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03.11.2019 •
Mathematics
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
d.
beijing
Solved
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Ответ:
Option C. Nanjing is the correct option.
Step-by-step explanation:
Jake is planning a trip to China. If we add expenses for all the cities then we get 557+681+595+728+449+534 = 3544, means Jake should have minimum 3544 yuan for the trip.
Unfortunately Jake has 2920 yuan available and he has to exclude one cheapest city from his trip.
Difference is 3544-2920 = 624 Therefore if he exclude Nanjing which has the expenditure of 681 yuan, his trip will remain under budget.
Option C. Nanjing is the answer.
Ответ: