Foundation June 2021 Paper 2 Q18
18 \(\mathscr{E}\) = {letters of the alphabet}
\(B\) = {b, r, a, z, i, l}
\(I\) = {i, r, e, l, a, n, d}
\(K\) = {k, e, n, y, a}
Cody writes down the statement \(B \cap K = \varnothing\)
Cody’s statement is wrong.
| Scheme | Marks |
|---|---|
| (i) other seen orders of letters: a, b, d, e, i, l, n, r, z b, r, I, a, e, z, l, n, d Answer: b, r, a, z, i, l, e, n, d | B1 |
| (ii) Answer: b, z | B1 |
| (2) |
Notes
B1: no repeats, letters can be in any order.
Condone capital letters rather than lower case letters.
(no need for commas)
B1: No repeats, letters can be in any order.
Condone capital letters.
(no need for commas)
| Scheme | Marks |
|---|---|
| correct explanation that shows they know the meaning of intersection and empty set | B1 |
| (1) | |
| (3 marks) |
Notes
B1: eg letter ‘a’ is in both sets
\(B \cap K = \{a\}\)
Set \(B\) and set \(K\) have an element (or letter) in common.
There is a letter that is in set \(B\) and in set \(K\)
There is an intersection so it isn’t the null set
There is a letter in common
(do not allow ‘letters’ or ‘elements’ (plural) in common)
(If students mention the letter that is in common, it must be the correct one (ie a))