<ol>
<li><p>What is the greatest possible area of a triangle with one side of length 7 and another side of length 10?
a 17 b 34 c 35 d 70 e 140</p></li>
<li><p>if p and n are integers such taht p>n>n and p^2-n^2 = 12, which of the following can be the value of p-n?
I.1
II.2
III.4</p></li>
<li><p>If j,k,n are consectutive integers such that 0<j<k<n and the units digit of the product jn is 9, what is the units digit of k?
a)0
b)1
c)2
d)3
e)4</p></li>
</ol>
<p>10/ area <= 7<em>10/2 = 35 –> c
15/ p^2-n^2= (p-n)</em>(p+n)
- p-n =1 –> p+n=12–> not integers –> I wrong
-p-n=2–> p+n=6–> p=4, n=2–> II
- p-n=4–>p+n=3–> not integers–>III wrong
20/ only the units digit of production of 1<em>9, 3</em>3, 7*7 is 9.
n-j=2 –> n =…1, j=…9 —> k= …0 ----> A</p>
<p>Where are these questions from? They seem far to hard to be SAT Questions.</p>
<p>I think i saw the first one from a practice sat i took either from online or the bluebook, but its definitely from collegeboard</p>
<p>what do you mean by not integers?</p>
<p>what do you mean by p>n>n?(question 15)</p>
<p>in this context, not integers = p-n<p+n</p>
<p>I hate math</p>
<p>bumppp…</p>
<p>oh i think i get it… it just means that the 3 integers that follow are greater than 0. not that they are 0123… i gett it.</p>
<p>15 is still confusing…</p>
<p>p^2-n^2=12; Find p-n</p>
<p>(p-n)(p+n)=12; (p-n)=1,2,3,4 [remember p>n]</p>
<p>So: I, II, III</p>
<p>not that hard… took about 30 seconds.</p>
<p>“what do you mean by p>n>n?(question 15)”</p>
<p>It’s a typo. It should read … p>n>0</p>
<p>greedisgood… the answer is only II… which is why i was confused.</p>
<p>The answer is II.
(p+n)(p-n)=12
if p-n=1, p+n=12. P and n will not be integers, because no numbers within the range of 0 to 12 can substitute p and n without changing them into fractions.
If p-n=4, then p+n=3, but n would be a negative integer, thus violating the p>n>0.
P-n could be 2, and p+n could be 6, because this pair is the only one that may satisfy all the conditions.
I hope my English will not hinder you from understanding this post. It’s hard to explain the concept. :)</p>
<p>ohhhh i sorta get it now… ur looking at the p-n and p+n as a whole number… i see. sorta…wow taht’s hard to recognize.</p>
<p>first one: C 35
triangular area=1/2 base<em>height
let the base of the triangle be 10 and let one side be 7. Now, only when the side 7 is perpendicular to the base 10 is the area greatest, because that’s when side 7 is the height. Consequently, when side 7 and 10 are not perpendicular, the height will be less than 7. Draw this on a piece of paper and you will understand. (</em> understanding what is the triangular height is very important)</p>
<p>second:
sorry, I don’t understand what this means, could you explain? –> p^2-n^2</p>
<p>third: Ok, I’m working on that too ‘’/</p>
<p>ok third: I’m going to give it a try, I think I’m right, but it seems a bit complicated.</p>
<p>since… product’s unit digit is 9
so… integers’ unit digits could be 1,3,7,9.
but since… 0<j<n
so… j=1, n=9</p>
<p>now we use the arithmetic progression formula: X2-X1=2d (X2,X1 are consecutive integers, d is the equal difference)</p>
<p>so… 2d could be: 8, 18, 28, 38, 48, …
so… d thus could be: 4, 9, 14, 19, 24…</p>
<p>thus k’s unit digit could be: 1+4= 5
1+9= 10 (0)
1+14=15 (5)
1+19=20 (0)
…</p>
<p>we can now see it alternates between 0 and 5.
since there is no “5”, the answer is “0”.</p>
<p>some math master please help correct my answer, thank you!</p>
<ol>
<li>If j,k,n are consectutive integers such that 0<j<k<n and the units digit of the product jn is 9, what is the units digit of k</li>
</ol>
<p>(J+2)(J)=NJ=J^2 + 2J; (By looking at calculator table): J=9</p>
<p>K=J+1=9+1=10</p>
<p>So answer is [A]</p>