Next Generation Arithmetic Study Guide for the ACCUPLACER Test

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When performing the four basic operations with fractions, you need to follow some additional rules to get the correct answer. Here are the basic procedures.

Addition and Subtraction of Fractions

When adding or subtracting fractions, there are some things you need to remember, especially if the denominators are not the same. We’ll start with the easiest process, though…when the denominators are the same.

Problems with “Like” Denominators

Problems with like denominators are pretty easy. Simply add or subtract the numerators and use the result as the numerator of the answer. The denominator remains the same.

Problems with “Unlike” Denominators

Fractions with unlike denominators must be converted to equivalent fractions with like denominators. The denominator of the “new” fractions will be the least common multiple of the original denominators. For example, to add \(\frac{3}{4} + \frac{1}{6}\), the least common multiple of \(4\) and \(6\) is \(12\).

The equivalent “like denominator” problem is \(\frac{9}{12} + \frac{2}{12} = \frac{11}{12}\).

Multiplication of Fractions

Multiplication of fractions is straightforward. Multiply the numerators of the fractions to get the numerator of the answer, and multiply the denominators of the fractions to get the denominator of the answer. Reduce to lowest terms if needed.

Division of Fractions

A division problem with fractions must first be converted to a multiplication problem. To do this, invert the fraction you are dividing by (the divisor, the fraction after the division sign) and then multiply this inverted fraction with the other fraction.

For example, to divide \(\frac {1}{3} \div \frac{1}{4}\), invert the \(\frac{1}{4}\) and multiply:

\[\frac{1}{3} \times \frac{4}{1} = \frac{4}{3}\]

Other Necessary Steps

When you have arrived at an answer for a fraction problem, you have not, necessarily, finished. There are two additional things you need to be able to do to really get a final answer, in most cases. Be sure you can:

Reduce to Lowest Terms (Simplify)

A fraction is in lowest terms, or simplified when the numerator and denominator have no common factor (there is no number that “goes into” both the numerator and denominator).

For example, \(\frac{8}{12}\) is not simplified because \(8\) and \(12\) can both be divided by \(4\) (\(4\) is a common factor for \(8\) and \(12\)). The simplified form is \(\frac{2}{3}\).

Create a “Proper Fraction” from an “Improper Fraction”

A proper fraction has a numerator that is less than the denominator. This means that the value of the fraction is less than one.

In the division example above, the result \(\frac{4}{3}\) is improper because the numerator is greater than the denominator, making the value more than one.

The proper way to write this is as a mixed number. Divide the numerator by the denominator. The quotient will be the integer part of the mixed number. The remainder is the numerator of the fraction portion. The denominator is the same as the original fraction.

In the example \(4 \div 3\) is \(1\) remainder \(1\). The mixed number is \(1\frac{1}{3}\).


Decimals are a way to extend place values to amounts less than one. Place values in whole numbers are in powers of \(10\), with decimals they are powers of \(\frac{1}{10}\).

For the decimal number 12345.6789, there are:

1 ten-thousands, worth 10,000
2 thousands, worth 2,000
3 hundreds, worth 300
4 tens, worth 40
5 ones, worth 5
6 tenths, worth \(\frac{6}{10}\)
7 hundreths, worth \(\frac{7}{100}\)
8 thousandths, worth \(\frac{8}{1000}\)
9 ten-thousandths, worth \(\frac{9}{10000}\)

Adding and Subtracting Decimals

When adding and subtracting decimals, it is vital that the problem is written with the decimal points lined up. This will ensure that all the place values are in the correct positions.

Align Decimal Points

For example, to add the numbers \(43.5\) and \(2.13\), the problem should be written with the decimals aligned:

\[\begin{align} 43.50& \\ \underline{+\quad 2.13}& \\ \end{align}\]

Notice that zeros can be added after the last decimal place so that the number of decimal places is the same in both numbers to be added or subtracted. Adding zeros does not affect the value of the number when added after the last numeral to the right of the decimal point.

Place the Decimal Point in the Answer

Next, place the decimal point in the answer right below where it is in the problem. Once this is done, the problem can be completed the same as for adding and subtracting integers.

\[\begin{align} 43.50& \\ \underline{+\quad 2.13}& \\ 45.63& \\ \end{align}\]

Multiplying Decimals

When multiplying decimals, the problem is written without aligning the decimals. The decimal will be placed in the answer as the last step. First, perform the multiplication as if the decimal points were not present:

\(\quad \quad 1\)
\(\quad \;\;1 1\)
\(\begin{align} \quad \; 43.5& \\ \underline{x\quad 2.13}& \\ \quad \; 1305& \\ \quad \; 4350& \\ \underline{ 87000}& \\ 92655& \\ \end{align}\)

Next, count the total number of decimal places in the problem. In the example, the first number has one place and the second number has two, for a total of three decimal places. Finally, place the decimal in the answer so that the answer has the same total number of decimal places. The answer for the example will be 92.655[.]

Dividing Decimals

To perform long division with decimals, first set up the problem with the original numbers. In our example, we will divide \(43.5\) by \(2.13\):

\[\require{enclose} \begin{array}{r} 2.13 \enclose{longdiv}{43.5} \\[-3pt] \end{array}\]

We need to adjust the decimal point in the divisor (outside the division sign) so that there are no decimal places. Then adjust the dividend (inside the division sign) by the same number of places. This way, the value of the result does not change.

\[\require{enclose} \begin{array}{r} 213. \enclose{longdiv}{4350.} \\[-3pt] \end{array}\]

Notice that we added a zero so that we could move the decimal point in the dividend the required two places to match the move in the divisor. Next, place the decimal point in the quotient directly above its new position.

\[\require{enclose} \begin{array}{r} . \\[-3pt] 213. \enclose{longdiv}{4350.} \\[-3pt] \end{array}\]

Now the problem can be executed in the same way that integer division is performed, with one modification. You can keep adding zeroes after the decimal point in the dividend to perform the division to as many decimal places as are needed. In this case, since the original divisor had two decimal places, we will perform the division to three places and then round to two places.

\[\require{enclose} \begin{array}{r} 20.422 \\[-3pt] 213 \enclose{longdiv}{4350.000} \\[-3pt] \underline{{426}\phantom{0.000}} \\[-3pt] 90.0\phantom{42} \\[-3pt] \underline{00.0}\phantom{42} \\[-3pt] 90.0\phantom{42} \\[-3pt] \;\;\underline{85.2} \phantom{00} \\[-3pt] 4.80\phantom{8} \\[-3pt] \underline{4.26}\phantom{0} \\[-3pt] \phantom{4.} 540 \\[-3pt] \underline{426} \\[-3pt] \end{array}\]

The answer is \(20.42\).

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