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Math Expert V
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A cafeteria offers seven types of sandwiches, each with a fixed price  [#permalink]

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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?

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

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Re: A cafeteria offers seven types of sandwiches, each with a fixed price  [#permalink]

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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 Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: http://www.gmatprepnow.com/articles/dat ... lug-values 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 Answer: B RELATED VIDEO FROM OUR COURSE _________________ Senior SC Moderator V Joined: 14 Nov 2016 Posts: 1348 Location: Malaysia A cafeteria offers seven types of sandwiches, each with a fixed price [#permalink] Show Tags 1 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 $$2.70$$. Official solution from Veritas Prep. 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 $$B$$. _________________ "Be challenged at EVERY MOMENT." “Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.” "Each stage of the journey is crucial to attaining new heights of knowledge." Rules for posting in verbal forum | Please DO NOT post short answer in your post! Advanced Search : https://gmatclub.com/forum/advanced-search/ Manager  B Joined: 03 Aug 2017 Posts: 64 A cafeteria offers seven types of sandwiches, each with a fixed price [#permalink] Show Tags 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. This is how i solved.... I am basically using what brent did but adding my test cases for even more simplification..... So as Brent mentioned " 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. Notice the Max cost = 2.7 and the average cost is 2.25 here the difference is 0.45 Let's assume all 7 sandwiches cost 2.25 each except the 7th one which cost = 2.70 so in order to maintain the avg we need to deduct 0.45 ( 2.70 - 2.25 ) from one of the other sandwiches... so let's test some values 2.25 - 0.45 = 1.80 So, in this case, no sandwich is below 1.75 Here the valus are 1.80 ,2,25, 2,25,2,25,2,25,2,25 and 2.70 = And overall avg of 2.25 But we can also deduct more from one sandwich eg 2.25 - 0.55 =1.70 and add also subtract the 0.10 from another sandwich ie 2.25- 0.10 = 2.15 So now the set is 1.70 , 2.15, 2.25 .2.25 2.25 ,2,25 and 2.70 and yet we are at the overall avg of 2.25 per sandwich,,, 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:

Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: Is perfectly explained by brent below

" 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 "

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Posts: 198
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GMAT 1: 780 Q51 V45 GRE 1: Q170 V167 A cafeteria offers seven types of sandwiches, each with a fixed price  [#permalink]

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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.

Analyzing the question:
Average price is 2.25, therefore the total is 7*2.25 = 14+1.4+0.35 = $15.75. Is there any less than$1.75? Note it is relatively straightforward to find a case where we answer "yes", so we should focus on finding the special "no" case. If there are none of "no" cases then this would be sufficient.

Statement 1:
The max price is 2.7, which is 2.7-2.25 = 0.45 above the median. We can have one sandwich at 1.8 which is 0.45 below the average, and the rest of the sandwiches at the average price so that all sandwiches have a price above $1.75. The key is there is enough room to have all sandwiches at a high enough price. We have found the "no" case so this is insufficient. Statement 2: Again we should focus on finding the "no" case, where all sandwiches cost above$1.75. We need at least three sandwiches above $2.7, we can assume all of the highest priced ones are 2.7 to maximize room for the low priced sandwiches. Then we have the three lowest-priced sandwiches adding up to = 15.75 - 2.7*4 = 15.75 - 10.8 =$4.95. That is not enough to give the rest of the low priced sandwiches $1.75 each, therefore we cannot have all sandwiches above$1.75 and this is sufficient.

Ans: B
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Re: A cafeteria offers seven types of sandwiches, each with a fixed price  [#permalink]

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hazelnut wrote:
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 $$2.70$$.

Official solution from Veritas Prep.

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 $$B$$.

I am surprisingly little confused with these sort of probolems, Can you please share a link with similar questions? Re: A cafeteria offers seven types of sandwiches, each with a fixed price   [#permalink] 08 Oct 2019, 01:36
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