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Of the vehicles in a parking garage, 54 are diesel,
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Updated on: 14 Jul 2022, 06:38
Updated on: 14 Jul 2022, 06:38
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Question Stats:
37% (01:48) correct
62% (02:53) wrong based on 24 sessions
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Of the vehicles in a parking garage, 54 are diesel, 42 are not blue, and 31 of the non-diesels are blue. Which of the following could be the number of vehicles in the parking garage?
Select all that apply!!
(A)60
(B)80
(C)100
(D)120
(E)140
(F)160
(G)180
Select all that apply!!
(A)60
(B)80
(C)100
(D)120
(E)140
(F)160
(G)180
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Official Answer
C,D,E,F,G
Originally posted by 2aam27 on 27 Mar 2017, 10:52.
Last edited by GreenlightTestPrep on 14 Jul 2022, 06:38, edited 2 times in total.
Last edited by GreenlightTestPrep on 14 Jul 2022, 06:38, edited 2 times in total.
UPDATED
Re: Of the vehicles in a parking garage, 54 are diesel,
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27 Mar 2017, 12:20
27 Mar 2017, 12:20
Expert Reply
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 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.
Re: Of the vehicles in a parking garage, 54 are diesel,
[#permalink]
25 Feb 2020, 21:55
25 Feb 2020, 21:55
1
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 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
