A car dealership carries only sedans and SUVs

Question

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.

(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.

Bunuel · Accepted 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?

(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.

Firstly, note that on Tuesday, the dealership sold 1/6 of the sedans it had, so the number of sedans at the beginning of Tuesday 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*(SUVs sold on Tuesday);
(sedans sold on Tuesday)/(SUVs sold on Tuesday) = 8/9.

Hence, the number of sedans sold on Tuesday must be a multiple of 8.

Now, since the dealership sold 1/6 of the sedans it had at the beginning of Tuesday, the number of sedans at the beginning of Tuesday equals 6*(sedans sold on Tuesday). Therefore, if sedans sold on Tuesday is a multiple of 8, then the number of sedans at the beginning of Tuesday must be a multiple of 48.

If the number of sedans sold on Tuesday was 8, and the number of SUVs sold on Tuesday was 9, then the number of sedans at the beginning of Tuesday was 48, and the dealership could have had fewer than 100 vehicles at the beginning of the day Tuesday. However, if the number of sedans sold on Tuesday was 16, and the number of SUVs sold on Tuesday was 18, then the number of sedans at the beginning of Tuesday was 96, and the dealership must have had more than 100 vehicles at the beginning of the day Tuesday (since it also had at least 18 SUVs at the beginning of 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*(sedans sold on Wednesday);
(sedans sold on Tuesday) = 17/20*(sedans sold on Wednesday);
(sedans sold on Tuesday)/(sedans sold on Wednesday) = 17/20.

Hence, the number of sedans sold on Tuesday must be a multiple of 17. Since the number of sedans at the beginning of Tuesday equals 6*(sedans sold on Tuesday), the number of sedans at the beginning of Tuesday must be a multiple of 6 * 17 = 102. Since the number of sedans at the beginning of Tuesday is positive, it must be at least 102, so the dealership definitely had more than 100 vehicles at the beginning of the day Tuesday. Sufficient.

Answer: B.

Hope it's clear.

mau5 · Answer

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.

SVaidyaraman · Answer

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.

ramannanda9 · Answer

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.

BillyZ · Answer

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.?

SVaidyaraman · Answer

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.

Rebaz · Answer

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!

Rebaz · Answer

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!

karanbadalia · Answer

a good and tricky question.

mollyweasley · Answer

KarishmaB   Bunuel   MartyMurray   gmatophobia , very basic question -
1. In statement 2, the equation formed is x/6=85%w , where w is the sales on wednesday.
2. The next step would be to reduce 85%, which would make the equation x/6=17w/20
3. This is where my question lies - is it wrong to cross multiply 6 with the RHS and make the equation x=17w*6/20? Because then it becomes x=51w/10; and the least value of x would be 51.
Would really appreciate a clarification on this, thanks!

KarishmaB · Answer

Nothing wrong in your algebra but you have not considered the constraint that x/6 is an integer. If x = 51, then it is not.

Solve it instead like this:

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

Assume number of sedans at the beginning of Tuesday was 6x and at the end 5x so x sedans were sold. Now x can be any positive integer.

Statement 2 says that
 x = \frac{85}{100} * W  which means
 x = \frac{17}{20} * W

Since  x is an integer, it must be a multiple of 17 which means 6x must be at least 102.

Answer (B)

rish_dutton · Answer

I will break it down such that you can explain this to a high-school student.

You just gotta keep one thing in your mind always:

Cars, in the GMAT world, cannot be sold in fractions.

Never seen half a car on the road unless there is something seriously wrong.

Let's go step by step. Let's start with the question.

On Tuesday morning, they have S sedans, U SUVs. No new stock is coming in. So the max number of cars at the end of the day could be S*(5/6) + U if no SUVs are sold.

On Tuesday, if they sell both sedans and SUVs, then the number of cars that are sold is

S*(1/6) + U*X

where X is the fraction of SUVs sold. X ranges >0 (when none of the SUVs are sold) to 1 where all the SUVs are sold.

