GMAT Math : Arithmetic

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Example Questions

Example Question #77 : Calculating Discrete Probability

Some balls are placed in a large box; the balls include one ball marked "A", two balls marked "B", and so forth up to twenty-six balls marked "Z". A ball is drawn at random.

Given a particular letter of the alphabet, does the probability that that ball will be marked with that letter exceed $\frac{1}{20}$ ?

Statement 1: The letter in question is in the word "pique".

Statement 2: The letter in question is a vowel.

(Note: for purposes of this problem, "Y" is a consonant.)

Possible Answers:

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

Correct answer:

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

Explanation:

The total number of balls in the box will be

1 + 2 + 3 + ... + 26 .

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 26= \frac{26 (26 + 1)}{2} = \frac{26 \cdot 27}{2} = 351$ .

The number of balls with a given letter of the alphabet is equal to the number of its position in the alphabet; the probability of a ball with that letter being drawn is that number divided by the total number of balls, 351. Therefore, for this probability to exceed $\frac{1}{20}$ , we must have the relation

$\frac{M}{351} > \frac{1}{20}$

$\frac{M}{351} \cdot 351 > \frac{1}{20} \cdot 351$

$M > \frac{351}{20} = 17 \frac{11}{20}$ .

Therefore, M > 17 .

The 17th letter of the alphabet is "Q", so in order to answer this question, it suffices to know whether the letter comes after "Q" in the alphabet.

Assume both statements to be true. There are three vowels in "pique"; "E" and "I" come before "Q" and "U", after "Q". Therefore, the two statements together are inconclusive.

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Example Question #111 : Outcomes

Aaron, Gary, Craig, and Boone are sitting down in a row of four chairs. What is the probability that Aaron and Gary will be seated beside each other?

Possible Answers:

$\frac{3}{4}$

$\frac{13}{24}$

$\frac{11}{12}$

$\frac{1}{2}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{2}$

Explanation:

Consider first all of the possible ways the men may be arranged, which is

4!=24

Now, consider all of the ways that Aaron and Gary could be seated beside each other; it may be easier to visualize by drawing it out:

A G _ _
G A _ _
_ A G _
_ G A _
_ _ A G
_ _ G A

As seen, there are six possibilities.

Finally, for each of these cases, Craige and Boone could be seated in one of two ways.

So the probability that Aaron and Gary will be seated beside each other is:

$\frac{2\cdot 6}{24}=\frac{12}{24}=\frac{1}{2}$

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Example Question #1 : Understanding Counting Methods

In how many ways can the 11th grade class elect a president, vice president, and treasurer from a class of 70 students?

Possible Answers:

$\dpi{100} \small 16,950$

$\dpi{100} \small 328,440$

$\dpi{100} \small 225,000$

$\dpi{100} \small 620,349$

$\dpi{100} \small 45,320$

Correct answer:

$\dpi{100} \small 328,440$

Explanation:

The president can be elected in 70 different ways. After a student is elected president, there are 69 students left to elect a vice president from. Similarly, there are then 68 students left for the spot of treasurer. So there are $\dpi{100} \small 70\times 69\times 68=328,440$ different arrangements.

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Example Question #321 : Arithmetic

What is the number of possible 4 letter code words that can be made from the alphabet, when all 4 letters must be different?

Possible Answers:

$\frac{26!}{22!}$

22!

$\frac{26!}{4!}$

26!

Correct answer:

$\frac{26!}{22!}$

Explanation:

This is a permutation of 26 objects (letters) taken 4 at a time. Here order matters, because for example, "abcd" is not the same code word as "bdca".

You must know the permutation formula! It is as follows:

$_{n}P_{r}=\frac{n!}{(n-r)!}$ , where n is the number of different objects taken r at a time.

Here we have $_{26}P_{4}=\frac{26!}{(26-4)!} = \frac{26!}{22!}$

Note: This is equivalent to 26 * 25 * 24 * 23.

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Example Question #1 : Counting Methods

There are 8 paths between places and and 5 paths betweeen places and . How many different routes are there between places and ?

