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At a restaurant, a group of friends ordered four main dishes [#permalink]

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11 Sep 2012, 06:10

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At a restaurant, a group of friends ordered four main dishes and three side dishes at a total cost of $89. The prices of the seven items, in dollars, were all different integers, and every main dish cost more than every side dish. What was the price, in dollars, of the most expensive side dish?

This was one of the questions that I encountered in MGMAT's tests.

MGMAT has provided a pretty good explanation for the answer. But I don't seem to understand it. Kindly explain. Also, I would like to know whether such time consuming questions appear on the real test or not. This question would take an average person like me almost 3 minutes to solve, even after I knew the underlying principle.

Explanation provided by MGMAT is in the spoiler. Please refer to it in case you can try and simply things down for me. Thanks..

There are seven different dishes (4 main dishes and 3 side dishes), each with a different integer cost; we cannot use just two variables to represent the two types of dishes. Each of the four main dishes is more expensive than any of the side dishes. The seven dishes add up to $89. The question asks for the price of the most expensive side dish.

(1) SUFFICIENT: Since this problem involves 7 different integers, some of which must be larger than others, it's a good idea to test extreme possibilities. We're told that the most expensive main dish is $16 so we'll also try maximizing the cost of the other three: $15, $14, and $13. In this case, the total cost of the four main dishes is $58, leaving $31 to split among the three side dishes. The prices of the three side dishes must be different integers and all must cost less than $13. Again, try maximum values first: if two side dishes cost $12 and $11, then the third must cost $8. What other possibilities exist that will still add to $31? 11 + 10 + 9 adds only to $30, so we have to keep $12 in the set. It turns out that there's one other possible combination: the three side dishes could cost $12, $10, and $9. In either case, the most expensive side dish costs $12.

Is it possible to get a value other than $12? Let's try to change the values for the main dishes. We can't make the main dishes more expensive, but we can take away $1 from the least expensive main dish; perhaps the most expensive side dish will become more expensive than $12. The next largest possible set of main dish prices are $16, $15, $14, and $12, for a total of $57. This leaves $32 for the three side dishes - but that value is impossible to achieve. The side dishes must cost less than the least-expensive main dish ($12), so the largest possible cost for the three side dishes is $11 + $10 + $9 = $30.

It's impossible, therefore, for the main dishes to cost anything other than $16, $15, $14, and $13. As a result, the most expensive side dish must cost $12.

(2) NOT SUFFICIENT: Since this problem involves 7 different integers, some of which must be larger than others, it's a good idea to test extreme possibilities. We're told that the least expensive side dish costs $9, so this time we'll try minimizing the cost of the other two: $10, and $11. The side dishes cost a total of $30, leaving $59 for the main dishes. That total can be achieved in many ways; for instance, the main dishes could cost $12, $13, $14, and $20. In this scenario, the most expensive side dish is $11. Are there any other possible values?

Let's try increasing the price of the most expensive side dish by $1: say the side dishes now cost $9, $10, and $12. That's a total of $31 for the side dishes, leaving $58 for the main dishes. That total can be achieved, exactly, if the main dishes cost $13, $14, $15, and $16. Therefore, the most expensive side dish can also cost $12. We have at least two possible values, $11 and $12, so this information is not sufficient to answer the question.

GmatPrep1 [10/09/2012] : 650 (Q42;V38) - need to make lesser silly mistakes. MGMAT 1 [11/09/2012] : 640 (Q44;V34) - need to improve quant pacing and overcome verbal fatigue.

At a restaurant, a group of friends ordered four main dishes and three side dishes at a total cost of $89. The prices of the seven items, in dollars, were all different integers, and every main dish cost more than every side dish. What was the price, in dollars, of the most expensive side dish?

(1) The most expensive main dish cost $16.

(2) The least expensive side dish cost $9.

The first thing I'd do when reading the question is quickly estimate the average cost of a dish: it's 89/7, which is just less than $13. If the highest price is only $16, as Statement 1 tells us, the prices will need to be very close together. So we could start by looking at an equally spaced list, decreasing by 1:

16, 15, 14, 13, 12, 11, 10

then the median of this list (13) is its average, so the sum of these numbers is (7)(13) = 91. So our sum is just slightly too high - we need to reduce it by $2. We can't reduce any of the higher prices, since that will make two of our numbers equal. We can do one of two things: we can lower the lowest price by $2:

16, 15, 14, 13, 12, 11, 8

or we can lower the lowest price by $1. Then the only other price we can also lower by $1 is the second-lowest price, since all of our prices need to be different:

16, 15, 14, 13, 12, 10, 9

Either way, the third-lowest price is $12, so Statement 1 is sufficient.

For Statement 2, we can again begin with an equally spaced list:

9, 10, 11, 12, 13, 14, 15

The sum of this list is (7)(12) = 84, so the sum is too low by $5. There are several possibilities now, but the simplest is to raise the price of the largest value by 5:

9, 10, 11, 12, 13, 14, 20

which makes the third-lowest price $11. But we also know that the third-lowest price can be $12 from our analysis of Statement 1. So Statement 2 is not sufficient, and the answer is A.
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Re: At a restaurant, a group of friends ordered four main dishes [#permalink]

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10 Jan 2013, 18:56

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Topic: Problems on which Algebra is not your friend. (by Ron Purewal - 10th Jan, 2013)

Question: At a restaurant, a group of friends ordered four entrées and three appetizers at a total cost of 89 dollars. The prices if the seven items, in dollars, were all different integers, and every entrée cost more than every appetizer. What was the price, in dollars, of the most expensive appetizer?

