Category Archives: Applications

Application of Rational Equations: Work Together (one pump stops working)

Problem:  Two pumps were required to pump the water out of a submerged area after a flood.  Pump A, the larger of the two pumps, can pump the water out in 24 hours, whereas it would take pump B 120 hours.  Both pumps were working for the first 8 hours until pump A broke down.  How long did it take pump B to pump the remaining water?

Solution:

Create a table.  List each individual and the time it takes to complete the job.  Also include a row for the pumps working together.  Use a variable to represent the unknown time to complete the job when the pumps are working together.

Time to complete the job
Pump A
 24 hours
Pump B
120 hours
Together x

Fill in the table with the portion of the job completed in one hour.

If it takes Pump A 24 hours to complete the whole job, one twenty-forth (1/24) of the job will be completed in one hour.

If it takes Pump B 120 hours to complete the whole job, one one hundred twentieth (1/120) of the job will be completed in one hour.

If it takes x number of hours when the pumps are working together 1/x portion of the job will be completed in one hour.

Time to complete the job
 Portion completed in 1 hour
Pump A
24 hours  1/24
Pump B
120 hours  1/120
Together x  1/x

From here an equation can be created with the portion of the job completed in one hour.

In one hour, the portion completed by pump A plus the portion completed by pump B should equal the portion when they are working together.

1/24+1/120=1/x

Solve the equation to find x.

 The original equation is a rational equation.
1/24+1/120=1/x
Multiply the least common multiple of the denominators by each term.  The least common multiple of 24, 120 and x is 120x. 
1/24+1/120=1/x
120x(1/24)+120x(1/120)=120x(1/x)
Simplify by canceling the common factors.
120x(1/24)+120x(1/120)=120x(1/x)
5x+x=120
Solve the remaining linear equation. 5x+x=120
6x=120
{6x}/6={120}/6
x=20

It would take both pumps working together 20 hours to pump out all of the water.

But the two pumps are only working together for 8 hours which means they only get eight-twentieths (8/20) of the job done.  This fraction reduces to two-fifths (2/5).

Three-fifths (3/5) of the water remains and pump B is working alone.

It takes pump B 120 hours to complete the whole job.  It will take 120(3/5) to pump the remaining water out.

120(3/5)= 72 hours.

It takes pump B 72 hours to pump the remaining water.

Application of Exponential Functions: Doubling Time

Example:

How long does it take for an investment to double if it is invested at 18% compounded continuously?

Solution:

Since this question involve continuous compound interest, we will use the associated formula.

A=Pe^{rt}

We are given that the interest rate is 18% or 0.18.  This tells me that when r=0.18 Since we are looking for the doubling time, A will be 2 times P.  I can write that in symbols A=2P.

Substitute these values into the continuous compound formula and solve for the interest rate.

Continuous compound formula
A=Pe^{rt}
Substitute the values of r and A into the formula
A=Pe^{rt}
2P=Pe^{0.18*t}
2P=Pe^{0.18t}
Solve for t by dividing both sides by P and simplifying
2P=Pe^{0.18t}
{2P}/P={Pe^{0.18t}}/P
2=e^{0.18t}
Solve for t by taking the log of both sides.
2=e^{0.18t}
ln 2=ln e^{0.18t}
Solve for t by using the power rule and simplifying
ln 2=ln e^{0.18t}
ln 2=0.18t ln e
ln 2=0.18t (1)
ln 2=0.18t
Solve for t by dividing both sides by 0.18 and simplifying
ln 2=0.18t
{ln 2}/0.18={0.18t}/0.18
{ln 2}/.018=t
Find the value in the calculator
{ln 2}/0.18=t
3.85081767=t
Write the answer rounded to two decimal places
t=3.85081767
t=3.85

It will take 3.85 years to double your money when interest is compounded continuously at 18%.

If you need to write this in years and months, you will need to convert the 0.85 to months.  Since there are 12 months in a year, multiply 0.85 by 12 to get 10.2.  I will round to the nearest months to get 10.

