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swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

At a ceratin upscale restaurant, there just two kinds of

by swerve » Sat Dec 28, 2019 10:31 am

At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrÃ©e 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

The OA is E

Source: Magoosh

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Source: Beat The GMAT — Problem Solving | Sat Dec 28, 2019 10:31 am

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pm; Followed by:5 members

by deloitte247 » Sun Dec 29, 2019 12:04 pm

Let the number of entrees = x
Then the number of appetizers = 15-x
Since each entree item cost $30 and appetizer $12
then entree = $30
appetizer = 12 (15-x) = 180 - 12x
Last year, total food items = 15
Average price = $18
$$Average\ price=\frac{Total\ price\ of\ food}{Total\ food\ items}$$
$$18=\frac{Total\ price\ of\ food}{15}$$
$$Total\ price\ of\ food=18\cdot15=270$$
i.e entree price + appetizer price = 270
30x + 180 - 12x = 270
18x = 270 - 180
18x = 90
x = 90/18
x = 5
Last year restaurant had 5 entrees and 10 appetizers
Let the number of appetizers added this year = n
number of entree = 5
number of appetizers = 10+n
Each entree item = $30 and appetizer = $12
Hence, entree = 5*30 = 150, and appetizer = (10+n)*12
Appetizer = 120 + 12n
Average price = 15
$$15=\frac{Total}{5+10+n}$$
$$Total=15\left(15+n\right)=225+15n$$
entree price + appetizer price = 225 + 15n
150 + 120 + 12n = 225 + 15n
270 - 225 = 3n
$$n=\frac{45}{3}=15$$
(i.e they added 15 appetizers)

Answer = option E

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Fri Jan 10, 2020 8:37 am

swerve wrote:At a certain upscale restaurant, there just two kinds of food items: entrees and appetizers. Each entrÃ©e 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

The OA is E

Source: Magoosh

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 entrÃ©es last year. Since the total number of items last year was 15, there were 10 appetizers and 5 entrÃ©es last year.

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

[30(5) + 12(10 + n)]/(15 + n) = 15

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

270 + 12n = 225 + 15n

45 = 3n

15 = n

Answer: E

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