1.) We can solve the problem by developing a linear model.
Let x represent the quantity to be used of the grade of coffee that sells for $70 per pound, and y represent the quantity to be used of the grade of coffee that sells for $80 per pound, then
![\left\begin{array}{c}x+y=80 . . . (1)\\70x+80y=76(80) . . . (2)\end{array}\right \\ \\ \Rightarrow \left\begin{array}{c}80x+80y=6400 . . . (3)\\70x+80y=6080 . . . (4)\end{array}\right \\ \\ \Rightarrow10x=320 \\ \\ \Rightarrow x=32 \\ \\ \Rightarrow y=80-32=48](https://tex.z-dn.net/?f=%20%20%5Cleft%5Cbegin%7Barray%7D%7Bc%7Dx%2By%3D80%20.%20.%20.%20%281%29%5C%5C70x%2B80y%3D76%2880%29%20.%20.%20.%20%282%29%5Cend%7Barray%7D%5Cright%20%20%5C%5C%20%20%20%5C%5C%20%5CRightarrow%20%20%5Cleft%5Cbegin%7Barray%7D%7Bc%7D80x%2B80y%3D6400%20.%20.%20.%20%283%29%5C%5C70x%2B80y%3D6080%20.%20.%20.%20%284%29%5Cend%7Barray%7D%5Cright%20%20%5C%5C%20%20%5C%5C%20%5CRightarrow10x%3D320%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20x%3D32%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20y%3D80-32%3D48)
Therefore, 32 pounds of the grade of coffee that sells for $70 per pound, and 48 pounds of the grade of coffee that sells for $80 per pound should be used.
2.) Volume of alcohol in the original mixture
![\frac{35}{100} \times 40=14](https://tex.z-dn.net/?f=%20%5Cfrac%7B35%7D%7B100%7D%20%5Ctimes%2040%3D14%20)
quarts
Let x be the number of quarts of alcohol to be added, then
Volume of alcohol in the new mixture
![\frac{48}{100} \times (40+x)=14+x \\ \\ \Rightarrow19.2+0.48x=14+x \\ \\ \Rightarrow0.52x=5.2 \\ \\ \Rightarrow x=10](https://tex.z-dn.net/?f=%5Cfrac%7B48%7D%7B100%7D%20%5Ctimes%20%2840%2Bx%29%3D14%2Bx%20%5C%5C%20%20%5C%5C%20%5CRightarrow19.2%2B0.48x%3D14%2Bx%20%5C%5C%20%20%5C%5C%20%5CRightarrow0.52x%3D5.2%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20x%3D10)
Therefore, 10 quarts of pure alcohol should be added
3.) Original proportion of white paint in the mixture
![=\frac{5}{5+11}=\frac{5}{16}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B5%7D%7B5%2B11%7D%3D%5Cfrac%7B5%7D%7B16%7D)
Let x be the number of gallons of white paint to be added, then
![\frac{5+x}{16+x}=\frac{2}{3} \\ \\ \Rightarrow3(5+x)=2(16+x) \\ \\ \Rightarrow15+3x=32+2x \\ \\ \Rightarrow x=32-15=17](https://tex.z-dn.net/?f=%5Cfrac%7B5%2Bx%7D%7B16%2Bx%7D%3D%5Cfrac%7B2%7D%7B3%7D%20%5C%5C%20%20%5C%5C%20%5CRightarrow3%285%2Bx%29%3D2%2816%2Bx%29%20%5C%5C%20%20%5C%5C%20%5CRightarrow15%2B3x%3D32%2B2x%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20x%3D32-15%3D17)
Therefore, 17 gallons of white paint should be added.
4a.) Let the three consecutive odd integers be x - 2, x and x + 2, then
![x - 2 + x + x + 2 = 25 \\ \\ \Rightarrow3x=25\Rightarrow x=8.33](https://tex.z-dn.net/?f=x%20-%202%20%2B%20x%20%2B%20x%20%2B%202%20%3D%2025%20%5C%5C%20%20%5C%5C%20%5CRightarrow3x%3D25%5CRightarrow%20x%3D8.33)
which is not an integer.
Hence, the sum of three consecutive odd integers cannot be 25.
4b.) Let the three consecutive odd integers be x - 2, x and x + 2, then
![x - 2 + x + x + 2 = 45 \\ \\ \Rightarrow3x=45\Rightarrow x=15](https://tex.z-dn.net/?f=x%20-%202%20%2B%20x%20%2B%20x%20%2B%202%20%3D%2045%20%5C%5C%20%5C%5C%20%5CRightarrow3x%3D45%5CRightarrow%20x%3D15)
Thus, the three consecutive odd intergers whose sum is 45 are 13, 15 and 17.
5.) If all pipes are open and the tank was initially empty, let t be the time it will take to fill the tank, then
![\frac{1}{t} = \frac{1}{9} + \frac{1}{12} - \frac{1}{15} = \frac{20+15-12}{180} = \frac{23}{180} \\ \\ \Rightarrow t= \frac{180}{23} =7.83](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bt%7D%20%3D%20%5Cfrac%7B1%7D%7B9%7D%20%2B%20%5Cfrac%7B1%7D%7B12%7D%20-%20%5Cfrac%7B1%7D%7B15%7D%20%3D%20%5Cfrac%7B20%2B15-12%7D%7B180%7D%20%3D%20%5Cfrac%7B23%7D%7B180%7D%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20t%3D%20%5Cfrac%7B180%7D%7B23%7D%20%3D7.83%20)
Therefore, it will take 7.83 hours to get the tank filled up.