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
HYC_Mock_1112_P2 (cont'd)
2015-01-11 11:51 pm
回答 (2)
2015-01-14 6:38 pm
✔ 最佳答案
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)
2015-01-12 1:56 am
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.
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.
參考: Myself, Myself
收錄日期: 2021-04-15 17:54:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150111000051KK00066