A car dealership carries only sedans and SUVs

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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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.

B.
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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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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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Question says If no new inventory arrived at any point on Tuesday -- couldn't new inventory have arrived on Wednesday making B insufficient?

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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Can someone please tell me where am i headed wrong?


Sent from my iPhone using GMAT Club Forum mobile app

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.

From question stem, we can infer the no of sedan will be 6k. statement 1 says, for 9 SUV, 8 Sedan were sold. So, the lowest no of Sedan sold can be 8 and in that case, 48 sedans were there in the beginning, but we don't know how many SUVs were there, so insufficient.

Statement 2 tells that, the lowest number of sedan sold on Tuesday was = 20 *8.5 = 17 as otherwise it won't be an integer. So we can say there were at least 17* 6 = 102 sedans in the inventory. No matter how many SUVs were in inventory at the beginning, this information is sufficient.

B is the answer.

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?

Firstly, note that on Tuesday, the dealership sold 1/6 of the sedans it had, so the number of sedans must be a multiple of 6.

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

This implies that:

(sedans sold on Tuesday) = 8/9*(SUV's sold on Tuesday);
(sedans sold on Tuesday)/(SUV's sold on Tuesday) = 8/9.

Hence, the number of sedans sold on Tuesday must be a multiple of 8. As it must also be a multiple of 6, it must be a multiple of the LCM of these numbers, which is 24. If the number of sedans sold on Tuesday was 24, and the number of SUVs sold on Tuesday was 27, then the total number of cars sold on Tuesday was 51, and the dealership might have had fewer than 100 cars at the beginning of the day Tuesday. However, if the number of sedans sold on Tuesday was 48, and the number of SUVs sold on Tuesday was 54, then the total number of cars sold on Tuesday was 102, and the dealership definitely had more than 100 cars at the beginning of the day Tuesday. Not sufficient.

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

This implies that:

(sedans sold on Tuesday) = 0.85*(SUV's sold on Wednesday);
(sedans sold on Tuesday) = 17/20*(SUV's sold on Wednesday);
(sedans sold on Tuesday)/(SUV's sold on Wednesday) = 17/20.

Hence, the number of sedans sold on Tuesday must be a multiple of 17. Since it also must be a multiple of 6, then it must be a multiple of the LCM of these numbers, which is 102. Since the dealership sold more than 100 sedans on Tuesday, it must have had more than 100 cars at the beginning of the day Tuesday. Sufficient.

Answer: B.

Hope it's clear.
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What a quant problem!

Interesting question.
let's get to it

lets assume number of sedans = x
number of SUVs = y
total cars = x+y
and we know that number of sedans sold = 1/6x

statement 1:
number of sedan sold = 8/9 of suv's sold
so 1/6x = 8/9y
the minimum value y can take is 9 (because you cannot sell fraction of any car, so y has to be multiple of 9)
if y=9, x=48, total cars = 57
if y = 36, x =192, total = 228

so we cant say whether total cars were more than 100

statement 2:
number of sedans sold on tuesday = 85% of that sold on wednesday
lets assume number of sedans sold on wednesday = W
1/6x=85/100w
so, 1/6x = 17/20w
least value w can take is 20, so x has to be 102 atleast.
hence inventory on tuesday was definitely greater than 100.

ans = b

Bunuel SVaidyaraman

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.



Can anyone please explain how you arrived at 20 for the number of sedan sold on wednesday?

Can you show it by setting up equations and explain even every mini step to show how you arrive at 20 cars sold on wednesday and atleast 102 Sedans sold on Tuesday?

Thanks in advance!

Bunuel

Thank you very much for your Clear explaination of this complex problem. Because of your detailed explaination, this question suddenly seems like a L15% question.

Again, Thank you!

­a good and tricky question.