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At a certain upscale restaurant, there just two kinds of food items: e

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At a certain upscale restaurant, there just two kinds of food items: e  [#permalink]

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New post 17 Mar 2015, 05:55
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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  [#permalink]

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New post 17 Mar 2015, 07:00
2
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  [#permalink]

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New post 17 Mar 2015, 23:17
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Answer= E = 15

Let the number of entrees = x, then number of appetizers = 15-x

Given that \(\frac{30x + 12(15-x)}{15} = 18\)

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  [#permalink]

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New post 18 Mar 2015, 04:13
PareshGmat wrote:
Answer= E = 15

Let the number of entrees = x, then number of appetizers = 15-x

Given that \(\frac{30x + 12(15-x)}{15} = 18\)

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 ?
\(\frac{30x + 12(15-x)}{15} = 18\)

Thanks
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Re: At a certain upscale restaurant, there just two kinds of food items: e  [#permalink]

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New post 18 Mar 2015, 08:29
PareshGmat wrote:
Answer= E = 15

Let the number of entrees = x, then number of appetizers = 15-x

Given that \(\frac{30x + 12(15-x)}{15} = 18\)

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  [#permalink]

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New post 18 Mar 2015, 09:26
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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=30-18:18-12=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=30-15:15-12=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  [#permalink]

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New post 18 Mar 2015, 10:10
1
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  [#permalink]

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New post 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 sums

First, 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 do-able, but we had to solve for many values.

Method II: proportional placement of the total average

Originally, 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 5-to-1 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  [#permalink]

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New post 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  [#permalink]

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New post 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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Re: At a certain upscale restaurant, there just two kinds of food items: e  [#permalink]

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Re: At a certain upscale restaurant, there just two kinds of food items: e   [#permalink] 07 Mar 2019, 19:04
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