SOLUTION: what is the sum of all integer values of x such that 31/90 < x/100 < 41/110 is true? The answer is 108. How?

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Question 1008112: what is the sum of all integer values of x such that 31/90 < x/100 < 41/110 is true?
The answer is 108. How?
Found 3 solutions by fractalier, MathLover1, MathTherapy:
Answer by fractalier(6550)   (Show Source): You can put this solution on YOUR website!
31/90 = .3444444
41/110 = .372727272
How many hundredths are in between them?
Well, 35/100, 36/100, and 37/100...
35 + 36 + 37 = 108
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

.............all terms multiply by




between and (including) we have :
, and and their sum is


Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

what is the sum of all integer values of x such that 31/90 < x/100 < 41/110 is true?
The answer is 108. How?




   ------------ Multiplying by LCD, 

It then becomes: 

 --------- Dividing by 99



Integers are therefore:  

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