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Beth s pizzeria offers x different toppings. What is the

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Beth s pizzeria offers x different toppings. What is the [#permalink] New post 02 Jul 2007, 02:06
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Beth’s pizzeria offers x different toppings. What is the value of x?

(1) There are an equal number of different pizzas that can be made with (x − 2) toppings as there are different pizzas with just 2 toppings.

(2) If the pizzeria were to add one additional topping, the number of different pizzas that could be made with 4 toppings would double.
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 [#permalink] New post 02 Jul 2007, 02:57
I found D.

Statement1) SUFF:

n-#of different pizzas that can be made with (x-2) and 2 toppings.
then (x-2)*n=2*n x=4.

Statement2) SUFF. p-#of pizzas that can be made with 4 toppings.
(x+1)C4 * p=2* [xC4]*p "combination". x+1=2*[x-3] x=7

I wonder if D then why X=4 and X=7 are different.
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 [#permalink] New post 02 Jul 2007, 08:50
I would go with D , but iten 2 I need help to explain...
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 [#permalink] New post 02 Jul 2007, 09:20
in order to solve this problem - we need to know that if we have x toppings then the sum of ways we can choose toppings on a certain pizza equal to 2^x.

this is a rule:

see also attached link:

http://www.gmatclub.com/phpbb/viewtopic.php?p=335466

statement 1


2^2 = 2^(x-2)

x = 4

sufficient

statement 2

2^4 = (2^(4+1))/2

x = 4

sufficient

the answer is (D)

vshaunak@gmail.com - did you write this problem yourself ? if so - well done on the initiative :-D

:-D
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 [#permalink] New post 02 Jul 2007, 15:37
your formula state that for 2 toppings one can have 2^2=4 different pizza
that is A, B, AB, BA
but pertaining to pizza, BA and AB is exactly the same thing so there must be only 3 ways to make pizza with 2 toppings and not 4.

I find D for the same reason of UMB, and i am also wondering why x=4 in one statement equal 7 in the other???
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 [#permalink] New post 02 Jul 2007, 20:55
boubi wrote:
your formula state that for 2 toppings one can have 2^2=4 different pizza that is A, B, AB, BA
but pertaining to pizza, BA and AB is exactly the same thing so there must be only 3 ways to make pizza with 2 toppings and not 4.

I find D for the same reason of UMB, and i am also wondering why x=4 in one statement equal 7 in the other???


Hello boubi:

A = topping 1
B = topping 2

way 1 = none
way 2 = A
way 3 = B
way 4 = AB (or BA it's the same)

2^2 = 4

this formula works for any number of toppings

A = topping 1
B = topping 2
C = topping 3

way 1 = none
way 2 = A
way 3 = B
way 4 = C
way 5 = AB
way 6 = AC
way 7 = BC
way 8 = ACB

2^3 = 8

:-D
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 [#permalink] New post 03 Jul 2007, 01:26
ok i will write down this formula. things are clearer now. Thanks KillerSquirrel
:)
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 [#permalink] New post 03 Jul 2007, 01:36
UMB wrote:
I found D.

Statement1) SUFF:

n-#of different pizzas that can be made with (x-2) and 2 toppings.
then (x-2)*n=2*n x=4.

Statement2) SUFF. p-#of pizzas that can be made with 4 toppings.
(x+1)C4 * p=2* [xC4]*p "combination". x+1=2*[x-3] x=7

I wonder if D then why X=4 and X=7 are different.


UMB! for statement 2, i think that our reasoning was wrong.
*C4 is the way to choose how many pizza can you make with a group of 4 topping and only 4. but we forgot that we can makee with 4 toppings *C3 *C2 an *C1 pizzas.
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 [#permalink] New post 03 Jul 2007, 02:57
I'm not english native speaker and don't know pizza cooking technology. Can anybody explain me what topping is and why for one pizza we can use 0, 1 or 2 toppings. Why not 3 or 4?

Thank you in advance.
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 [#permalink] New post 03 Jul 2007, 03:11
Andrey2010 wrote:
I'm not english native speaker and don't know pizza cooking technology. Can anybody explain me what topping is and why for one pizza we can use 0, 1 or 2 toppings. Why not 3 or 4?

