What mass of zinc sulfate must be added to 100 g of a solution of this salt with a mass fraction

What mass of zinc sulfate must be added to 100 g of a solution of this salt with a mass fraction of 5% zinc sulfate to obtain a 10% solution.

w (ZnSO4) in vitriol = 65/161 = 0.4;
m (ZnSO4) in solution = 1000 * 0.005 = 5g;
m (ZnSO4 * 10H2O) = 5 / 0.4 = 12.5g;
m (solution) = 100

w1 = 5%
w2 = 10%

m (add salt) -?

m (salt) = m (p-pa) * w1 = 100 * 0.05 = 5 g
w = m (salts) / m (p-pa)
m2 (p-pa) = m1 (p-pa) + m (add salt)
w2 = (m (salt) + m (add salt)) / (m (p-pa) + m (add salt))
Let, m (add salt) = x / Then:
(5 + x) / (100 + x) = 0.1
5 + x = 0.10 (100 + x)
5 + x = 10 + 0.10x
1.10x = 15