During a week-long sale at a car dealership, the most number of cars s

Given
1. Max no of cars sold on any day = 12
2. Min no of cars sold on any day >=2

No we have to determine whether the Avg no of cars sold during the week > 6
=> is total no of cars sold > 6*7 =42 ?

Statement 1
Possibility 1- No of cars sold on 7 days = 2, 4, 4, 4, 4, 4, 4 => Total Cars sold = 26 <42
Possibility 2- No of cars sold on 7 days = 2, 4, 10, 10, 10, 10, 12 => Total cars sold = 58 >42

Not sufficient.

Statement 2
Taking the minimum no of cars sold on 7 days = 2, 2, 2, 10 (median), 10, 10, 12 = 48 >42

Sufficient

Answer - B

Hi,

Statement 1 tells us that the second smallest number of cars sold is 4.
Is is possible to derive that the no. of cars sold could also be (2, 2, 2, 4, 12)? Second smallest does not mean to me that 2 cars were sold only once!

In any way I also arrived at B with this method, so its just for my personal clarification.

Thanks in advance!!

Yes, you are right. This is definitely a possibility and there would always by different possibilities to evaluate for any statements. Normally, you need to consider the extremities since they may challenge or give different results.

Yes, you are right. This is definitely a possibility and there would always by different possibilities to evaluate for any statements. Normally, you need to consider the extremities since they may challenge or give different results.

From Question stem , min car sold each day \(x\geq{2}\) and Max sold in particular day of week 12

Is weekly Avg > 6 or total > 42 ?


Stmt 1 : During that week, the second smallest number of cars sold on any one day was 4.

it can be 2, 4, 5, 5, 5, 5 ,12 = 38 (avg <6)
it can be 2, 4, 10, 10, 10, 10, 12 ~ (avg >6)
Not sufficient.

Stmt 2 : During that week, the median number of cars sold was 10.

it can be 2, _, _, 10, _, _ ,12
lets fill minimum values it can be 2, 2, 2, 10, 10, 10, 12 = 48 (avg >6)
Sufficient.

Ans : B

MANHATTAN GMAT OFFICIAL SOLUTION:

First, do you see why I described this as a “scenario” problem? All these different days… and some number of cars sold each day… and then they (I!) toss in average and median… and to top it all off, the problem asks for a range (more than 6). Sigh.

Okay, what do we do with this thing?

Because it’s Data Sufficiency, start by establishing the givens. Because it’s a scenario, Draw It Out.

Let’s see. The “highest” day was 12, but it doesn’t say which day of the week that was. So how can you draw this out?

Neither statement provides information about a specific day of the week, either. Rather, they provide information about the least number of sales and the median number of sales.

The use of median is interesting. How do you normally organize numbers when you’re dealing with median?

Bingo! Try organizing the number of sales from smallest to largest. Draw out 7 slots (one for each day) and add the information given in the question stem:
Now, what about that question? It asks not for the average, but whether the average number of daily sales for the week is more than 6. Does that give you any ideas for an approach to take?

Because it’s a yes/no question, you want to try to “prove” both yes and no for each statement. If you can show that a statement will give you both a yes and a no, then you know that statement is not sufficient.

Try this out with statement 1

(1) During that week, the least number of cars sold on any one day was 4.

Draw out a version of the scenario that includes statement (1):
Can you find a way to make the average less than 6? Keep the first day at 2 and make the other days as small as possible:
The sum of the numbers is 34. The average is 34 / 7 = a little smaller than 5.

Can you also make the average greater than 6? Try making all the numbers as big as possible:
(Note: if you’re not sure whether the smallest day could be 4—the wording is a little weird—err on the cautious side and make it 3.)

You may be able to eyeball that and tell it will be greater than 6. If not, calculate: the sum is 67, so the average is just under 10.

Statement (1) is not sufficient because the average might be greater than or less than 6. Cross off answers (A) and (D).

Now, move to statement (2):

(2) During that week, the median number of cars sold was 10.

Again, draw out the scenario (using only the second statement this time!).
Can you make the average less than 6? Test the smallest numbers you can. The three lowest days could each be 2. Then, the next three days could each be 10.
The sum is 6 + 30 + 12 = 48. The average is 48 / 7 = just under 7, but bigger than 6. The numbers cannot be made any smaller—you have to have a minimum of 2 a day. Once you hit the median of 10 in the middle slot, you have to have something greater than or equal to the median for the remaining slots to the right.

The smallest possible average is still bigger than 6, so this statement is sufficient to answer the question. The correct answer is (B).

Oh, and the OG question is DS #121 from OG13. If you think you’ve got the concept, test yourself on the OG problem.

Key Takeaway: Draw Out Scenarios

(1) Sometimes, these scenarios are so elaborate that people are paralyzed. Pretend your boss just asked you to figure this out. What would you do? You’d just start drawing out possibilities till you figured it out.

(2) On Yes/No DS questions, try to get a Yes answer and a No answer. As soon as you do that, you can label the statement Not Sufficient and move on.

(3) After a while, you might have to go back to your boss and say, “Sorry, I can’t figure this out.” (Translation: you might have to give up and guess.) There isn’t a fantastic way to guess on this one, though I probably wouldn’t guess (E). The statements don’t look obviously helpful at first glance… which means probably at least one of them is!

While I’ve seen the explanations and know the solution, I have a very specific query that I hope someone can resolve- as per manahattan one of the cases for statement 1 is 4,4,12,12,12,12,12. How’s that even possible? The statement clearly says 4 is the second lowest! It’s just for my clarification I want to know and I do understand we can choose any case. But my main concern is how come they have given this as one of the cases?

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