<p>I need some help.I don't understand one or two questions and when ambiguity rises,i will post it here...</p>
<p>I hope some of you help me.I will truly appreciate it. :)</p>
<p>"At a certain hospital, 89 children were born in the month of June. If more children were born on the fifteenth of June than on any other day in June,what is the least number of children that could have been born on the fifteenth of June?"
i wrote 45, but answer was 4 =))</p>
<p>"when x is expressed as a decimal,the hundredths digit is 8.What is the greatest possible value of 1/x ?
1) 0.00125...
2)0.010101...
3)0.9899...
4)11.111...
5)12.5</p>
<p>I don't understand how to solve it regardless i know an answer which is 12.5.</p>
<p>Please tell me how to solve it...Thanks a lot!</p>
<p>The first one is an example of the “pigeonhole principle.” We’re trying to “evenly disperse” the students throughout the days of the month. We can roughly put 3 students on each day (one day will only have 2 students, but that doesn’t matter for this question). Now if we do this, then the fifteenth will have the SAME number of students as most of the other days. In order to ensure it has MORE than every other day we need to put one more student on the 15th. So the answer is 4.</p>
<p>For example, we can have 4 students on the 15th, 3 students on the 1st through 14th, and 3 students on the 16th through 29th, and 1 student on the 30th.</p>
<p>For the second question, because x takes place as the denomenator, so the smaller the value of x is, the larger value of 1/x becomes. x must be 0.08 so that 1/x is maximum. 1/0.08 =12.5</p>
<p>But i still don’t understand the first question:
If 4 children were born in the 15 of june, there is a possibility, that there somewhere in the beginning of the June, 30 children were born.
Information presented here doesn’t exclude this variation.It doesn’t say that every day at least 1 or 2 children were born.
That’s why i chose 45,because even if only 2 days were the days of fertility “second day”(other then 15th) would have 44 children…</p>
<ol>
<li>June has 30 days, 89/30 = 2.966… < 3. The answer’s not 3, because the distribution would be 2,3,3,3,3… which doesn’t satisfy the condition of the problem. However, we can change the 3 that corresponds to June 15 to a 4 and another 3 to a 2, and it works.</li>
</ol>
<p>45 children could have been born on the 15th of June, but that is not the LEAST possible answer. The question is asking for the least number that works - 45 is just one of many other numbers that also work.</p>
<p>Kristiuna, you’re misreading the question. You aren’t being asked for the smallest number of babies that will *ensure *that there were more babies born on the 15th than on any other day (which would be half the babies, plus 1). You’re being asked for the smallest number that *makes it possible *that there were more babies born on the 15th than on any other day of the month.</p>
<p>Imagine this :
Because there are 89 born children in a month of 30 days, we have the least children born each day is 2. So the rest is 29 children.
Distribute 29 into 30 days, 2 is needed to add to least greatest number of children born in June.
So 2+2 = 4.
Does it make sense?
Hope this help</p>
<p>1) substitute 1 for p and get 90n+23=4523
reduce and get 90n=4500<br>
(Lookie here! see how perfectly the numbers work out? 4500 is a multiple of 90)
reduce and get n = 50
n+p = 50 + 1 = 51</p>
<p>2) The nature of the question implies that every choice of n will lead to the same answer (see remark below for an exception). So choose n = 1. Then we have x^(1/2)=5, so that x=25. Then 1/x^n = 1/x = 1/25 or .04</p>
<p>Exception: We cannot choose k = 0 because the left side would then become 1 making the given equation 1 = 5 which is false. </p>
<p>Complete algebraic solution:: ((x^(1/2))^n) = x^(n/2), so that the equation is x^(n/2) = 5. We square both sides to get x^n = 25. So 1/x^n = 1/25.</p>
<p>On the number line above,the tick marks are equally spaced.Which of the lettered points represents y?
1)A
2)B
3)C
4)D
5)E</p>
<p>2)A large solid cube is assembled by gluing together identical unpainted small cubic blocks.All six faces of the large cube are then painted red.If exactly 27 of the small cubic blocks that make up the large cube have no red paint on them,how many small cubic blocks make up the large cube?
This grid-in question…</p>
<p>I was thinking so much to solve this…But really could not.Thanks in advance!</p>
<p>1) For the sake of simplicity let the length of an increment be 1. Then y=2 since x+y is two marks further along the line than x. Notice that x+y<0 since the average of x and y is greater than their some. So a,b,and c must be negative since they are <x+y, which leaves us with D and E. D can not be right since the mark (x+y)/2 is negative and we can not go from a negative to 2 (at least three units) in the space of two tick marks. This leaves us with E.</p>
<p>2) Let x be the number of blocks along a side of the cube. The cubes left unpainted are those in the inner cube with side length x-2. This cube has volume (x-2)^3=27 so x-2=3 and x=5. If the large cube is 5x5x5, the total number of blocks is 125.</p>
<ol>
<li>The fact that (x+y)/2 is greater than x+y should immediately suggest, “maybe x+y is negative!” It turns out that point D corresponds to 0 on the number line because that’s the only point where (x+y)/2 is exactly halfway between x+y and 0.</li>
</ol>
<p>To get from “x” to “x+y” we move right two tick marks. Hence, to get from “0” to “y” we can also move right two tick marks, to land on point E. The answer is E.</p>
<ol>
<li>Cubes have volume.</li>
</ol>
<p>We can assume that the small cubes have side length 1. In an n<em>n</em>n cube, the cubes that don’t have paint on them will be the ones contained in the cube of length n-2. (NOT n-1 – visualize it and you’ll see why). Therefore, (n-2)^3 cubes don’t have paint on them, so (n-2)^3 = 27 → n = 5. Therefore there are 5^3 = 125 unit cubes.</p>
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