# Moore Math

<p>more questions:
15. A cartaon contains b boxes of paper plates and each box contains n plates. If carton costs c dollars, what is the cost per paper plate, in dollars, when the plates are bought by the carton?
a.cbn
b.c/bn
c.bn/c
d.cn/b
e.n/cb</p>

<p>2.In the figure above, lines k,l and m are parallel. if y = 125, what is the value of x +z?
answer = 110, i have no idea cause the picture looks like its obtuse angles... sorry i don't have a scanner... this is from test 5 i believe.</p>

<ol>
<li><p>There are k students in a class, only one of whom is a junior. If 5 seniors are added to the class, how many students in the class will not be juniors?
a.k+1
b.k+2
c.k+3
d.k+4
e.k+5</p></li>
<li><p>The quadratic function graphed above has equation f(x)=a-x^2. Which of the following must be true. correct answer is a>0. The parabola is downwards, and it is raised from the origin.</p></li>
<li><p>Its a figure of a triangle with hypotenuse 20 and sides a and b. Which of the following inequalities is true about the lengths a and b of the sides of the triangle above?
a.0<=(a+B)^2<20
b.20<=(a+b)^2<40
c.40<=(a+b)^2<100
d.100<=(a+b)^2<400
e.400<=(a+b)^2</p></li>
</ol>

<ol>
<li>the answer is C I guess. </li>
all you have to do in this question is to translate to math… so k students… only one of those students is a junior… so k-1 is all of the other students excluding the junior and then you add the other new 5 seniors… so k-1+5 is k+4</li>
</ol>

<p>15 is b… i thought it was c as well but i was wrong…</p>

<p>Yes, 15 should be b., c/bn.
Let’s use simple substitutions to figure out this problem.
Let’s say, a plate costs \$1 each, there are 20 plates to a box, and 5 boxes in a carton. Therefore, there are 100 plates in a carton, which a carton would cost \$100. In order to find the cost of each plate, in this given situation, we would use \$100 (cost of a carton) divide by however many plates there are in a carton, or 20 (n plates in a box)*5 (b boxes in a carton), which would yield the result of \$1 per plate. Thus, by using known numbers, we can derive the correct answer b.) c/bn
Theoretically, we know that c/bn = c/b/n, whereas c/b=cost per box and c/b then divide by n=cost per plate in each box.
Hope it helps! :D</p>

<p>Plugging in does wonders for question 6.</p>

<p>And I’m not quite sure I understand #8, is the triangle a right triangle? And by “(a+B)^2” do you mean a^2 + b^2?</p>

<p>yeah i guess. i think it has to do with triangle inequality theorem.</p>

<p>and angelfish… according to your example of plugging in, if you do bn/c, its also 100/100…</p>

<p>Haha true, sorry didn’t notice that! I just realized I made up bad numbers. You can use the exact substituting way to solve this problem except just change the original price for each plate to \$2 each. So, basically, each plate is \$2, 20 plates in a box and 5 boxes in a carton. Thus, a carton would cost \$200. Using c/bn, (200/(20<em>5)), you’ll get \$2. But using bn/c ((20</em>5)/200), you’ll get \$.5.
Again, it makes sense theoretically. As what I said previously, “Theoretically, we know that c/bn = c/b/n, whereas c/b=cost per box and c/b then divide by n=cost per plate in each box.” On the other hand, theoretically, bn/c means how many plates you can buy for a dollar by using the total number of plates divide by the total cost.
Thank you!</p>

<p>oh right… that makes sense. my answer was basically a ratio for how many plates you could get for a certain amount of money. therefore you ahve to flip it to get the price per…</p>

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<p>The carton has B boxes in it.
The box has N plates it it.
The carton cost C dollars.</p>

<p>C/B = cost of carton per plate
B/N = cost of plate per box</p>

<p>C/B/N

<p>Method #2 - Plug Numbers In:
Let C = \$100
Let B = 10
Let N = 5</p>

<p>5 * 10 = 50 (50 plates)
\$100 / 50 = \$2</p>

<p>100/(10*5) = \$2</p>

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<p>Without the picture we cannot solve the problem.</p>

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<p>k + 5 (seniors added) = total class, but there is one junior in the class, so subtract 1.

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<p>Once again we need the graph…</p>

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<p>Once again we need the picture…</p>

<p>Avid student, did you not read my comments? And for 8 its just a right triangle with hypotenuse 20 and sides a,b… i think you can figure that one out… i appreciate the effort but those have been addressed. If anyone has access to the online course and take a look at 2, i would appreciate it.</p>

<p>and Congrats on Dartmouth!!!</p>

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<p>The problem seems relatively straightforward, but you failed to clarify until this post that the problem is a right triangle.</p>

<p>Actually, though it is a right triangle, that actually does not mattter (and I think is designed to be a distraction). It’s just that a + b > 20 bc of the triangle inequality, then square both sides. (a + b) ^2 > 400</p>

<p>yeah^^ pckeller rescuing me from a potential blunder… but yeah triangle inequality.</p>