Students evaluate the harder problem of the total value of the coins rather than the number of coins, i.e., (2 × $1) + (9 × 50 cents) + (8 × 10 cents) = $7.30

Gives the value of the three coins shown, i.e., 1 + 0.50 + 0.10 = 1.60

Diagnostic and formative information:

Common error

Likely reason

Next steps

a)
b)
c)
d)

5 (or 6)
14 (or 13)
18 (or 17)
11 (or 12)

Uses "inclusive counting"
For example, for 16 - 7, starts counting back at 16 rather that 15
i.e., e.g., 16, 15, 14, 13, 12, 11, 10 (instead of 15, 14, 13, 12, 11, 10, 9)

Get student to do the problems on materials. Starting with 16, get them to say "Take 1 away and I have 15".

c)

$7.30
(or 1.60)

Students evaluate total value of the coins rather than the number of coins,

Get the students to re-read the problem.

d)

11

Uses subtraction but takes away the bigger digit from the smaller.

Expose students to similar subtractions where the strategy does not work (e.g., 13 - 4, which should be 9 rather than 1 or 11).
Get students to show their working on the number line.