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If the reciprocals of two consecutive positive integers are added toge

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Bunuel
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If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   02 Jun 2015, 06:51
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If the reciprocals of two consecutive positive integers are added together, what is the sum in terms of the greater integer x?

A. x/3
B. x^2 - x
C. 2x - 1
D. (2x - 1)/(x^2 + x)
E. (2x - 1)/(x^2 - x)
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E
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   02 Jun 2015, 07:41
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Let two consecutive positive integers be x and x-1 (Greater integer is x)
So,
(1/x) + [1/(x-1)]
= (2x-1)/x(x-1)
=(2x-1)/(x^2- x)


Answer E
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   02 Jun 2015, 08:59
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I do have to say I was puzzled when they stated in terms of the greater x.
But after some thought I was able to come up with this.

so we have X and X-1 as two consecutive positive integers.
X-1 because X is the greater number.

The reciprocals of X and X-1 are.
1/x and 1/(x-1)

The sum is 1/x + 1/(x-1)
To come up with a common denominator we multiply x and (x-1), which is x^2 - x
And the first numerator will be (x-1) times 1, which is x-1, and the other 1 times x, which is x
So the numerator added together (since we have a common denominator this is allowed) will be 2x-1

Therefore our result is (2x-1)/(X^2-X), hence answer E
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   03 Jun 2015, 16:30
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Hi All,

This question is perfect for TESTing VALUES.

We're told to take the sum of the reciprocals of two CONSECUTIVE POSITIVE INTEGERS....

Let's TEST 2 and 3....

The sum of the reciprocals would be...

1/2 + 1/3 =
3/6 + 2/6 =
5/6

We're asked how this sum relates to the greater integer X. In simple terms, we're looking for an answer that equals 5/6 when X = 3...

Answer A: X/3 = 3/3 = 1 NOT a match
Answer B: X^2 - X = 3^2 - 3 = 6 NOT a match
Answer C: 2X - 1 = 6 - 1= 5 NOT a match
Answer D: (2X - 1)/(X^2 + X) = 5/12 NOT a match
Answer E: (2X - 1)/(X^2 - X) = 5/6 This IS a MATCH

Final Answer:
Show Spoiler
E


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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   03 Jun 2015, 20:43
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tjerkrintjema wrote:
I do have to say I was puzzled when they stated in terms of the greater x.
But after some thought I was able to come up with this.

so we have X and X-1 as two consecutive positive integers.
X-1 because X is the greater number.

The reciprocals of X and X-1 are.
1/x and 1/(x-1)

The sum is 1/x + 1/(x-1)
To come up with a common denominator we multiply x and (x-1), which is x^2 - x
And the first numerator will be (x-1) times 1, which is x-1, and the other 1 times x, which is x
So the numerator added together (since we have a common denominator this is allowed) will be 2x-1

Therefore our result is (2x-1)/(X^2-X), hence answer E


"in terms of" should also make you think of "number plugging". If I want x in terms of y, I assume a value for y, find out what x should be and then plug in the y value in the options.
So here, I should take x = 2. So two consecutive positive integers such that the greater one is 2 are 1 and 2.
Sum of reciprocal = 1 + 1/2 = 3/2
Put x = 2 in all the options and check. Only (E) gives you 3/2. Hence (E) is the answer.
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   04 Jun 2015, 00:15
E
If greater integer is x then smaller one is x-1
Thus,
1/x + 1/(x-1)
= (2x-1)/x(x-1)
=(2x-1)/(x^2- x)
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   08 Jun 2015, 04:45
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Bunuel wrote:
If the reciprocals of two consecutive positive integers are added together, what is the sum in terms of the greater integer x?

A. x/3
B. x^2 - x
C. 2x - 1
D. (2x - 1)/(x^2 + x)
E. (2x - 1)/(x^2 - x)


MANHATTAN GMAT OFFICIAL SOLUTION:

If the greater of the two integers is x, then the two integers can be expressed as x – 1 and x. The sum of the reciprocals would therefore be:

\(\frac{1}{x-1} + \frac{1}{x} = \frac{x+(x-1)}{(x-1)x}=\frac{2x-1}{x^2-x}\).

The correct answer is E.
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   17 Jul 2016, 04:50
Hi!
I tried using x nd x+1
kindly help me with this solution as using this i am unable to reach to the correct answer choice.
thanks
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   17 Jul 2016, 04:56
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Celestial09 wrote:
Hi!
I tried using x nd x+1
kindly help me with this solution as using this i am unable to reach to the correct answer choice.
thanks


You won't be able because the question says that the greater integer is x. So to get correct answer you have to use x-1 and x.
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   18 Jul 2016, 12:12
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Celestial09 wrote:
Hi!
I tried using x nd x+1
kindly help me with this solution as using this i am unable to reach to the correct answer choice.
thanks


Hi Celestial09,

From your post, it's not clear what work you did when attempting to answer this question. However, if you look a few posts 'up', you'll see how it's relatively easy to answer this question by TESTing VALUES.

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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   18 Jul 2016, 18:38
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Option E
The greater integer is x and therefore the lesser integer is x-1.
Plug numbers or solve algebraically...
1/x+1/(x-1)
2x-1/x^2 -x


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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink] New post   02 Dec 2022, 04:32
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Re: If the reciprocals of two consecutive positive integers are added toge [#permalink]
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