+5 What is the sum of all integer values of \(x\) such that \(\frac{31}{90} < \frac{x}{100} < \frac{41}{110}\) is true?
+729 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.
+10 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.
