Let's walk through each step of solving the problem.
1. Identify the costs for each city Walter planned to visit:
- Marrakech: MAD 965
- Fes: MAD 478
- Kenitra: MAD 778
- Oujda: MAD 683
2. Calculate the total cost of the original itinerary:
[tex]\( 965 + 478 + 778 + 683 = 2,904 \)[/tex]
The original itinerary costs MAD 2,904, which exceeds Walter’s budget of MAD 2,500.
3. Determine the different itinerary changes to reduce the cost:
- Remove Marrakech:
[tex]\[ 2,904 - 965 = 1,939 \][/tex]
- Remove Fes:
[tex]\[ 2,904 - 478 = 2,426 \][/tex]
- Remove Kenitra:
[tex]\[ 2,904 - 778 = 2,126 \][/tex]
- Remove Oujda:
[tex]\[ 2,904 - 683 = 2,221 \][/tex]
- Remove Marrakech and Fes:
[tex]\[ 2,904 - 965 - 478 = 1,461 \][/tex]
- Remove Marrakech and Kenitra:
[tex]\[ 2,904 - 965 - 778 = 1,161 \][/tex]
- Remove Marrakech and Oujda:
[tex]\[ 2,904 - 965 - 683 = 1,256 \][/tex]
- Remove Fes and Kenitra:
[tex]\[ 2,904 - 478 - 778 = 1,648 \][/tex]
- Remove Fes and Oujda:
[tex]\[ 2,904 - 478 - 683 = 1,743 \][/tex]
- Remove Kenitra and Oujda:
[tex]\[ 2,904 - 778 - 683 = 1,443 \][/tex]
4. Identify which of these changes keep the total cost within the budget of MAD 2,500:
- Remove Marrakech: MAD 1,939
- Remove Fes: MAD 2,426
- Remove Kenitra: MAD 2,126
- Remove Oujda: MAD 2,221
- Remove Marrakech and Fes: MAD 1,461
- Remove Marrakech and Kenitra: MAD 1,161
- Remove Marrakech and Oujda: MAD 1,256
- Remove Fes and Kenitra: MAD 1,648
- Remove Fes and Oujda: MAD 1,743
- Remove Kenitra and Oujda: MAD 1,443
Given these calculations, the possible changes that allow Walter to stay within his budget of MAD 2,500 are:
- Remove Marrakech: MAD 1,939
- Remove Fes: MAD 2,426
- Remove Kenitra: MAD 2,126
- Remove Oujda: MAD 2,221
- Remove Marrakech and Fes: MAD 1,461
- Remove Marrakech and Kenitra: MAD 1,161
- Remove Marrakech and Oujda: MAD 1,256
- Remove Fes and Kenitra: MAD 1,648
- Remove Fes and Oujda: MAD 1,743
- Remove Kenitra and Oujda: MAD 1,443
Thus, each of the listed changes individually would allow Walter to stay within his budget.
