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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!

31%2F90+%3C+x%2F100+%3C+41%2F110.............all terms multiply by 100
%28100%2A31%29%2F90+%3C+100x%2F+100%3C+%28100%2A41%29%2F110
%2810%2A31%29%2F9+%3C+x%3C+%2810%2A41%29%2F11
310%2F9%3Cx%3C410%2F11
34%3C+x+%3C+37
between 34 and 37 (including37) we have :
35,36 and 37 and their sum is
35%2B36%2B37=108

Answer by MathTherapy(10552) About Me  (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?
31%2F90+%3C+x%2F100+%3C+41%2F110



%283410%2F9900%29+%3C+%2899x%2F9900%29+%3C+%283690%2F9900%29   ------------ Multiplying by LCD, 

It then becomes: 3410+%3C+99x+%3C+3690

%283410%2F99%29+%3C+%2899x%2F99%29+%3C+%283690%2F99%29 --------- Dividing by 99

34.444+%3C+x+%3C+37.27273

Integers are therefore: highlight_green%2835+%2B+36+%2B+37+=+108%29