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Re: Of the vehicles in a parking garage, 54 are diesel, [#permalink]
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prototypevenom wrote:
sandy wrote:
Hi,

Let number of non diesel cars be \(x\).

Number of diesel cars= \(54\)

Total number of cars are = \(54 +x\)

Given that 31 of the cars non diesel type are blue so \(x \geq 31\).

Thus minimum number of cars= \(54+31\)= \(85\).

Let number of non diesel cars that are blue = \(y\).

Total number of blue cars = \(31 + y\). Blue cars =\(42\).

So even if total number of cars 180. (54 Diesel 126 Non diesel) no constraints are being violated.

Hence every option Greater than 85 is correct.

Please share the source of the question, because this is particularly peculiar as a GRE question.


Let number of non diesel cars that are blue = \(y\). (isnt it 31?)

Total number of blue cars = \(31 + y\). Blue cars =\(42\).

What ? shouldn't total of blue car be 31+y
non blue cars=42

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Re: Of the vehicles in a parking garage, 54 are diesel, [#permalink]
2aam27, sandy, the correct answer is C. However, all answer choices work except A when I plug into table above. How to check that only C is the correct option?
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Re: Of the vehicles in a parking garage, 54 are diesel, [#permalink]
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tkorzhan18 wrote:
2aam27, sandy, the correct answer is C. However, all answer choices work except A when I plug into table above. How to check that only C is the correct option?


The diagram provided by Albs perfectly describes the situation.

The correct answers are C, D, E, F and G.

I have edited the official answer accordingly.
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Re: Of the vehicles in a parking garage, 54 are diesel, [#permalink]
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