Great question!

First, let x and x+1 be two consecutive integers.
So, 1/x and 1/(x+1) are the reciprocals.
So, their difference = 1/x - 1/(x+1)
To subtract these fractions we must find common denominators.
So, 1/x - 1/(x+1) = (x+1)/(x)(x+1) - x/(x+1)(x)
= 1/(x)(x+1)

Notice that the denominator is the product of two consecutive integers AND the numerator is 1
Check the denominators of the answer choices:
A. 1/100 We can't write 100 as the product of two consecutive integers. ELIMINATE.
B. 1/121 We can't write 121 as the product of two consecutive integers. ELIMINATE.
C. 1/45 We can't write 45 as the product of two consecutive integers. ELIMINATE.
D. 1/56 We CAN write 56 as the product of two consecutive integers (7x8). KEEP
E. 3/72 We CAN write 72 as the product of two consecutive integers (8x9). 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
Answer: D

ASIDE: 1/56 = 1/(7)(8) = 1/(x)(x+1)
So, x = 7 and x+1 = 8
Notice that 1/7 - 1/8 = 8/56 - 7/56 = 1/56!
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=1/n - 1/(n+1)

=1/n(n+1)

Only answer choice that satisfies above equation is D

1/7*8


D


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Must be D. 1/56 since 56=7x8. The difference will be of the form 1/a X (a+1)


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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

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

1/56 = 1/7-1/8

IMO D

Posted from my mobile device
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\(\frac{1}{x}-\frac{1}{(x+1)}=\frac{1}{x*(x+1)}\)

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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