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I, …, E, …, M, …, S, …, K, …, W, … Z The frequencies of the appearance

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

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New post 22 Dec 2017, 06:45
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

56% (01:59) correct 44% (01:31) wrong based on 52 sessions

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[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 \(\frac{1}{26}\).

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

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New post 22 Dec 2017, 09:01
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 \(\frac{s}{10^5}>\frac{1}{27} => s>\frac{10^5}{27} ?\)

Statement 1: implies \(\frac{m}{10^5}>\frac{1}{27} =>m>\frac{10^5}{27}\)

Also frequency of M is greater than S, so we have \(m>s\)

But from this we can have two scenarios \(m>s>\frac{10^5}{27}\) or \(m>\frac{10^5}{27}>s\). Hence Insufficient

Statement 2: implies \(\frac{k}{10^5}>\frac{1}{26} => k>\frac{10^5}{26}\)

Also \(s>k=> s>k>\frac{10^5}{26}\)

Now \(\frac{10^5}{26}>\frac{10^5}{27}\) so we can say that

\(s>k>\frac{10^5}{26}>\frac{10^5}{27}\). Sufficient

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

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New post 25 Dec 2017, 18:02
=>

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.

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

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New post 25 Dec 2017, 23:10
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.

Thus B answer.
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Re: I, …, E, …, M, …, S, …, K, …, W, … Z The frequencies of the appearance &nbs [#permalink] 25 Dec 2017, 23:10
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