Brine worksheet

1. Salt: 100.8 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $1\text{\hspace{0.17em}US gal}$$=\mathrm{3,785.41}\text{\hspace{0.17em}g}$

2. Add the food's weight to the water's, because in this kind of brine the salt spreads into both

   $\mathrm{3,785.41}\text{\hspace{0.17em}g}+\mathrm{1,814.37}\text{\hspace{0.17em}g}$$=\mathrm{5,599.78}\text{\hspace{0.17em}g}$

3. Take 1.8% of that combined weight for the salt

   $\mathrm{5,599.78}\times \frac{1.8}{100}$$=100.796063\text{\hspace{0.17em}g}$

2. Salt: 204 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $1\text{\hspace{0.17em}US gal}$$=\mathrm{3,785.41}\text{\hspace{0.17em}g}$

2. Find the water's share of the finished brine: whatever is left after the salt and sugar shares

   $1-0.05-0.022$$=0.928$

3. Multiply the water by the salt share, then divide by the water's share. That makes salt the chosen share of the whole brine

   $\frac{\mathrm{3,785.41}\times 0.05}{0.928}$$=203.955376\text{\hspace{0.17em}g}$

4. Work out the sugar the same way, using the sugar share

   $\frac{\mathrm{3,785.41}\times 0.022}{0.928}$$=89.740366\text{\hspace{0.17em}g}$

5. Check the answer: the salt divided by the whole brine's weight gives back your chosen percentage

   $\frac{203.96}{\left(\mathrm{3,785.41}+203.96+89.74\right)}\times 100$$=5\text{\hspace{0.17em}\%}$

3. Salt: 205.7 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $1\text{\hspace{0.17em}US gal}$$=\mathrm{3,785.41}\text{\hspace{0.17em}g}$

2. Find the water's share of the finished brine: whatever is left after the salt and sugar shares

   $1-0.05-0.03$$=0.92$

3. Multiply the water by the salt share, then divide by the water's share. That makes salt the chosen share of the whole brine

   $\frac{\mathrm{3,785.41}\times 0.05}{0.92}$$=205.728901\text{\hspace{0.17em}g}$

4. Work out the sugar the same way, using the sugar share

   $\frac{\mathrm{3,785.41}\times 0.03}{0.92}$$=123.437341\text{\hspace{0.17em}g}$

5. Check the answer: the salt divided by the whole brine's weight gives back your chosen percentage

   $\frac{205.73}{\left(\mathrm{3,785.41}+205.73+123.44\right)}\times 100$$=5\text{\hspace{0.17em}\%}$

4. Salt: 162.5 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $1\text{\hspace{0.17em}US gal}$$=\mathrm{3,785.41}\text{\hspace{0.17em}g}$

2. Find the water's share of the finished brine: whatever is left after the salt and sugar shares

   $1-0.04-0.028$$=0.932$

3. Multiply the water by the salt share, then divide by the water's share. That makes salt the chosen share of the whole brine

   $\frac{\mathrm{3,785.41}\times 0.04}{0.932}$$=162.464025\text{\hspace{0.17em}g}$

4. Work out the sugar the same way, using the sugar share

   $\frac{\mathrm{3,785.41}\times 0.028}{0.932}$$=113.724818\text{\hspace{0.17em}g}$

5. Check the answer: the salt divided by the whole brine's weight gives back your chosen percentage

   $\frac{162.46}{\left(\mathrm{3,785.41}+162.46+113.72\right)}\times 100$$=4\text{\hspace{0.17em}\%}$

5. Salt: 84.6 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $1\text{\hspace{0.17em}US gal}$$=\mathrm{3,785.41}\text{\hspace{0.17em}g}$

2. Add the food's weight to the water's, because in this kind of brine the salt spreads into both

   $\mathrm{3,785.41}\text{\hspace{0.17em}g}+\mathrm{2,721.55}\text{\hspace{0.17em}g}$$=\mathrm{6,506.97}\text{\hspace{0.17em}g}$

3. Take 1.3% of that combined weight for the salt

   $\mathrm{6,506.97}\times \frac{1.3}{100}$$=84.590558\text{\hspace{0.17em}g}$

6. Salt: 496.4 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $2\text{\hspace{0.17em}US gal}$$=\mathrm{7,570.82}\text{\hspace{0.17em}g}$

2. Find the water's share of the finished brine: whatever is left after the salt and sugar shares

   $1-0.06-0.025$$=0.915$

3. Multiply the water by the salt share, then divide by the water's share. That makes salt the chosen share of the whole brine

