All GMAT Math Resources
Example Questions
Example Question #1901 : Problem Solving Questions
A pizza parlor is offering a five-topping large pizza for $14.59. The toppings must be different, and they must include two meats and two vegetables; the fifth can be either. The meat toppings are pepperoni, beef, and sausage; the vegetable toppings are mushrooms, olives, onions, and green peppers.
How many possible ways can the toppings be chosen?
Apply combinations and the multiplication principle here.
One possibility is to order three of three meats, which can only be done one way, and two of four vegetables, which can be done
ways
The number of possible pizzas with three meats and two vegetables is .
The other possibility is to order two of three meats, which can be done
ways,
and three of four vegetables, which can be done
ways.
The number of possible pizzas with two meats and three vegetables is ways.
This is a total of ways to build a pizza to the given specifications.
Example Question #32 : Understanding Counting Methods
A college cafeteria offers sizes of pizza - small, medium, and large. With a small pizza, up to one topping is included without additional charge; for a medium or large, up to two toppings are included without additional charge. The cafeteria allows a double topping (i.e. double mushrooms) to count as two toppings.
The cafeteria offers three meat toppings - pepperoni, beef, and sausage. If the cafeteria offers other toppings, then how many ways can someone order a pizza -choosing a size and up to the maximum number of toppings- without having to pay extra?
There are toppings overall.
For a small pizza, since one topping is allowed, there are possible choices.
For a medium pizza, there are
ways to order two different toppings, and ways to order a double topping. The same numbers hold for a large pizza.
Add these:
Example Question #33 : Understanding Counting Methods
What is the 58th digit of the following repeating decimal?
What is the 58th digit of the following repeating decimal?
Begin this problem by realizing there is an easier way than counting out to the 58th digit.
The easier way requires us to divide 58 by the number of digits in the repeating decimal. In this case, we have 6.
So, perform the following:
9 remainder 4 means that the pattern will repeat 9 times before the 58th digit, and will then go 4 more. That means our answer must be 8, because 8 is the 4th digit in the sequence.