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A cafeteria offers seven types of sandwiches, each with a fixed price
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08 Feb 2017, 00:51

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A cafeteria offers seven types of sandwiches, each with a fixed price and a fixed set of ingredients (no additions or substitutions). If the average price among the seven sandwiches is $2.25 do any sandwiches cost less than $1.75?

Re: A cafeteria offers seven types of sandwiches, each with a fixed price
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08 Feb 2017, 08:21

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Bunuel wrote:

A cafeteria offers seven types of sandwiches, each with a fixed price and a fixed set of ingredients (no additions or substitutions). If the average price among the seven sandwiches is $2.25 do any sandwiches cost less than $1.75?

(1) The maximum price of a sandwich is $2.70.

(2) The median price of a sandwich is $2.70.

Great question!

Target question:Do any sandwiches cost less than $1.75?

Given: The average price among the seven sandwiches is $2.25 So, (TOTAL cost of all 7 sandwiches)/7 = $2.25 So, TOTAL cost of all 7 sandwiches = (7)($2.25) = $15.75

Statement 1: The maximum price of a sandwich is $2.70 This statement doesn't FEEL sufficient, so I'll TEST some values. There are several scenarios that satisfy statement 1 and the given information (that the TOTAL cost of all 7 sandwiches = $15.75). Here are two: Case a: the sandwich prices numbers are {$1.00, $1.25, $2.70, $2.70, $2.70, $2.70, $2.70}, in which case there IS a sandwich that costs less than $1.75 Case b: the sandwich prices numbers are {$2.05, $2.20, $2.20, $2.20, $2.20, $2.20, $2.70}, in which case NO sandwich costs less than $1.75 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The median price of a sandwich is $2.70 Notice that, in our analysis of statement 1, we came up with a scenario that happens to meet the condition set out in statement 1. That is, {$1.00, $1.25, $2.70, $2.70, $2.70, $2.70, $2.70} meets the condition that the median price of a sandwich is $2.70. So, we can see that it IS POSSIBLE to have a situation in which there IS a sandwich that costs less than $1.75

Is it also possible to have a situation in which there are NO sandwiches that cost less than $1.75? Let's find out. Let's arrange the 7 sandwich prices in ASCENDING order with a median price of $2.70. We get: _ _ _ $2.70 _ _ _

We're trying to create a scenario in which there are NO sandwiches that cost less than $1.75. This means we're trying to MAXIMIZE the values to the LEFT of the median. To do this, we must MINIMIZE the values to the RIGHT of the median.

Since the values to the RIGHT of the median must be $2.70 or greater, we can MINIMIZE these values by making them all $2.70. We get: _ _ _ $2.70 $2.70 $2.70 $2.70

4 x $2.70 = $10.80, so we have already accounted for $10.80 of the total $15.75 $15.75 - $10.80 = $4.95, so the remaining 3 prices must add to $4.95 At this point, we might recognize that it is IMPOSSIBLE to achieve our goal of creating a scenario in which there are NO sandwiches that cost less than $1.75 Here's why. If all 3 remaining prices were $1.75, then we would need $5.25 (3 x $1.75 = $5.25). However, we only have $4.95 left. So, at least one of the 3 remaining prices will be LESS THAN $1.75

So, statement 2 guarantees that there IS a sandwich that costs less than $1.75 Since we can answer the target question with certainty, statement 2 is SUFFICIENT

A cafeteria offers seven types of sandwiches, each with a fixed price
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17 Feb 2017, 16:31

Bunuel wrote:

A cafeteria offers seven types of sandwiches, each with a fixed price and a fixed set of ingredients (no additions or substitutions). If the average price among the seven sandwiches is \($2.25\) do any sandwiches cost less than \($1.75\)?

B. In this Min/Max Data Sufficiency problem, it's important to remember the threshold for sufficiency: if the answer is "only yes" the information is sufficient; if the answer is "only no" the information is sufficient; and if the answer is "yes or no" the information is not sufficient. So your goal should be to use each statement to see if you can get both a "yes" and a "no" - Min/Max Data Sufficiency is almost always about "prove insufficiency".

In this case, statement 1 allows for multiple answers. You could certainly have one sandwich at \($2.70\), one at \($1.80\) (the min and max averaging out to \($2.25\)), and the rest all at \($2.25\) (for an answer of "no"). Or you could set the low price at \($1.70\) ("yes"), the high at \($2.70\), and slightly shift one of the middle values to average out to \($2.25\). Statement 1 should quickly prove to be insufficient.

Statement 2 is somewhat "sneaky sufficient" alone. If you try to maximize the value of the lowest-priced sandwich to see if it can be above \($1.74\), you should see that your goal, then, is to minimize the value of the higher-priced sandwiches. That would mean that four sandwiches are priced at \($2.70\) (to keep the median price \($2.70\) but leave as much of the \(7($2.25)\) average for the lowest priced sandwich) and similarly minimize the other two sandwiches that are below the median (again, saving as much of that value as possible for the lowest-priced item to see if you can get it to \($1.75\)). If you do that, then, all the lowest-priced items would have the same value, so you'd have the equation:

\(3x + 4(2.70) = 7(2.25)\)

\(3x + 10.80 = 15.75\)

\(3x = 4.95\)

\(x = 1.65\), meaning that even at its highest price, the lowest priced sandwich is less than \($1.75\).

This makes statement 2 sufficient, and the answer \(B\).
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Re: A cafeteria offers seven types of sandwiches, each with a fixed price
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