ＧＷＤ18-13和18-6

本人老文科,求解题思路:

Q13:

Each of 20 parents chose one of five days from Monday through Friday to attend parent-teacher conferences.  If more parents chose Monday than Tuesday, did at least one of the parents choose Friday?

(1)     None of the five days was chosen by more than 5 parents.

(2)     More parents chose Monday than Wednesday.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.

Answer:

Q6:

The function f is defined for each positive three-digit integer n by f(n) = 2^x 3^y 5^z, where x, y and z are the hundreds, tens, and units digits of n, respectively.  If m and v are three-digit positive integers such that f(m) = 9 f(v), then m - v =

A.       8

B.       9

C.     18

D.     20

E.      80

Answer:

最好给出答案，这样解答时可以避免错误，呵呵。

Q13答案B。已知条件中，知道选M的人比T的人多，而条件1说每天不超过5人，所以假设M是5人，那么T最多4人，剩下的周三，周四都是5人，那么还是有一人必须选周五，所以是充分的。

条件2只是说周一比周三人多，但没法限制每天的人数上限，所以就没法判断，是不充分的。

Q6答案是D

设M=xyz,V=x'y'z'，由已知条件f(m)=9f(v)可以得到y-y'=2，x=x',z=z'，所以m-v时，百位数，个位数都是0，只有十位数为2，所以就是20。

還是看不懂!可以再講仔細點嗎?我無法看出""得到y-y'=2，x=x',z=z'，所以m-v时，百位数，个位数都是0，只有十位数为2，所以就是20""多謝!

f(m)=2^x3^y5^z,f(v)=2^x'3^y'5^z'

f(m)=9f(v)=>2^x3^y5^z=3^2*2^x'*3^y'*5^z'=2^x'3^(y'+2)5^z'

所以y=y'+2

因为y和y'均代表该数的十位，所以相当于差为20

caesar講的很有理...

但Q13答案不是應該選A嗎?...