Mixture with Money Answers to 1 - 12
1. Yolanda has dimes and quarters totaling $5.25. If she has 33 coins in all how many of each does she have?
a. Read the problem several times. Guess.
This time we are going to do some sophisticated guessing.
# of dimes # of quarters Value of dimes Value of Q’s Total value
15 18 $1.50 $4.50 $6.00 (too big)
23 10 $2.30 $2.50 $4.70 (too small)
If you continue with this type of organized guessing, you can get the answer soon.
b. Define x, usually to answer the question.
Let x = the number of dimes.
c. Label other unknown quantities in terms of x.
Let 33 – x = the number of quarters.
Let 10x = the value of the dimes.
Let 25(33 – x) = the value of the quarters.
d. Form an equation.
10x + 25(33 – x) = 525 (This problem is organized in cents to avoid decimals but you could also use dollars: .10x + .25(33 – x) = 5.25.)
e. Solve the equation.
10x + 25(33 – x) = 525
10x +825 – 25x = 525
-15x = -300
x = 20 dimes
f. Answer the question.
33 – x = 33 – 20 = 13 quarters
2. Tony has 39 bills in fives and tens. If the total value is $285 how many of each does he have?
a. Read the problem several times. Guess.
b. Define x, usually to answer the question.
Let x = the number of fives.
c. Label other unknown quantities in terms of x.
Let 39 – x = the number of tens
Let 5x = the value of the fives.
Let 10(39 – x) = the value of the tens.
d. Form the equation.
5x + 10(39 – x) = 285
e. Solve the equation.
5x + 10(39 – x) = 285
5x + 390 –10x = 285
-5x = -105
x = 21 fives
f. Answer the question.
39 – x = 39 – 21 = 18 tens
3. The Drama Club sold 500 tickets to their fall performance. The adult tickets were $5 each and the student tickets were $3 each. If they took in $2080, how many of each did they sell?
a. Read the problem several times. Guess.
Make sure that in your guess, the total number of tickets is 500.
b. Define x, usually to answer the question.
Let x = the number of adult tickets.
c. Label other unknown quantities in terms of x.
Let 500 – x = the number of student tickets
d. Form the equation.
5x + 3(500 – x) = 2080
e. Solve the equation.
5x + 3(500 – x) = 2080
5x + 1500 – 3x = 2080
2x = 580
x = 290 adult tickets
f. Answer the question.
500 – x = 500 – 290 = 210 student tickets
4. Edie has 27 coins in dimes and quarters. If the total value is $3.75 how many of each does she have?
a. Read the problem several times. Guess.
Try guessing and checking and guessing again until you get it numerically (without algebra).
b. Define x, usually to answer the question.
Let x = the number of dimes.
c. Label other unknown quantities in terms of x.
Let 27 – x = the number of quarters.
Let 10x = the value of the dimes.
Let 25(27 – x) = the value of the quarters.
d. Form the equation.
10x + 25(27 – x) = 375 (This equation is organized in cents, not dollars.)
e. Solve the equation.
10x + 25(27 – x) = 375
10x + 675 – 25x = 375
-15x = -300
x = 20 dimes
f. Answer the question.
27 – x = 27 – 20 = 7 quarters
5. Venus bought 40 stamps for $12.40. Some of the stamps were 33 cent stamps and some were 23 cent stamps. How many of each did she buy?
a. Read the problem several times. Guess.
b. Define x, usually to answer the question.
Let x = the number of 33 cent stamps.
c. Label other unknown quantities in terms of x.
Let 40 – x = the number of 23 cent stamps.
Let 33x = the value of the 33 cent stamps.
Let 23(40 – x) = the value of the 23 cent stamps.
d. Form the equation.
33x + 23(40 – x) = 1240
e. Solve the equation.
33x + 920 – 23x = 1240
10x = 320
x = 32 stamps at 33 cents each
f. Answer the question.
40 – x = 40 – 32 = 8 stamps at 23 cents each
6. Sonia has 26 bills in ones and fives. If their total value is $50 how many of each does she have?
a. Read the problem several times. Guess.
You are not going to need many fives in this guess. Check your first guess to see why.
b. Define x, usually to answer the question.
Let x = the number of ones
c. Label other unknown quantities in terms of x.
Let x = the value of the ones. It is unusual that x can be both the number of bills and the value of the bills. This will only happen with $1 bills, one cent stamps, $1 tickets, etc.
Let 26 – x = the number of fives.
Let 5(26 – x) = the value of the fives.
d. Form the equation.
x + 5(26 – x) = 50
e. Solve the equation.
x + 5(26 – x) = 50
x + 130 – 5x = 50
-4x = - 80
x = 20 one dollar bills
f. Answer the question.
26 – x = 26 – 20 = 6 five dollar bills
7. Mr. Planter wants to mix 20 lb of macadamia nuts that cost $8.10 per pound with pecans that cost $5.40 per pound. How many pounds of pecans should he use if the mixture is to cost $6.48 per pound?
a. Read the problem. Guess.
The final price is closer to the pecan price than to the macadamia price so more
pecans are in the mixture than macadamias. Your guess?
b. Define x.
