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Take a test score represented by the fraction ${a\over b}$. This test score could be curved by removing a wrong answer to get ${a\over b-1}$ or adding a correct answer to get ${a+1\over b+1}$. Sometimes the former will cause a greater increase to the fraction, sometimes the latter will cause a greater increase, and sometimes they will be equal. For example

${99\over 100}$ would become either ${99\over 99}$ or ${100\over 101}$ (a is better)

${1\over 3}$ would become either ${1\over 2}$ or ${2\over 4}$, (same)

${0\over 3}$ would become either ${0\over 2}$ or ${1\over 4}$ (b is better)

Is there a formula for at which points or intervals one way is better than the other? (Equal would be at $-a^2 + ab + a = 1$)

Michael Hardy
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  • in the second line, do you mean $a \over b-1$? – Guy Mar 25 '14 at 19:46
  • @Sabyasachi thanks – Y     e     z Mar 25 '14 at 19:47
  • Your formula for both being the same can't work. $-a^2 + ab + a = 1$ means that $a$ is a factor of $1$, therefore $a$ can only be $1$. Since $2\over 5$ gives the same result either way, this doesn't work. In this case, $-a^2 + ab + a = 8$ which is a multiple of $a=2$. – CJ Dennis May 26 '18 at 20:46

2 Answers2

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$${a\over b-1}\gt \frac{a+1}{b+1}$$

$$ab+a\gt ab+b-a-1\implies b\lt 2a+1$$

If $b$ is less than $2a+1$ the first is better. If it is equal, both options are equal. If it is more, then, the second option is better.

Guy
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We need to solve $$\frac{a}{b-1}>\frac{a+1}{b+1} $$or $$ {a}(b+1)> (a+1)(b-1),$$which is $$ 2a > b -1.$$

TZakrevskiy
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