Cost Per Use worksheet

1. Cost per use: $0.26

1. Uses per year: 3 per week × 52

   $3\times 52$$=156$

2. Multiply by the years it lasts to get total uses over 4 years

   $156\times 4$$=624$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$160.00}}{624}$$=\mathrm{\$0.26}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$46.00}}{104\times 0.5}$$=\mathrm{\$0.88}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.26}-\mathrm{\$0.88}$$=\mathrm{-\$0.63}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$160.00}}{\mathrm{\$0.88}}$$=180.8696$

2. Cost per use: $0.23

1. Uses per year: 4 per week × 52

   $4\times 52$$=208$

2. Multiply by the years it lasts to get total uses over 4 years

   $208\times 4$$=832$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$190.00}}{832}$$=\mathrm{\$0.23}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$52.00}}{104\times 0.62}$$=\mathrm{\$0.81}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.23}-\mathrm{\$0.81}$$=\mathrm{-\$0.58}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$190.00}}{\mathrm{\$0.81}}$$=235.6$

3. Cost per use: $0.32

1. Uses per year: 3 per week × 52

   $3\times 52$$=156$

2. Multiply by the years it lasts to get total uses over 4 years

   $156\times 4$$=624$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$200.00}}{624}$$=\mathrm{\$0.32}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$69.00}}{208\times 0.8}$$=\mathrm{\$0.41}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.32}-\mathrm{\$0.41}$$=\mathrm{-\$0.09}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$200.00}}{\mathrm{\$0.41}}$$=482.3188$

4. Cost per use: $0.71

1. Uses per year: 2 per week × 52

   $2\times 52$$=104$

2. Multiply by the years it lasts to get total uses over 3 years

   $104\times 3$$=312$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$220.00}}{312}$$=\mathrm{\$0.71}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$70.00}}{156\times 0.76}$$=\mathrm{\$0.59}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.71}-\mathrm{\$0.59}$$=\mathrm{\$0.11}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$220.00}}{\mathrm{\$0.59}}$$=372.6171$

5. Cost per use: $0.26

1. Uses per year: 3 per week × 52

   $3\times 52$$=156$

2. Multiply by the years it lasts to get total uses over 3 years

   $156\times 3$$=468$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$120.00}}{468}$$=\mathrm{\$0.26}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$68.00}}{156\times 0.7}$$=\mathrm{\$0.62}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.26}-\mathrm{\$0.62}$$=\mathrm{-\$0.37}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$120.00}}{\mathrm{\$0.62}}$$=192.7059$

6. Cost per use: $0.32

1. Uses per year: 3 per week × 52

   $3\times 52$$=156$

2. Multiply by the years it lasts to get total uses over 4 years

   $156\times 4$$=624$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$200.00}}{624}$$=\mathrm{\$0.32}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$60.00}}{104\times 0.74}$$=\mathrm{\$0.78}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.32}-\mathrm{\$0.78}$$=\mathrm{-\$0.46}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$200.00}}{\mathrm{\$0.78}}$$=256.5333$

7. Cost per use: $0.34

1. Uses per year: 3 per week × 52

   $3\times 52$$=156$

2. Multiply by the years it lasts to get total uses over 3 years

   $156\times 3$$=468$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$160.00}}{468}$$=\mathrm{\$0.34}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$60.00}}{104\times 0.55}$$=\mathrm{\$1.05}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.34}-\mathrm{\$1.05}$$=\mathrm{-\$0.71}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$160.00}}{\mathrm{\$1.05}}$$=152.5333$

8. Cost per use: $0.58

1. Uses per year: 3 per week × 52

   $3\times 52$$=156$

2. Multiply by the years it lasts to get total uses over 2 years

   $156\times 2$$=312$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$180.00}}{312}$$=\mathrm{\$0.58}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$55.00}}{156\times 0.62}$$=\mathrm{\$0.57}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.58}-\mathrm{\$0.57}$$=\mathrm{\$0.01}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$180.00}}{\mathrm{\$0.57}}$$=316.5382$

9. Cost per use: $0.30

1. Uses per year: 4 per week × 52

   $4\times 52$$=208$

2. Multiply by the years it lasts to get total uses over 3 years

   $208\times 3$$=624$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$190.00}}{624}$$=\mathrm{\$0.30}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$62.00}}{156\times 0.58}$$=\mathrm{\$0.69}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.30}-\mathrm{\$0.69}$$=\mathrm{-\$0.38}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$190.00}}{\mathrm{\$0.69}}$$=277.2774$

10. Cost per use: $0.42

1. Uses per year: 2 per week × 52

   $2\times 52$$=104$

2. Multiply by the years it lasts to get total uses over 4 years

   $104\times 4$$=416$

3. Divide the cost by the number of uses to spread it over every use

   $\frac{\mathrm{\$175.00}}{416}$$=\mathrm{\$0.42}$

4. Do the same for the other option: its price divided by its own total uses

   $\frac{\mathrm{\$70.00}}{156\times 0.5}$$=\mathrm{\$0.90}$

5. Subtract to compare. A negative difference means the first item is cheaper per use

   $\mathrm{\$0.42}-\mathrm{\$0.90}$$=\mathrm{-\$0.48}$

6. Divide the first item's cost by the other option's cost per use to see how many uses it needs to match

   $\frac{\mathrm{\$175.00}}{\mathrm{\$0.90}}$$=195$