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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?

 May 15, 2024
 #1
avatar+418 
-1

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.

 May 15, 2024
 #2
avatar+10 
+1

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.

 May 15, 2024
edited by nerdboy4000  May 15, 2024

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