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A car dealership carries only sedans and SUVs

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A car dealership carries only sedans and SUVs  [#permalink]

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New post 06 Sep 2013, 08:19
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A car dealership carries only sedans and SUVs, and on Tuesday it sold 1/6 of the sedans that it had in stock at the beginning of the day. If no new inventory arrived at any point on Tuesday, and the only change in inventory was that some vehicles were sold, did the dealership have more than 100 vehicles in inventory at the beginning of the day Tuesday?

(1) By the end of the day, the dealership had sold 8/9 as many sedans as SUVs.

(2) The dealership sold 85% as many sedans on Tuesday as it did on Wednesday.

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Re: A car dealership carries only sedans and SUVs  [#permalink]

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New post 06 Sep 2013, 19:08
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Practicegmat wrote:
A car dealership carries only sedans and SUVs, and on Tuesday it sold 1/6 of the sedans that it had in stock at the beginning of the day. If no new inventory arrived at any point on Tuesday, and the only change in inventory was that some vehicles were sold, did the dealership have more than 100 vehicles in inventory at the beginning of the day Tuesday?

(1) By the end of the day, the dealership had sold 8/9 as many sedans as SUVs.

(2) The dealership sold 85% as many sedans on Tuesday as it did on Wednesday.


To find , Is sedans(inv on Tue) + suvs(inv on Tue) >100

Given : sedans(sold on Tue) = sedans(inv on Tue) /6

Statement 1:

sedans(sold on Tue)=suvs(sold on Tue) * 8/9

We cannot find either sedans(inv on Tue) or suvs(inv on Tue) based on the statement and what is given. Not sufficient

Statement 2:

sedans(sold on wed) * 85/100 = sedans(sold on Tue)

From the above we can say that the minimum possible number of sedans (sold on Wed) was 20 because that is the minimum value when sedans(sold on wed) * 85/100 will be an integer and consequently the minimum sedans(sold on Tue) = 17. Therefore sedans(inv on Tue) = sedans(sold on Tue) *6= 102. Hence sufficient to answer the question.
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Re: A car dealership carries only sedans and SUVs  [#permalink]

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New post 06 Sep 2013, 12:07
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Practicegmat wrote:
A car dealership carries only sedans and SUVs, and on Tuesday it sold 1/6 of the sedans that it had in stock at the beginning of the day. If no new inventory arrived at any point on Tuesday, and the only change in inventory was that some vehicles were sold, did the dealership have more than 100 vehicles in inventory at the beginning of the day Tuesday?

(1) By the end of the day, the dealership had sold 8/9 as many sedans as SUVs.

(2) The dealership sold 85% as many sedans on Tuesday as it did on Wednesday.


From F.S 1, as the # of cars sold can only assume integral values, we know that the # of SUV's sold at the end of the day were of the form 9k(k is a a positive integer).
Thus, the # of sedans sold at the end of the day = \(\frac{8}{9}*9k\) = 8k = This represents \(\frac{1}{6^{th}}\)of the total sedans present at the start of the day. Thus, the # of sedans at the start of the day = 8k*6 = 48k. The no of SUV's sold can be minimum 9(for k=1) and the # of sedans present at the start of the day can be minimum(48). Still, no information about the actual # of SUV's present. Insufficient.

From F.S 2, Let the # of sedans sold on Wednesday be S. We know that # of Sedans sold on Tuesday = 85% of S = \(\frac{17}{20}*S\) = The minimum value can be when S=20, i.e. the # of sedans sold on Tuesday = 17. Also, as this is \(\frac{1}{6^{th}}\) of the total stock of sedans at the start of the day, the total no of sedans at the start = 17*6 = 102>100(Irrespective of the no of SUV's sold). Sufficient.

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Re: A car dealership carries only sedans and SUVs  [#permalink]

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New post 07 Sep 2013, 06:14
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My way is a bit different

A is not sufficient
B As we know that only the integer number of sedans can be sold on any day.

Number of sedans sold on wednesday =85% of that sold on tuesday=85%*1/6(sold on tuesday)*m(no of sedans on tuesday)=k(some integer)

=> m=120/17*k => k must be a multiple of 17 => let k be 17 then m=120
therefore even in this bare minimum case we have 120 sedans on tuesday which implies that total number of cars will be more than 100.
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Re: A car dealership carries only sedans and SUVs  [#permalink]

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New post 21 Aug 2016, 07:26
Practicegmat wrote:
A car dealership carries only sedans and SUVs, and on Tuesday it sold 1/6 of the sedans that it had in stock at the beginning of the day. If no new inventory arrived at any point on Tuesday, and the only change in inventory was that some vehicles were sold, did the dealership have more than 100 vehicles in inventory at the beginning of the day Tuesday?

