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What is the sum of all integer values of $x$ such that $\frac{31}{90} < \frac{x}{100} < \frac{41}{110}$ is true?

MEXICO May 15, 2024

eramsby1O1O · Answer 1 · 2024-05-15T07:28:29

To find the integers satisfying the inequality, we can rewrite it to isolate x:

Multiply both sides by 100 (the least common denominator):

3100 < 100x < 4100

Divide both sides by 100:

31 < x < 41

Since we only care about integer values of x, we need to find all the integers between 31 (not including) and 41 (including). These integers are:

32, 33, 34, 35, 36, 37, 38, 39, 40

Finally, add all the listed integers to find the sum:

32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 = 334

Therefore, the sum of all integer values of x satisfying the inequality is 334.

nerdboy4000 · Answer 2 · 2024-05-15T07:29:44

Erams, you made a simple mistake, this would actually be the answer.

$\frac{31}{90}=0.3444...$

$\frac{41}{110}=0.372727...$

$\frac{35}{100}>\frac{31}{90}$

$\frac{37}{100}<\frac{41}{100}$

$35+36+37=108$

The answer would be 108.