Jake is planning a trip to China. He has made a list of cities he would like to visit, as well as the approximate amount of money he plans to spend in each as a result of travel, lodging, shopping, and so on. All costs are listed in renminbi (¥).

| City | Cost (¥) |
| ----------- | -------- |
| Tianjin | 557 |
| Nanjing | 681 |
| Zhengzhou | 595 |
| Beijing | 728 |
| Wuhan | 449 |
| Chengdu | 534 |

Unfortunately, Jake only has ¥2,920 available. What is the cheapest city that Jake can remove from his travel plans and still stay under budget?
a. Tianjin
b. Wuhan
c. Nanjing

Jake needs to remove cities from his travel plans to stay under his budget of ¥2920. After calculating the costs, removing Nanjing is the only option that allows him to stay within budget, despite it being the most expensive city of the options considered. The answer is Nanjing.
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Answered by Anonymous | 2025-07-04

Calculate the total cost of all cities: Y 3544.
Determine that the total cost exceeds Jake's budget of Y 2920.
Calculate the new total cost after removing each of the cities: Tianjin ( Y 2987), Wuhan ( Y 3095), and Nanjing ( Y 2863).
Identify that removing Nanjing is the only option that keeps Jake within his budget. Therefore, the answer is Nanjing.
The final answer is c. Nanying N anjin g ​

Explanation

Calculate Total Cost First, let's calculate the total cost of visiting all the cities on Jake's list. We have: 557 + 681 + 595 + 728 + 449 + 534 = 3544 The total cost is Y 3544.

Compare Total Cost to Budget Jake's budget is Y 2920. Since the total cost Y 3544 exceeds his budget, he needs to remove at least one city from his itinerary. The question asks us to find the cheapest city he can remove to stay within his budget. We are given three options: Tianjin, Wuhan, and Nanjing. Let's examine each option.

Removing Tianjin If Jake removes Tianjin, the new total cost would be: 3544 − 557 = 2987 Since Y 2987 is still greater than Y 2920, removing Tianjin alone is not sufficient.

Removing Wuhan If Jake removes Wuhan, the new total cost would be: 3544 − 449 = 3095 Since Y 3095 is greater than Y 2920, removing Wuhan alone is not sufficient.

Removing Nanjing If Jake removes Nanjing, the new total cost would be: 3544 − 681 = 2863 Since Y 2863 is less than Y 2920, removing Nanjing is sufficient to stay within budget.

Determine the Cheapest City to Remove Now, we need to determine which of the three cities (Tianjin, Wuhan, Nanjing) is the cheapest. Their costs are:



Tianjin: Y 557
Wuhan: Y 449
Nanjing: Y 681 Wuhan is the cheapest at Y 449. However, removing Wuhan alone is not sufficient to stay within budget. Nanjing is the most expensive of the three, but removing it is sufficient to stay within budget. The question asks for the cheapest city that Jake can remove and still stay under budget. Since removing Nanjing allows Jake to stay within budget, we need to check if removing the cheaper cities (Tianjin or Wuhan) would also allow Jake to stay within budget. We already know that removing Tianjin or Wuhan alone is not sufficient. Therefore, the correct answer is Nanjing, as it is the cheapest of the three cities that, when removed, allows Jake to stay within his budget.


Final Answer Therefore, the cheapest city that Jake can remove from his travel plans and still stay under budget is Nanjing.

Examples
Imagine you're planning a road trip with friends, and each city you want to visit has associated costs like gas, food, and activities. You have a limited budget, so you need to decide which cities to cut from your itinerary to stay within your financial constraints. This problem is similar to figuring out which cities to skip to minimize the overall cost while still having a great trip. By calculating the costs associated with each city and comparing them to your budget, you can make informed decisions about which destinations to remove, ensuring you don't overspend and can still enjoy your vacation.

Answered by GinnyAnswer | 2025-07-04