Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

51% (01:00) correct 49% (00:52) wrong based on 41 sessions

HideShow timer Statistics

[GMAT math practice question]

I, …, E, …, M, …, S, …, K, …, W, … Z The frequencies of the appearance of the letters from ‘A’ to ‘Z’ in a book are listed in decreasing order as shown above. ‘I’ is the most frequently occurring letter. The book contains a total of 100,000 alphabetic characters. Is the probability of randomly selecting an ‘S’ from the alphabetic characters appearing in the book greater than \(\frac{1}{27}\)?

1) The probability of selecting ‘M’ is greater than \(\frac{1}{27}\). 2) The probability of selecting ‘K’ is greater than \(\frac{1}{26}\).

I, …, E, …, M, …, S, …, K, …, W, … Z The frequencies of the appearance [#permalink]

Show Tags

22 Dec 2017, 08:01

MathRevolution wrote:

[GMAT math practice question]

I, …, E, …, M, …, S, …, K, …, W, … Z The frequencies of the appearance of the letters from ‘A’ to ‘Z’ in a book are listed in decreasing order as shown above. ‘I’ is the most frequently occurring letter. The book contains a total of 100,000 alphabetic characters. Is the probability of randomly selecting an ‘S’ from the alphabetic characters appearing in the book greater than \(\frac{1}{27}\)?

1) The probability of selecting ‘M’ is greater than \(\frac{1}{27}\). 2) The probability of selecting ‘K’ is greater than \(\frac{1}{26}\).

let the frequencies of appearance of each alphabet be represented by that alphabet.

so the frequency of "S" will be \(s\)

Hence probability of selecting S \(= \frac{s}{100,000}=\frac{s}{10^5}\)

we need to find IS \(\frac{s}{10^5}>\frac{1}{27} => s>\frac{10^5}{27} ?\)

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables which are the probability of S, P(S), that of M, P(M) and that of K, P(K) and we have 1 equation which is P(M) > P(S) > P(K), C is most likely to be the answer and so we should consider both conditions 1) & 2) together first.

Conditions 1) & 2) Since \(P(M) > P(S) > P(K) > \frac{1}{26} > \frac{1}{27}\), both conditions together are sufficient.

Since this is an inequality question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1) If \(P(M) > P(S) = \frac{1}{27}\), the answer is ‘no’. If \(P(M) = \frac{1}{25} and P(S) = \frac{1}{26} > \frac{1}{27}\), the answer is ‘yes’. Thus, condition 1) is not sufficient.

Condition 2) Since \(P(S) > P(K) > \frac{1}{26} > \frac{1}{27}\), condition 2) is sufficient.

Therefore, B is the answer.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

Re: I, …, E, …, M, …, S, …, K, …, W, … Z The frequencies of the appearance [#permalink]

Show Tags

25 Dec 2017, 22:10

MathRevolution wrote:

[GMAT math practice question]

I, …, E, …, M, …, S, …, K, …, W, … Z The frequencies of the appearance of the letters from ‘A’ to ‘Z’ in a book are listed in decreasing order as shown above. ‘I’ is the most frequently occurring letter. The book contains a total of 100,000 alphabetic characters. Is the probability of randomly selecting an ‘S’ from the alphabetic characters appearing in the book greater than \(\frac{1}{27}\)?

1) The probability of selecting ‘M’ is greater than \(\frac{1}{27}\). 2) The probability of selecting ‘K’ is greater than \(\frac{1}{26}\).

(1) Probability of selecting M is > 1/27, there are a lot of possibilities > 1/27. And since S appears less often than M, its possible that Probability of M occurring is > 1/27, it could be = 1/27 or it could be < 1/27. We cant say. Insufficient.

(2) Probability of selecting K is > 1/26.. Now 1/26 is greater than 1/27. Now, since S appears more frequently than K, Probability of selecting S will be more than that of K. So Probability of selecting S > Probability of selecting K > 1/26. Thus probability of selecting S > 1/26, so obviously its greater than 1/27. Sufficient.