The Mathematics 11 Competency Test, Base, and Rate Sorting out a percent problem always involves correctly attaching numbers to three fundamental quantities: the percent (rate), the value to which the percent rate is applied, and the resulting amount of material or amount of change. Here we will use the full words to represent each quantity, but when you ve developed some skill in algebra, you could use shorter symbols (such as A for amount, b for, and r for rate ). The percent rate is just the percent value with which we are dealing. So when we say the metal alloy is 10% zinc, the percent rate of zinc in the alloy is equal to 10% Hank has received a 5% salary increase, the percent rate of change in Hank s salary is equal to 5%. When it comes time to do calculations involving the percent rate, we always use the fractional or decimal value corresponding to the percent rate. The value is the value of the quantity to which the percent rate is applied. Every percent rate is always associated with a very specific apart from this value, a percent rate has no meaning. The easiest way to identify the quantity is to ask percent of what? The what here must be a numerical quantity. So when we say the metal alloy is 10% zinc, the that the percent rate refers to is the total mass of metal alloy. (Ask: Zinc is 10% of what? Answer: It is the total mass of the metal alloy that zinc makes up 10%.) Hank has received a 5% salary increase the to which the percent rate is applied is his previous salary. (Ask: The increase is 5% of what? Answer: It is Hank s previous salary that the increase is 5% of.) Sometimes when people are speaking of percent change, they use the term original value instead of. The two terms mean exactly the same thing. The amount is the actual quantity of stuff or quantity of change that is represented by the stated percent of the. The amount and the, as these words are used here, always have the same units (and in fact, are almost always quantities that have units of measurement). So, when we say the metal alloy is 10% zinc, the amount of zinc is the actual number of kilograms (say) of zinc in the kilograms of alloy that is present. Hank has received a 5% salary increase, the amount of this increase is the additional dollars and cents he receives for each hour, say, of work. The relationship between these three fundamental characteristics of a percent problem is very simple: = rate David W. Sabo (2003), Base, and Rate Page 1 of 5
If you know a bit of algebra already, you know that this formula can easily be rearranged to give = rate and Thus, if we know any two of these three quantities, it is easy to calculate the third one. In all of these formulas, rate stands for the fractional or decimal form of the percent value. Example 1: Calculate 67% of 586. Clearly 67% is the rate here. Since the value 586 follows the word of in 67% of 586, 586 must be the to which the percent refers. Then, the question is asking for the amount corresponding to 67% of 586. So = x rate = 586 x 0.67 = 392.62. Note that we used 0.67, the decimal equivalent of 67%. So, to respond to the instruction, 67% of 586 is 392.62. You might get a better sense of the way language is being used here if you note that 67% is equivalent to 67/100 or about two-thirds as a fraction. When we say, calculate two-thirds of 586, we mean multiply 586 by two-thirds. In a similar way, since 67% is equivalent to the decimal value 0.67, to calculate 67% of 586 is to calculate 0.67 of 586, and indicates the result of multiplying 586 by 0.67. Example 2: Hank buys a DVD player on sale. Before tax, he is given a 30% discount which reduces the price by $68.99. What was the original price of the DVD player? The rate here is 30% or 0.30. This 30% discount is applied to the original price of the DVD player (so the original price must be the corresponding to the rate of 30% -- this means that the question is asking us to find a value). The $68.99 is the amount of dollars that the 30% of the original () price amounted to. So, we are told the rate and the amount, and we are asked to find the. Using = rate = 68.99 $ 0.30 = 229.96666 $ David W. Sabo (2003), Base, and Rate Page 2 of 5
Now, prices of DVD players do not normally include fractions of cents. So, at the least we should round our final answer off to the nearest cent. To respond to the question, then we can state that it appears that the original price of the DVD player must have been $229.97. (Note that you can check this answer. Just calculate how much money that 30% of $229.97 amounts to. You should get $68.99 if we haven t made a mistake here.) Example 3: Hank buys a house for $227,000. One month later he sells the house for an additional $15,500 more than he paid for it. By what percent did the price of the house increase? The question asks us for a percent, which means we are being asked to calculate a rate. The formula we have for a rate is So, to be able to calculate a rate, we need to identify the values of the relevant and amount. Recalling that the is the original value of whatever is changing, we conclude that the appropriate value here must be the original price of the house, $227,000. The is the amount by which the values changes. Here the original price of the house increased by $15,500, so this is the in the formula. Substituting these values into the formula gives $15,500 = 0.06828 $227,000 rounded to five decimal places. To get the percent equivalent of this decimal value, we need to multiply by 100: 0.06828 x 100 6.828% Thus, the price of the house increased by 6.828%. Example 4: If the price of a new car is $27,545, and tax is payable on this car at a rate of 14.5%, how much tax must the purchaser actually pay? The rate of tax is 14.5%. This rate of taxation is d on the actual price of the car. The tax paid, the amount of tax paid in dollars is then = x rate = $27,545 x 0.145 = $3994.03 Example 5: One day Hank spent $42.50 on gasoline and $123.75 on a textbook. What percent of his purchases that day was for gasoline? David W. Sabo (2003), Base, and Rate Page 3 of 5
There s a bit of a twist here, but it s not too hard to sort things out. Clearly, we are asked for a percent, so we need to calculate a rate, and the formula for that is The part of his expenses for which a rate is requested is the part Hank spent on gasoline. So here = $42.50 The value to which this percent applies must then be the total amount of money Hank spent that day: So = $42.50 + $123.75 = $166.25 $42.50 = $166.25 = 0.2556 rounded to four decimal places. This decimal fraction corresponds to 25.56%. So, 25.56% of Hank s expenses for that day were for gasoline. Example 6: The label on a 1.5 litre bottle of wine that Hank buys states that it is 11.7% alcohol by volume. How many millilitres of alcohol are present in that bottle of wine. Here we have a straightforward request to calculate the. When we are told the alcohol is 11.7% of the total of 1.5 litres of wine in the bottle, we get that the amount of alcohol present must be = x rate = (1.5 litres)(0.117) = 0.1755 litres Notice that we carried units along where appropriate (the quantity here was a number of litres of wine, and hence the we calculated carries with it units of litres). But the question asked for the answer as millilitres of alcohol, so we must do a unit conversion here: David W. Sabo (2003), Base, and Rate Page 4 of 5
1000 millilitres = 0.1755 litres = 175.5 millilitres 1litre Thus, this bottle of wine contains 175.5 millilitres of alcohol. NOTE: A method for carrying out unit conversions correctly and reliably is described in detail in other documents on this site. If you haven t studied that method, you can still easily calculate the answer above because if involves a very simple unit conversion. If then 1 litre = 1000 millilitres (by definition) 0.1755 litres = 0.1755 x 1000 millilitres = 175.5 millilitres This rather informal approach will do here for more involved unit conversions you ll definitely want to use the more systematic method described in the documents on unit conversion elsewhere in these notes. David W. Sabo (2003), Base, and Rate Page 5 of 5