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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If k is an integer such that 56 < k < 66, what is the value of k ?

(1) If k were divided by 2, the remainder would be 1. (2) If k + 1 were divided by 3, the remainder would be 0.

In the original condition, there is 1 variable(k), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), k=2t+1=57,59,61...., which is not unique and not sufficient. For 2), k+1=3m, k=3m-1=59, 62, 65, which is not unique and not sufficient. When 1) & 2), n=59,65, which is not unique and not sufficient. Therefore, the answer is E.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

Re: If k is an integer such that 56 < k < 66, what is the value [#permalink]

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20 Jan 2016, 23:55

1

This post received KUDOS

k is an integer such that 56 < k < 66 which means k=(57,58,59,60,61,62,63,64,65) Now Statement 1 If k were divided by 2, the remainder would be 1. which means k is odd so possible values of k=(57,59,61,63,65), not sufficient

Statement 2 If k + 1 were divided by 3, the remainder would be 0 so the possible values of k+1 are 60,63,66 so possible values of k=(59,62,65), insufficient

Combining 1 and 2 we still get 2 values 59 and 65 so insufficient

Answer is E

Please consider a kudos if you get the explanation

If k is an integer such that 56 < k < 66, what is the value of k ?

(1) If k were divided by 2, the remainder would be 1. (2) If k + 1 were divided by 3, the remainder would be 0.

Solution:

We are given that k is between 56 and 66. We must determine the value of k.

Statement One Alone:

If k were divided by 2, the remainder would be 1.

This means k must be an odd number. However, since 56 < k < 66, k can be any odd integer between 56 and 66, which means that k could be 57, 59, 61, 63, or 65. Statement one is not sufficient to determine a value of k. We can eliminate answer choices A and D.

Statement Two Alone:

If k + 1 were divided by 3 the remainder would be 0.

This means k + 1 is a multiple of 3. However, since 56 < k < 66, there is more than one value between 56 and 66 that, when increased by 1, will be a multiple of 3. For example, since 60, 63 and 66 are all multiples of 3, we see that k could be 59, 62 or 65. Since we have three possible values for k, statement two is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

From statement one we know that k can be 57, 59, 61, 63, or 65, and from statement two we know that k could be 59, 62 or 65. From the combined statements, then, k could be either 59 or 65; thus, we can’t determine a unique value for k. Statements one and two together are not sufficient to answer the question.

The answer is E.
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

Re: If k is an integer such that 56 < k < 66, what is the value [#permalink]

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05 Oct 2017, 06:23

The 2 options state that K and K+1 being consecutive numbers are not divisible by 2 and 3 respectively. K not divisible by 2 but K+1 is divisible by 3. This is not possible and hence the answer is straight E.

Re: If k is an integer such that 56 < k < 66, what is the value [#permalink]

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13 Dec 2017, 22:21

somesh86 wrote:

The 2 options state that K and K+1 being consecutive numbers are not divisible by 2 and 3 respectively. K not divisible by 2 but K+1 is divisible by 3. This is not possible and hence the answer is straight E.

Hi Somesh86

The answer is indeed E but I have doubt in your explanation. Why is it not possible to have an integer K that is not divisible by 2, but simultaneously (K+1) to be divisible by 3. Eg, we can take K=5, here K is not divisible by 2, but K+1 = 6, IS divisible by 3.