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.

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.

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.

Solve the equation to find x.

The original equation is a rational equation.
Multiply the least common multiple of the denominators by each term.  The least common multiple of 24, 120 and x is 120x.
Simplify by canceling the common factors.
Solve the remaining linear equation.

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.

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.

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.

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.

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.

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

For the 6% account:

For the 5% account:

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.

Solve the equation.
Use the distributive property and combine like terms to simplify each side of the equation.
Solve for x by subtracting 1550 on both sides and simplifying.
Solve for x by dividing both sides by 0.01 and simplifying.

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:

Here is a handout about consecutive numbers.

MAT0024_2.5CW Consecutive