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

Question

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. 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.

The correct answer is A.

IanStewart · Accepted Answer

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.

goutamread · Answer

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 https://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.
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goutamread · Answer

Yeap! they are good. Visit, for more info, https://www.manhattangmat.com/gmat-prep-programs.cfm And for more feedback, manhattan-gmat-class-discount-everything-about-mgmat-84213.html All the best!

mau5 · Answer

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.

A.

srcc25anu · Answer

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

psdhali · Answer

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

mbaapp1234 · Answer

This problem is best solved by simply listing values and doing the math. So if we start with statement (1), then we can work our way backward:

$16 $15 $14 $13 $12 $11 $10

We know that the sum of these prices is $91. We can't increase the maximum price of the most expensive side dish ($12) because we would start a ripple effect that would make the most expensive dish more expensive than the given information. We can't reduce the price of the most expensive side dish because then the total would not add up to $91. We can't make the second most-expensive side dish compensate by increasing its price by 1 and decreasing the most expensive side dish's price by 1, because then the second-most would become the most-expensive side dish. Only one possible value exists.

For statement (2), we can use a similar approach. If we start at $10 and work our way upwards, then we notice that there is only one possible value for the most expensive side dish. If we increase the price of the most expensive side dish, then we cannot reduce the prices elsewhere to make up for it. There is no "wiggle room." If we increase the price of every other dish by 1 (or even if we increase the price of 1 dish by 1), then the total would be greater than $91. Sufficient.

PrayasT · Answer

Bunuel  Can you suggest a better approach to this question?

AvidDreamer09 · Answer

Most expensive side dish = ???

Clues from the question stem:
1) 4 main dishes and 3 side dishes
2) Total cost of $89
3) The prices of the seven items, in dollars, were all different integers
4) Every main dish costs more than every side dish

(1) The most expensive main dish cost $16.
We know the total cost of the bill is $89 and each Main dish is more expensive than every side dish

MD 1 = 16 ( from 1)
MD 2 = 15 (cost of all dishes are different) 
MD 3 = 14
MD 4 = 13
Total = 58

SD1=12 Every main dish costs more than every side dish. Side Dish Amount has to be less than 13 because 
SD2 = 10
SD3 = 9
Total = 31

MD + Sd = 89 Sufficient

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

There is no upper limit
Sd may cost, 9, 10 , 11 or 9 11 12

Two possibilities, Not sufficient

warrior1991 · Answer

AjiteshArun   VeritasKarishma

Why is option B not sufficient??

My reasoning:-

Least expensive dish is 9$ and all the dish amount are different integer values. We have to find highest value of side dish . So we have to minimize all other dish price.

Side dish price after minimization: 9,10,x (Least expensive dish is 9$ given as per St2 and X is most expensive price of side dish)
Main dish price after minimization:-X+1,X+2,X+3,X+4

So 9+10+X+X+1+X+2+X+3+X+4=89
5X+29=89
5X=60
X=12

Therefore,X=12 is the most expensive price of the side dish .

Where am I going wrong??

AvidDreamer09 · Answer

I hope i can clear your doubt for you.

When you assume price of each dish to be (X),(X+1),(X+2),(X+3),(X+4) , you assume price of each dish to be   consecutive   integers.

But the question stem says different integers.
Which is why the following two possibilities are allowed

Possibility 1 => 9+10+12+13+14+15+16=89

Possibility 2 => 9+10+11+13+14+15+17=89

warrior1991 · Answer

chetan2u   VeritasKarishma   AjiteshArun

Please help.

chetan2u · Answer

Hi,

You are going wrong by taking this as the ONLY possibility.
We just know that the least is 9. Nothing more

If all were equal, then the median would be 89/7 or between 12 and 13. The three side dishes will be just less than the median, so  the most expensive side dish will be surely less than 13 ..
 The least is 9, so most expensive dish cannot be 9 or 10. . Possible values 11 or 12

Check for 11..
so the side dishes are 9, 10, 11... 9+10+11=30.....main dishes can be 12, 13, 14, and (89-30-12-13-14=20) OR 12, 13, 15, 19 and so on
check for 12..
so the side dishes are 9, 10, 12... 9+10+12=31.....main dishes can be 13, 14, 15 and (89-31-15-13-14=16) ..

So both 11 and 12 are possible. Hence not sufficient.

KarishmaB · Answer

The question does not ask for the maximum possible value of the most expensive side dish. It asks for the actual cost of the most expensive side dish. 
So only if $89 can be distributed under the given constraints in only such ways that the most expensive side dish can have only one value, then the statement would be sufficient.

If the least expensive side dish costs $9, the cost of most expensive side dish can take multiple values:

9, 10,  11 , 12, 13, 14, 20 (total = $89)
9, 10,  12 , 13, 14, 15, 16 (total = $89)

So the most expensive side dish cost is not known uniquely using stmnt 2 alone.

mangamma · Answer

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.

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

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

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 $91. The question asks for the price of the most expensive side dish. This is a statistics question, so it will be useful to consider different cases.
 
(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 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 $33 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: $12, $11 and $10 are the largest possible prices for the three side dishes and these indeed sum to $33. However, note that this maximized case is the only case that will work – otherwise the 3 side dishes will not sum to $33 (and the total will not sum to $91). If we back up and try something other than the maximum values for the four main dishes, the main dishes will then sum to less than $58 and the lowest price of these main dishes will be something less than $13. This will mean we'll need the sum of the three side dishes to be greater than $33, and we'll have to achieve that with numbers smaller than 12, 11, and 10. That's impossible! Thus, $12 is the price of the most expensive side dish. 
 
Conceptually, we could have started with the average of these 7 dishes that sum to $91, that is $91/7 or $13. With the greatest statistic at $16 (only $3 above the mean), we might have understood that the other 6 prices can’t be that far away from the mean (the mean is the balancing point for all of the statistics in the set). The set of consecutive integers from $16 to $10 would have been the first to try. After that worked (13 is the mean of these 7 integers and 7 x 13 = 91), we could have proved that all other cases would decrease the sum to less than $91.
 
(2) 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 $10, so this time try minimizing the cost of the other two: $11, and $12. The side dishes would then cost a total of $33, leaving $58 for the main dishes. Since the minimum values for each of the four side dishes, i.e. $13, $14, $15, and $16 already sum to $58 there is no way to choose any larger values for these prices, which also means that we also couldn’t have chosen any larger values for the price of the side dishes. Thus, $12 is the price of the most expensive side dish.
 
Conceptually, we could have started with the average of the 7 dishes that sum to $91, that is $91/7 or $13. With the smallest statistic at $10 (only $3 below the mean), we might have understood that the other 6 prices can’t be that far away from the mean (the mean is the balancing point for all of the statistics in the set). The set of consecutive integers from $16 to $10 would have been the first to try. After that worked (13 is the mean of these 7 integers and 7 x 13 = 91), we could have proved that all other cases would increase the sum to greater than $91.

Note that this is a C-trap. Knowing that the greatest and smallest values of the set of 7 different integers were only 6 apart would imply that they must be consecutive integers. However, it was possible to get the answer with each statement alone.

nj23598 · Answer

Please find the compiled .docx note enclosed.

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

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