<p>Number 14 on the last math section:</p>
<p>The average of three numbers is 15. Add 20 to one number and the new average is 19. What is the sum of the original 3 numbers?</p>
<p>60
65
70
75
80</p>
<p>Reading the forum here, I’m getting that 75 was the correct answer.
How do you do it?</p>
<p>I keep seeing 1008 pop up. Anyone remember the exact question?</p>
<p>It never said 3 numbers. You had to figure out how many numbers there were.</p>
<p>1008 - First integer greater than 1000 that is divisible by 6,8,9.</p>
<p>did anyone get the question about the calculus students of years 1998 1999 2000? where 2000 was 20% of 1999 or something like that? 
was that the experimental section for math? section 7?</p>
<p>@ keekee
I got that question, and it wasn’t on experimental. Sorry :(</p>
<p>It wasnt an experimental. The answer was 10, or A.</p>
<p>I just posted it above you lawls</p>
<p>@ jollygood</p>
<p>You had to figure out how many numbers. I just used the answer choices to figure out what it was:</p>
<p>Test Choice A: 60. If the original sum was 60 then there had to be four numbers since 60/4 is 15. Add 20 to 60 and you get 80. 80/4 is definitely not 19.
You keep doing that and you figure out 75 is the correct sum.</p>
<p>I’m pretty sure there’s a mathematical way to do it, but I chose to do it this way haha (:</p>
<p>LOLOLOL GENSIS.</p>
<p>sorry x)</p>
<p>what was the answer the the question like whats a+b=
something about set A set B not having certain x and y numbers and n is neutral numbers?</p>
<p>The quadrilateral question wasn’t that difficult.</p>
<p>
<a href=“http://i55.■■■■■■■.com/2a0sgn.png[/img]”>http://i55.■■■■■■■.com/2a0sgn.png
</a>
AB was 12, CD was 10</p>
<p>Essentially you had two triangles split across the y-axis and you needed to find the area of both of them, and then add them together.</p>
<p>Area of a triangle = .5(b/h)</p>
<p>The problem with this question was that it gave you the length of AB and CD, with no intermediaries. You basically just have to realize that there are going to be two equations which give you the area of each triangle, and consequentially, the area of the entire quadrilateral.</p>
<p>Area of left triangle = .5(the entire base, because remember they’re split across the y axis / part of the height)
Area of right triangle = .5(the entire base / the remaining part of the height)</p>
<p>Basically I realized that however you split up 12 between the two equations, the answer would always be 60. i.e. plug in 6 in the first equation and 6 in the second equation, or 4 in the first, 8 in the second, and vice versa. It all works out to 60.</p>
<p>How do you work that problem with the right triangular prism? It was like equilateral triangles for the base with length 6, the height of the prism was 8. Then you had to find the perimeter of a triangle made up of 2 points on the bottom of the base and one point on the back of the prism, and the point on the back was 3 down from the top. Nobody has mentioned it yet… Was it experimental?</p>
<p>@keekee2tiffany: Don’t remember that but I do recall a question about sets n and p. Forgot the rest of the question though.</p>
<p>@chasel.
I did it that way too just to double check myself, but it worked physically too! 
Since it didn’t say ‘figure not drawn to scale’ I made myself a mini ruler and marked ‘10’ and ‘12’ (and added increments of 2), measured the neccessary sides, and calculated it…
It worked. o_______o</p>
<p>@yudawg.
definitely experimental cause I didn’t get it and I didn’t see it on this forum. yay for you!</p>
<p>You do it like this lol.</p>
<p>The average of numbers is 15. Add 20 to one number and the new average is 19. What is the sum of the original numbers?</p>
<p>So…</p>
<p>Let us say x is number of numbers.</p>
<p>15x + 20 (sum of old numbers) = 19x (sum of new ones)</p>
<p>x=5</p>
<p>Therefore 5x15=75</p>
<p>Also:</p>
<p>Actually easy way to solve the quadrilateral question is to draw a huge box around it. try it out!</p>
<p>can anyone tell me the question where the answer was 109?</p>
<p>What? ***, that doesn’t help at all.</p>
<p>I am I remembering the question correctly?</p>
<p>The average of three numbers is 15. Add 20 to one number and the average becomes 19. What is the sum of the original three numbers.</p>
<p>The sum of the original three must equal 45. Why wasn’t 45 listed.</p>
<p>How can the average of the original three increase from 15 to 19 when 20 is added. The average is supposed to increase by 20/3, from 15, to 21.666</p>
<p>The question was like:</p>
<p>The sequence of terms is a, 2a + 1, 2(2a +1) where you have multiply the preceding term by 2 and add 1. What is the first term of a sequence where exactly 4 numbers are less than 1000?</p>
<p>Definitely not worded like that though…</p>
<p>@gensis: OH THAT’S THE WAY TO DO IT. thanks!
@jolly: it didn’t work because it never stated there were three numbers. you had to figure out how many numbers there were</p>
<p>does anybody remember that question? the 15, three numbers average one</p>
<p>Did anyone have the question where it asked how many balls of radius 3 cm you could fit in a 12x12x12 box? I put 8.</p>
<p>It never said there were three numbers.</p>
<p>eta: pchs10 beat me to the punch :)</p>
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