<p>oh sorry i wont say the answer next time lol</p>
<p>Or:</p>
<p>Cathy answered 1/6 of the questions incorrectly.
The number of her wrong answers can't be lower than 9 (only 27-18=9 "bad" questions on the full test).
For 9 incorrect answers being 1/6 of the total number of questions, that number is 9x6=54.
More incorrect answers -> more questions on the test (for 10 it'd be 10x6=60).
Hence (E) 54.</p>
<p>thanks so much guys, especially dchow you're so good at explaining things lol. sorry for the long response couldn't find this thread for a while >.<</p>
<p>Oki here's a Q </p>
<pre><code> ZWYX
</code></pre>
<p>The above sequence may be changed in either two ways. Either two adjacent letters may be interchanged or the entire sequence may be reversed. What is the LEAST number of changes needed to put the letters in alphabetical order from left to right?</p>
<p>Thnx in advance</p>
<p>It tells us to put the letters in alphabetical order that is WXYZ. My approach is that, since the sequence can be reversed, try to put them in the order ZYXW.
Therefore, W and Y are interchanged (1 change) and W and X are interchanged (2 changes) and at last the whole sequence is reversed (3 changes).
My advice is that just try to be very simple on this problem. Don't get this too hard.</p>
<p>Thnx tsengunn
:)</p>
<p>Q. If a card is pulled at random from an ordinary 52-card deck of playing cards, what is the probability of pulling a face card (Jack, Queen, or King)?
(A) 6/25
(B) 3/13
(C) 2/15
(D) 1/10
(E) 3/52</p>
<ol>
<li> Susanne has eleven different medals from her two years of competitive swimming. Unfortunately, the mounting frame she wishes to place them in has room for only two. How many different combinations of medals can Susanne place in her frame?
(A) 21
(B) 33
(C) 55
(D) 66
(E) 110 </li>
</ol>
<p>Many thanks</p>
<ol>
<li>You have 12 face cards: Jack Queen King = 3 * Diamond Heart Spade Club = 4 3*4 = 12
In total there are 52 cards so the probability of pulling a face card is 12/52 = 3/13</li>
<li>Let's name the two rooms of the mounting frame. Let's say we have room A and room B.
Starting from room A we have 11 different choices . After placing a medal in room A, there are 10 different choices for room B, so in total there are 10*11=110 combinations we can do.</li>
</ol>
<h1>7 is ambiguous: are spots for the medals in the frame fixed, or Susanne can place the medals in the frame in any place she wants?</h1>
<p>In the latter is true, there is just one more step to be added to baphomet's solution.</p>
<p>Medals A and B can go in order AB and BA - 2 different ways. If the order is not important (that's combinations), the number of arrangements should be divided by 2: 11x10/2 = 55.</p>
<p>Incidentally, if there were room for three medals, the number of combinations (not arrangements a.k.a. permutations!) would be 11x10x9/(3x2) = 165.</p>