   $\frac{\mathrm{7,570.82}\times 0.06}{0.915}$$=496.447447\text{\hspace{0.17em}g}$

4. Work out the sugar the same way, using the sugar share

   $\frac{\mathrm{7,570.82}\times 0.025}{0.915}$$=206.853103\text{\hspace{0.17em}g}$

5. Check the answer: the salt divided by the whole brine's weight gives back your chosen percentage

   $\frac{496.45}{\left(\mathrm{7,570.82}+496.45+206.85\right)}\times 100$$=6\text{\hspace{0.17em}\%}$

7. Salt: 204.4 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $1\text{\hspace{0.17em}US gal}$$=\mathrm{3,785.41}\text{\hspace{0.17em}g}$

2. Find the water's share of the finished brine: whatever is left after the salt and sugar shares

   $1-0.05-0.024$$=0.926$

3. Multiply the water by the salt share, then divide by the water's share. That makes salt the chosen share of the whole brine

   $\frac{\mathrm{3,785.41}\times 0.05}{0.926}$$=204.395885\text{\hspace{0.17em}g}$

4. Work out the sugar the same way, using the sugar share

   $\frac{\mathrm{3,785.41}\times 0.024}{0.926}$$=98.110025\text{\hspace{0.17em}g}$

5. Check the answer: the salt divided by the whole brine's weight gives back your chosen percentage

   $\frac{204.4}{\left(\mathrm{3,785.41}+204.4+98.11\right)}\times 100$$=5\text{\hspace{0.17em}\%}$

8. Salt: 120 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $1\text{\hspace{0.17em}US gal}$$=\mathrm{3,785.41}\text{\hspace{0.17em}g}$

2. Find the water's share of the finished brine: whatever is left after the salt and sugar shares

   $1-0.03-0.024$$=0.946$

3. Multiply the water by the salt share, then divide by the water's share. That makes salt the chosen share of the whole brine

   $\frac{\mathrm{3,785.41}\times 0.03}{0.946}$$=120.044771\text{\hspace{0.17em}g}$

4. Work out the sugar the same way, using the sugar share

   $\frac{\mathrm{3,785.41}\times 0.024}{0.946}$$=96.035817\text{\hspace{0.17em}g}$

5. Check the answer: the salt divided by the whole brine's weight gives back your chosen percentage

   $\frac{120.04}{\left(\mathrm{3,785.41}+120.04+96.04\right)}\times 100$$=3\text{\hspace{0.17em}\%}$

9. Salt: 325.6 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $2\text{\hspace{0.17em}US gal}$$=\mathrm{7,570.82}\text{\hspace{0.17em}g}$

2. Find the water's share of the finished brine: whatever is left after the salt and sugar shares

   $1-0.04-0.03$$=0.93$

3. Multiply the water by the salt share, then divide by the water's share. That makes salt the chosen share of the whole brine

   $\frac{\mathrm{7,570.82}\times 0.04}{0.93}$$=325.62682\text{\hspace{0.17em}g}$

4. Work out the sugar the same way, using the sugar share

   $\frac{\mathrm{7,570.82}\times 0.03}{0.93}$$=244.220115\text{\hspace{0.17em}g}$

5. Check the answer: the salt divided by the whole brine's weight gives back your chosen percentage

   $\frac{325.63}{\left(\mathrm{7,570.82}+325.63+244.22\right)}\times 100$$=4\text{\hspace{0.17em}\%}$

10. Salt: 241.1 g

1. Change the water into grams so everything is weighed the same way. 1 mL of water weighs about 1 g

   $2\text{\hspace{0.17em}US gal}$$=\mathrm{7,570.82}\text{\hspace{0.17em}g}$

2. Find the water's share of the finished brine: whatever is left after the salt and sugar shares

   $1-0.03-0.028$$=0.942$

3. Multiply the water by the salt share, then divide by the water's share. That makes salt the chosen share of the whole brine

   $\frac{\mathrm{7,570.82}\times 0.03}{0.942}$$=241.109031\text{\hspace{0.17em}g}$

4. Work out the sugar the same way, using the sugar share

   $\frac{\mathrm{7,570.82}\times 0.028}{0.942}$$=225.035095\text{\hspace{0.17em}g}$

5. Check the answer: the salt divided by the whole brine's weight gives back your chosen percentage

   $\frac{241.11}{\left(\mathrm{7,570.82}+241.11+225.04\right)}\times 100$$=3\text{\hspace{0.17em}\%}$