Final answer:
Hamza can make 20 combinations with odd digits, while Grace can make 12 combinations with even digits. Hamza can thus make 8 more combinations than Grace, so the value of x is 8.
Step-by-step explanation:
The maths problem involves Hamza and Grace making 2-digit numbers with specific conditions. Hamza uses only odd digits while Grace uses only even digits, and the digit 0 is not used. Additionally, digits cannot be repeated in each 2-digit number they create. To find the value of x, which represents how many more codes Hamza can make than Grace, let's go through the possible digit combinations for each person.
For the tens place, Hamza has 5 possibilities (1, 3, 5, 7, 9), and for the ones place, he has 4 remaining options after choosing a digit for the tens place. Thus, Hamza can make 5 × 4 = 20 combinations.
Grace has fewer options. For the tens place, there are 4 possibilities (2, 4, 6, 8), and for the ones place, there are 3 remaining options. Therefore, Grace can make 4 × 3 = 12 combinations.
To find out how many more combinations Hamza can make than Grace, we subtract Grace's combinations from Hamza's:
Hamza's combinations - Grace's combinations = 20 - 12 = 8
Therefore, Hamza can make 8 more combinations than Grace. So, the value of x is 8.