2500 – 2000 = 500

The amount of increase is 500. The percent increase is found by dividing the amount increase by the original amount (2000).

500/2000 = 25/100 = 25%

Begin by translating your sentence into an equation.  The original number can be given the value "\(\displaystyle x\)."  When you increase that number by \(\displaystyle 25\%\), you can write it one of two ways:

Way 1: \(\displaystyle x + 0.25 * x = 16\)

Way 2: \(\displaystyle 1.25x = 16\)

Note that the second equation is merely a simplification of the first.  Now, just divide both sides by \(\displaystyle 1.25\) to get \(\displaystyle x\). \(\displaystyle x = 12.8\).

To find the percentage difference, you merely subtract the start value from the end value and divide by the start.  The absolute change was \(\displaystyle 575 - 450\) cubic centimeters or \(\displaystyle 125\) cubic centimeters.  Divide this by the start value, \(\displaystyle 450\), and you get \(\displaystyle 0.2778\) or \(\displaystyle 28\%\).

The product must make another \(\displaystyle \$4\) in order to break even.  Therefore, it must be marked up to a total price of \(\displaystyle \$22\).  The question is really asking for a percentage change.  The percentage change from \(\displaystyle \$18\) to \(\displaystyle \$22\) is:

\(\displaystyle \frac{22-18}{18} = \frac{4}{18} = 0.2222222 \text{ or } 22.23%\).  NOTE: It must be marked up at least slightly more than 22.22% in order to break even.

If Sam is \(\displaystyle 145\) lbs with a body fat percentage of \(\displaystyle 18\%\), this means that she has \(\displaystyle 145 * 0.18 \text{ or } 26.1 lbs\) of body fat.  Before losing the weight, she would have had \(\displaystyle 26.1 + 6 \text{ or } 32.1 lbs\) of body fat.

If Pat is \(\displaystyle 157\) lbs. with \(\displaystyle 21\%\) body fat, he has \(\displaystyle 157 * 0.21 \text{ or } 32.97\) lbs. of body fat.  

When he was \(\displaystyle 12\) lbs. heavier, he weighed \(\displaystyle 169\) (that is, \(\displaystyle 157 + 12\)) lbs., and would have had \(\displaystyle 44.97\) (that is, \(\displaystyle 32.97 + 12\)) lbs. of body fat on his body.  

This would mean that his body fat percentage is \(\displaystyle \frac{44.97}{169}\) or approximately \(\displaystyle 26.61\%\).

\(\displaystyle .08\cdot 2800=224\)

\(\displaystyle 2800+224=3024\)

Between 2003 and 2004, the number of people enrolled increases by 12%:

\(\displaystyle 3000\cdot 1.12=3360\)

Between 2004 and 2005, 65 people unenroll:

\(\displaystyle 3360-65=3295\)

\(\displaystyle \textup{To find a }15\%\textup{ increase, multiply the original number by 1.15}\)

\(\displaystyle 25,000\times1.15=28,750\)

Do not overthink this question.  Remember that if you have consecutive increases in value, you can multiply them in a row.  For instance, if you increase something costing \(\displaystyle x\) dollars by 50% and then again by 50%, you merely multiply twice by 1.5:

\(\displaystyle New Price = 1.5 * 1.5 * x = 2.25 * x\)

For our data, this would be:

\(\displaystyle 1.1 * 1.2 * 20 = 1.32 * 20\)

Note that we DO NOT want to multiply everything out.  Instead, we can see immediately that we have 132% of 20 as the new value.  This means that it has been increased by 32% from its original price ($20).