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

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Intern
Joined: 17 Nov 2013
Posts: 39
A parking garage has spaces for cars and vans only.  [#permalink]

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18 Sep 2018, 03:41
00:00

Difficulty:

25% (medium)

Question Stats:

78% (01:18) correct 22% (01:07) wrong based on 72 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
Intern
Joined: 26 Aug 2018
Posts: 7
Re: A parking garage has spaces for cars and vans only.  [#permalink]

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18 Sep 2018, 04:59
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.

Intern
Joined: 17 Nov 2013
Posts: 39
A parking garage has spaces for cars and vans only.  [#permalink]

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18 Sep 2018, 06:24
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.

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?
Intern
Joined: 06 Oct 2017
Posts: 38
GPA: 3.6
Re: A parking garage has spaces for cars and vans only.  [#permalink]

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18 Sep 2018, 18:03
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.

For some reason Im missing the reasoning behind this part:

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

Am I missing something obvious?
Intern
Joined: 20 Jun 2018
Posts: 40
Re: A parking garage has spaces for cars and vans only.  [#permalink]

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19 Sep 2018, 00:53
I chose the answer D. Can someone explain how to go about Statement 2?
Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 678
Re: A parking garage has spaces for cars and vans only.  [#permalink]

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17 Nov 2018, 22:26
Let's say there are V vans and C cars already in the garage. We need to find V.

(1)Sufficient as we are given some more info about V & C other than V + C= 118 from the question stem. Hence we have two equations and two unknowns and can solve for V.

(2)Insufficient no additional info is given as x and y are unknowns and does not reveal anything.

Note we don't need to find out what is V but just realize that we can do so.

Best,
harish1986 wrote:
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

_________________

Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)

Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 678
Re: A parking garage has spaces for cars and vans only.  [#permalink]

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17 Nov 2018, 22:29
Hi Mike03,

Statement two gives us info about the relationship between Vans and Cars, however, it uses more unknowns x and y. Hence we cannot determine a unique number for Vans.

In DS, we can say a statement is sufficient only if it gives a unique answer to the original question asked in the Q-stem.

Hope this helps.

Best,

Mike03 wrote:
I chose the answer D. Can someone explain how to go about Statement 2?

_________________

Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)

DS Forum Moderator
Joined: 21 Aug 2013
Posts: 1435
Location: India
Re: A parking garage has spaces for cars and vans only.  [#permalink]

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18 Nov 2018, 04:06
1
aanjumz92 wrote:
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.

For some reason Im missing the reasoning behind this part:

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

Am I missing something obvious?

Hello

The statement says that if 14 more vans are driven, there will be twice as many vans as cars.
Now if 14 more vans turn up, total no of vehicles will be 118+14 or 132.

and if earlier the no of vans was x, now the no of vans will be x+14, and since this number is twice the number of cars, the no of cars will be (x+14)/2.
Now no of vans x+14, no of cars (x+14)/2, and this total should be equal to 118+14.

Hope this is clear now.
DS Forum Moderator
Joined: 21 Aug 2013
Posts: 1435
Location: India
Re: A parking garage has spaces for cars and vans only.  [#permalink]

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18 Nov 2018, 04:09
harish1986 wrote:
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.

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?

Hello

The second statement does NOT say that equal number of cars and vans are driven into the garage. It says 'x' vans and 'y' cars are driven.

Also the new no of vehicles will NOT be 118, but now it will be (118+x+y)
Re: A parking garage has spaces for cars and vans only. &nbs [#permalink] 18 Nov 2018, 04:09
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