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02 Jun 2015, 06:51
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E
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Difficulty:
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79% (01:34) correct
21%
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based on 370
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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)
A. x/3
B. x^2 - x
C. 2x - 1
D. (2x - 1)/(x^2 + x)
E. (2x - 1)/(x^2 - x)
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
So,
(1/x) + [1/(x-1)]
= (2x-1)/x(x-1)
=(2x-1)/(x^2- x)
Answer E
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
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
