One-sixth of the orders placed at a take-out restaurant on Friday nigh

100mitra
For the case of 17<$15
Don't you think it would cover orders <$17 as well..

My wrong, I didn't got that one possible situation can be consider
Thank you

This question is a part of Are You Up For the Challenge: 700 Level Questions collection.­

It's a very good task, so please let me share my way of thinking here.

Step 1: from the Statement 1 we get that 1/6x <= 5, which gives us x<=30. Since both number of orders <$15 and <$17 are integers, only 5 values are possioble: x= 6, 12, 18, 24, 30. It could be that all orders under $17 were also under $15, but we do not know. Thus, insufficient.

Step 2: from statement 2 it's not clear whether there were any orders in $15-$20 range -> insufficient.

Step 3: combine 2 statements. Here it starts to be really interesting.

First, remember, that we could only have some integral values for x from statement 1. Second, we can see that only 30 matches two our conditions: 1/6x <= 5 and that there are at least 26 orders (5 less than $17 AND 21 more than $20). It clearly shows that it's only possible to have 30 orders in total, hence, sufficient.

Bunuel, may I kindly ask you to correct my reasoning if needed.

1/6th of all orders have Value < $15


(1) The number of orders with a value of less than $17 was 5

5 orders have value < $17

Then 5/4/3/2/1 orders could have value less than $15. In respective cases, total number of orders would be 30, 24, 18, 12 and 6 (because 1/6th of total orders are less than $15 value)
Insufficient alone.


(2) The number of orders with a value of more than $20 was 21

21 orders have value > $20.
We could have any number of orders in the range $15 to $20.
We get no information on how many orders were below $15 and how many total orders there are.

Using both, we know that total number of orders could be 30, 24, 18, 12 or 6.

Assume total = 30; Since 21 orders have value above $20, 9 orders would be at $20 or lower. Then 5 orders (1/6th of 30) could have been below $15 and that is fine. It all works.

Assume total = 24; Since 21 orders have value above $20, 3 orders would be at $20 or lower. Then at most 3 orders could have been below $15 but that is less than 1/6th of 24. It doesn't work. We cannot have 1/6th total orders with value less than $15. Not possible.

Total number of orders cannot be lower than 21 since we have 21 orders with value more than $20. Hence, 18, 12 or 6 total number of orders are not possible.

Hence total number of orders must be 30.

Answer (C)

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Please provide an expert response. Thanks!


We are told that 1/6 of the orders were less than $15, which means the total number of orders must be a multiple of 6.

Statement One Alone:
\(\Rightarrow\) The number of orders with a value of less than $17 was 5

It follows that the maximum number of orders with a value of less than $15 is 5 (if we assume all of the 5 orders less than $17 were also less than $15), which means the total number of orders can be at most 6 * 5 = 30. For instance, if there were 5 orders with a value of $10 and 25 orders with a value of $18, then there are a total of 30 orders, the number of orders with a value of less than $17 is 5, and 1/6 of all orders have a value of less than $15; so one possible answer to the question is 30. However, if we only assume this statement, then the total number of orders can also be some multiple of 6 less than 30, such as 18. For instance, there could be 3 orders with a value of $10, 2 orders with a value of $16, and 13 orders with a value of $18. In this scenario, the number of orders with a value of less than $17 is 3 + 2 = 5, and 1/6 of all orders are less than $15. Thus, another possible answer to the question is 18. Since we have more than one possible answer to the question, this statement is not sufficient on its own.

Eliminate answer choices A and D.

Statement Two Alone:
\(\Rightarrow\) The number of orders with a value of more than $20 was 21

Since the smallest multiple of 6 greater than 21 is 24, assuming this statement implies the total number of orders is at least 24. However, if there were 24 orders in total, then 1/6 * 24 = 4 of them would have to be less than $15, which means the number of orders with a value of more than $20 can be at most 24 - 4 = 20. Since we are told that there were 21 such orders, the total number of orders cannot equal 24. On the other hand, if we assume only this statement, the total number of orders can still be 30 or any multiple of 6 greater than 30. If there were 5 orders with a value of $10, 4 orders with a value of $18, and 21 orders with a value of $25, then 1/6 of all orders are below $15 and the number of orders with a value of more than $20 is 21. So 30 is one possible answer. If there were 10 orders with a value of $10, 29 orders with a value of $18, and 21 orders with a value of $25, then the total number of orders is 10 + 29 + 21 = 60, and 1/6 of all orders have a value of less than $15, and the number of orders with a value of more than $20 is 21. So we see that 60 is another possible answer. Since there are more than one possible answers, statement two alone is not sufficient on its own.

Eliminate answer choice B.

Statements One and Two Together:

From our analysis of statement one, we know the total number of orders can be at most 30. From our analysis of statement two, we know the total number of orders must be at least 30. Combined, it follows that the total number of orders must be exactly 30. Statements one and two together are sufficient.

Answer: C­