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Simple one :)

by theCodeToGMAT » Sat Sep 28, 2013 4:48 am
A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option to add extra cheese, mushrooms, or both to any kind of pizza, how many different pizza varieties are available?

a) 12
b) 16
c) 8
d) 20
c) 10
OA {B}
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by vinay1983 » Sat Sep 28, 2013 5:35 am
theCodeToGMAT wrote:A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option to add extra cheese, mushrooms, or both to any kind of pizza, how many different pizza varieties are available?

a) 12
b) 16
c) 8
d) 20
c) 10
OA {B}
Is it 16? 12 with the cheese and and stuff and the regular 4
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by theCodeToGMAT » Sat Sep 28, 2013 5:37 am
Yes.. correct..
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by Brent@GMATPrepNow » Sat Sep 28, 2013 6:39 am
theCodeToGMAT wrote:A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option to add extra cheese, mushrooms, or both to any kind of pizza, how many different pizza varieties are available?

a) 12
b) 16
c) 8
d) 20
c) 10
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 ([spoiler]= 16 ways[/spoiler])

Cheers,
Brent

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by Brent@GMATPrepNow » Sat Sep 28, 2013 6:50 am
theCodeToGMAT wrote:A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option to add extra cheese, mushrooms, or both to any kind of pizza, how many different pizza varieties are available?

a) 12
b) 16
c) 8
d) 20
c) 10
OA {B}
Since the 5 answer choices are all relatively small, we can also consider listing all of the possible options as a viable approach.

P = pepperoni
C = chicken
H = hawaiian
V = vegetarian
E = extra cheese
M = mushrooms

All possible pizzas
- P
- C
- H
- V
- PE
- CE
- HE
- VE
- PM
- CM
- HM
- VME
- PME
- CME
- HME
- VME

Total = [spoiler]16 = B[/spoiler]

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by Rich.C@EMPOWERgmat.com » Sat Sep 28, 2013 12:01 pm
Hi All,

This question is more about organization and thoroughness than math skills. The various explanations present different way to organize the work, so I'll add one more option:

With 4 types of pizzas (pepperoni, chicken, Hawaiian and vegetarian) and 4 options (AS IS, extra cheese, extra mushrooms, BOTH extra cheese and mushrooms).

4 x 4 = 16

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by nikhilgmat31 » Tue Oct 06, 2015 11:33 pm
Brent@GMATPrepNow wrote:
theCodeToGMAT wrote:A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option to add extra cheese, mushrooms, or both to any kind of pizza, how many different pizza varieties are available?

a) 12
b) 16
c) 8
d) 20
c) 10
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 ([spoiler]= 16 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Hi Brent
I think Answer should be 12

since 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian.
3 Options -Add extra cheese, mushrooms, or both to any kind of pizza,

Total Options - 4 * 3 = 12

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by theCEO » Wed Oct 07, 2015 2:13 am
nikhilgmat31 wrote:
Brent@GMATPrepNow wrote:
theCodeToGMAT wrote:A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option to add extra cheese, mushrooms, or both to any kind of pizza, how many different pizza varieties are available?

a) 12
b) 16
c) 8
d) 20
c) 10
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 ([spoiler]= 16 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Hi Brent
I think Answer should be 12

since 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian.
3 Options -Add extra cheese, mushrooms, or both to any kind of pizza,

Total Options - 4 * 3 = 12
There are actually 4 options: Add extra cheese, mushrooms, both to any kind of pizza or add no toppings (cheese or muchroom)

The question states that the customer has the option to add the toppings. This means that he does not have to add any toppings
Last edited by theCEO on Wed Oct 07, 2015 2:18 am, edited 1 time in total.

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by Rich.C@EMPOWERgmat.com » Wed Oct 07, 2015 9:05 am
Hi nikhilgmat31,

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

Total = 4+4+4+4 = 16 varieties

Final Answer: B

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by patco » Wed Oct 07, 2015 1:28 pm
vinay1983 wrote:
theCodeToGMAT wrote:A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option to add extra cheese, mushrooms, or both to any kind of pizza, how many different pizza varieties are available?

a) 12
b) 16
c) 8
d) 20
c) 10
OA {B}
Is it 16? 12 with the cheese and and stuff and the regular 4
At first glance I would say just 12.. But, you gave the right answer extremely fast and with a logic! :)
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by nikhilgmat31 » Wed Oct 07, 2015 9:04 pm
Thanks Rich, I missed the option of No Toppings.