Anonymous
Anonymous asked in 社會與文化語言 · 1 decade ago

GMAT數學問題

For any positive integer n, the sum of the first n positive intergers equals n(n+1)/2. What is the sum of all the even intergers between 99 and 301?

正確答案是20200

請問該如何解?

下面這2題是資料充分性

An interger greater than 1 that is not prime is called composite. If the two-digit interger n is greater than 20, is n composite?

1.The tens digit of n is a factor of the units digit of n

2.The tens digit of n is 2

答案是A

Each of the 45 boxes on shelf J weights less than each of the 44 boxes on shelf K. What is the median weight of the 89 boxes on these shelves?

1.The heaviest box on shelf J weight 15 pounds.

2.The lightest box on shelf K weight 20 pound.

答案是A

這幾題怎麼想都想不通~拜託會的人幫忙解釋一下!!!!

Update:

不好意思~本人對數學真的很頓

請問下面這段是怎麼來的???

2×(150×151)/2-2×(49×50)/2

Update 2:

請問從哪裡得知j44 < j45和k1 < k2???

2 Answers

Rating
  • Peter
    Lv 6
    1 decade ago
    Favorite Answer

    樓上第一題答得太妙了.

    在下提供第二題解答, 請參考:

    An interger greater than 1 that is not prime is called composite. If the two-digit interger n is greater than 20, is n composite?

    1.The tens digit of n is a factor of the units digit of n

    2.The tens digit of n is 2

    設n=(10a+b), 且a>1(因為n>20), 依題目所示, 何者條件可以說明n為"合數"?

    1. The tens digit of n is a factor of the units digit of n

    >> a為b的因數, 代表b可以被整除, b/a仍為一正整數

    >> n=(10a+b) = a(10+b/a) = 整數a x 整數(10+b/a)

    >> 質數只能被1和自己整除, 但整數a和整數(10+b/a)

    皆大於1, 又不等於n, 所以為合數, 條件絕對成立.

    2.The tens digit of n is 2

    >> 2字頭的2位數中, 23,29即為質數, 所以條件非絕對成立.

    2009-03-27 09:55:53 補充:

    第三題:

    若將J和K內所有箱子依重量排序,

    可得集合 J[j1<= j2<= j3<= j4<= ...

    2009-03-27 10:18:46 補充:

    第三題:

    若將J和K內所有箱子依重量排序,

    可得集合 J[j1<= j2<= j44< j45] < K[k1 < k2 <= k43 <= k44]

    2009-03-27 10:20:44 補充:

    其中:

    1. "<=" 為 "小於, 等於"

    2. 套上題目給的2個條件, j45為J中最大數字,

    2009-03-27 10:23:30 補充:

    j44 < j45, 已排除j44=j45之可能性. 同理 k1 < k2, 已排除k1=k2之可能性

    2009-03-27 10:23:47 補充:

    3. j1為第1數, j2為第2數, j45為第45數, k1為第46數, k2為第47數, k44為第89數.

    中數median= 第(1+89)/2數=第45數=j45=J櫃中最重的箱子.

    2009-03-27 10:25:30 補充:

    由於版面限制, 第3題請看意見欄

    2009-03-29 04:21:48 補充:

    1. 因為只取100~300之間所有偶數之總和 >>100+102+……+300....(式1)

    2. 因為偶數, 可以將上述式子中提2出來 >>2×(50+51+……+150)...(式2)

    3. (式2)中的(50+51+...150)為50~150之總和,

    而(50~150之總和)=(1~150之總和)-(1~49之總和)

    4. 因此分別套上n=150及n=49至題目已知的公式: 1+2+3+...+n=n(n+1)/2,

    可得2×(50~150之總合)=2×(150×151)/2-2×(49×50)/2=20200

    2009-03-29 04:52:27 補充:

    請問從哪裡得知j44 < j45和k1 < k2???

    >>從題目給的條件:

    1."The heaviest box on shelf J" weight 15 pounds....

    J櫃中共有45個箱子, 依輕至重排序, 最重為j45. 而因為叫"45箱中最重", 必只會有一個, 不會有兩箱以上最重, 所以設定j44

    2009-03-29 04:53:05 補充:

    所以設定j44 < j45, 而不是j44 <= j45.

    舉一個反證:

    若J櫃有3箱最重, 則j43=j44=j45, "the heaviest box on shelf J"= j43 or j44 or j45...(式A)

    題目選項一"the heaviest box on shelf J" = median(式B)

    由(式A)和(式B), 可得median=j43 or j44 or j45>>這是不對的

    j1~j89的median只能為j45, 所以 "the heaviest box on shelf J"只能為j45, 所以j43, j44皆不等於45.

    2009-03-29 04:53:47 補充:

    同理可證:

    2.The lightest box on shelf K weight 20 pound >> k1 < k2

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  • 1 decade ago

    For any positive integer n, the sum of the first n positive intergers equals n(n+1)/2. What is the sum of all the even intergers between 99 and 301?

    Ans: 100+102+……+300=2×(50+51+……+150)=2×(150×151)/2-2×(49×50)/2=20200

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