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A pizza can be assembled from given choices, with customers'

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A pizza can be assembled from given choices, with customers'

by BTGmoderatorDC » Sun Aug 18, 2019 5:09 pm

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A pizza can be assembled from given choices, with customers' preferences. The crust is available in 3 varieties - cheese burst, classic, and wheat thin; toppings are available in 4 variants - bacon, black olives, pepperoni and sausage; and finally cheese available in 3 types - Cheddar, American, and Swiss. How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

A. 3
B. 9
C. 12
D. 36
E. 60

OA E

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by deloitte247 » Sat Aug 24, 2019 6:16 am
Given that:
Number of varieties for pizza crust = 3
Number of varieties for toppings = 4
Number of varieties for cheese = 3
Customer cannot choose either of toppings or cheese alone
Possible ways in which the pizza can be made :
- Pizza made with crust only = 3C1 = 3*1 = 3
- Pizza made with crust and toppings = 3C1 * 4C1
= (3*1) * (4*1)
= 3*4 = 12
- Pizza made with crust and cheese = 3C1 * 3C1
=(3*1) * (3*1)
= 3*3 = 9
- Pizza made with crust, cheese and toppings = 3C1 * 3C1 * 4C1
=(3*1) * (3*1) * (4*1)
= 3 * 3 * 4
= 9*4 = 36
Various types of pizzas that can be made = pizza with crust only + pizza with crust and toppings + pizza with crust and cheese + pizza with crust, cheese and toppings
= 3 + 12 + 9 + 36
= 60 various types of pizzas
Answer = option E
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by Ian Stewart » Sat Aug 24, 2019 7:40 am
It's not clear what the question is asking - does the customer have the option to omit either toppings or cheese, or can she omit both? How many toppings are allowed on a pizza?

Based on the answer choices, I can guess what they intend: a customer can pick one crust, one cheese, and one topping, but for toppings and for cheese, the customer can also choose none. So the customer has 3 choices for crust, 5 choices for topping (one of the 4 available, or none), and 4 choices for cheese (one of the 3 available, or none), for a total of (3)(5)(4) = 60 choices.

But if you notice that a customer choosing one item for each group would have (3)(4)(3) = 36 options, you can tell instantly that the answer must be 60, the only answer choice larger than 36, since with additional options (omitting toppings, etc) there will be more possibilities in total.
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by Scott@TargetTestPrep » Sun Aug 25, 2019 5:31 pm
BTGmoderatorDC wrote:A pizza can be assembled from given choices, with customers' preferences. The crust is available in 3 varieties - cheese burst, classic, and wheat thin; toppings are available in 4 variants - bacon, black olives, pepperoni and sausage; and finally cheese available in 3 types - Cheddar, American, and Swiss. How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

A. 3
B. 9
C. 12
D. 36
E. 60

OA E

Source: e-GMAT
The customer has 3 choices of crust, 5 choices of toppings (either one of the 4 toppings or none) and 4 choices of cheese (either one of the 3 types or none). Thus, a total of 3 x 5 x 4 = 60 different pizzas can be ordered.

Answer: E

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by Brent@GMATPrepNow » Mon Aug 26, 2019 5:56 am
BTGmoderatorDC wrote:A pizza can be assembled from given choices, with customers' preferences. The crust is available in 3 varieties - cheese burst, classic, and wheat thin; toppings are available in 4 variants - bacon, black olives, pepperoni and sausage; and finally cheese available in 3 types - Cheddar, American, and Swiss. How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

A. 3
B. 9
C. 12
D. 36
E. 60

OA E

Source: e-GMAT
ASIDE: On the GMAT, "or" is an INCLUSIVE OR, which means both options can occur (unless stated otherwise).
By the looks of the OA, we are supposed to assume that the pizza has only ONE topping.

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

Stage 1: Select a crust
There are 3 crusts from which to choose, so, we can complete stage 1 in 3 ways

Stage 2: Select a topping
There are 5 options: bacon, black olives, pepperoni, sausage and no topping
So, we can complete this stage in 5 ways.

Stage 3: Select a cheese
There are 4 options: Cheddar, American, and Swiss and no cheese
So we can complete this stage in 4 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus build a pizza) in (3)(5)(4) ways (= 60 ways)

Answer: E

--------------------------

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