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

Question

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.

A. 3 
B. 6 
C. 9 
D. 12 
E. 15

Barnal · Answer

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

PareshGmat · Answer

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

Lucky2783 · Answer

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
Lucky

mr7183 · Answer

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?

chetan2u · Answer

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

Kudos for a correct solution.

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..

rdoshi · Answer

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)

Bunuel · Answer

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.

Hawkmoon · Answer

I directly solved for X the number of appetizers added.
We know that   / (15 + X) = 15
We can solved directly for X without caring what was the number of appetizers last year.

douglasvg · Answer

Total of last year = 15*18 = 270

this year -> (270 + 12x)/(15+x) = 15

x= 15

ScottTargetTestPrep · Answer

We see that 30 is 12 away from 18, and 12 is 6 away from 18. Since 12/6 = 2, there were twice as many appetizers as entrées last year. Since the total number of items last year was 15, there were 10 appetizers and 5 entrées last year.

If we let n = the number of appetizers added this year, we can create the equation:

/(15 + n) = 15

150 + 120 + 12n = 15(15 + n)

270 + 12n = 225 + 15n

45 = 3n

15 = n

Answer: E

siglon · Answer

For those aiming at 700+ who really want to understand averages and weighted averages: 
This might not be the best way to solve this question, but it is very important to understand why the following is possible.
It is important to understand how a set's term's difference from the mean is relevant when working with averages.

Finding the distribution of food items before adding appetizers: 
We know the average (arithmetic mean) price, the price for each food item, and the total number of food items.
In a set, the sum of the differences between the mean and the terms are 0. In other words, we can find the distribution by multiplying the differences from the mean by variables for each food item type, and adding these differences together.

e=entrees 
 a=appetizers  
  -12e+6a=0   (-12 and 3 are the differences from the mean of 18)
  e+a=15   (The total number of food items)

From this we get   e=5   and   a=10  .

Then finding the distribution of food items after adding appetizers 
The amount of entrees stay the same, but the number of appetizers increase.
We know the average (arithmetic mean) price, the price for each food item, and the number of entrees (5).

-12(5)+3a=0  
  5+a=?

From the first equation we get   a=25  . So the number of appetizers increase from 10 to 25.

bumpbot · Answer

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

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