anceer wrote:

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.

\(? = x\,\,\)

(1) In this statement we must implicitly assume that x is at least 3, because no pizzas can be made with x-2 toppings if x is less than that.

> If there are exactly x=3 toppings, with 3-2=1 topping we can make 3 pizzas, exactly the same number of different pizzas we can make with just 2 toppings: C(3,2) = 3.

> If there are exactly x=4 toppings, with 4-2=2 toppings we can make exactly the same number of different pizzas that with just 2 toppings... sure!

(2) In this statement

we must implicitly assume that x is at least 4 (*), because no pizzas could be made with 4 toppings if x is less than that.

(If, for instance, we had only x+1=3 toppings after 1 topping was added, the number 0 would be equal to twice itself, but we are not dealing with this scenario as viable.)

The number of different pizzas that can be made with 4 toppings among x+1 toppings (x at least 4) is

\(C\left( {x + 1,4} \right)\,\, = \,\,\frac{{\left( {x + 1} \right)!}}{{\,4!\,\,\left( {x - 3} \right)!\,}}\,\,\, = \,\,\,\frac{{\left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 2} \right)}}{{4!}}\)

and the number of different pizzas that can be made with 4 toppings among x toppings (x at least 4) is

\(C\left( {x,4} \right) = \frac{{x!}}{{\,4!\,\,\left( {x - 4} \right)!\,}} = \frac{{x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{4!}}\)

Therefore, using statement (2), we know that

\(\frac{{\left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 2} \right)}}{{4!}} = 2 \cdot \frac{{x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{4!}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\frac{{x\left( {x - 1} \right)\left( {x - 2} \right)}}{{4!}}\,\, \ne \,0\,\,\,\left( * \right)} \,\,\,\,\left( {x + 1} \right) = 2\left( {x - 3} \right)\)

\(\left( {x + 1} \right) = 2\left( {x - 3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 7\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\,\,\,\,\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,

Fabio.

_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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