求解：2数学－GWD最新13套第二套－Q18,22

zsx

    

Q18.
On a certain transatlantic crossing, 20% of a ship's passengers held
round-trip tickets and also took their cars abroad the ship. If 60% of
the passengers w/ round-trip tickets did NOT take their cars abroad the
ship, what % of the ship's passengers held round-trip tickets?

A. 33 1/3%

B. 40%

C. 50%

D. 60%

E. 66 2/3%

        

我觉得这一题好像条件不足以解答。是这样吗？这题我很不解。答案给出的是C。

        

Q22. A construction company was paid a total of $500,000 for a
construction project. The company's only costs for the project were for
labor and materials. Was the company's profit for the project greater
than $150,000?

(1) The company's total cost was 3 times its cost for materials

(2) The company's profit was greater than its cost for labor

        

My considerations: profit = 500,000 - (labor + material); according to
statement (1), total cost = labor + material = 3 x materials' cost
--> 500,000 - 3 x materials > 150,000 ---> materials' cost
< 350,000/3 ---> labor's cost < 700,000/3, which is greater
than the profit and conflicts w/ statement (2) ... very confused ...
这一题，我很不懂。答案给出的是C。不解 ...

        

麻烦大家帮忙解决一下！感激ing ...

zl35

第1题，设总的PASSENGERS有X人，那么根据题意可得，X20%既有往返票又带车；再设PASSENGERS中有R人有往返票，题目已知R60%没有带车，即R40%带了车。由此可得X20%=R40%，（这时等式两边表示既有往返票又带了车上船的人）即R/X=1/2，所以选C

第2题，很显然条件1和条件2单独都不能求解，合并一起考虑

设公司的总成本为X，MATERIAL成本为M，LABOR成本为L，利润为R

由条件1可以知道，X=3M；又因为M+L=X，那么L=2M

由条件2可以知道，P要大于L

反证法，假设M=100000，则L=200000，P就=500000-100000-200000=200000=L，不符合条件2，所以M不=100000；

再假设M=110000，则L=220000，P就=500000-110000-220000=170000，小于L，不符合条件2，所以M也不=110000；

再假设M=99999，则L=199998，P就=500000-99999-199998=200003，大于L，更加大于题目问的150000，所以可以得出M一定小于100000，即P一定会大于150000，所以选C

zsx

非常之感谢！！！现在我终于弄懂了！恩人 ...