(1) The most expensive entrée cost 16 dollars. (2) The least expensive appetizer cost 9 dollars.

Check out more of similar kind of questions, on which Algebra is not your friend, in this session (below).

Problems on which you need not to proceed with algebraic equations (by 'the' Ron Purewal)

Note: The content/knowledge was not produced by me; it was merely assimilated from Ron's sessions. ( Please visit http://www.manhattangmat.com/thursdays-with-ron.cfm for more such lessons). I have just compiled various notes , for my convenience,from his videos. You could keep this doc for your revision once you attend the video session.

In this session, the 'Ron' showed how complex algebraic problems could be solve easily, even without touching algebra (algebraic equations).

Please find the compiled .docx note enclosed.

Thanks!

Attachments

File comment: In this session (once again, please visit http://www.manhattangmat.com/thursdays-with-ron.cfm to check out this session), the 'Ron' showed how complex algebraic problems could be solve easily, even without touching algebra (algebraic equations).Here is the compiled .docx note. Problems on which Algebra is not your friend (by Ron Purewal).docx [1.03 MiB]
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Re: At a restaurant, a group of friends ordered four main dishes [#permalink]

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05 Mar 2013, 06:40

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thebigr002 wrote:

At a restaurant, a group of friends ordered four main dishes and three side dishes at a total cost of $89. The prices of the seven items, in dollars, were all different integers, and every main dish cost more than every side dish. What was the price, in dollars, of the most expensive side dish?

(1) The most expensive main dish cost $16.

(2) The least expensive side dish cost $9.

From the F.S 1, we know that the price of the most expensive side dish is 16. Let me assume that the price of the costliest side dish(call it r) is 15. Is this possible? No as because every main dish cost more than every side dish. Thus, if I were to have a side dish costing 15, this will contradict the given statement. Now lets assume that the side dish costs 14. Again, not possible. Thus, the costliest side dish can have a value only below 13. Now, we know that 16+15+14+13 = 58. Thus, the sum of the remaining three dishes is 89-58 = 31. Now, again, the price of the cheapest and the second cheapest side dishes be x and y respectively.Ans we know that x,y and r are different integers.Also, 31-(x+y) = r has to be less than 13. Thus, 31-(x+y)<13 or (x+y)>18.

For r = 31-(x+y),put all values for (x+y) more than 18.

For x+y = 19, r = 12.

As you keep on increasing the value for (x+y), the value of r will decrease. Thus, the value of r = 12. Also, for any other value of (x+y), the highest value will always come out to be 12. Sufficient.

From F.S 2, we have that x = 9. We can have y=10, r = 11. The price of the main dishes can be 17,15,14,13 to give a total of 89 dollars.Again, for y=8, r= 12, we have the price of the other dishes to be 18,15,14,13. Not Sufficient as the value of r is not fixed.

Re: At a restaurant, a group of friends ordered four main dishes [#permalink]

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24 May 2013, 10:27

each dish was a different integer. each main dish cost more than any side dish

St1: Most expensive main dish = 16 lets say other 3 main dish cost 15, 14 and 13 that leaves 31 to be divided between 3 side dishes. maximum price of a side dish can be 12 as 13 is the least expensive main dish so side dishes could be priced 12, 11 and 8

Consider another scenario: lets say most exp MD costs 16 and three other MD costs 14,12 and 10 that leaves 37 for 3 SD. This cannot be the case since all side dished must cost less than 10 (minimum cost of any main dish)

and 1 more scenario: so its established that the 4 MD cost 16, 15, 14 and 13 leaving 31 for three SD. lets say most exp side dish SD3 cost 11 that leaves 20 for SD1 and SD2 To get 20, one price must be greater than 10 and it cant be 11 (since no 2 dish cost the same) and in that case it could be 12 and 8 or 13 and 7 BUT then 11 wont be the most expensive.

So the only option left for the most expensive side dish = 12. Sufficient. Ans A

Re: At a restaurant, a group of friends ordered four main dishes [#permalink]

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03 Jul 2014, 05:28

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Re: At a restaurant, a group of friends ordered four main dishes [#permalink]

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25 Aug 2015, 23:24

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Re: At a restaurant, a group of friends ordered four main dishes [#permalink]

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26 Jun 2016, 04:23

I understood that most problems are built using some core concepts. This one might be based on sum of consecutive integers

If we look at 91 = 13 * 7 and problem talks about 7 numbers and if we are able to connect the consecutive integers with this formula (I know lots of ifs)

Then 13 * 7 = (a1+an)/2 * n (number of terms in the sequence of consecutive integers where a1 is the first term and an is the last term)

Now if we take statement one and plug in 16 for an , this gives a1= 10; So if any answer would bound either the first term in the series or the last term in the series (10 or 16), then the second extreme and hence all numbers get fixed.

Hence D as both 1 and 2 bind the upper and lower extremes of the consecutive integers sequence

This was the only way I could think of to complete this in less than 2 minutes

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Re: At a restaurant, a group of friends ordered four main dishes
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