BGCSE Mathematics 0563 common mistakes
BGCSE Mathematics 0563 · Understand the traps and how to avoid them.
When ordering 0.4, 0.15 and 0.105, compares the digits after the decimal point as if they were whole numbers, so writes 0.105 < 0.15 < 0.4 as 0.4 < 0.15 < 0.105 or claims 0.15 is the largest.
The learner is applying whole-number place-value rules to decimal fractions and reads 15 as bigger than 4; the shorter decimal also looks smaller because it has fewer digits.
What to do instead
Give every decimal the same number of decimal places by annexing zeros: 0.400, 0.150, 0.105. Compare column by column from the tenths place, then locate each on a number line between 0 and 1. Practise with pairs that break the whole-number intuition (0.4 and 0.15, 0.09 and 0.1).
A jacket is reduced by 20% in a sale and now costs P640. The learner finds the original price by adding 20% of P640, obtaining P768.
The percentage is applied to the only number visible. The learner does not recognise that the 20% was calculated on the original price, not on the sale price, so the operation must be reversed rather than repeated.
What to do instead
Set the problem up as original × 0.8 = 640, then divide by 0.8. Insist on a checking line: 20% of the answer must give back P128 and P768 − P128 must equal P640. Teach the general rule that reverse percentage means dividing by the multiplier, never adding the percentage back on.
Expands 5 − 2(x − 3) as 5 − 2x − 6 and simplifies to −2x − 1, instead of 5 − 2x + 6 = 11 − 2x.
The minus sign is attached only to the 2 and is not carried into the bracket, so the second term keeps the sign it had inside the bracket. Learners often treat the sign in front of a bracket as punctuation rather than as part of the multiplier.
What to do instead
Rewrite the expression as 5 + (−2)(x − 3) and draw an arrow from −2 to each term inside the bracket before writing anything. Then substitute x = 1 into both the original and the simplified form to confirm they match. Repeat with brackets preceded by a negative sign until the arrow routine is automatic.
Finds the gradient of the line through (2, 5) and (6, 11) as (6 − 2)/(11 − 5) = 2/3, using change in x over change in y or reversing the order of subtraction.
The formula is memorised as a string of letters without meaning, so the learner cannot tell which difference belongs on top. Points are also subtracted in inconsistent orders, which flips the sign.
What to do instead
Label the points (x1, y1) and (x2, y2) on the diagram first, then write gradient = change in y ÷ change in x with the numbers substituted. Check against the sketch: a line rising from left to right must have a positive gradient, and a steep line must give a gradient greater than 1.
For a circle of diameter 10 cm, calculates the area as π × 10² = 314 cm², or the circumference as 2π × 10.
The only number given is 10, and the formula is recalled as πr² without checking whether the question supplies a radius or a diameter. The learner substitutes the visible number into the formula.
What to do instead
Make 'd = 10, so r = 5' the compulsory first line of working, before any substitution. Keep a formula card where each formula is written with the meaning of its symbols, and circle in the question whether radius or diameter is stated. Check the answer against an estimate: the area should be a little over 3 × 25.
Draws a bearing of 120° as an angle of 120° measured anticlockwise from the positive x-axis, or from a south line, producing a direction of about 060° or 300°.
Bearings are confused with standard angles measured from the x-axis in coordinate geometry. The learner does not draw a north line because the diagram in the question may not show one.
What to do instead
Before measuring, draw a north arrow at the point and mark the angle turning clockwise from it. State the three-figure convention (000° to 360°) and test the answer by asking whether the direction lies in the correct quadrant. Practise reverse bearings by adding or subtracting 180° and adjusting into the 000–360 range.
For A(2, 7) and B(5, 3), calculates the vector AB as (3, 4) or (7, −4) instead of (3, −4); or writes 2(3, −1) as (6, −1).
Coordinates are subtracted in the order A minus B, and the y-components are subtracted the wrong way round. For scalar multiplication, only the first component is multiplied because the bracket is read as a label rather than as two separate numbers.
What to do instead
Use the routine 'destination minus start' and write it out as (5 − 2, 3 − 7) before simplifying. Sketch the arrow on a grid to check the direction. For scalars, draw arrows from the multiplier to each component inside the bracket and multiply both.
With 40 values in a cumulative frequency table, finds the median by reading across at 40 on the cumulative frequency axis rather than at 20; or divides the total of midpoint × frequency by the number of classes instead of the total frequency.
The cumulative frequency axis is treated as the data axis, and the total frequency is mistaken for the position of the median. In grouped data, the number of rows looks like the number of items.
What to do instead
Write n = 40, so the median is the 20th value, and draw the horizontal line at 20 before reading down to the curve. For the mean, display the column headings and show that the divisor is the sum of the frequencies, then check that the mean lies inside the modal class.
A bag holds 5 red and 3 blue counters. For two counters taken without replacement, calculates P(red then red) as (5/8) × (5/8).
The two draws are treated as independent because multiplication of probabilities has just been learned. The learner does not update the contents of the bag after the first counter is removed.
What to do instead
Draw the bag twice, writing the counts before each draw: 8 counters, then 7. Write the second fraction as 4/7 because one red has gone, changing both numerator and denominator. Compare directly with the with-replacement case so the difference in the second fraction is visible.
Describes a rotation as 'a 90° turn' without giving the centre or direction, or reflects the point (3, 2) in the line y = x and writes (−3, 2).
Only part of a transformation is perceived, usually the angle or the mirror line. For reflection in y = x, the learner applies a sign change instead of swapping coordinates because the two rules are mixed up.
What to do instead
Require a four-part description for every rotation: rotation, angle, direction (clockwise or anticlockwise), and centre. For reflections, test the rule on a point lying on the mirror line, which must map to itself: (3, 2) in y = x must give (2, 3), and only swapping does that.