Higher June 2018 Paper 3 Q17
17 Here are two methods to make a 4-digit code.
Codes can have repeated digits.
Method A
For the first two digits use an odd number between 30 and 100
For the last two digits use a multiple of 11
Method B
Use four digits in the order \(\quad\) even \(\quad\) odd \(\quad\) even \(\quad\) odd
Do not use the digit zero
Which method gives the greater number of possible codes?
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(7 \times 5\) (\(\times\) 9) or \((100 - 30) \div 2\) (\(\times\) 9) or 35 (\(\times\) 9) or \(99 \div 11\) or 9 or \(4 \times 5 \times 4 \times 5\) | M1 | First two digits of Method A Last two digits of Method A Complete for Method B |
| 315 or 400 | A1 | |
| 315 and 400 with Method B identified | A1 | Method B can be implied by choosing 400 |
Additional guidance
| 315 and 400 and B with no working | M1A1A1 |
| 315 and 400 with 400 circled | M1A1A1 |
| Beware \(40 \times 10 = 400\) (for Method A) is incorrect working |