<ol>
<li>If n is postive and 50/n = n percent of 50, what is the value of n?
a.1
b.5
c.10
d.50
e.100</li>
</ol>
<p>30.If each of 12 artists submitted either one or two paintings for an exhibition, and a total of 17 paintings were submitted, how many artists submitted only one painting?</p>
<p>I had to guess and check on both these so im looking for the best way to do these problems. and an algebraic way too</p>
<p>30)
Perhaps:
{ x + y = 12
{ x + 2y =17</p>
<p>Multiply the first equation by -1, and then add the two equations (x’s cancel out, leaving you with y=5). Plug your new y=5 back into the first equation, and you get x=7. </p>
<p>So 7 artists bought 1 painting in, and 5 bought 2 paintings in.</p>
<p>mmm anyone care to verify? and how bout 27?</p>
<p>You could use algebra or just plug your way to victory.
For 27 plug the answer choices for n and see what works or…
50/n=(n/100)*50
50/n=n/2
Cross multiply
n^2=100
n=10</p>
<p>30
I thought it would be easier just to try and think about this problem conceptually…
So 17 total paints. 12 artists. Each give 1 so 17-12=5. 5 would be the number of extra paintings. problem says you can either do 1 or 2 paintings. (not 3 or 4…) so you know that 5 dudes did extra paintings. 12-5=7 which gives us the number of people too lazy to do 2 paintings.</p>
<p>B. </p>
<p>50/n = 10% of 50
50/n = 50(n/100)
50/n = n/2
100 = n^2
n = 10 (n > 0)</p>
<p>I think. If you can’t do it algebraically, plug it in ;).</p>
<p>Was beat to it, but glad that we did it the same way xD.</p>
<ol>
<li><p>50/n = 50n/100
50(n^2)=5000
n^2 = 100
((n^2)-100) = 0
Difference of squares: (n-10)(n+10) = 0
n = 10 or -10
Reject “-10” because n>0
n = 10</p></li>
<li><p>We know that all 12 submitted at least 1. That leaves us with 5 more paintings to get to 17. Thus, 5 submitted two paintings, and the rest (7) entered one.
The answer is 7.</p></li>
</ol>