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At a certain upscale restaurant, there just two kinds of food items: e
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17 Mar 2015, 05:55
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70% (03:06) correct 30% (03:07) wrong based on 101 sessions
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At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrée item costs $30, and each appetizer item costs $12. Last year, it had a total of 15 food items on the menu, and the average price of a food item on its menu was $18. This year, it added more appetizer items, and the average price of a food item on its menu dropped to $15. How many appetizer items did it add this year? A. 3 B. 6 C. 9 D. 12 E. 15 Kudos for a correct solution.
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Re: At a certain upscale restaurant, there just two kinds of food items: e
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17 Mar 2015, 07:00
It is just about resolving some equations :
X is the number of entrees Y is the number of appetizers
1) X+Y = 15 2) (30X+12Y)/15=18
1 and 2 give X=5 and Y = 10
Then next year, Z is the number of added appetizers 3) (30*5+12*(10+Z))/(15+Z)=15
I found Z = 15



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Re: At a certain upscale restaurant, there just two kinds of food items: e
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17 Mar 2015, 23:17
Answer= E = 15 Let the number of entrees = x, then number of appetizers = 15x Given that \(\frac{30x + 12(15x)}{15} = 18\) Solving we get x = number of entrees = 5 & number of appetizers = 10 Say number of appetizers added = y New equation \(= \frac{30*5 + 12(10+y)}{15+y} = 15\) y = 15
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Re: At a certain upscale restaurant, there just two kinds of food items: e
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18 Mar 2015, 04:13
PareshGmat wrote: Answer= E = 15
Let the number of entrees = x, then number of appetizers = 15x
Given that \(\frac{30x + 12(15x)}{15} = 18\)
Solving we get x = number of entrees = 5 & number of appetizers = 10
Say number of appetizers added = y
New equation \(= \frac{30*5 + 12(10+y)}{15+y} = 15\)
y = 15 Hi Paresh, One question : The current price is as follows: Each entrée item costs $30, and each appetizer item costs $12. Last year the breakup was > Last year, it had a total of 15 food items on the menu, and the average price of a food item on its menu was $18. Then why are we setting up a equation which takes current prices and average prices from last year ? \(\frac{30x + 12(15x)}{15} = 18\) Thanks Lucky



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Re: At a certain upscale restaurant, there just two kinds of food items: e
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18 Mar 2015, 08:29
PareshGmat wrote: Answer= E = 15
Let the number of entrees = x, then number of appetizers = 15x
Given that \(\frac{30x + 12(15x)}{15} = 18\)
Solving we get x = number of entrees = 5 & number of appetizers = 10
Say number of appetizers added = y
New equation \(= \frac{30*5 + 12(10+y)}{15+y} = 15\)
y = 15 This is pretty straightforward and I took the same approach, but it still took me over 3 minutes to work out the equations and solve. 1. Is this too long to take for this type of question? 2. Any tips to increase speed?



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At a certain upscale restaurant, there just two kinds of food items: e
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18 Mar 2015, 09:26
mr7183 wrote: This is pretty straightforward and I took the same approach, but it still took me over 3 minutes to work out the equations and solve.
1. Is this too long to take for this type of question? 2. Any tips to increase speed? hi, 1. 3 minutes is slightly more and specially when you feel it is straightforward.. 2. ofcourse you need to tackle different type of Q in different ways to lower the time..  few Q are best by choosing/putting appropriate value for variable and then check the answer  few are best solved by putting the value given in choices  few through proper mathematical way... in this Q, best method is weighted average but you require to practice these.. At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrée item costs $30, and each appetizer item costs $12. Last year, it had a total of 15 food items on the menu, and the average price of a food item on its menu was $18. This year, it added more appetizer items, and the average price of a food item on its menu dropped to $15. How many appetizer items did it add this year? A. 3 B. 6 C. 9 D. 12 E. 15 first lets find initial ratio of E:A... there average values are 12 and 30 and combined average brings it to 18... so A:E=3018:1812=12:6=2:1 or A=10 and E=5 as total is 15.. let y appetizers be added so number of A=10+y.... and now combined average is 15 so A:E=3015:1512=5:1=10+y:5..... \(\frac{5}{1}=\frac{(10+y)}{5}\) 10+y=25... so y=15... ans E..
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Re: At a certain upscale restaurant, there just two kinds of food items: e
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18 Mar 2015, 10:10
The total cost of appetizers + entrees last year = $18 * 15
Let there be 'x' appetizers added this year, then total cost of food items = $15 * (15+x)
Since, cost of each appetizer is $12, we have
18*15 + 12x = 15*15 + 15x 3x = 3*15
Therefore, no. of appetizers added this year, x = 15
Answer (E)



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Re: At a certain upscale restaurant, there just two kinds of food items: e
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23 Mar 2015, 04:10
Bunuel wrote: At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrée item costs $30, and each appetizer item costs $12. Last year, it had a total of 15 food items on the menu, and the average price of a food item on its menu was $18. This year, it added more appetizer items, and the average price of a food item on its menu dropped to $15. How many appetizer items did it add this year?
A. 3 B. 6 C. 9 D. 12 E. 15
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:Method I: using sumsFirst, last year. Let x be the number of entrees. Then (15 – x) is the number of appetizers. The sums are: entrees = 30x appetizers = (15 – x)*12 = 12*15 – 12x = 6*30 – 12x = 180 – 12x total = 15*18 = 30*9 = 270 Notice the use of the Doubling and Halving trick in the second and third lines. The two individual sums should add up to the total sum. 30x + 180 – 12x = 270 18x = 90 x = 5 They start out with 5 entrees and 10 appetizers. Let N be the number of appetizers added, so now there are 5 entrees and (10 + N) appetizers. We need to solve for N. Again, the sums: entrees = 5*30 = 150 appetizers = (10 + N)*12 = 120 + 12N total = (15 + N)*15 = 225 + 15N Again, the two individual sums should add up to the total sum. 150 + 120 + 12N = 225 + 15N 270 = 225 + 3N 45 = 3N 15 = N They added 15 more appetizers. Answer = (E) Method I was doable, but we had to solve for many values. Method II: proportional placement of the total averageOriginally, the entrée price was 30 – 18 = 12 from the total average, and the appetizer price was 18 – 12 = 6. This means there must have been twice as many appetizers as entrees. Therefore , with 15 items, there must have been 10 appetizers and 5 entrees. The number of entrees doesn’t change. The average drops to $15, so the distance from the entrée price is now 30 – 15 = 15, and the distance from the appetizer price is now 15 – 12 = 3. That’s a 5to1 ratio, which means there must be 5x as many appetizers as entrees. Since there still are 5 entrees, there must now be 25 appetizers, so 15 have been added. Answer = (E) If you know how to employ this method, it is much more elegant.
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Re: At a certain upscale restaurant, there just two kinds of food items: e
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21 Dec 2015, 02:35
I directly solved for X the number of appetizers added. We know that [(18 x 15) + 12 x X ] / (15 + X) = 15 We can solved directly for X without caring what was the number of appetizers last year.



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Re: At a certain upscale restaurant, there just two kinds of food items: e
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21 Dec 2015, 04:15
Total of last year = 15*18 = 270
this year > (270 + 12x)/(15+x) = 15
x= 15



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