MSCE Mathematics common mistakes
MSCE Mathematics · Understand the traps and how to avoid them.
When solving a linear equation such as 5(2x - 3) = 4x + 9, the candidate expands only the first term inside the bracket, writing 10x - 3 = 4x + 9, or drops the minus sign when the bracket is subtracted, e.g. 7 - (x - 2) becomes 7 - x - 2.
The distributive law is applied mechanically to the first term only, and the negative sign in front of a bracket is treated as belonging to the first term rather than to every term inside. Under time pressure the candidate's eye jumps from the sign to the first symbol after the bracket.
What to do instead
Before expanding, draw an arrow from the multiplier or minus sign to each term inside the bracket and write the intermediate line with every product shown, e.g. 7 - (x - 2) = 7 - x + 2. Then check by substituting a simple value such as x = 1 into the original and simplified forms and confirming they agree.
The candidate collects unlike terms, for example simplifying 3a + 4b to 7ab, or cancels terms across a plus sign, reducing (x + 6)/6 to x + 1 or (2x + 8)/2 to x + 8.
Visual similarity between letters is mistaken for algebraic compatibility, and cancellation is treated as deleting whatever appears on both sides of a fraction bar. This usually traces back to learning simplification as a set of symbol-moving rules rather than as operations on quantities.
What to do instead
Replace letters with numbers to test any simplification: 3(2) + 4(3) = 18 while 7(2)(3) = 42, so the two are not equal. For fractions, state the rule explicitly: only factors may be cancelled, never terms joined by + or -. Practise by factorising the numerator first, e.g. (2x + 8)/2 = 2(x + 4)/2 = x + 4.
On a negative index or fractional index, the candidate writes 2^-3 = -8 or 16^(1/2) = 8, or evaluates 27^(2/3) by multiplying 27 by 2/3. Related errors appear when a negative number is squared, writing -3^2 = 9 instead of -9.
The minus sign in an exponent is confused with a minus sign in front of a number, and a fractional exponent is read as multiplication by a fraction. The order of operations for unary minus versus exponentiation is also frequently reversed.
What to do instead
Learn the two definitions as sentences: a^-n means 1/a^n, and a^(m/n) means the nth root of a raised to the mth power. For -3^2, apply the rule that exponentiation binds tighter than the leading minus, so the square applies to 3 only; write (-3)^2 when the minus is inside. Do a short drill converting between forms until each conversion is automatic.
The candidate computes a percentage change by dividing the change by the new value, or answers 'a price rises 20% then falls 20%, so it returns to the original' without checking. Similarly, a 15% increase is applied by multiplying by 0.15 rather than 1.15.
Percentage is treated as an amount to add rather than as a multiplier, and the base of the percentage is not identified. Successive changes feel reversible because the two percentages match, so the candidate stops before testing with numbers.
What to do instead
Always write the multiplier first: an increase of p% multiplies by (1 + p/100), a decrease by (1 - p/100). For successive changes multiply the multipliers, e.g. 1.20 x 0.80 = 0.96, a 4% net fall. Test with a concrete starting value such as 100 whenever a question involves more than one percentage change.
When solving x^2 = 16 or a quadratic by taking square roots, the candidate gives only x = 4 and omits x = -4; when solving an inequality such as -2x > 6, the candidate divides by -2 and keeps the direction, writing x > -3.
Square roots are habitually associated with the positive answer only, since the square-root key on a calculator returns the principal root. The reversal rule for inequalities is memorised as a phrase without a reason, so it is forgotten under pressure.
What to do instead
Write 'x = +/- sqrt(16)' as a standard step whenever solving x^2 = k with k positive. For inequalities, justify the flip with a number line: -2x > 6 means x < -3 because x = -4 satisfies the original (-2 times -4 = 8 > 6). Test one value on each side of the boundary to confirm the direction.
In coordinate geometry, the candidate substitutes points into the gradient formula in inconsistent order, computing (y2 - y1)/(x1 - x2), or finds the perpendicular gradient by negating the original slope to -2 instead of taking the negative reciprocal, -1/2.
The formula is recalled as a shape rather than as 'change in y over change in x', so the order of subtraction is chosen arbitrarily for each coordinate. The perpendicular condition is remembered as 'flip the sign', losing the reciprocal half of the rule.
What to do instead
Label the two points (x1, y1) and (x2, y2) on the page before substituting, and keep the same point first in both numerator and denominator. State the perpendicular rule as m1 x m2 = -1, then derive the second gradient by division rather than by memory. Verify by multiplying the two gradients to confirm the product is -1.
The candidate uses the basic sine or cosine ratio in a triangle that is not right-angled, or applies Pythagoras to a non-right triangle. In right-triangle work, the opposite and adjacent sides are chosen relative to the wrong angle, or the candidate uses the hypotenuse as the adjacent.
SOHCAHTOA is practised so heavily on right triangles that it becomes the default response to any triangle problem. The labels opposite and adjacent are defined relative to a chosen angle, and candidates often fix them to the diagram rather than to the angle in the question.
What to do instead
Before choosing a method, mark the right angle on the diagram or note explicitly that none exists. If there is no right angle and you know a matching side-angle pair, use the sine rule; if you know two sides and the included angle or all three sides, use the cosine rule. When labelling, put a finger on the angle in question and name each side aloud before writing a ratio.
When finding an angle with the sine rule, the candidate reports the acute answer from the calculator and stops, missing the obtuse possibility. In radian work, the candidate evaluates sin(30) in radian mode, or gives a trig solution outside the requested interval, such as omitting the second solution of sin x = 0.5 in 0 to 360 degrees.
The inverse sine function on a calculator returns only values between -90 and 90 degrees, and the candidate does not connect this to the symmetry of the sine curve. Calculator mode is rarely checked because the display looks plausible either way.
What to do instead
After any inverse-sine result, immediately write the second candidate as 180 degrees minus the answer and test whether it fits the triangle or interval. Sketch the relevant curve over the required interval and mark all intersections before finalising. Check the calculator's degree/radian indicator at the start of every trigonometry question.
The candidate confuses mean, median and mode when a question asks which average best represents a data set, or computes the mean of grouped data by averaging the class midpoints without weighting by frequency. In probability, the candidate adds probabilities of overlapping events without subtracting the intersection.
The three averages are learned as parallel procedures rather than as measures with different sensitivities, so outliers and skew are not considered. The addition rule is recalled as P(A) + P(B) without the overlap correction, often because Venn diagrams are not drawn.
What to do instead
For each average, state what it is sensitive to: the mean uses every value and is pulled by outliers, the median depends on position, the mode is the most frequent value. For grouped data, write the frequency-times-midpoint column and divide the total by the total frequency. For combined events, draw a two-circle Venn diagram and apply P(A or B) = P(A) + P(B) - P(A and B) explicitly.
On mensuration, the candidate uses the diameter in place of the radius in area and volume formulas, or confuses the formulas for the circumference and area of a circle, or uses the slant height of a cone as its perpendicular height in the volume formula.
Diagrams frequently label the full width, and the candidate copies the given number into the formula without converting. Circle formulas are similar in appearance and are often memorised as a pair without the dimensional check that would separate a length from an area.
What to do instead
Write 'r = d/2' as a visible line before substituting any diameter. Check units dimensionally: circumference is a length, area is a length squared, so a formula giving the wrong power signals a mix-up. For cones, identify the perpendicular height and the slant height separately on the diagram and use the slant height only in the curved surface area, with the perpendicular height in the volume.