\(\textbf{1)}\) How many pints of an 30% and a 50% alcohol solution must be mixed to create 8 pints of a 35% solution?
6 pints of 30% solution and 2 pints of 50% solution
Let \(x\) = pints of 30% solution, and \(y\) = pints of 50% solution.
We know:
\(x + y = 8\) (total mixture)
\(0.30x + 0.50y = 0.35(8)\) (alcohol content equation)
Simplify the second equation:
\(0.30x + 0.50y = 2.8\)
Substitute \(y = 8 – x\):
\(0.30x + 0.50(8 – x) = 2.8\)
\(0.30x + 4 – 0.50x = 2.8\)
\(-0.20x = -1.2\)
\(x = 6\)
Then \(y = 8 – 6 = 2\).
\(6\) pints of 30% solution and \(2\) pints of 50% solution.
\(\textbf{2)}\) Gummy worms cost $1.50 a pound and sour apple candy cost $2.00 a pound. How many pounds of each would you mix if you want to create a 4 pound candy bag that costs $1.70 a pound?
2.4 pounds of gummy worms and 1.6 pounds of sour apple candy
Let \(x\) = pounds of gummy worms at $1.50/lb, \(y\) = pounds of sour apple at $2.00/lb.
Total weight: \(x+y=4\).
Cost equation: \(1.50x+2.00y=1.70(4)=6.8\).
Substitute \(y=4-x\): \(1.50x+2(4-x)=6.8\).
\(1.50x+8-2x=6.8 \Rightarrow -0.50x=-1.2 \Rightarrow x=2.4\).
\(y=4-2.4=1.6\).
Answer: \(2.4\) lb gummy worms, \(1.6\) lb sour apple.
\(\textbf{3)}\) How may pounds of grapes for $2.50 per pound must be mixed with 10 lbs of melon selling for $1.60 per pound to make a mix that costs $2.10 per pound?
12.5 pounds of grapes
Let \(x\) = pounds of grapes at $2.50/lb. Melon is fixed at \(10\) lb at $1.60/lb.
Target: \(2.10(x+10)\) dollars.
Equation: \(2.50x+1.60(10)=2.10(x+10)\).
\(2.5x+16=2.1x+21 \Rightarrow 0.4x=5 \Rightarrow x=12.5\).
Answer: \(12.5\) lb grapes.
\(\textbf{4)}\) Megan has $5.50 in quarters and nickels. She has 30 coins. How many quarters does Megan have?
20 quarters and 10 nickles
Let \(q\) = quarters, \(n\) = nickels.
Count: \(q+n=30\).
Value (in cents): \(25q+5n=550\).
From count, \(n=30-q\). Substitute:
\(25q+5(30-q)=550 \Rightarrow 25q+150-5q=550\).
\(20q=400 \Rightarrow q=20\); then \(n=10\).
Answer: \(20\) quarters, \(10\) nickels.
\(\textbf{5)}\) Jake has $10.45 in nickels and dimes. He has 109 coins. How many nickles does Jake have?
9 nickles and 100 dimes
Let \(n\) = nickels, \(d\) = dimes.
Count: \(n+d=109\).
Value (in cents): \(5n+10d=1045\).
Use \(n=109-d\):
\(5(109-d)+10d=1045 \Rightarrow 545-5d+10d=1045\).
\(5d=500 \Rightarrow d=100\); then \(n=9\).
Answer: \(9\) nickels, \(100\) dimes.
\(\textbf{6)}\) Carrie has $4.00 in quarters, nickles and dimes. She has 38 coins. She has twice as many nickles as dimes. How many quarters does Carrie have?
8 quarters, 20 nickles, and 10 dimes
Let \(q\) = quarters, \(n\) = nickels, \(d\) = dimes. Given \(n=2d\).
Count: \(q+n+d=38 \Rightarrow q+2d+d=38 \Rightarrow q=38-3d\).
Value (in cents): \(25q+5n+10d=400\). Substitute:
\(25(38-3d)+5(2d)+10d=400\).
\(950-75d+10d+10d=400 \Rightarrow 950-55d=400\).
\(-55d=-550 \Rightarrow d=10\); then \(n=20\), \(q=38-3(10)=8\).
Answer: \(8\) quarters, \(20\) nickels, \(10\) dimes.
\(\textbf{7)}\) You have a tub of a 60% alcohol solution and a tub of an 82% alcohol solution. Your chemistry experiment requires 8 cups of a 72% alcohol solution. How many cups of each solution do you need? \( \)
3.64 cups 60% solution and 4.36 cups 82% solution\( \)
Let \(x\) = cups of 60% solution, \(y\) = cups of 82% solution.
Total: \(x+y=8\).
