Which of the following fractions can be written as the difference of

Question

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

PS04867

WoundedTiger · Accepted Answer

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.

PareshGmat · Answer

Difference of reciprocals means 1 has to be in the numerator;

So option B & D can be ignored

For option A, B & E, we require to search two consecutive numbers whose product is the denominator

24 = 4 * 6 ......... Ignore

45 = 5 * 9 ........ Ignore

72 = 8 * 9 ................ Answer

Answer = E

PiyushK · Answer

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.

desaichinmay22 · Answer

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

generis · Answer

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.

RashedVai · Answer

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

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aborja87 · Answer

In addition to this, it has to be even so you can eliminate 45 (option c) with ease as well...

Basshead · Answer

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!

Fish181 · Answer

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)

DanTheGMATMan · Answer

Look for a pattern wrt the difference between reciprocals of consecutives:

https://www.youtube.com/watch?v=pJQtlcMyFZo

MarkGMATMentor · Answer

This question tests an obscure number property involving the reciprocals of two consecutive integers.  On questions that test obscure number properties, it's helpful to test small numbers and look for patterns.
 
First, make sure you’re clear on what the question is asking. You need to find two consecutive integers, take their reciprocals, and subtract one from the other to get one of the answer choices.

Should you simply try random pairs of consecutive integers? A better method is to start small and look for patterns. For example, you might start with consecutive integers 2 and 3. Then, their reciprocals are  \frac{1}{2}  and  \frac{1}{3} , and their difference is:
 \frac{1}{2} - \frac{1}{3} = \frac{3}{6} – \frac{2}{6} = \frac{1}{6
} 
Then, try 3 and 4:
 \frac{1}{3} – \frac{1}{4} = \frac{4}{12} – \frac{3}{12} = \frac{1}{12}

Now look for the pattern. In both cases, the result is 1 over the product of the integers.

What answer choices fit that pattern? B and D are out: they don’t have 1 in the numerator (and can’t be reduced).

The denominator of A is 24, which is the product of 4 and 6, 3 and 8, 2 and 12, or 1 and 24 – none of which are consecutive.

The denominator of C is 45, which is the product of 5 and 9, among other pairs, none of which are consecutive.

The denominator of E is 72, which is the product of 8 and 9, two consecutive integers. Let’s confirm that this works:
 \frac{1}{8} – \frac{1}{9} = \frac{9}{72} – \frac{8}{72} = \frac{1}{72}

The answer is E.

By starting small and recognizing the pattern, you can organize your solution and avoid trying random cases. This strategy is particularly helpful on problems that test unusual number properties.

Which of the following fractions can be written as the difference of

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GMAT Problem Solving (PS) Questions

Which of the following fractions can be written as the difference of

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