Let x = the number of pounds of pecans in the mixture.
c. Define unknown quantities in terms of x.
20 – x = the number of pounds of macadamias
5.40 x = the value of x pounds of pecans
8.10 (20 – x) = the value of (20 – x) pounds of macadamias
d. Form the equation. Read the problem again. The final value will be 6.48 (20).
5.40 x + 8.10(20 – x) = 6.48(20)
e. Solve the equation.
5.40 x + 162 – 8.10 x = 129.60
-2.70 x = - 32.4
x = 12 pounds of pecans
f. Answer the question.
20 – x = 20 – 12 = 8 pounds of macadamias
8. Mr. Planter wants a 40 lb mixture of pecans and walnuts that will sell for $4.80 per pound. If the pecans cost $5.40 per pound and the walnuts cost $4.40 per pound, how many of each should he use?
a. Read the problem. Guess.
b. Define x, usually to answer the question.
Let x = the number of pounds of pecans.
c. Define other unknown quantities in terms of x.
Let 40 – x = the number of pounds of walnuts.
Let 5.40x = the value of the pecans.
Let 4.40(40 – x) = the value of the walnuts.
d. Form an equation.
5.40x + 4.40(40 – x) = 4.80(40)
e. Solve the equation.
5.40x + 4.40(40 – x) = 4.80(40)
5.40x + 176 – 4.40x = 192
x = 16 pounds of pecans
f. Answer the question.
40 – x = 40 – 16 = 24 pounds of walnuts.
9. Eddie invested $2400 in two simple interest accounts. The annual rate in one account is 8% and the annual rate in the other is 11%. How much did he invest in each account if the annual interest totaled $240?
a. Read the problem. Guess.
Make sure that the amounts in each account add to be $2400.
b. Define x, usually to answer the question.
Let x = the amount invested at 8%.
c. Define other unknown quantities in terms of x.
Let 2400 – x = the amount invested at 11%.
Let .08x = the interest earned in the 8% account. (I = PRT, but here, T = 1.)
Let .11(2400 – x) = the interest earned in the 11% account.
d. Form an equation.
.08x + .11(2400 – x) = 240
e. Solve the equation.
.08x + .11(2400 – x) = 240
.08x + 264 - .11x = 240
-.03x = -24
x = $800 invested at 8%.
f. Answer the question.
2400 – x = 2400 – 800 = $1600 invested at 11%.
10. Fred and Irma invested $12,500 in two simple interest accounts. One account earns annual simple interest of 7% while the other earns 6%. How much was invested at each rate if each account earned the same interest? Round your answer to the nearest dollar.
a. Read the problem. Guess.
b. Define x, usually to answer the question.
Let x = the amount invested at 7%
c. Define other unknown quantities in terms of x.
Let 12500 – x = the amount invested at 6%.
Let .07x = the amount of interest earned at 7%.
Let .06(12500 – x ) the amount of interest earned at 6%.
d. Form an equation.
Since the interest in each account is the same, .07x = .06(12500 – x)
e. Solve the equation.
.07x = .06(12500 – x)
.07x = 750 - .06x
.13x = 750
x = $5769 invested at 7%
f. Answer the question.
12500 – x = 12500 – 5769 = $6731 invested at 6%.
11. Bubba has two part time jobs. One pays $7.50 per hour and the other pays $9.00 per hour. Last week he earned $427 while working 50 hours. How many hours did he work at each job?
a. Read the problem. Guess.
b. Define x, usually to answer the question.
Let x = the number of hours worked at the $7.50 rate.
c. Define other unknown quantities in terms of x.
Let 50 – x = the number of hours worked at the $9 rate.
Let 7.50x = the amount of money earned at the $7.50 rate.
Let 9.00(50 – x) = the amount of money earned at the $9 rate.
d. Form an equation and solve the equation.
7.50x + 9.00(50 – x) = 427.00
7.50x + 450 – 9.00x = 427
-1.50x = -23
x = 15.33 hours (or 15 1/3 hours) at the $7.50 rate.
e. Answer the question.
50 – x = 50 – 15.33 = 34.67 hours at the 9$ rate.
12. Kim had to pay $1200 in labor for some repairs to the house. The carpenter charged $18 an hour and the painter charged $15 per hour. If she paid for 73 hours of work, how much work was done by each?
a. Read the problem. Guess.
b. Define x, usually to answer the question.
Let x = the number of carpenter hours.
c. Define other unknown quantities in terms of x.
Let 73 – x = the number of painter hours.
Let 18x = the money paid to the carpenter.
Let 15(73 – x) = the money paid to the painter.
d. Form an equation.
18x + 15(73 – x) = 1200
e. Solve the equation.
18x + 15(73 – x) = 1200
18x + 1095 – 15x = 1200
3x = 105
x = 35 hours of work by the carpenter
f. Answer the question.
73 – x = 73 – 35 = 38 hours of work by the painter.Created by: Marcia Tharp
Revised on: 10.14.02 by Dr. Marcia Tharp
Send comments to: mtharp@tcc.edu or rgill@tcc.edu