Alex 發問於 科學及數學數學 · 5 年 前

HYC_Mock_1112_P2 (cont'd)

David joined a 10-km cycling competition. David rode the bicycle with the speed xkm/h from the starting point to the checking point and he went the speed (x-5)km/h from the checking point to the ending point. It is known that the checking point is in the middle between the starting point and the ending point. Finally, David spent 30 minutes to finish the whole competition. Find x

A. 23

B. (25 + 5sqt17)/2

C. 87/20

D. (32 + sqt409)/12

Given that a>b>0. Which of the following must be true?

I. log(b)a > 1

II. log(a)b > 0 (I dunno this)

III. log(b)ab = log(b)a + 1

() is the base

A bag contains 15 cards, each of the numbered 2 to 16. If 2 cards are drawn without replacement. find the probability that the sum of the numbers on the cards are even

A. 3/15

B. 7/15

C. 8/15

D. 11/15

更新:

Why the equation of question 1 is = 5??? 5 is ???

2 個解答

評分
  • 5 年 前
    最佳解答

    1.) Time required for the first half is 5/x hr, for the second half is 5/(x-5) hr.As total time taken is 1/2 hr, so,5/x+5/(x-5)=1/2==> 10(x-5)+10x=x(x-5)==> x²-25x+50=0==> x=(25+5√17)/2 or (25-5√17)/2 (rej, as x > 5)Ans:(B)

    2.) log(b) a=log a / log bIf a > 1 > b, then log a > 0 > log b, and, log a / log b < 0. ie. (I) is not true.log(a) b=log b / log aSame as (I), if a > 1 > b, then log a > 0 > log b, and, log b / log a < 0.Therefore, (II) is also not true.log(b) ab=log(b) a+log(b) b=log(b) a+1Therefore, only (III) is true.

    3.) P(the sum of the numbers on the cards are even)=P(two cards are even)+P(two cards are odd)=8C2/15C2+7C2/15C2=4/15+3/15=7/15 ⋯⋯ (B)

  • 5 年 前

    The time consumed before reaching the checkpoint is 5/x hours

    The time consumed from the checkpoint to the end of race is (1/2 - 5/x) hours

    Hence,

    (x - 5)(1/2 - 5/x) = 5

    x/2 - 5 - 5/2 + 25/x = 5

    x/2 -25/2 = -25/x

    x^2 -25x = -50 (as x not equals to 0)

    (x - 25/2)^2 = -50 + 625/4 = 425/4

    x = (425/4)^(1/2) + 25/2 or -(425/4)^(1/2) + 25/2 (rejected as x > 0)

    = (425/4)^(1/2) + 25/2

    = 5(17)^(1/2)/2 + 25/2

    = [25 + 5(17)^(1/2)]/2

    Consider when a > b > 1, for any constant c > 0,

    log(c)a > log(c)1 = 0 and

    log(c)b < log(c)1 = 0

    i.e. log(c)a > 0 and log(c)b < 0

    Now

    log(a)b = log(c)b / log(c)a

    log(b)a = log(c)a / log(c)b

    which are negative when b < 1.

    On the other hand,

    log(b)ab

    = log(b)a + log(b)b

    = log(b)a + 1

    even + even = even

    odd + even = odd

    odd + odd = even

    There are 8 even-number cards and 7 odd-number cards

    i.e. P(sum are even)

    = P(2 evens) + P(2 odds)

    = (8/15)(7/14) + (7/15)(6/14)

    = 7/15

    2015-01-11 17:59:06 補充:

    Opps.

    There's a mistake.

    For the logarithm problem, consider a > 1 > b.

    2015-01-15 00:41:34 補充:

    5(km) = 10/2 is half distance of the whole race.

    資料來源: Myself, Myself
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