(1) By the end of the day, the dealership had sold 8/9 as many sedans as SUVs.

(2) The dealership sold 85% as many sedans on Tuesday as it did on Wednesday.


What a quant problem!
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Re: A car dealership carries only sedans and SUVs  [#permalink]

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New post 09 Mar 2017, 19:06
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Practicegmat wrote:
A car dealership carries only sedans and SUVs, and on Tuesday it sold 1/6 of the sedans that it had in stock at the beginning of the day. If no new inventory arrived at any point on Tuesday, and the only change in inventory was that some vehicles were sold, did the dealership have more than 100 vehicles in inventory at the beginning of the day Tuesday?

(1) By the end of the day, the dealership had sold 8/9 as many sedans as SUVs.

(2) The dealership sold 85% as many sedans on Tuesday as it did on Wednesday.


Official solution from Veritas Prep.

The question doesn't put any restriction on the number of SUVs not sold, so you can make this number as big as you want to prove the answer "yes".

Statement 2, however, is sufficient. While it may not look so at first, it combines with the given information to guarantee that the dealership had a minimum of 102 sedans. Knowing that:

-The number of sedans sold was 17/20 (85%) of the number sold on another day tells us that the number of sedans sold must be a multiple of 17.

-The number of sedans in stock is equal to 6(the number sold) tells us that the number of sedans in stock must be 17 x a multiple of 6. The lowest such number is 102, guaranteeing that at least 102 vehicles were in stock at the beginning of the day.

Strategically, this problem is a good opportunity to "Leverage Assets" and ask "Why Are You Here?". The fact that statement 2 talks about Wednesday when everything else in the problem talks only about Tuesday might seem like a total throwaway, but remember: the GMAT rarely, if ever, completely throws away a DS statement. If a statement seems irrelevant upon first glance, it's a good opportunity for you to ask why it's there:

-Could it be a clue that helps you better understand the other statement? -Could it be a trap, a piece of information that you thought you might need but you actually don't? -Could it be, as it is in this case, a cleverly-worded piece of information that you can make useful by Leveraging Assets - manipulating algebra, applying a number property or divisibility rule, etc.?
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Re: A car dealership carries only sedans and SUVs  [#permalink]

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New post 18 Apr 2017, 21:58
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Question says If no new inventory arrived at any point on Tuesday -- couldn't new inventory have arrived on Wednesday making B insufficient?
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A car dealership carries only sedans and SUVs  [#permalink]

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New post 09 Aug 2017, 21:57
Given:x/6 sedans sold by the end of Tuesday where x is the inventory of sedans at the beginning of the day.

Statement 1: Suv sold on Tuesday = 9/8*x/6 = 3x/16. Total vehicles sold = x/6+3x/16= 17x/48. We cannot answer from statement (1) as x can be 48 or a multiple of 48 and no information about suvs

Statement 2: Sedans sold on Wednesday is (100/85)*x/6. x has to be at least 102, otherwise the number of sedans sold on Wednesday will not be an integer. So B is sufficient.
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Re: A car dealership carries only sedans and SUVs  [#permalink]

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New post 09 Apr 2018, 11:46
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Can someone please tell me where am i headed wrong?


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Re: A car dealership carries only sedans and SUVs  [#permalink]

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New post 13 Sep 2018, 21:09
Assume total number of sedans at the beginning of Tuesday as x and number of SUV's as y.
Sold on Tuesday (Sedans) - x/6
Stock at the end of Tuesday (Sedans) - x-x/6 = 5x/6.
SUV's we have no information. We also no new vehicles added.
Question: Is x+y>100?
St. 1 - Suppose SUVs sold on Tue is K. We have, x/6(Sedans sold on Tue) = 8/9*K (SUVs sold on Tuesday). Minimum value of K could be 9, i.e K>=9.
x/6>=8/9*9 which implies, x>=48. x could be less than 100, equal to 100 or more than 100. Not sufficient.
St.2 - We have x/6 (Sedans sold on Tue) = 85/100*(SUVs sold on Wed), 17/20*(SUVs sold on Wed)
Now, Minimum value of SUVs sold on Wed is 20.
Therefore, x/6>=17/20*20 implies to x>=102. Sufficient.

Answer B.
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Re: A car dealership carries only sedans and SUVs &nbs [#permalink] 13 Sep 2018, 21:09
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