Concentration: \(0.60x+0.82y=0.72(8)=5.76\).
Use \(y=8-x\): \(0.60x+0.82(8-x)=5.76\).
\(0.60x+6.56-0.82x=5.76 \Rightarrow -0.22x=-0.80 \Rightarrow x=\frac{80}{22}\approx3.64\).
\(y=8-3.64\approx4.36\).
Answer (rounded to hundredths): \(3.64\) cups of 60% and \(4.36\) cups of 82%.
\(\textbf{8)}\) You have a tub of a 40% alcohol solution and a tub of an 32% alcohol solution. Your chemistry experiment requires 10 quarts of a 34% alcohol solution. How many quarts of each solution do you need?
2.5 quarts 40% solution and 7.5 quarts 32% solution
Let \(x\) = quarts of 40%, \(y\) = quarts of 32%.
Total: \(x+y=10\).
Concentration: \(0.40x+0.32y=0.34(10)=3.4\).
Use \(y=10-x\): \(0.40x+0.32(10-x)=3.4\).
\(0.40x+3.2-0.32x=3.4 \Rightarrow 0.08x=0.2 \Rightarrow x=2.5\).
\(y=10-2.5=7.5\).
Answer: \(2.5\) qt of 40% and \(7.5\) qt of 32%.
\(\textbf{9)}\) Walnuts cost $2.50 a lb and brazil nuts cost $8.00 a pound. How many pounds of each would you mix if you want to create a 6 pound bag that costs $4.50 a pound?
3.82 pounds walnuts and 2.18 pounds brazil nuts
Let \(x\) = pounds of walnuts at $2.50/lb, \(y\) = pounds of brazil nuts at $8.00/lb.
Total weight: \(x+y=6\).
Cost: \(2.50x+8y=4.50(6)=27\).
Use \(y=6-x\): \(2.5x+8(6-x)=27\).
\(2.5x+48-8x=27 \Rightarrow -5.5x=-21 \Rightarrow x=\frac{21}{5.5}\approx3.82\).
\(y=6-3.82\approx2.18\).
Answer (rounded): \(3.82\) lb walnuts, \(2.18\) lb brazil nuts.
\(\textbf{10)}\) Aliyah invested \$180,000 in two accounts paying 7% and 3% annual interest. After one year she earned \$8,400 in interest altogether. How much was invested at each rate?
\$75,000 at 7% and \$105,000 at 3%.
Let \(x\) = dollars at 7%, \(y\) = dollars at 3%.
Total principal: \(x+y=180{,}000\).
Interest: \(0.07x+0.03y=8{,}400\).
Use \(y=180{,}000-x\):
\(0.07x+0.03(180{,}000-x)=8{,}400\).
\(0.07x+5{,}400-0.03x=8{,}400 \Rightarrow 0.04x=3{,}000 \Rightarrow x=75{,}000\).
\(y=180{,}000-75{,}000=105{,}000\).
Answer: $\(75{,}000\) at 7% and $\(105{,}000\) at 3%.
\(\textbf{11)}\) Brandon placed \$250,000 into two funds: one earns 5% annually and the other earns 2.5%. In the first year the total interest was \$10,000. How much did he put in each fund?
\$150,000 at 5% and \$100,000 at 2.5%.
Let \(x\) = dollars at 5%, \(y\) = dollars at 2.5%.
Total: \(x+y=250{,}000\).
Interest: \(0.05x+0.025y=10{,}000\).
Use \(y=250{,}000-x\):
\(0.05x+0.025(250{,}000-x)=10{,}000\).
\(0.05x+6{,}250-0.025x=10{,}000 \Rightarrow 0.025x=3{,}750 \Rightarrow x=150{,}000\).
\(y=250{,}000-150{,}000=100{,}000\).
Answer: $\(150{,}000\) at 5% and $\(100{,}000\) at 2.5%.
\(\textbf{12)}\) Nina split \$90,000 between municipal bonds paying 6% and a CD paying 4.5%. Her total interest after one year was \$4,650. How much was invested in each?
\$40,000 at 6% and \$50,000 at 4.5%.
Let \(x\) = dollars at 6%, \(y\) = dollars at 4.5%.
Total: \(x+y=90{,}000\).
Interest: \(0.06x+0.045y=4{,}650\).
Use \(y=90{,}000-x\):
\(0.06x+0.045(90{,}000-x)=4{,}650\).
\(0.06x+4{,}050-0.045x=4{,}650 \Rightarrow 0.015x=600 \Rightarrow x=40{,}000\).
\(y=90{,}000-40{,}000=50{,}000\).
Answer: $\(40{,}000\) at 6% and $\(50{,}000\) at 4.5%.
See Related Pages\(\)
