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If the sum of the reciprocals of two consecutive odd integers

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If the sum of the reciprocals of two consecutive odd integers [#permalink]

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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
[Reveal] Spoiler: OA

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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]

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New post 23 Jul 2017, 12:19
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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New post 25 Jul 2017, 03:52
Or algebraically: 1/x + 1/y=12/35
(x+y)/xy=12/35
You can see from the answer choices that only 7 can be the correct answer.

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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]

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New post 25 Jul 2017, 09:39
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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 [#permalink]

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New post 26 Jul 2017, 16:30
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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 let the first odd integer = x and the next odd integer = x + 2; thus, the reciprocals are 1/x and 1/(x + 2). Thus:

1/x + 1/(x+2) = 12/35

Multiplying by 35x(x+2), we have:

35(x + 2) + 35x = 12x(x + 2)

35x + 70 + 35x = 12x^2 + 24x

12x^2 - 46x - 70 = 0

6x^2 - 23x - 35 = 0

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

Thus, x = -7/6 or x = 5.

Since x is an integer, x must be 5 and the greater integer is x + 2 = 7.

Alternate solution:


We are given that the sum of the reciprocals of two consecutive odd integers is 12/35. We see that the denominator is 35. It’s not difficult to conjecture that the integers have to be 5 and 7 since 5 x 7 = 35. Finally, we can check the sum of 1/5 and 1/7 to see if they sum to 12/35:

1/5 + 1/7 = 7/35 + 5/35 = 12/35

Since they do add up to 12/35, the larger integer is 7.

Answer: C
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Re: If the sum of the reciprocals of two consecutive odd integers is 12/35 [#permalink]

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New post 19 Oct 2017, 01:35
Assuming the two integers are x and y.

1/x + 1/y = 12/35, so

(x+y)/xy=35

xy=5 x 7 = 35

Therefore, the greater integer is 7.

Answer is C.

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

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New post 19 Oct 2017, 01:44
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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New post 19 Oct 2017, 07:07
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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:
[Reveal] Spoiler:
C


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If the sum of the reciprocals of two consecutive odd integers [#permalink]

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New post 19 Oct 2017, 10:08
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


let x=greater integer
x+(x-2)=12
x=7
C

Last edited by gracie on 05 Dec 2017, 16:21, edited 2 times in total.

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Re: If the sum of the reciprocals of two consecutive odd integers [#permalink]

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New post 05 Dec 2017, 12:43
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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:
[Reveal] Spoiler:
C


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Re: If the sum of the reciprocals of two consecutive odd integers   [#permalink] 05 Dec 2017, 12:43
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