Question 1195878
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The question is incomplete. I'll give an example of how to solve these types of problems.


Let's say the true volume is 7.5 liters.
Furthermore, let's say you erroneously measured it to be 8 liters.


The difference is 8 - 7.5 = 0.5 liters
Divide this difference over the true value
0.5/7.5 = 0.0667 approximately
There's about a 6.67% error 


Another example:
True value = 7.5 liters
measured value = 5 liters
difference = 7.5 - 5 = 2.5 liters
difference/trueValue = 2.5/7.5 = 0.3333 = 33.33% error approximately


Note: When computing the difference, use the positive version. Use absolute value if needed.
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