A couple "difficult" mathematics questions.

<p>18: A 50-foot wire runs from the roof of a building to the top of a 10 -foot pole 14 feet across the street. How much taller would the pole have to be if the street were 16 feet wider and the wire remained the same length?</p>

<p>(A): 2 feet
(B): 8 feet
(C): 14 feet
(D): 16 feet
(E): 18 feet</p>

<p>I know that I have to use the Pythagorean Theorem, but am unsure as to where to label the sides. Also, would I have to use any of the trigonometry functions?</p>

<p>Another...</p>

<p>20: Graham walked to school at an average speed of 3mph and jogged back along the same route at 5mph. If his total traveling time was one hour, what was the total number of miles round trip?</p>

<p>(A): 3
(B): 3 1/8
(C): 3 3/4
(D): 4
(E): 5</p>

<p>Also, these are problems from TPR’s “Cracking The SAT” 2009 Edition.</p>

<p>Also, one more question:</p>

<p>If the average of x, y, and 80 is 60 more than the average of y, z, and 80, what is the value of x - z?</p>

<p>Would I set it up like this?</p>

<p>x + y + 80<br>
------------ = 6 +
3</p>

<h2>y + z +80</h2>

<pre><code> 3
</code></pre>

<p>bump. Anyone?</p>

<p>For the first one the answer is B. To solve, simply set up a pythagorean theorem using 14, 50, and solving for the unknown value x. The 50 is the hypotenuse and represents the wire coming across. The 14 is for the width of the street. After solving we find that x is equal to 48. Thus, we now know the house is 58 ft tall. Then we set up another equation, this time using 30 instead of 14. Using pythagorean theorem once again we get the x value as 40. This means the pole has to be 8 feet higher as 48-40 means the pole is 8 ft short.</p>

<p>I thought that 50 was the left leg…</p>

<p>Is B the right answer?</p>

<p>Second one is simple too: just do (2 * 3 * 5)/(3 + 5).</p>

<p>Let</p>

<p>distance = x
time for outgoing trip = t</p>

<p>For outgoing trip
x = 3t</p>

<p>For return trip,
x = 5 ( 1 - t)</p>

<p>As distance is the same for outgoing and return trips, </p>

<p>3t = 5 - 5t</p>

<p>8t = 5</p>

<p>t = 5/8 for outgoing trip</p>

<p>Use to calculate distance</p>

<p>x = 3 * (5/8) = 15/8</p>

<p>Roundtrip </p>

<p>2x = 15/4 So the answer is (C)</p>

<p>They’re not hard, but I definitely did something wrong (or seriously misread the question…). Thanks, guys. Can anyone help me out with the last one?</p>

<p>For the last one set up your equation as:</p>

<p>(x+y+80)/3 = (z+y+80)/3 + 60</p>

<p>Multiplying by 3 results in</p>

<p>x+y+80 = z+y+80 + 180</p>

<p>Subtract y, 80, and z from both sides</p>

<p>x-z = 180</p>

<p>^ Thank you, sir.</p>