Possible Answers:

336

170

Correct answer:

Explanation:

Multiple the number of routes for each piece of the trip: $8 \times 5 = 40$

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Example Question #1 : Counting Methods

How many subsets does a set with 12 elements have?

Possible Answers:

144

$\small 12^{12}$

132

4096

Correct answer:

4096

Explanation:

The number of subsets in a set of size is $\small 2^{n}$ . If n=12 , then the set has $\small \small \small \small 2^{12} = 4,096$ subsets.

Alternatively, each subset of this twelve-element set is essentially a sequence of 12 independent decisions, one per element - each decision has two possible outcomes, exclusion or inclusion. By the multiplication principle, this is 2 taken as a factor 12 times, or

$\small \small \small \small 2^{12} = 4096$

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Example Question #3 : Understanding Counting Methods

How many ways can a president, a vice-president, a secretary-treasurer, and three Student Senate representatives be selected from a class of thirty people? You may assume these will be six different people.

Possible Answers:

729,000,000

427,518,000

71,253,000

593,775

121,500,000

Correct answer:

71,253,000

Explanation:

This can be seen, without loss of generality, as choosing each officer in turn.

There are 30 ways of choosing the president; there are then 29 ways of choosing the vice-president, and 28 ways of choosing the secretary-treasurer. Then 3 Student Senate representatives are chosen from the remaining 27 students; this is a combination of 3 elements from 27 - that is, $C_{3}^{27}$ . By the multiplication principle, the number of possible selections of the officers is:

$N = 30 \cdot 29\cdot 28\cdot C_{3}^{27}$

$N= 30 \cdot 29\cdot 28\cdot \frac{ 27!}{(27-3)!\; 3!}$

$N= 30 \cdot 29\cdot 28\cdot \frac{ 27!}{24!3!}$

$N= \frac{30 \cdot 29\cdot 28\cdot 27 \cdot26\cdot25}{3\cdot 2\cdot 1}$

N= 71,253,000

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Example Question #2 : Counting Methods

How many ways can you select three different prime numbers between 1 and 20?

Possible Answers:

336

6,720

210

Correct answer:

Explanation:

There are eight prime numbers between 1 and 20:

$\left \{ 2,3,5,7,11,13,17,19\right \}$

The number of ways to select three of them, without regard to order, is the number of combinations of three out of eight:

$C(8,3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{6\cdot 7\cdot 8}{1\cdot 2\cdot 3}= 56$

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Example Question #4 : Understanding Counting Methods

A= 1-3+5-7+9-11+...+9,997-9,999

B = 2-4+6-8+10-12...+9,998 -10,000

Which of the following statements is true?

Possible Answers:

B= A + 2,500

B= A - 5,000

B= A - 2,500

B= A

B= A + 5,000

Correct answer:

B= A

Explanation:

The finite series is obtained from by increasing each term by 1; since is an alternating series, this results in adding to :

(1-1) + (1-1) + (1-1) + ...(1-1) = 0

so B = A + 0 = A

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Example Question #5 : Understanding Counting Methods

Define set $A = \left \{ 1,2,3,4,5,6,7,8\right \}$ .

How many four-element subsets of include at least three even numbers?

Possible Answers:

Correct answer:

Explanation:

Only one subset of has four even numbers - that is the subset $\left \{ 2,4,6,8\right \}$ .

Forming a subset with three even numbers and one odd number can be restated as choosing which even number to leave out and which odd number to include. There are 4 choices each way, so the number of sets fitting this description is $4\times 4 = 16$ .

Therefore, the number of subsets with at least three even elements is 16 + 1 = 17 .

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GMAT Math : Arithmetic

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #77 : Calculating Discrete Probability

Example Question #111 : Outcomes

Example Question #1 : Understanding Counting Methods

Example Question #321 : Arithmetic

Example Question #1 : Counting Methods

Example Question #1 : Counting Methods

Example Question #3 : Understanding Counting Methods

Example Question #2 : Counting Methods

Example Question #4 : Understanding Counting Methods

Example Question #5 : Understanding Counting Methods

All GMAT Math Resources

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