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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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New post 23 Jul 2017, 11:17
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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  [#permalink]

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New post 23 Jul 2017, 12:19
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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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New post 25 Jul 2017, 03:52
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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 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
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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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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:

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

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New post Updated on: 11 Aug 2018, 12:13
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


35=7*5
7/35+5/35=12/35
7
C

Originally posted by gracie on 19 Oct 2017, 10:08.
Last edited by gracie on 11 Aug 2018, 12:13, edited 3 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.

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

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New post 11 Aug 2018, 11:01
ScottTargetTestPrep 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 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


I used the alternative route for this one but I was wondering if you could elaborate on whether there are any tips/tricks to dealing with messy quadratics, those that have a constant (for my purposes, I'm referring to a constant >1) in front of the x^2 term, since it's less straight forward and more time-consuming to figure out the roots in those situations?

I get that for quadratics with no constant in front of the x^2 term, we can use the "sum" and "product" rules to quickly figure out the form of the two roots in our head but the "sum" rule doesn't hold when we have a constant in front of the x^2 term, only the "product" rule does....I know you could technically plug values into the quadratic equation to figure out the roots but that's not a viable gmat approach so just trying to figure out what I'm missing that is keeping me from quickly solving these sorts of quadratics such as 6x^2 - 23x - 35 = 0.
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Re: If the sum of the reciprocals of two consecutive odd integers  [#permalink]

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New post 18 Aug 2018, 17:39
Hi,
well this can be solved assuming integers as x and y
then 1/x+1/y= x+y/xy
x+y/xy= 12/35
5,7 are the integers
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Re: If the sum of the reciprocals of two consecutive odd integers  [#permalink]

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Re: If the sum of the reciprocals of two consecutive odd integers   [#permalink] 03 Sep 2019, 06:58
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