<PRE><B>Question 77427</B>
&lt;pre&gt;&lt;font size = 4&gt;&lt;b&gt;
(1) Two circles of radius 1 and 4 units are tangent
to each other externally. the length of the common 
tangent line between the points of tangency is 
closest to which of the following?
(a) 4.8 (b) 4.5 (c) 3.9 (d) 3.6 

Draw Circle A with radius 1 externally tangent to 
Circle B with radius 4.

Let CD be their common tangent, where CD is tangent 
to circle A at C and tangent to circle B at D.
  
Connect the two centers with segment AB

Draw radii AC and BD.

AC and BD are parallel because they are both
perpendicular to CD.

Draw AE parallel and equal to to CD where E is on BD.

Here is the figure, I can&#39;t put in the letters here but 
you can on your paper:

{{{graph( 500,300,0,10,-2,4,sqrt(1-(x-1)^2),(-4/3)((x-1)-5)*sqrt(5-(x-1))/sqrt(5-(x-1))*sqrt((x-1)-2.6)/sqrt((x-1)-2.6),(-4/3)(x-1)*sqrt(-(x-1))/sqrt(-(x-1))*(sqrt((x-1)+.6)/sqrt((x-1)+.6)),sqrt(16-((x-1)-5)^2),(3/4)*(x-1)*(sqrt((x-1))/sqrt((x-1)))(sqrt(3.2-(x-1))/sqrt(3.2-(x-1))),-sqrt(16-((x-1)-5)^2),(.75(x-1)+1.25)*(sqrt((x-1)+.6)/sqrt((x-1)+.6))*(sqrt(2.6-(x-1))/sqrt(2.6-(x-1)))   )}}}



ABE is a right triangle.

AB is its hypotenuse and equals 1+4 or 5

AC=DE=1, and BD = 4, so BE = 3

By the Pythagorean theorem
BE² + AE² = AB²
3² + AE² = 5²
9 + AE² = 25
    AE² = 16
     AE = 4

and CD = 4 since CD=AE and AE=4

Therefore the common tangent line is 4.
So the answer is (c) because 3.9 is 
nearest of the choices to 4.

------------------------------------- 

(2)If the diameter of a cylindrical jar is 
increased by 25% without altering the volume 
then its height must decrease by:
(a) 10% (b)25% (c)36% (d)50% 

Let the answer be p% 

Volume of a cylinder = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(radius)²(height)
Radius = Diameter/2
Volume of cylinder = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(diameter/2)²(height) = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(diameter)²height/4

Diameter before = d
Diameter after = d + .25d = 1.25d
Height before = h
height after = h - .01ph = h(1-.01p)
Volume before = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(d)²h/4
Volume after = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(1.25d)²[h(1-.01p)]/4

Volume after = Volume before

&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(1.25d)²[h(1-.01p)]/4 = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(d)²h/4

&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(1.25)²d²h(1-.01p)/4 = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;d²h/4

&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(1.5625)d²h(1-.01p)/4 = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;d²h/4

Multiply both sides by 4

&lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;(1.5625)d²h(1-.01p) = &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;d²h

Divide both sides by &lt;font face = &quot;symbol&quot;&gt;p&lt;/font&gt;d²h

1.5625(1-.01p) = 1
1.5625-.015625p = 1
      -.015625p = -.5625
              p = (-.5625)/(-.015625)
              p = 36   

So the height decreased by 36%, choice (c) 

----------------------------------------

Interesting note: for some reason, statistics have shown that 
multiple-choice test-makers make the correct choice
(c) more often than (a), (b), or (d). So if a student does
not know an answer and cannot eliminate (c), and must guess, 
they should always guess (c), because their chances of guessing 
correctly is greater.  You will notice that both correct choices
here were indeed (c), bearing out the theory.  :-)

Edwin&lt;/pre&gt;

 </PRE>