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by [email protected] » Sun Nov 10, 2013 4:52 pm
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.

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by GMATGuruNY » Sun Nov 10, 2013 7:29 pm
[email protected] wrote: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.
Let the 4 entrées, in descending order, be A, B, C and D.
Let the 3 appetizers, in descending order, be x, y, and z.
Since every entrée costs more than every appetizer, D>x.

What is the value of x?

Statement 1: A=16
Here, the maximum possible sum for A+B+C+D = 16+15+14+13 = 58.
Implication:
The minimum possible sum for x+y+z = 89-58 = 31.
Options for x, y and z:
12, 11, 10, 9, 8, 7...

Case 1: x=12
Here, y+z = 31-12 = 19, implying that y=11 and z=8 or that y=10 and z=9.

Case 2: x=11
Here, y+z = 31-11 = 20.
Not possible, since the maximum possible sum for y+z in this case = 10+9 = 19.

Thus, when A+B+C+D = 16+15+14+13 = 58, x=12.

If the value of A+B+C+D decreases, then the value of x must compensate by INCREASING to a value greater than 12.
The next greatest possible sum for A+B+C+D = 16+15+14+12.
Not possible:
If D=12, then the value of x CANNOT increase to a value GREATER THAN 12, since it is required that D>x.

Thus, there is only one possible option for x:
x=12.
SUFFICIENT.

Statement 2: z=9
Case 1 can work here: A+B+C+D = 58, x=12, y=10, z=9.

Case 3: If A+B+C+D = 17+15+14+13 = 59, then it's possible that x=11, y=10, and z=9.

Since x can be different values, INSUFFICIENT.

The correct answer is A.
Last edited by GMATGuruNY on Mon Nov 11, 2013 6:55 am, edited 2 times in total.
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by Uva@90 » Sun Nov 10, 2013 10:16 pm
GMATGuruNY wrote:
[email protected] wrote: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.
Let the 4 entrées, in descending order, be A, B, C and D.
Let the 3 appetizers, in descending order, be x, y, and z.
Since every entrée costs more than every appetizer, D>x.

What is the value of x?

Statement 1: A=16
Here, the maximum possible sum for A+B+C+D = 16+15+14+13 = 58.
Implication:
The minimum possible sum for x+y+z = 89-58 = 31.
Options for x, y and z:
12, 11, 10, 9, 8, 7...

Case 1: x=12
Here, y+z = 31-12 = 19, implying that y=11 and z=8 or that y=10 and z=9.

Case 2: x=11
Here, y+z = 31-11 = 20.
Not possible, since the maximum possible sum for y+z in this case = 10+9 = 19.

Thus, when A+B+C+D = 16+15+14+13 = 58, x=12.

If the value of A+B+C+D decreases, then the value of x must compensate by INCREASING to a value greater than 12.
The next greatest possible sum for A+B+C+D = 16+14+13+12.
Not possible:
If D=12, then the value of x CANNOT increase to a value GREATER THAN 12, since it is required that D>x.

Thus, there is only one possible option for x:
x=12.
SUFFICIENT.

Statement 2: z=9
Case 1 can work here: A+B+C+D = 58, x=12, y=10, z=9.

Case 3: If A+B+C+D = 17+15+14+13 = 59, then it's possible that x=11, y=10, and z=9.

Since x can be different values, INSUFFICIENT.

The correct answer is A.
Mitch,
I have a doubt.
In Question they mentioned as
What was the price, in dollars, of the most expensive appetizer?
Doesn't it mean to find the actual or original value of Expensive Appetizer ?

so answer can be 12, or 11 or something else. Which can't be said as-surely.

If they mentioned as "what CAN/MAY be the price, in dollars, of the most expensive appetizer?"

Then we can come to conclusion as you solved.
12 wil

Please tell me am I understanding something wrong.

Thanks in advance.

Regards,
Uva.
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by [email protected] » Sun Nov 10, 2013 10:53 pm
Hi Uva@90,

DS questions always ask you a specific question; it's up to you to determine if there's just one possible answer (based on the given information) or if there's MORE than one possible answer.

Mitch's explanation proves that Fact 1 would lead to just one answer, so Fact 1 is SUFFICIENT. Fact 2 gives us information that leads to more than one possible answer, so it's INSUFFICIENT.

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by sana.noor » Sun Nov 10, 2013 11:57 pm
it can not be 11 as 89-58 = 31. one has to divide this 31 into 3 appetizers in such a way that all three must be different. if u choose 11 the most expensive appetizer then 31-11= 20. and you re left with highest most prices as 10 and 9. the other two appetizers cant be 10 and 9 as they two combined 19 which is one less than 20. so 12 is for sure the only answer.
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by truongynhi » Fri Feb 19, 2016 5:35 am
GMATGuruNY wrote: Statement 1: A=16
Here, the maximum possible sum for A+B+C+D = 16+15+14+13 = 58.
Implication:
The minimum possible sum for x+y+z = 89-58 = 31.
Options for x, y and z:
12, 11, 10, 9, 8, 7...
Hi Mitch,
x+y+z = 89-58 = 31 is the maximum possible sum for x+y+z, isn't it?

I thought it was a typo in your post. If it is not, please help me understand why 31 is the minimum sum.

Thank you very much!
Nhi
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by GMATGuruNY » Fri Feb 19, 2016 6:06 am
truongynhi wrote:
GMATGuruNY wrote: Statement 1: A=16
Here, the maximum possible sum for A+B+C+D = 16+15+14+13 = 58.
Implication:
The minimum possible sum for x+y+z = 89-58 = 31.
Options for x, y and z:
12, 11, 10, 9, 8, 7...
Hi Mitch,
x+y+z = 89-58 = 31 is the maximum possible sum for x+y+z, isn't it?

I thought it was a typo in your post. If it is not, please help me understand why 31 is the minimum sum.

Thank you very much!
Nhi
(A+B+C+D) + (x+y+z) = 89.
If A+B+C+D = 58, then x+y+z = 89-58 = 31.
If A+B+C+D = 57, then x+y+z = 89-57 = 32.
If A+B+C+D = 56, then x+y+z = 89-56 = 33.
As illustrated by the option in red, the MAXIMUM VALUE for A+B+C+D yields the MINIMUM VALUE for x+y+z.
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by Matt@VeritasPrep » Fri Feb 19, 2016 2:12 pm
Here's another approach to S1:

Let's treat the entrees and the appetizers as groups, i.e. (sum of the other three entrees) = m and (sum of the three appetizers) = n.

Since m > n and m + n = 73, we know that m ≥ 37. Since 16 > each of our other entrees, we know that the max of those three = 15 + 14 + 13 = 42. So 42 ≥ m ≥ 37.

We also know that the max of n = 12 + 11 + 10 = 33. So 12 is a possible answer. Now suppose the most expensive appetizer = 11. That gives us 11 + 10 + 9 = 30. But then we need the entrees to cost 43, which is higher than our maximum! So the most expensive appetizer must = 12.
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