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Re: a-1/b+1 vs a/b [#permalink]
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a−1/b+1
Considering a/b as out reference, conceptually, as a,b>0, if the value of numerator decreases (a-1) or as the value of denominator increases (b+1) the obtained fraction is always lesser than the original fraction (a/b) pick B

Another way would be by representing a-1 as a- (a- is a value that is less than a)
representing b+1 as b+ (b+- is a value that is greater than b)
clearly a/b will be the greater fraction
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Re: a-1/b+1 vs a/b [#permalink]
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Hi,

You can assume a and b to be some number substitute and the find the answer.

For eg. a=2 and b=1

Qty A: (2-1)/1+1=1/2
Qty B:=2/1

Here, A<B
Also, try solving with number in which a<b
For eg. a=2 & y=3
Qty A: (2-1)/(3+1)=1/4
Qty B: 3/4

Thus A<B

IMO B

Hope this helps!
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Re: a-1/b+1 vs a/b [#permalink]
GreenlightTestPrep wrote:
a and b are both positive.

Quantity A
Quantity B
\(\frac{a-1}{b+1}\)
\(\frac{a}{b}\)


A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Since \(a\) and \(b\) are both positive, we can cross-multiply their denominators

Col. A: \((a - 1)b\)
Col. B: \((b + 1)a\)

Col. A: \(ab - b\)
Col. B: \(ab + a\)

Col. A: \(-b\), a negative number
Col. B: \(a\), a positive number

So, Col. A < Col. B

Hence, option B
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Re: a-1/b+1 vs a/b [#permalink]
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Re: a-1/b+1 vs a/b [#permalink]
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