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We can represent two consecutive numbers as n and n+1
Where n < n+1

Reciprocate them = \(\frac{1}{n}> \frac{1}{n+1}\)

Subtract small from big..

\(\frac{1}{n} - \frac{1}{n+1}\)
\(\frac{1}{n^2+n}\)

So we must have 1 in numerator and denominator term can can be expressed in the from of n^2+1.

5 + 25 = 30
6+ 36 = 42
7+ 49 = 56
8 + 64 = 72 <<< matched << we have one answer choice (E) with 72 in denominator.
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let a and b be the consecutive integers.
1/b-1/a=(a-b)/ab

inverse of (a-b)/ab can be compared with inverse of each option

option 1 : ab/(a-b) = 24/1 .... no consecutive numbers have product as 24
option 2 : ab/(a-b) = 30/7.... denominator has to be 1 as a and b are consecutive numbers
option 3 : ab/(a-b) = 54/1 ..... no consecutive numbers have product as 54
option 4 : ab/(a-b) = 56/13 .... denominator has to be 1 as a and b are consecutive numbers
option 5 : ab/(a-b) = 72/1 ..... 8 & 9 satisfy the equation.

Ans- E
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NoHalfMeasures
Which of the following fractions can be written as the difference of reciprocals of two consecutive integers?

A. 1/24
B. 7/30
C. 1/45
D. 13/56
E. 1/72
Look only at the denominators to start.

Which denominators are a product of two consecutive integers?

B) 7/30 (5*6 = 30)

D) 13/56 (7*8 = 56)

E 1/72 (8*9 = 72)

Now numerators. There is only one choice: a numerator of 1. The difference between two consecutive integers is 1.
(9-8) = 1.

That is Answer E.
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Two consecutive integers have no common factor except 1.
If we subtract the reciprocals of two consecutive integer, Then we actually multiply them in the denominator. So, we want their product in the denominator.

again, if we subtract the reciprocals of two consecutive integer, their difference will be in the numerator, which will be always 1.

Only (E), satisfies both of the restrictions.
9*8=72 and 9-8=1

Posted from my mobile device
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which of the following fractions can be written as the difference of reciprocals of two consecutive integers?

a) 1/24
b) 7/30
c) 1/45
d) 13/56
e) 1/72

Let the 2 consecutive integers be "a" and "a+1"

So we have \(\frac{1}{a}\) - \(\frac{1}{(a+1)}\)

Simplify and we get \(\frac{1}{a(a+1)}\)....So the answer needs to have 1 in numerator and denominator should be product of consecutive integers...

Option B and D can be ruled out

24=6*4 (Not consecutive)
45= 9*5 (not consecutive)
72=8*9 (Consecutive) and hence your answer.

In addition to this, it has to be even so you can eliminate 45 (option c) with ease as well...
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Two consecutive integers = \(x, x+1\)

\(\frac{1}{x} - \frac{1}{x-1} = \frac{1}{x(x-1)}\)

At this point we can look at the answer choices

\(\frac{1}{72} = \frac{1}{8(9)}\)

Perfect match!
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strategy: write what it's asking algebraically and then plug and play the answer choices. You know that the number has to be 1/x(x+1) therefore 1 / (multiplication of two consecutive integers)
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­Look for a pattern wrt the difference between reciprocals of consecutives:

­
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