Problem Solving

Hi danielheulensen,

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.

We're asked to figure out the SUM of the RECIPROCALS of the integers from 201 to 300, inclusive. Since 1/300 < 1/201, the sum of those 100 terms would be.... (100)(1/300) = 1/3.... at the MINIMUM. Thus, the only answer that's possible... The extra work that you might do to calculate the maximum value of the sum would be unnecessary.

Final Answer: A

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.

GMAT assassins aren't born, they're made,
Rich

Hi danielheulensen,

Since the Quant section of the GMAT is NOT a 'math test', you will never face a question on Test Day that requires lots of complex calculations to solve (so if you think that's what you need to do to get to the correct answer, then there's almost certainly some other approach that is far easier and faster). It's also important to remember that GMAT questions are not 'randomly' put together - the wording, the numbers involved and the 5 answer choices are all carefully designed to test your critical thinking skills (and in many cases, to provide 'shortcuts' that will help you save time). Thus, part of your training for this Test has to ultimately focus on Tactics and pattern-matching, so that you can take full advantage of all of the opportunities that come your way on Test Day.

GMAT assassins aren't born, they're made,
Rich

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

Here is some general advice on this question: https://www.beatthegmat.com/m-is-the-sum ... tml#793900

danielheulensen wrote:100. 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 100 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 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 = ½. This value is the high estimate of M.

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

Answer: A