It will take 3 years and 10 months to double your money when interest is compounded continuously at 18%.

Here is a video that is similar except that you are looking for the investment to triple.

Application of Quadratic Equation: Translation

Example:  The product of some negative number and 5 less than twice that number is 273.  Find the number.

Solution:  Translate the statement into an equation and then solve the equation.

“The product of some negative number…”

Let x be some negative number and multiply that number by the expression that comes next in the statement.

x  (expression that comes next)

“…and 5 less than twice that number…”

5 less than means take 5 away from what follows.  Twice that number refers to 2 times the negative number that was described before.  (2x-5)

x  (2x-5)

“…is 273”

This translates to equals 273.

x (2x-5)=273

Solve the equation.

 

The translated equation
x (2x-5)=273
Use the distributed property to rewrite the equation (recognize that the equation is a quadratic equation)
2x^2-5x=273
Choose the method for solving the quadratic equation (factoring, square root method, completing the square or quadratic formula)  I will demonstrate factoring so I need to move all terms to the same side to have zero on one side.
2x^2-5x=273
2x^2-5x-273=273-273
2x^2-5x-273=0
Factor
2x^2-5x-273=0
(       )(       )=0
(2x       )(x       )=0
(2x      21)(x     13)=0
(2x    +  21)(x  -   13)=0
Use the zero product property and set each factor equal to zero and solve.
(2x    +  21)(x  -   13)=0
2x    +  21=0 or x  -   13=0
2x    +  21-21=0-21 or x  -   13+13=0+13
2x =-21 or x =13
{2x}/2 ={-21}/2 or x =13
x ={-21/2} or x =13

Since the statement says that the number is negative, the number is {-21/2}.

Application of Linear Equation (Simple Interest)

Example:  Larry invested part of his $31,000 advance at 6% annual simple interest and the rest at 5% annual simple interest.  If his total yearly interest from both accounts was $1,760, find the amount invested at each rate.

Solution:  This question involved simple interest.  We will use the simple interest formula.

I=Prt

The I stands for interest, the P stands for principal (initial investment), r stands for the interest rate and t stands for time in years.

When starting an application problem like this it can be helpful to organize the information in a table using the formula.

I P r t
6% Account
5% Account

Begin filling in the information from the problem.  The interest rate is the most obvious and you should use the decimal version of the percent by moving the decimal place left two places.  Since the problem state that $1760 is the yearly interest for both accounts, we will use a time of 1 year.

I P r t
6% Account .06 1
5% Account .05 1

The goal in this question is to find the amount invested in each account.  I will let x be the amount invested in the 6% account.  The problem states that the rest of the $31,000 will be invested in the 5% account.  We can represent the rest of the money as 31000-x.

I P r t
6% Account x .06 1
5% Account  31000-x .05 1

You can fill in the interest column using the simple interest formula.

For the 6% account: I=Prt=(0.06)(x)(1)=0.06x

For the 5% account: I=Prt=(0.05)(31000-x)(1)=0.05(31000-x)

I P r t
6% Account 0 .06x x 0.06 1
5% Account 0.05(31000-x) 31000-x 0.05 1

You can now create an equation with this information.  The interest from the first account plus the interest from the second account should equal the total interest of $1760.

0.06x+0.05(31000-x)=1760

Solve the equation.
0.06x+0.05(31000-x)=1760
Use the distributive property and combine like terms to simplify each side of the equation.
0.06x+1550-0.05x=1760
0.01x+1550=1760
Solve for x by subtracting 1550 on both sides and simplifying.
0.01x+1550-1550=1760-1550
0.01x=210
Solve for x by dividing both sides by 0.01 and simplifying.
{0.01x}/0.01=210/0.01
x=21000

Since x represents the amount invested in the 6% account, $21000 is invested in the 6% account and the rest is invested in the 5% account.  The rest is $31,000-$21,000=$10,000.  $10,000 is invested in the 5% account.

Video Example: