Possible percent problem

Algebra Level 3

Consider numbers $a$ and $b$ such that $\frac{a}{b}=24\%$ when rounded to the nearest whole percent.

$b$ could be 63 since if $a=15$ we would have $\frac{15}{63}\approx .2381$ which rounds to $24\%$

$b$ could not be 10 however since no fraction of the form $\frac{a}{10}$ rounds to $24\%$

Let $x=$ the least possible value of $b$ for which there is a value of $a$ that makes $\frac{a}{b}=24\%$ when rounded to the nearest whole percent.

Let $y=$ the greatest possible value of $b$ for which there is no value of $a$ that makes $\frac{a}{b}=24\%$ when rounded to the nearest whole percent.

Give the value of $x+y$ .

The answer is 98.

1 solution

Jeremy Galvagni
Apr 16, 2018

We seek $\frac{23.5}{100}<\frac{a}{b}<\frac{24.5}{100}$ or

$.235b < a < .245b$

The upper bound can then be found by $[.245b]$ and the lower bound by $-[-.235b]$

If the difference between these is -1 there are no values of $a$ but if the difference is zero, we have found a value that works.

The first such value is $x=17$ where both are $4$ and indeed $\frac{4}{17}=.23529$

$y$ must be below 100 because the gaps become too small for $b$ to miss.

The last -1 is at $y=81$ and indeed $\frac{19}{81}=.234568$ and $\frac{20}{81}=.246914$

Solution: $x+y=17+81=\boxed{98}$

Incidentally, this table can be used to find other things. Such as: The smallest $b=119$ where $a$ could be either of two values: ${28 or 29}$

Possible percent problem

The answer is 98.

1 solution

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