CLEP College Mathematics common mistakes
CLEP College Mathematics · Understand the traps and how to avoid them.
Solving a polynomial or rational equation, then reporting every root produced by the algebra without checking whether it makes a denominator zero or makes an even root's radicand negative.
Students are trained to 'solve for x' and treat the derived values as automatic answers. Multiplying both sides by an expression containing the variable, or squaring both sides, can introduce extraneous solutions, and the habit of stopping at the last algebraic step carries over from simpler linear equations where no check was ever needed.
What to do instead
Adopt a fixed closing routine for every equation with a variable in a denominator or under an even radical: substitute each candidate back into the original equation, or first record the excluded values from the denominators and discard any candidate that matches. Practise on equations such as 1/(x-2) + 3 = x/(x-2) and sqrt(x+6) = x, where one candidate survives and one does not, so the check becomes automatic rather than an afterthought.
Treating combinations and permutations as interchangeable, or forgetting to divide by factorials for repeated letters when counting distinguishable arrangements.
Both formulas are quotients of factorials and look similar, so candidates pattern-match on keywords like 'choose' or 'arrange' rather than asking whether order matters and whether items repeat. Word-arrangement problems with duplicate letters are especially prone to this because the raw factorial feels like the natural answer.
What to do instead
Before computing, answer two questions in writing: does swapping two selected items create a different outcome, and are any items identical? If order is irrelevant, divide by r!; if items repeat, divide by the factorial of each repeat count. Drill contrasting pairs, for example the number of 4-person committees from 9 people (126) versus the number of ways to seat 4 of those 9 in a row (3024), and the arrangements of the letters of MISSISSIPPI (34650, not 11!).
Reading a histogram bars' heights as the raw data values, or reporting the mean of grouped data by averaging the class midpoints without weighting by frequency.
Histograms and frequency tables present intervals, not individual observations, and candidates often visually estimate a 'typical' value or average the midpoints equally. The visual pull of a tall bar near the middle reinforces the wrong answer, and the unweighted midpoint average can coincidentally sit close to the true mean, hiding the error.
What to do instead
Write out the frequency-weighted computation explicitly: multiply each midpoint by its frequency, sum, then divide by the total frequency. Check with a quick sanity test: if one class holds most of the data, the mean must be pulled toward that class's midpoint. Also distinguish mean, median, and mode questions before computing, since a skewed histogram makes these three answers visibly different and the exam often tests exactly that contrast.
Computing simple interest when the problem describes compounding, or applying the compound formula with the annual rate and an annual period count when interest compounds monthly or quarterly.
The words 'interest rate of 6% per year' trigger the familiar I = Prt formula, and the compounding frequency is buried in a phrase like 'compounded monthly'. Candidates also substitute the annual rate directly into A = P(1+r)^t without converting r to r/n and t to nt, which produces an answer that looks plausible but is too small.
What to do instead
Whenever compounding is mentioned, immediately rewrite the given rate and time into per-period units: r/n for the rate and nt for the number of periods, then apply A = P(1 + r/n)^(nt). Build a comparison table for one principal, say 2000 at 6% for 3 years, computed as simple interest, annually, quarterly, and monthly, so the direction and size of the difference become intuitive and a wrong-frequency answer is easy to spot.
Using the wrong formula for the area of a triangle or parallelogram, or mixing perimeter and area when a problem asks for fencing versus covering.
Formulas for triangles, trapezoids, and circles blur together under time pressure, and the factor of one-half is easily dropped. Perimeter and area both involve the side lengths, so a problem about 'border' or 'trim' versus 'covering' or 'paint' is misread as a cue to multiply lengths rather than add them.
What to do instead
Before calculating, label what the question wants in units: a linear unit (feet, metres) signals perimeter or circumference, a squared unit signals area, a cubed unit signals volume. Write the formula from a personal reference sheet each time, including the one-half for triangles and the average-of-parallel-sides for trapezoids. Verify with a rough estimate, such as a triangle inside a rectangle having half its area.
Negating a quantified statement incorrectly, writing 'not all A are B' as 'no A are B', or misreading 'only if' and 'unless' as plain 'if'.
Everyday speech treats 'not all' and 'none' loosely, and conditional phrasing in English does not map cleanly onto the arrow. Candidates translate the words one at a time rather than identifying the logical structure, so the converse and the negation of a conditional get confused with the original.
What to do instead
Translate each sentence into symbols before judging it: 'All A are B' becomes A implies B, whose negation is 'some A is not B', not 'no A is B'. For 'P only if Q' and 'P unless Q', both yield P implies Q; test by asking what the sentence rules out. Practise by writing the negation and the converse of ten short statements and checking each against a concrete counterexample.
Adding or subtracting fractions and radicals by operating on numerators and denominators separately, for example writing 1/2 + 1/3 = 2/5 or sqrt(9) + sqrt(16) = sqrt(25).
The rules for multiplication of fractions and for products under a radical are learned more firmly than the addition rules, and the symmetry of the notation invites applying the same operation to top and bottom. The error often survives because the wrong answer looks tidy.
What to do instead
Rehearse the addition rules aloud with a numeric check: 1/2 + 1/3 must exceed 1/2, so 2/5 is impossible; sqrt(9) + sqrt(16) is 3 + 4 = 7, while sqrt(25) = 5. For radicals, only identical radicands combine, so simplify first (sqrt(50) = 5 sqrt(2)) and then add like terms. Estimate the decimal value of every answer before selecting it.
Determining the domain or range of a function from its algebraic form alone, missing that a square root restricts the input or that a denominator excludes a value, or reading the range off the wrong axis.
Domain and range are often taught as vocabulary rather than as consequences of the operations present. Candidates scan for x-values that 'look fine' and forget that the output of an even root cannot be negative, and on a graph they confuse horizontal position with output value.
What to do instead
Run a fixed checklist on every function: denominators cannot be zero, even roots need non-negative radicands, and logarithms need positive arguments; solve the resulting inequalities. For range, ask what outputs the operation can produce, for example sqrt(x-3) has range y >= 0. On graphs, state 'domain reads left to right along x, range reads bottom to top along y' before answering, and confirm with a test point.
Computing the probability of 'at least one' by adding the individual probabilities instead of using the complement, giving answers greater than 1 or double-counting overlaps.
The phrase 'at least one' sounds additive, and candidates add P(A) + P(B) without subtracting the intersection. The complement method requires a mental flip that is easy to skip under time pressure, especially when the events are described in words rather than symbols.
What to do instead
When you see 'at least one', immediately write P(at least one) = 1 - P(none) and compute the none case as a product for independent trials. Verify the result lies between 0 and 1 and is larger than any single event's probability. Compare both methods on a two-dice or two-coin example to see the double-counting appear, then use the complement as the default.
Interpreting a z-score or a probability from a normal distribution without converting to standard units, or treating a percentile as a raw score.
The normal curve is often memorised as a picture with percentages attached, so candidates answer from the shape rather than from the standardisation formula. Percentile language is also misread as a score value, when it is a position in the distribution.
What to do instead
Always compute z = (x - mean)/standard deviation before consulting any table or rule, and state in words what the z-score means: how many standard deviations above or below the mean. For percentile questions, translate 'the 90th percentile' into 'the value with 90% of the area to its left' and find it by working backward from the table. Check that a percentile answer is expressed in the same units as the original data.