If the sum of the reciprocals of two consecutive

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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

The OA is C.

Should I solve the equation 1/(2k+1) + 1/(2k+3) = 12/35? Or there is another way to solve it?

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Let us assume that two consecutive odd integers are a and b where b = a+2. Then, sum of their reciprocal is (a+b)/ab. We have been given that sum of two consecutive odd integers is 12/35.

As GCD of two consecutive integers is always 1, sum of these two is 12 and multiplication of these two integers is 35.

Obviously, these two odd integers are 5 and 7.

Thus, larger integer is 7. So, answer is C.
Last edited by pannalal on Sun Sep 24, 2017 7:22 am, edited 1 time in total.

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by Brent@GMATPrepNow » Sun Sep 24, 2017 7:01 am
Vincen 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: C

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by [email protected] » Sun Sep 24, 2017 10:34 am
Hi Vincen,

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: C

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by Jeff@TargetTestPrep » Tue Dec 19, 2017 7:14 am
Vincen 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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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