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A university cafeteria offers 4 flavors of pizza – pepperoni, chicken,
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Updated on: 12 Oct 2014, 01:43

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A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option (but not the obligation) to add extra cheese, mushrooms or both to any kind of pizza, how many different pizza varieties are available ?

Re: A university cafeteria offers 4 flavors of pizza – pepperoni, chicken,
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12 Oct 2014, 01:47

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jrymbei wrote:

A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option (but not the obligation) to add extra cheese, mushrooms or both to any kind of pizza, how many different pizza varieties are available ?

(A) 4 (B) 8 (C) 12 (D) 16 (E) 32

There are 4 flavors of pizza and each can be: 1. Without cheese and mushrooms; 2. With cheese; 3. With mushrooms; 4. With cheese and mushrooms.

So, the total number of pizza varieties is 4*4 = 16.

Re: A university cafeteria offers 4 flavors of pizza – pepperoni, chicken,
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13 Feb 2018, 21:45

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Hi All,

From the answer choices, we know that the number of total possibilities is limited, so we can 'map them' all out:

4 types of pizza and 4 options per pizza:

Pepperoni - as is, extra cheese, mushrooms, BOTH cheese/mushrooms = 4 options Chicken - as is, extra cheese, mushrooms, BOTH cheese/mushrooms = 4 options Hawaiian - as is, extra cheese, mushrooms, BOTH cheese/mushrooms = 4 options Vegetarian - as is, extra cheese, mushrooms, BOTH cheese/mushrooms = 4 options

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Re: A university cafeteria offers 4 flavors of pizza – pepperoni, chicken,
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12 Oct 2014, 01:53

4 flavours * 4 choices = 4C1*4C1 = 4*4=16

Note: The question says in that among the options of cheese, mushroom or both, they have the choice but not the obligation of picking the option. So we have a 4 options - cheese, mushroom, both or *none*.

Re: A university cafeteria offers 4 flavors of pizza – pepperoni, chicken,
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15 Feb 2018, 10:04

jrymbei wrote:

A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option (but not the obligation) to add extra cheese, mushrooms or both to any kind of pizza, how many different pizza varieties are available ?

(A) 4 (B) 8 (C) 12 (D) 16 (E) 32

We are given 4 flavors for pizza and there are 4 options for each flavor: no toppings, cheese-only, mushrooms-only, and both toppings. Thus, we have 4 x 4 = 16 different pizzas possible.

Re: A university cafeteria offers 4 flavors of pizza – pepperoni, chicken,
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06 Dec 2018, 16:13

Top Contributor

jrymbei wrote:

A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option (but not the obligation) to add extra cheese, mushrooms or both to any kind of pizza, how many different pizza varieties are available ?

(A) 4 (B) 8 (C) 12 (D) 16 (E) 32

Take the task of building a pizza and break it into stages.

Stage 1: Choose one of the flavors There are 4 flavors of pizza (pepperoni, chicken, Hawaiian and vegetarian), so we can complete stage 1 in 4 ways

Stage 2: Choose whether to add extra cheese We can either add extra cheese or not add extra cheese, so we can complete stage 2 in 2 ways.

Stage 3: Choose whether to add mushrooms We can either add mushrooms or not mushrooms, so we can complete stage 3 in 2 ways.

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus build a pizza) in (4)(2)(2) ways (= 16 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

Re: A university cafeteria offers 4 flavors of pizza – pepperoni, chicken,
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10 Dec 2019, 03:00

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