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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
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carcass wrote:
If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is

A. 3

B. 5

C. 7

D. 9

E. 11


We can directly use the options as that will be the fastest method to this problem.
(1/5 + 1/7) = 12/35
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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
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carcass wrote:
If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is

A. 3

B. 5

C. 7

D. 9

E. 11

\(x = 5\) & \(y = 7\)

\(\frac{1}{x} + \frac{1}{y} = \frac{1}{5} + \frac{1}{7}\) = \(\frac{12}{35}\)

Thus, the greater of the two integers is 7 answer must be (C) 7
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Re: If the sum of the reciprocals of two consecutive odd integers is 12/35 [#permalink]
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jedit wrote:
If the sum of the reciprocals of two consecutive odd integers is 12/35, then the greater of the two integers is

A. 3
B. 5
C. 7
D. 9
E. 11

Searched this question before I posted but nothing showed up.


I always prefer to do such questions by using the options.

Three things to keep in mind:

1. We have odd integers
2. We need to find out the largest integer
3. Integers are consecutive.

Okay, let's take Option A.

If largest is 3, the smallest has to be 1. 1/1 + 1/3 is not equal to 12/35. OUT

Option B: 1/3 + 1/5 is NOT equal to 12/35. OUT

Option C: 1/5 + 1/7 = 12/35. It worked. Answer.
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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
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carcass wrote:
If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is

A. 3

B. 5

C. 7

D. 9

E. 11


Since the denominator of the sum is 35, and since 35 equals the product of 5 and 7 (two consecutive odd integers), let's start by testing whether 5 and 7 are the integers in question.

So, the RECIPROCALS are 1/5 and 1/7
1/5 + 1/7 = 7/35 + 5/35
= 12/35
Voila!!

So, the odd integers are 5 and 7
The greater of the two integers is 7
Answer:

Cheers,
Brent
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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
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Hi All,

We're told that the sum of the reciprocals of two CONSECUTIVE ODD integers is 12/35. We're asked for the GREATER of the two integers. This question can be solved by TESTing THE ANSWERS.

Let's TEST Answer B: 5
IF.... the values are 3 and 5....
then the sum of the reciprocals is 1/3 + 1/5 = 5/15 + 3/15 = 8/15
This is clearly not the correct answer, but the denominator ends in a '5', so it's likely that 12/35 will ALSO include a 5...

Let's TEST Answer C: 7
IF.... the values are 5 and 7....
then the sum of the reciprocals is 1/5 + 1/7 = 7/35 + 5/35 = 12/35
This is an exact match for what we were told, so this MUST be the answer.

Final Answer:

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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
ScottTargetTestPrep wrote:
6x^2 - 23x - 35 = 0

(6x + 7)(x - 5) = 0



I've always struggled with finding the zeroes when the coefficient is > 1 for the x^2 figure. Can anyone explain how you go from the 1st line to the second line?
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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
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Hi thesip2323,

To start, most GMAT questions can be approached in more than one way - and it's worth noting that you do NOT need to approach this question Algebraically to solve it (and by extension, if you are choosing methods that involve more complicated math than is necessary, then you might just be making everything harder than it needs to be).

That all having been said, when dealing with a Quadratic in which the first term includes a co-efficient that is greater than 1, a bit of 'brute force' Arithmetic is usually required to factor the Quadratic. Thankfully, there are usually context clues for what the possible values will be (and there won't be that many). Here, the clues are:

1) The "-35" at the end of the equation; this will be the PRODUCT of the two numbers, so we know that one number will be negative and one will be positive. Second, there are only a couple of ways (that involve integers) to get to 35: 1 x 35 and 5 x 7.

2) Similarly, the "6X^2" at the beginning is likely just one of two options: (X)(6X) or (2X)(3X).

3) The "-23X" in the middle will be the result of adding a positive and a negative (remember that "-35" at the end: the product of one positive and one negative, so the two terms that sum to make that middle term will be one positive and one negative). Since that term is not bigger than 35 (and not that close to 35 either), it's likely that we're NOT going to be dealing with 1 x 35 (it'll probably be a 5 x 7).

Now it's just a matter of doing the 'brute force' work. How long will it take to multiply and add the various possible calculations to get to "-23X"? It's important to do all of that work on your pad, since even if one of your calculations doesn't hit that total, you might spot a pattern that will help you to find the actual terms that will.

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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
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carcass wrote:
If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is

A. 3

B. 5

C. 7

D. 9

E. 11


The approach of looking at the denominator, noticing that you're probably going to want 5 and 7, and then testing the answer choice sure seems like a great one here! Here's another way.

12/35 is basically 1/3. We need two numbers that are each really close to half of that, one higher and one lower. Half of 1/3 is 1/6, so 1/5 and 1/7 sounds good.

Answer choice C.
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If the sum of the reciprocals of two consecutive odd integers [#permalink]
Consecutive odd will be of form (x-1) and (x+1)

sum of reciprocals -> 2x / (x^2 - 1) , but note x is the even between the two consecutive odds

Now if 3 is highest number then x = 2 , (i.e 1, 2, 3 are the consecutive integers and 1, 3 the consecutive odds)

putting the options in our equation will give

A) 2*2 = 4 in numerator
B) 2*4 = 8 in numerator
C) 2*6 = 12 numerator , just to validate also check denominator 6^2 - 1 = 35
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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
praveen8047 wrote:
carcass wrote:
If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is

A. 3

B. 5

C. 7

D. 9

E. 11


We can directly use the options as that will be the fastest method to this problem.
(1/5 + 1/7) = 12/35


does that mean that i need to combine 2 possibilites of the answer choices? like 3 and 5, 5 and 7, 7 and 9, 9 and 11; and see which one yield 12/35?
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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
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Thib33600 wrote:
praveen8047 wrote:
carcass wrote:
If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is

A. 3

B. 5

C. 7

D. 9

E. 11


We can directly use the options as that will be the fastest method to this problem.
(1/5 + 1/7) = 12/35


does that mean that i need to combine 2 possibilites of the answer choices? like 3 and 5, 5 and 7, 7 and 9, 9 and 11; and see which one yield 12/35?


Hi Thib33600

Yes if we are checking the options we need to plug in the pairs and check : 1/3 + 1/5 , 1/5 + 1/7 , .. for all the options.

Standard solving process would be to consider an even number -> 2a
Then the consecutive odd integers must be 2a - 1, 2a +1

1/(2a+1) + 1/(2a-1) = 4a/(4a^2 - 1) = 12/35

eliminate the 4 in numerator from left and right then we get:
a/(4a^2 - 1) = 3/35
We may see that the numerator is 3 and there is no common factor between 3 and 35 so a must be 3 !
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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]
35 has to be a product of two integers: factors of 35 is just 1, 5, 7, 35. So has to be 5 and 7
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