SAT I Math problem...help would be appreciated

Test Preparation
1

<p>I’m having trouble figuring out what to do with this problem. It’s a grid-in from the SAT I from May 2000, Section 4, #25 (I’m getting this from the 10 Real SATs book, which doesn’t appear to have answer explanations anywhere).</p>

<li>“A flock of geese on a pond were being observed continuously. At 1:00pm, 1/5 of the geese flew away. At 2:00pm, 1/8 of the geese that remained flew away. At 3:00pm, 3 times as many geese as had flown away at 1:00pm flew away, leaving 28 geese on the pond. At no other times did any geese arrive or fly away. How many geese were in the original flock?”</li>
</ol>

<p>The answer is 280, but I’m not sure how to arrive at that answer besides random guess and check, which is obviously unfeasible for the real test.</p>

<p>Any help would be greatly appreciated, thanks.</p>

2

<p>this has been answered many times, try serching the forums,this would not be a typical sat rephrase question so dont worry about it anyway</p>

3

<p>Denote the total amount of geese as x.</p>

<p>At 1:00pm, 1/5 of the geese flew away. </p>

<p>so (1/5)x geese flew away</p>

<p>At 2:00pm, 1/8 of the geese that remained flew away. </p>

<p>that means (1/8) of the remaining, or (x - x/5), so (1/8)(x - x/5)</p>

<p>At 3:00pm, 3 times as many geese as had flown away at 1:00pm flew away</p>

<p>That means three times as much as in step one so 3 * (1/5)x or (3/5)x</p>

<p>...leaving 28 geese on the pond. </p>

<p>So your final equation, you must get a remainder of 28 after subtracting all the flown-away geese from x:</p>

<p>x - (1/5)x - (1/8)(x - x/5) - (3/5)x = 28</p>

<p>Solving for x will get you the total amount of geese</p>

<p>x - x/5 - x/8 + x/40 - 3x/5 = 28
40x/40 - 8x/40 - 5x/40 + x/40 - 24x/40 = 28
4x/40 = 28
4x=1120
x = 280</p>

4

<ol>
<li>"A flock of geese on a pond were being observed continuously. At 1:00pm, 1/5 of the geese flew away. At 2:00pm, 1/8 of the geese that remained flew away. At 3:00pm, 3 times as many geese as had flown away at 1:00pm flew away, leaving 28 geese on the pond. At no other times did any geese arrive or fly away. How many geese were in the original flock?"</li>
</ol>

<p>SOLUTION:</p>

<p>Since we're looking for the number of geese that were in the original flock, we'll call that number x.</p>

<p>At 1pm, 1/5 fly away, so at that time there are </p>

<p>x - (1/5)x ... or (4/5)x geese left on the pond</p>

<p>Then, at 2pm, 1/8 of those remaining 4/5 fly away, so there are</p>

<p>(7/8)(4/5)x ... or (7/10)x geese left on the pond</p>

<p>At 3pm, three times the amount that flew away at 1pm, which was (1/5)x, flew away, so there are</p>

<p>(7/10)x - 3(1/5)x ... or 1/10x geese left on the pond</p>

<p>(1/10)x represents the number of geese left in relation to the original number of geese on the pond (x). The problem tells us at there are 28 geese left on the pond, so we can set this number equal to 28 and then solve for x:</p>

<p>(1/10)x = 28 <-- multiply each side by 10</p>

<p>x = 280</p>

5

<p>I actually feel though that this question would be much easier by just guessing and checking until your remainder gets closer and closer to 28.</p>

6

<p>No way.... I just did it in a few seconds.</p>

<p>The only trick is the 1/8 of the remaining geese part.
28=x-[(1/5x)+(4/40x)+(3/5x)]</p>

7

<p>zach447 and legendofmax, thanks a bunch, i understand it now, although i can see that this problem would be quite time-consuming whatever way you attempt it...</p>

<p>justperfect, i didn't know that this forum had a search function (never looked up i guess...lol). i'll do it next time.</p>