. Yellow is 14 less than Red which means number of Yellow is 5*14 = 70

Blue = (4/5)Yellow = (4/5)*70 = 56

Answer (D)
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Quote:
In a certain parking lot, the number of blue cars is \(\frac{2}{3}\) the number of red cars, and the number of yellow cars is \(\frac{1}{4}\) greater than the number of blue cars. If there are 14 more red cars than yellow cars, how many blue cars are in the parking lot?

A) 12

B) 28

C) 42

D) 56

E) 70
Let the number of red cars = the LCM of the two denominators = 12

Since there are 2/3 as many blue cars as red cars, the number of blue cars \(= \frac{2}{3} * 12 = 8\)
Since there are 1/4 more yellow cars than blue cars, the number of yellow cars \(= 8 + \frac{1}{4}(8) = 10\)
In this case, red - yellow = 12-10 = 2
Since the actual difference must be 14 -- seven times as great -- all of the values above must increase by a factor of 7, with the result that blue = 7*8 = 56

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