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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...

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Re: If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is...

by Jay@ManhattanReview » Sun May 24, 2020 11:49 pm
AAPL wrote: ↑
Sun May 24, 2020 9:59 am
 Official Guide

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

OA C
Let's solve this question by selecting values from the option. Let's first pick Option B.

Option B: So, we have the greater of the two integers = 5; thus, the smaller of the two integers = 3

=> 1/3 + 1/5 = 8/15

Since 8/15 > 12/35, the two integers must be greater than 3 and 5 for the sum of their reciprocal would
be lesser.

Let's try Option D:

Option D: So, we have the greater of the two integers = 9; thus, the smaller of the two integers = 7

=> 1/7 + 1/9 = 16/63

Since 16/63 < 12/35, the two integers must be smaller than 7 and 9 for the sum of their reciprocal would be greater.

The correct answer must be C. There's no need to even try Option C for, with Option B, the sum of reciprocal was greater than 12/35 and with Option D, the sum of reciprocal was smaller than 12/35. So only Option C would work.

Let's check this.

Option C: So, we have the greater of the two integers = 7; thus, the smaller of the two integers = 5

=> 1/5 + 1/7 = 12/35 = given (12/35)

The correct answer: C

Hope this helps!

-Jay
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Re: If the sum of the reciprocals of two consecutive odd integers is 12/35 then the greater of the two integers is...

by Scott@TargetTestPrep » Sat May 30, 2020 1:46 pm
AAPL wrote: ↑
Sun May 24, 2020 9:59 am
 Official Guide

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

OA C
Solution:

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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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