The first thing to do is work out the error interval (bounds) for a and b:
6.425 ≤ a < 6.435
5.5135 ≤ b < 5.5145
As we want to work out v by considering these bounds, we have to find both the upper and lower limit of the value of v. To find the upper value, we must take the highest possible value of a and divide it by the lowest possible value of b. To find the lower value, we must take the lowest possible value of a and divide it by the highest possible value of b.
Upper bound of v = √(6.435 ÷ 5.5135) = 1.080340323
Lower bound of v = √(6.425 ÷ 5.5145) = 1.079402689
Now to find out what the suitable degree of accuracy is, you find the most accurate value that both bounds round to. It sounds complicated, but it’s easy once you understand it.
To one digit, they both round to 1. To one decimal place, they both round to 1.1. To two decimal places, they both round to 1.08. To three decimal places, the upper value rounds to 1.080, however the lower value rounds to 1.079. These values are now different, therefore the most accurate value that you can take is 1.08.
So v = 1.08 because all values round to 1.08 ( 2dp).
I hope this helps anyone reading this! I am also studying for GCSEs and I have a feeling the other responses on this question are looking at it from too advanced a level. My answer will get you full marks on this question at GCSE. Good luck to everyone! X