Thank you in advance.


see:

http://www.pizzahut.com/Menu.aspx?tab=pizzaToppings

:-D
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 [#permalink] New post 03 Jul 2007, 03:27
Sorry don't agree with you guys
My answer is B
Please post OA
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 [#permalink] New post 03 Jul 2007, 04:32
1) not sufficient as xCy always equal to xC(x-y)
2) sufficient: 2*xC4 = (x+1)C4 , x=7

Squirrel, you shouldn't take into consideration option of 1,2, 3 etc toppings.
Question asks about combinations - how many pizzas can we make using (x-2) toppings if there are totally x toppings.

The answer is B
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Re: [#permalink] New post 10 Jan 2008, 03:16
boubi wrote:
your formula state that for 2 toppings one can have 2^2=4 different pizza
that is A, B, AB, BA
but pertaining to pizza, BA and AB is exactly the same thing so there must be only 3 ways to make pizza with 2 toppings and not 4.

I find D for the same reason of UMB, and i am also wondering why x=4 in one statement equal 7 in the other???



it is not obligatory that two answers must be equal, when you state that each answer is sufficient you mean that they are sufficient isolated from each other, information in statement two is not related to information in statement 1.
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Re: DS Perm/Comb [#permalink] New post 10 Jan 2008, 07:59
Quote:
Beth’s pizzeria offers x different toppings. What is the value of x?

(1) There are an equal number of different pizzas that can be made with (x − 2) toppings as there are different pizzas with just 2 toppings.

(2) If the pizzeria were to add one additional topping, the number of different pizzas that could be made with 4 toppings would double.


1. (x-2) = 2 so x = 4
There's no other number of toppings that will give you exact number of options as having 2 toppings except for 2. Even though they're disagreeing over there being 3 or 4 options, it doesn't really matter. The fact is, to get the same number of options as with 2 toppings you need to have 2 toppings.

and you can have 4 options with 2 toppings (call them x and y) assuming neither is an option.

1. x only
2. y only
3. both x and y
4. neither x nor y

2. The number of options for a 4 topping pizza:
4C4 = 1
5C4 = 5

5C4 = 5
6C4 = 15

6C4 = 15
7C4 = 35

7C4 = 35
8C4 = 70 WORKS

Only adding 1 more topping to a current list of 7 toppings will double the number of 4 topping pizzas.


Unless I'm mistaken we are getting sufficient answers from each statement and the answers are different from each other. I've never seen this before. Is this allowed on the GMAT?
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Re: DS Perm/Comb [#permalink] New post 10 Jan 2008, 08:20
statement 1 and 2 cannot lead to contradicting answers or several answers.
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Re: DS Perm/Comb [#permalink] New post 10 Jan 2008, 08:28
bmwhype2 wrote:
statement 1 and 2 cannot lead to contradicting answers or several answers.


do you see any flaws in how I approached the problem? I'm getting contradicting answers and am pretty confident in my work.

Maybe it's a made up question?
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Re: DS Perm/Comb [#permalink] New post 10 Jan 2008, 09:54
eschn3am wrote:
bmwhype2 wrote:
statement 1 and 2 cannot lead to contradicting answers or several answers.


do you see any flaws in how I approached the problem? I'm getting contradicting answers and am pretty confident in my work.

Maybe it's a made up question?


1. is pretty clear and I came to the same solution
2. i tought pC(x+1)=2pC4...then, for any value of p would find the value of x....anyway it would be a very tough equation, although we don't need to calculate it.

I didn't actually catch the rule about 2^n toppings ! can someone explain that to me clearly.
I appreciate.
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Re: DS Perm/Comb [#permalink] New post 10 Jan 2008, 10:27
for n toppings there are 2^n ways to make a pizza. Example:

1 topping 2^1 = 2 (choices are with topping or without
2 toppings 2^2 = 4 (topping 1, topping 2, both or neither)
3 toppings 2^3 = 8 (t1, t2, t3, t1 and t2, t1 and t3, t2 and t3, t1, t2, and t3, no toppings)

This happens because you have two choices for each topping, either you put it on the pizza or you don't. The more toppings you have the more choices you end up with.

this isn't the same thing we're doing for statement 2 though because we can only use 4 toppings, no more no less when making our pizza. In that case it's NC4 (where N = number of available toppings). order doesn't matter since it's all going on the pizza.
Re: DS Perm/Comb   [#permalink] 10 Jan 2008, 10:27
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