Spring constant, should be easy?

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Homework Statement


A spring is suspended vertically parallel to a meterstick. When a 170-g mass is attached to the bottom of the spring, the spring stretches until its bottom is adjacent to the 80-cm mark on the meterstick. When a 760-g mass is then used to replace the 170-g mass, the bottom of the spring is measured to be at the 30-cm mark. Based on these measurements, the value of the spring constant is


Homework Equations


k=mg/x


The Attempt at a Solution


Seemed simple enough, (0.17*9.8)/.2 = 8.33 N/m
then (0.76*9.8)/.7=10.64 N/m

But neither of these are correct, and if i take the sum of the masses and distance i get:
(.93*9.8)/0.9=10.126~

None of these can possibly be the correct answer.. and I am sort of lost =/
 
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New question: A spring is suspended vertically parallel to a meterstick. When a 130-g mass is attached to the bottom of the spring, the spring stretches until its bottom is adjacent to the 20-cm mark on the meterstick. Based on this measurement, the value of the spring constant is
6.37 kg/s2.
637 kg/s2.
0.00637 kg/s2.
unknown.
0.650 kg/s2.
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Where i thought it was 6.37; apparently it is unknown because you don't know that the spring's bottom was at 0cm when it started =/ Still not sure how it will help me with the first one though [the first question doesn't state that either]
 
For the first one:

Take the two points (use the weight) and plot them on a graph... your gradient is the spring constant.

Alternatively k = (m2 - m1)g / (x2 - x1)
 
Ugh, so it was the difference in the mass and distance; no idea why i took the sum. Thank you.