Download 1 04.0 5000 = = = t r P 200 )1)(04.0)( 5000( Pr = = = I I t I

M098
Carson Elementary and Intermediate Algebra 3e
Section 3.5
Objectives
1.
2.
Use a table to solve problems involving two investments.
Use a table to solve problems involving mixtures.
Prior Knowledge
Converting from a percent to a decimal: To convert from a percent to a decimal, move the decimal point
two places to the left.
Clearing decimals: To clear decimals in an equation, multiply each term by an appropriate power of 10
(e.g. 10, 100, 1000, etc.)
Vocabulary
Principal
APR
Concentration
An amount of money invested into (or borrowed from) an account
Annual Percentage Rate
Amount of a substance per unit of volume
New Concepts
Use a table to solve problems involving two investments
When solving problems involving investments, it is helpful to remember the formula for simple interest: I = Prt
In this formula, “I” is the interest earned on account where a principal “P” is invested at a given rate “r” for a
certain time (in years) “t.”
Example 1
You invest $5000 into an account that earns a 4% APR. How much interest will you earn after 1 year?
First, identify the key information about the principal, rate and time, then use the formula:
Notice, the
interest rate
of 4% was
converted to
a decimal
P  5000
I  Pr t
r  0.04
I  (5000)(0.04)(1)
t 1
I  200
This means that after 1 year, the account earns $200 in interest.
In this section, we will be dealing with interest earned between two accounts for one year. Since the “t” will
always be 1, our formula for interest will be just I = Pr
We can use a table to organize the information like we did when we solved problems involving distances.
When solving problems involving investments:
 Identify the key words indicating the principal, interest rate and total interest earned.
 Organize information in a table.
 Write the equation and solve.
Example 2
Suppose you invested $8000 into two separate accounts. The APR for the first account was 4% and the APR
for the second account was 8%. After one year the total amount of interest earned was $600. How much did
you invest in each account?
C.Stewart 2011
page 1
M098
Carson Elementary and Intermediate Algebra 3e
Section 3.5
Before we fill in the table, let’s break this down into its components:
Principal – the total principal invested into the two accounts is $8000. To figure out how much is invested into
the two accounts, look at the following table:
Amount Invested
First Account
1000
1500
3700
p
Now, we can fill in the table:
Accounts
Interest Rate
First Account
0.04
Second Account
0.08
(totals)
Amount Invested
Second Account
8000 – 1000 = 7000
8000 – 1500 = 6500
8000 – 3700 = 4300
8000 – p
Principal
p
8000 - p
In general
terms
Since we do not know
the interest for each
account, we will use the
formula R•P = I
Interest
0.04p
0.08(8000 – p)
600
0.04 p  0.08(8000  p)  600
In general, you will need to add
the interest from the first account
(I1) and the interest from the
second account (I2) to find the
total interest (It).
Equation to solve: I1 + I2 = It
Clear decimals as necessary.
(100)[0.04 p  0.08(8000  p)]  (100)600
4 p  8(8000  p)  60000
4 p  64000  8 p  60000
 4 p  64000  60000
 4 p  4000
 4 p  4000

4
4
p  1000
Looking back at the table, the answer p = 1000 represents the amount invested into the first account, therefore
the amount invested into the second account is 8000 – p or $7000.
Example 3
Marvin invests $12,000 in two plans. Plan 1 is at an APR of 6% and Plan 2 is at an APR of 9%. If the total
interest earned after one year is $828, what principal was invested in each plan?
Accounts
Plan 1
Plan 2
(totals)
Interest Rate
0.06
0.09
Principal
p
12000 - p
0.06 p  0.09(12000  p)  828
6 p  9(12000  p)  82800
Equation to solve: I1+I2 = It
Interest
0.06p
0.09(12000 – p)
828
Clear decimals by
multiplying by 100
6 p  72000  9 p  82800
 3p  7200  82800
 3p  10800
p  3600
Therefore, $3600 was invested into the first plan and $8400 was invested into the second account.
C.Stewart 2011
page 2
M098
Carson Elementary and Intermediate Algebra 3e
Section 3.5
Use a table to solve problems involving mixtures
Sometimes, it is necessary to create a mixture. The process for solving problems involving mixtures is the
same as solving investment problems; the only difference is the terminology:
Investments: Rate x Principal = Total Interest (PR = I)
Chemical Mixtures: Concentration of Chemical x Volume of Solution = Total Amount of Chemical (CV = A)
Other Mixtures:
Price x Quantity = Total Cost
Notice how similar all of these formulas are. IN GENERAL: Value x Quantity = Total
Example 4
Suppose you buy a 250 mL bottle of hydrogen peroxide at your local pharmacy with a concentration of 3%.
How much of that bottle is pure hydrogen peroxide?
First, identify the key information about the concentration and volume, then use the formula: CV = A
The concentration of
3% was converted
to a decimal.
V  250
250(0.03)  A
7.5  A
C  0.03
This means that there are 7.5 mL of pure hydrogen peroxide in a 250 mL bottle.
Again, the process for solving a mixture problem is much like solving an investment problem. You can still use
a table to organize your information. The main difference is that usually, two solutions are mixed together to
achieve a third solution.
Example 5
Ken has 80 milliliters of 15% acid solution. How much of a 20% acid solution must be added to create a
solution that is 18% acid? What will the total volume of the mixture be?
Solutions
First Solution
Second Solution
Mixture
Concentration
0.15
0.20
0.18
Volume of Sol
80
x
x + 80
Amount of Sol
0.15(80)
0.20x
0.18(x + 80)
Since we do not know
the amount for each
solution, we will use the
formula C•V = A
The problem tells us that he is using 80 mL of the
first solution but asks “how much” of the second
solution, so we will call that “x.” Since we are adding
these two solutions together to get the third solution
we will call the volume of the mixture “x + 80.”
In general, you will need to add
the amount from the first solution
and the amount from the second
solution to find the total amount.
Equation to solve: A1 + A2 = At
Clear decimals as necessary.
0.15(80)  0.20 x  0.18(x  80)
15(80)  20 x  18(x  80)
1200  20 x  18 x  1440
2 x  240
x  120
Therefore, he must use 120 mL of the second 20% solution. The total amount of the new mixture is “x + 80” or
200 mL.
C.Stewart 2011
page 3
M098
Carson Elementary and Intermediate Algebra 3e
Section 3.5
Example 6
The Candy Shoppe wants to mix 115 pounds of candy to sell for $0.80 per pound. How many pounds of $0.60
candy must be mixed with a candy costing $1.20 per pound to make the desired mix?
It’s helpful to
keep the
mixture on the
bottom row.
Types
$0.60 candy
$1.20 candy
$0.80 Mixture
Price
0.60 → 0.6
1.20 → 1.2
0.80 → 0.8
0.6 x  1.2(115  x)  0.8(115)
6 x  12(115  x)  8(115)
6 x  1380  12 x  920
Quantity
x
115 - x
115
Total Cost
0.6x
1.2(115 – x)
0.8(115)
Finding the
quantity here is
much like
example 2
 6 x  1380  920
 6 x  460
x  76.6 6
115  x  38.34
This means that 76.66 pounds of the $0.60 candy should be mixed with 38.34 pounds of the $1.20 candy to
make 115 pound mixture that sells for $0.80 per pound.
C.Stewart 2011
page 4