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M is the sum of the reciprocals of the consecutive integers

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barksdale Junior | Next Rank: 30 Posts Default Avatar
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M is the sum of the reciprocals of the consecutive integers

Post Thu Jun 01, 2017 9:21 pm
Totally confused about the process of getting to a solution in the 2 minute average time. Guess on this question?

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

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Post Mon Jun 05, 2017 11:37 pm
Also, when in doubt, try one (or more) of the following:

1) Try a smaller version of the same pattern to get a feel for what's going on;

2) Approximate;

3) Backsolve from the answers

It's amazing how many GMAT problems respond to one of those three approaches.

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Post Tue Jun 13, 2017 7:53 am
barksdale wrote:
M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9
Let's first analyze the question. We are trying to find a potential range for M in which M is the sum of the reciprocals from 201 to 300 inclusive. Thus, M is:

1/201 + 1/202 + 1/203 + … + 1/300

Since we probably would not be expected to do such time-consuming arithmetic (i.e., to add 100 fractions, each with a different denominator), that is exactly why the answer choices are in the form of an inequality. Thus, we do not need to know the EXACT value of M. The easiest way to determine the RANGE of values for M is to use easy numbers that can be quickly manipulated.

Notice that 1/200 is greater than each of the addends and that 1/300 is less than or equal to each of the addends. Therefore, instead of trying to add together 1/201 + 1/202 + 1/203 + … + 1/300, we are going to add 1/200 one hundred times and 1/300 one hundred times. These two sums will give us a high estimate of M and a low estimate of M. Again, we are adding 1/200 one hundred times and 1/300 one hundred times because there are 100 numbers from 1/201 to 1/300.
Instead of actually adding each one of these values one hundred times, we will simply multiply each value by 100:

1/300 x 100 = 1/3. This value is the low estimate of M.

1/200 x 100 = 1/2. This value is the high estimate of M.

We see that M is between 1/3 and 1/2.

Answer: A

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Matt@VeritasPrep GMAT Instructor
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Post Mon Jun 05, 2017 11:37 pm
Also, when in doubt, try one (or more) of the following:

1) Try a smaller version of the same pattern to get a feel for what's going on;

2) Approximate;

3) Backsolve from the answers

It's amazing how many GMAT problems respond to one of those three approaches.

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Post Tue Jun 13, 2017 7:53 am
barksdale wrote:
M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9
Let's first analyze the question. We are trying to find a potential range for M in which M is the sum of the reciprocals from 201 to 300 inclusive. Thus, M is:

1/201 + 1/202 + 1/203 + … + 1/300

Since we probably would not be expected to do such time-consuming arithmetic (i.e., to add 100 fractions, each with a different denominator), that is exactly why the answer choices are in the form of an inequality. Thus, we do not need to know the EXACT value of M. The easiest way to determine the RANGE of values for M is to use easy numbers that can be quickly manipulated.

Notice that 1/200 is greater than each of the addends and that 1/300 is less than or equal to each of the addends. Therefore, instead of trying to add together 1/201 + 1/202 + 1/203 + … + 1/300, we are going to add 1/200 one hundred times and 1/300 one hundred times. These two sums will give us a high estimate of M and a low estimate of M. Again, we are adding 1/200 one hundred times and 1/300 one hundred times because there are 100 numbers from 1/201 to 1/300.
Instead of actually adding each one of these values one hundred times, we will simply multiply each value by 100:

1/300 x 100 = 1/3. This value is the low estimate of M.

1/200 x 100 = 1/2. This value is the high estimate of M.

We see that M is between 1/3 and 1/2.

Answer: A

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Jeffrey Miller Head of GMAT Instruction

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Post Mon Jun 05, 2017 7:20 pm
barksdale wrote:
Totally confused about the process of getting to a solution in the 2 minute average time. Guess on this question?

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9
I think the biggest issue that students have with questions like this is that they grab their pen and start trying to solve before figuring out what is really going on in the question.

As the others have pointed out, this one isn't about actually adding reciprocals; it's about benchmarking and using common sense. But the only way to figure that out is to look at the answer choices and ask yourself: "why are they all ranges, and not an exact sum?" Clearly there is an exact value answer to this question, but that's not what the GMAT is looking for. If you grab your pen and start adding things before even looking at the answer choices, you would miss that point entirely!

Here's some good general advice for all PS: put the pen down and read the whole problem + answer choices before you start solving!

More on how and why to do that here: https://www.manhattanprep.com/gmat/blog/2016/03/10/want-to-do-better-on-gmat-quant-put-your-pen-down/

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Harvard Graduate School of Education


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Post Fri Jun 02, 2017 4:11 am
Quote:
M is the sum of the reciprocals of the consecutive integers from 201 to 300 inclusive. Which of the following is true?
A) 1/3 B)1/5 C)1/7 D) 1/9 < M < 1/7
E) 1/12
We want to find 1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300

NOTE: there are 100 fractions in this sum.

Let's examine the extreme values (1/201 and 1/300)

First consider a case where all of the values are equal to the smallest fraction (1/300)
We get: 1/300 + 1/300 + 1/300 + ... + 1/300 = 100/300 = 1/3
So, the original sum must be greater than 1/3

Now consider a case where all of the values are equal to the biggest fraction (1/201)
In fact, let's go a little bigger and use 1/200
We get: 1/200 + 1/200 + 1/200 + ... + 1/200 = 100/200 = 1/2
So, the original sum must be less than 1/2

Combine both cases to get 1/3 < M < 1/2 = A

Cheers,
Brent

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