We are not gonna use this too much but this is important to understand. Clear?

Let's move forward.

We are concerned with finding out whether S is greater than 100 or not.

Or you can ask: is the minimum value of S is less than or equal to 100?

Now let's push the question.

At the end of Tuesday, dealership has sold S/6 sedans.

Cars, in the GMAT world, cannot be sold in fractions.

Cars can't be sold in fractions, so this means that the sedans on that day could only be multiples of 6. What does this say about the number of sedans as of this point? It could be 6, 12, 18 etc.

As of now, do we know anything about the number of SUVs sold? No. It could be 0, it could be 1 million.. we don't know.

If the number of SUVs sold is zero, and number of sedans sold is 6, we have a "NO" answer to our main question.

If number of SUVs is zero, number of sedans is 120, then we have a "YES" answer.

This proves that the data in the question part is not sufficient to answer the question.

(This is super important IMO in extreme difficulty DS questions.)

Because, only after proving this can we start with a blank slate before going to the statements.

Now we are ready for statement 1:

Statement 1:

Number of sedans sold is 8/9 the number of SUVs sold.

That simply means

S*(1/6) = (8/9) (U*X)
S*(1/6) = (8/9) (U*X)

where, from before, X is the fraction of SUVs sold.

Pause here.

If we want to find the minimum number of sedans sold, S, how would we go about doing that?

Cars cannot be sold in fractions.

(8/9) (U*X)

Lets us observe that U*X is divisible by 9

And it also lets us observe that S is a multiple of 8

What is the least such quantity where S is divisible by 6 and 8? LCM.

You can say that one way to get to the lowest value is when S is divisible by 8 and divisible by 6. This is possible when S = 24

This means
S
= 8*3
= 24

So the minimum value of S is 24 when the value of U is 27 and X = 1 because.

That is when the value of the total number of cars is 24 + 27 = 51

However, the maximum value of S has no bounds as of this equation

Remember, U*X just needs to be divisible by 9

That means it could be any multiple above 9

It could be 9, 18, 180 or even 9 million

In the second case, S would also scale up because the value of S*(1/6) and thus S is dependent proportionally on U*X

So although the minimum is 51, the maximum could be anything. That means the maximum could be greater than 100

So we cannot come to a YES or NO definitive answer.

So statement 1 is insufficient, and we can rule out options A and D.

Now let's look at statement 2 alone.

Statement 2:

According to statement 2,

S/6 is equal to 85% of the number of sedans sold on Wednesday

No details are given about how many sedans were sold on Wednesday.

Pause here.

What does this statement say about S?

Cars cannot be sold in fractions.

It first says that 85% of some quantity yields a value of S/6 such that S/6 is a whole number.

What is the minimum such quantity of which 85% is a whole number?

85%
= 85/100
= (17*5)/(20*5)
= 17/20

Because we can't reduce the fraction any further, we can say 17/20 of the sale on Wednesday is a whole number

The lowest such number of which 17/20 is a whole number is 20

This means the lowest S/6 can go is when the sale on Wednesday is 20

Then, to solve for minimum quantity of S,

S/6 = 85% of minimum sale of sedans on Wednesday

S/6 = (85/100) *20
S/6 = (17/20) *20
S/6 = 17
S = 17*6

AHA! Hence, minimum S is 102.

Thus we come to a YES answer that S is indeed greater than 100.

And that

Statement B is sufficient to answer. Hence answer is B.

Now clap.

PriteshS · Answer

Where are the options for the questions present? kindly help

Bunuel · Answer

Hi,

This is a data sufficiency question. Options for DS questions are always the same and usually omitted on the site.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Hope this helps.

PriteshS · Answer

Thank you. Grateful

Adit_ · Answer

Ahh for statement 2 I felt Wednesday more vehicles could have been bought and thus we couldn't make the inference and missed out on the LCM idea there. Great question!

A car dealership carries only sedans and SUVs

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GMAT Data Sufficiency (DS) Questions

A car dealership carries only sedans and SUVs

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