5道9月下JJ数学难题 -确认答案！

gustavechi

我明白了，一下子昏了头，谢谢各位！

bwainecho

1. A.
2. E.
Think this case,  A = (1, 2 x 13, 3), B = (1, 3 x 16, 4).  Move 1 from B to A.

3. A
From 1, X + 10 = 37*k => X = 37 *(k-1) + 27.  
From 2. 10 | (63 + 37)      10 | (163+37) , but 63 and 163 have different remainder when divided by 37.
4. E
5. E.
(1) Think 100 and 999.   (2) 10000 < m^2 < 100000 => lg(10000) < 2*lg(m) < lg(100000) => 2 < lg(m) < 2.5 => 6 < 3*lg(m) < 7.5 => 10^6 < m^3 < 10^7.5.   M^3 can be 6 or 7 digits.

petron12

bwainecho , according to your example. answer is C, corrent?

gustavechi

茅塞顿开，谢谢！

bwainecho

which one are you talking about?

petron12

    bwainecho , according to your example. answer is C, corrent?

seven7bob

4题　答案是Ｂ把

只要Ｘ是大于零的　Ｘ＋１就应该是别人的平方把　毕竟也没说Ｘ＋１是整数

bwainecho

2.
                        一个集合m 有15个数，一个集合k有19 个数，从k中拿走一个数加到m ，问新的m平均数是不是大于k 平均数

1）以前m的中位数小于k 的中位数;   2）以前m的平均数小于k 的中位数

1, Think m = (1 x 15) k = (2x18, 10000000)

In this case, Both 1) and 2) are correct,

move 2 from K to M   => avg(M + {2}) < avg (K - {2})

move 10000000 from K to M => avg (M + {1000000}) > avg (K - {10000000})

So both 1) and 2) are NOT sufficient => E

bwainecho

Another similar question, but different answer
13. 一个集合m 有15个数，一个集合k有19 个数，从k中拿走一个数加到m ，问新的m平均数是不是大于k

平均数
  1）n 的中位数保持不变
  2）m,n的平均数都增加了


By moving k' from K to M, from 2) we have that   [sum(M) + k']/16 > sum(M)/15 => sum(M)/15 < k'
                                                                                    [sum(K) - k']/18 > sum(K)/19  =>  k' < sum (K)/19;

[sum(M) + k']/16 < [sum(M)+sum(K)/19]/16 (replace k' with sum(K)/19 ) < [15*sum(K)/19 + sum(K)/19] /16 (replace sum(M) with 15*sum(K)/19) = sum(K)/19 < [sum(K)-k']/18

So 2 is sufficient  => choose B

小土狗

谢谢楼上的，这道题貌似蛮难的，我也推了很久才推出来。 

考试的时候那么紧张，时间也紧，如果遇到肯定乱猜E了。