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stonecold
Which of the following fractions can be written as the difference of reciprocals of two consecutive integers?

A. 1/100
B. 1/121
C. 1/45
D. 1/56
E. 3/72
Must be D. 1/56 since 56=7x8. The difference will be of the form 1/a X (a+1)


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stonecold
Which of the following fractions can be written as the difference of reciprocals of two consecutive integers?

A. 1/100
B. 1/121
C. 1/45
D. 1/56
E. 3/72


This is how I solved

let The no be x and x+1 .
Which means the reciprocals = 1/x + 1/( X+1) which also means 2 consecutive integer should add upto the no.

Check ans choice and see which no is the multiplication of two consecutive nos
eg : a) 100 = 10 * 10 so rule out
c) 45 = 7*6 = 42+3 so rule out
d) 56 = 7*8 = 56 keep
e) 72 = 9*8 = keep

We're left with D and E.
Since E is not in the correct format (with 1 in the numerator), we can ELIMINATE E
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Similar question to: https://gmatclub.com/forum/which-of-the ... 72318.html

I feel it is wrong to eliminate E based on having 3 in the numerator. Simplify the fraction and see that it is 1/24, now it can't.

Very similar to terminating decimal questions where it has 3/30 and the numerator and denominator need to be simplified to see that only 2s and 5s are in the denominator
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stonecold
Which of the following fractions can be written as the difference of reciprocals of two consecutive integers?

A. 1/100
B. 1/121
C. 1/45
D. 1/56
E. 3/72

1/56 = 1/7-1/8

IMO D

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stonecold
Which of the following fractions can be written as the difference of reciprocals of two consecutive integers?

A. 1/100
B. 1/121
C. 1/45
D. 1/56
E. 3/72


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

note that the consecutive integers x and x+1 product will always be even. So straight away eliminate B and C. Now look for other options. E reduce to the simplest form. Now you can see that the only D satisfies the condition, a product of 7 and 8
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