Increasing/Decreasing by a Percent - 4
A box has a square bottom. The height has not yet been determined, but the bottom is 10 inches by 10 inches. The volume formula is V = lwh, because each of the length and width is 10, lw becomes 10 – 10 = 100.
Increasing/Decreasing by a Percent - 3
For some word problems, nothing more will be required of you than to substitute a given value into a formula, which is either given to you or is readily available. The most difficult part of these problems will be to decide which variable the given quantity will be. For example, the formula might look like R = 8q and the value given to you is 440. Is R = 440 or is q = 440? The answer lies in the way the variables are described In R ¼ 8q, it might be that R represents revenue (in dollars) and q represents quantity (in units) sold of some item. ‘‘If 440 units were sold, what is the revenue?’’ Here 440 is q. You would then solve R = 8(440). ‘‘If the revenue is $440, how many units were sold?’’ Here 440 is R, and you would solve 440 = 8q.
Increasing/Decreasing by a Percent - 2
At times the percent is the unknown. You are given two quantities and are asked what percent of one is of the other. Let x represent the percent as a decimal number.
Increasing/Decreasing by a Percent - 1
Many word problems involving percents fit the above model—that is, a quantity being increased or decreased. Often you can solve these problems using one of the following formats:
Increasing/Decreasing by a Percent
Increasing/Decreasing by a Percent
As consumers, we often see quantities being increased or decreased by some percentage. For instance, a cereal box boasts ‘‘25% More.’’ An item might be on sale, saying ‘‘Reduced by 40%.’’ When increasing a quantity by a percent, first compute what the percent is, then add it to the original quantity. When decreasing a quantity by a percent, again compute the percent then subtract it from the original quantity.
