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Estimation Question

by saadishah » Thu Dec 03, 2015 1:05 am
This is an OG question for which explanation is given. I tried applying the explanation on a smaller set however it doesn't work (for me). So either I do not follow the explanation or am trying to apply it to a wrong set. Please help me learn the principles that are used in the explanation so that I can solve a similar question myself next time.

Q. 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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by GMATGuruNY » Thu Dec 03, 2015 3:16 am
M is the sum of 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
If all of the 100 values were 1/300, the result would be the following sum:
100(1/300) = 1/3.
Since all but one of the values are actually GREATER THAN 1/300, the value of M must be GREATER THAN 1/3.

If all of the 100 values were 1/200, the result would be the following sum:
100(1/200) = 1/2.
Since all of the values are actually LESS THAN 1/200, the value of M must be LESS THAN 1/2.

Thus:
1/3 < M < 1/2.

The correct answer is A.

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by [email protected] » Thu Dec 03, 2015 10:04 am
Hi saadishah,

In this prompt, the answer choices are "ranges"; this usually means that there's a way to avoid doing lots of math and instead use patterns and logic to save you time.

Mitch's already pointed out the easiest way to figure out the minimum and maximum values of the sum of the reciprocals. You can actually stop working once you figure out the minimum though:

Since 1/300 < 1/201 and the sum of those 100 terms would be 1/3 AT THE MINIMUM, the only answer that's possible would be A. The extra work that Mitch did just confirms the maximum value of the sum, but it's unnecessary.

As you continue to study, be mindful of how the answer choices are written - they can sometimes provide a huge hint into the fastest way to answer the question.

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by Brent@GMATPrepNow » Thu Dec 03, 2015 10:15 am
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
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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by bhoglund » Fri Nov 04, 2016 1:23 pm
All the explanations provided so far have been very helpful! Another way to solve this problem is as follows:

- Numerator: There are 100 fractions in the set. All of them have a numerator of 1, so 1 x 100 = 100.
- Denominator: The average value of all the denominators in the set is 250.5. (201+300)/2 = 250.5.

Then simplify 100/250.5 to 1/2.505.

1/3 < 1/2.505 < 1/2

The answer is A.
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by Scott@TargetTestPrep » Mon Nov 07, 2016 10:55 am
saadishah wrote:

Q. 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 difficult math (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.

This gives us:

1/300 x 100 = 1/3

1/200 x 100 = 1/2

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

Answer: A

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by Matt@VeritasPrep » Fri Nov 11, 2016 3:37 pm
bhoglund wrote:All the explanations provided so far have been very helpful! Another way to solve this problem is as follows:

- Numerator: There are 100 fractions in the set. All of them have a numerator of 1, so 1 x 100 = 100.
- Denominator: The average value of all the denominators in the set is 250.5. (201+300)/2 = 250.5.

Then simplify 100/250.5 to 1/2.505.

1/3 < 1/2.505 < 1/2

The answer is A.
This is a little dangerous. It does give you a ballpark, but denominators don't average in the same way that numerators would: (1/2 + 1/3)/2 ≠ 1/2.5, (1/2 + 1/3 + 1/4)/3 ≠ 1/3, etc. The average is "close" in some sense, but whether that close is close enough depends on the answers given.
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