The given alloy contains 30% tin and another 70% tin, how many kg is needed to get 100 kg from 46%.

Let’s take for x and y the number of the 1st and 2nd alloys, respectively, then:

0.3 * x – the amount of pure tin in the 1st alloy;

0.7 * y – quantity in the 2nd.

Then we get the equation:

(0.3x + 0.7y) / (x + y) = 0.46

Since the total mass of the alloys is 100 kg, we get the 2nd equation:

x + y = 100

Express x from the 2nd equation and substitute it into the 1st:

x = 100 – y

0.3 * (100 – y) + 0.7y = 0.46 * (100 – y) + 0.46y

30 + 0.4y = 46

0.4y = 16

y = 64.

Then x = 100 – 64 = 32.

Answer: you need 32 kg of 30% and 64 kg of 70%