<p>1-The integers 1 through 5 are written on each of five cards. The
cards are shuffled and one card is drawn at random. That card
is then replaced, the cards are shuffled again and another card
is drawn at random. This procedure is repeated one more time
(for a total of three times). What is the probability that the sum
of the numbers on the three cards drawn was between 13 and
15, inclusive?</p>
<p>Answer:2/25 or .08</p>
<p>2-Last month Joe the painter painted many rooms. He used 3
coats of paint on one third of the rooms he painted. On two
fifths of the remaining rooms he used 2 coats of paint, and he
only used 1 coat of paint on the remaining 24 rooms. What was
the total number of coats of paint Joe painted last month?</p>
<p>Answer:116</p>
<p>3-A group of 286 parents is to be divided into committees with 3
or more parents on each committee. If each committee must
have the same number of parents and every parent must be on
a committee what is the maximum number of committees
possible?</p>
<ol>
<li><p>The only ways to obtain 13 thru 15 are {5,5,5}, {5,5,4}, {5,5,3}, and {5,4,4} (up to permutations). Therefore there are 1+3+3+3 = 10 ways to obtain a sum of 13-15. There are 5^3 = 125 total ways to select 3 cards, 10/125 = 2/25 = 0.08.</p></li>
<li><p>1 - 1/3 - 2/5 = 4/15, 24 rooms is 4/15 of the total #, so Joe painted 90 rooms total. Therefore, 30 rooms had three coats of paint, 36 had two coats, 24 had one coat. 30(3) + 36(2) + 24 = 188. Uhh. You might want to check that to see if there are any errors.</p></li>
<li><p>Look for factors of 286. 286 = 2<em>11</em>13, the smallest factor of 286 greater than 3 is 11, so there are 11 parents on each committee, and 2*13 = 26 committees.</p></li>
</ol>
<p>Your error in the second one is that it says “2/5 of the REMAINING rooms.” So your first computataion should be 1 - 1/3 - (2/5)(2/3) = 1 - 1/3 - 4/15 = 2/5. So 24 is 2/5 of the total giving a total of 60 rooms (not 90).</p>
<p>Number 2 is problem that is quite tricky to do algebraically. It may take quite a bit of thought to see why we’re subtracting off (2/5)(2/3) = 4/15. </p>
<p>I would actually recommend doing this one with the most basic of strategies: taking a guess.</p>
<p>Now in this case, an informed guess will get you to the answer quicker than a random guess. And guessing the number of rooms will be easier than guessing the number of coats of paint.</p>
<p>A moment’s thought should convince you that the number of rooms must be a multiple of 15 larger than 30 (at the very least you should see that it has to be larger than 24).</p>
<p>So, for example if we guess that there are 45 rooms, then (1/3)(45) = 15 have 3 coats of paint, and (2/5)(30) = 12 have 2 coats of paint. Adding up we get 15 + 12 + 24 = 51 rooms which is not equal to 45. So our guess was wrong.</p>
<p>Guessing 60 rooms next, we get (1/3)(60) = 20 have 3 coats and (2/5)(40) = 16 have 2 coats. Adding we get 20 + 16 + 24 = 60. It matches. So we get 20(3) + 16(2) + 24(1) = 116 coats of paint.</p>
<p>Dr.Steve
Really thank you for your efforts on this beautiful book !
But Please Can you explain no.2 and 3 as I didn’t understand them !
Thank You</p>
<p>Once you get 60 rooms, the rest is easy. There are 1/3<em>60 3 coat-paintings for a total of 60/3</em>3=60. Similarly, 4/15<em>60</em>2=32. 60+32+24=116</p>
<p>Ok. Once you know there are 60 rooms, go back to the beginning of the question and take it line by line.</p>
<p>So first it says “he used 3 coats of paint on one third of the rooms he painted.” Since there are 60 rooms, and 1/3 of 60 is 20, he used 3 coats of paint on 20 of the rooms. That’s 3*20 = 60 coats of paint.</p>
<p>Next it says “on two fifths of the remaining rooms he used 2 coats of paint.” Now, how many rooms are remaining? Well we already took care of 20, and there are 60 total, so there are 40 remaining. 2/5 of 40 is 16. So he used 2 coats of paint on 16 of the rooms. That’s 2*16 = 32 coats of paint.</p>
<p>Finally, we have 24 rooms with 1 coat of paint.</p>
<p>Now just add up all the coats: 60 + 32 + 24 = 116.</p>
<p>I do like guesses. The only time, I do not like the guessing games on SAT problems is when they are timesinks and are superfluous. </p>
<p>In this case, I do NOT think that guessing is really the best option. And there is really no need to build a long formula neither. All that is needed is working through the problem … logically. </p>
<p>Here’s what my paper could have shown – in case I wanted to document the steps. </p>
<p>
</p>
<p>In reality all my paper had was this, and it took about 10 seconds. What is needed is to start with the right place, and that is a CONCRETE number. In this case, all you need is 24 and understand that it represents 3/5 of the rooms with 1 and 2 coats! </p>
<p>24 x 5/3 = 40 </p>
<p>24
2 x16
3 x 20 … >> 24 32 30 > 116</p>
<p>And ten seconds is about what it would take to come up with a legitimate guess. As in about every case, guesses ought to be the last option in your bag of tricks. Reasoning and thinking, the … first ones!
I</p>
<p>I like what Xiggi did here - it is quite clever. </p>
<p>Let me try to add a bit more explanation. He worked from the end of the question to the beginning (this works quite often by the way), adding in one piece of information at a time.</p>
<p>So 24 of the rooms have 1 coat of paint. </p>
<p>Next, 2/5 of the rooms with 1 or 2 coats have 2 coats. So 3/5 have 1 coat. In other words, 24 is 3/5 of these rooms (24 = 3/5 x, so x = 24(5/3) = 40). So there are a total of 40 rooms with 1 or 2 coats of paint.</p>
<p>Working backwards one more step, we have 1/3 of the rooms with 3 coats, so that 40 is 2/3 of the total. So there are a total of 40(3/2) = 60 rooms.</p>
<p>We have seen 3 methods for finding the number of rooms:</p>
<p>(1) By guessing (the safest method in my opinion - also my favorite overall because this method can be applied quite easily to lots of problems)</p>
<p>(2) Logically (this is Xiggi’s solution above which is a nice, clever solution - certainly the quickest method for this particular problem)</p>
<p>(3) Algebraically (not recommended - very often leads to errors on a question of this difficulty level).</p>
