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A parking garage has spaces for cars and vans only.

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A parking garage has spaces for cars and vans only.  [#permalink]

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New post 18 Sep 2018, 04:41
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Difficulty:

  25% (medium)

Question Stats:

79% (01:29) correct 21% (00:58) wrong based on 42 sessions

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A parking garage has spaces for cars and vans only. If there is a total of 118 vehicles in the garage, how many of them are vans?

(1) If 14 more vans are driven into the garage, there will be twice as many vans as cars in the garage.

(2) If x vans and y cars are driven into the garage, there will be an equal number of vans and cars in the garage.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT SUFFICIENT
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Re: A parking garage has spaces for cars and vans only.  [#permalink]

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New post 18 Sep 2018, 05:59
Being x the number of vans in the garage and y the number of cars in the garage:
x + y = 118

(1) If 14 more vans are driven into the garage, there will be twice as many vans as cars in the garage.

y = (x+14)/2, thus

(x+14)+(x+14)/2 = 118+14

(2x+28)+(x+14) = 264

3x+42 = 264

x = 222/3 = 74

Initially 74 vans. Sufficient.

(2) If x vans and y cars are driven into the garage, there will be an equal number of vans and cars in the garage.

We can't find relation between additional x and y, so 2 unknown factors. Not sufficient.

Answer A
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A parking garage has spaces for cars and vans only.  [#permalink]

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New post 18 Sep 2018, 07:24
iamzyw wrote:
Being x the number of vans in the garage and y the number of cars in the garage:
x + y = 118

(1) If 14 more vans are driven into the garage, there will be twice as many vans as cars in the garage.

y = (x+14)/2, thus

(x+14)+(x+14)/2 = 118+14

(2x+28)+(x+14) = 264

3x+42 = 264

x = 222/3 = 74

Initially 74 vans. Sufficient.

(2) If x vans and y cars are driven into the garage, there will be an equal number of vans and cars in the garage.

We can't find a relation between additional x and y, so 2 unknown factors. Not sufficient.

Answer A


I don't understand option B. If an equal number of cars and vans are driven into the garage, why should not we do 118/2? 54 cars + 54 vans? What am i misinterpreting?
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Re: A parking garage has spaces for cars and vans only.  [#permalink]

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New post 18 Sep 2018, 19:03
iamzyw wrote:
Being x the number of vans in the garage and y the number of cars in the garage:
x + y = 118

(1) If 14 more vans are driven into the garage, there will be twice as many vans as cars in the garage.

y = (x+14)/2, thus

(x+14)+(x+14)/2 = 118+14

(2x+28)+(x+14) = 264

3x+42 = 264

x = 222/3 = 74

Initially 74 vans. Sufficient.

(2) If x vans and y cars are driven into the garage, there will be an equal number of vans and cars in the garage.

We can't find relation between additional x and y, so 2 unknown factors. Not sufficient.

Answer A



For some reason Im missing the reasoning behind this part:


(x+14)+(x+14)/2 = 118+14

Am I missing something obvious?
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Re: A parking garage has spaces for cars and vans only.  [#permalink]

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New post 19 Sep 2018, 01:53
I chose the answer D. Can someone explain how to go about Statement 2?
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Re: A parking garage has spaces for cars and vans only. &nbs [#permalink] 19 Sep 2018, 01:53
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A parking garage has spaces for cars and vans only.

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