Let
and
be the amounts of available alloy (AA) and pure magic steel that are used, so we also have
![x + y = 2.7](https://tex.z-dn.net/?f=x%20%2B%20y%20%3D%202.7)
DA13 is supposed to contain 7% gold, 3% silver, and 90% steel, so that 2.7 kg of it is made up of
![0.07 \times 2.7 \,\mathrm{kg} = 0.189 \,\mathrm{kg} \text{ gold}](https://tex.z-dn.net/?f=0.07%20%5Ctimes%202.7%20%5C%2C%5Cmathrm%7Bkg%7D%20%3D%200.189%20%5C%2C%5Cmathrm%7Bkg%7D%20%5Ctext%7B%20gold%7D)
![0.03 \times 2.7 \,\mathrm{kg} = 0.081 \,\mathrm{kg} \text{ silver}](https://tex.z-dn.net/?f=0.03%20%5Ctimes%202.7%20%5C%2C%5Cmathrm%7Bkg%7D%20%3D%200.081%20%5C%2C%5Cmathrm%7Bkg%7D%20%5Ctext%7B%20silver%7D)
![0.90 \times 2.7 \,\mathrm{kg} = 2.43 \,\mathrm{kg} \text{ magic steel}](https://tex.z-dn.net/?f=0.90%20%5Ctimes%202.7%20%5C%2C%5Cmathrm%7Bkg%7D%20%3D%202.43%20%5C%2C%5Cmathrm%7Bkg%7D%20%5Ctext%7B%20magic%20steel%7D)
For each kg of the available alloy (AA), there is a contribution of 0.21 kg of gold, 0.09 kg of silver, and therefore 0.70 kg of steel;
kg of it will contain
kg of gold,
kg of silver, and
kg of steel. Each kg of magic steel of course contributes 1 kg of steel;
kg of it will contribute
kg of steel.
Then the dwarves need
• total gold: ![0.21x = 0.189](https://tex.z-dn.net/?f=0.21x%20%3D%200.189)
• total silver: ![0.09x = 0.081](https://tex.z-dn.net/?f=0.09x%20%3D%200.081)
• total steel: ![0.70x + y = 2.43](https://tex.z-dn.net/?f=0.70x%20%2B%20y%20%3D%202.43)
Solve for
and
. The first two equations are consistent and give
, and substituting this into the third we find
. So the dwarves must combine 0.90 kg of AA and 1.80 kg of magic steel.
1.) 11,000(0.04 or 4 percent)= 440
2.) 440(15 years)= 6,600
3.) The population will be about 6,600 after 15 years.
Answer:
................................
Answer:
5
Step-by-step explanation: