SAT Math problems help

Archive SAT Subject Test Preparation
1

<p>EXPLAIN how you got the answers please:

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  3. <a href=“http://i1150.■■■■■■■■■■■■■■■/albums/o606/triassiic/Image24.jpg[/url]”>http://i1150.■■■■■■■■■■■■■■■/albums/o606/triassiic/Image24.jpg&lt;/a&gt;
  4. <a href=“http://i1150.■■■■■■■■■■■■■■■/albums/o606/triassiic/Image25.jpg[/url]”>http://i1150.■■■■■■■■■■■■■■■/albums/o606/triassiic/Image25.jpg&lt;/a&gt;&lt;/p&gt;
2

<ol>
<li>Even if you had no idea how to solve this problem (which can be solved by finding a,b), this problem is pretty guess-able. The average speed is simply d(t)/t, or a + (1/2)bt. This eliminates choices B,C,E because the answer must be in the form (constant) + (constant)t.</li>
</ol>

<p>Also, A doesn’t work, because that would imply d(t) = 2t^2. Clearly, when t = 2, d(t) would equal 8, but it equals 6 according to the diagram. For D, d(t) = t + t^2, and when t = 2, d(t) = 6. The answer is D.</p>

3

<ol>
<li><p>b-25 is between -1/4 and 1/4, so |b-25| < 1/4, E.</p></li>
<li><p>The area of ABCD is 100. This is also equal to the area of sector ABC + sector ADC - shaded part (due to an inclusion-exclusion principle argument). Thus,</p></li>
</ol>

<p>100 = 25pi + 25pi - (shaded part) = 50pi - (shaded part). Therefore (shaded part) = 50pi - 100 = 50(pi - 2), B.</p>

<ol>
<li>OP is a 60-degree arc, because triangle OPR is equilateral. Therefore the length of OP is (1/6)(4pi) = 2pi/3. Because the shaded region has three arcs equivalent to OP, the perimeter of the shaded region is simply 3OP, or 2pi, D.</li>
</ol>

4

<p>I’m not sure if rspence meant this, but:</p>

<ol>
<li>Let the shaded area above the straight line AC (not drawn) be area x and the shaded area below the line AC be area y. The question is asking for the entire shaded area, which would be area z. Because area z is the sum of both the shaded area above and below the line AC, z=x+y.</li>
</ol>

<p>That being said, the area of ABCD is 100 because it is a square and AB=BC=CD=AD=10. Because the line AC divides ABCD into 2 equal right triangles, the two newly made right triangles ABC and ADC both have area 50. </p>

<p>You will realize that area x is all the area BENEATH the UPPER arc (section of a circle; the curved line that sets the upper boundary for area x) henceforth E minus the area of triangle ADC, and area y is all the area above the LOWER arc henceforth F minus the area of triangle ABC.
To clarify, E AND F ARE AREAS such as 25ft^2, NOT LINES, such as 5 ft.
In other words, E-ADC=x and F-ABC=y</p>

<p>You can make a circle with radius AD=10 and center D. The area of circle D is (10)^2(pi)=100pi. Because 90 degrees is 1/4 of 360 degrees(360/90=4), you can divide circle D up into 4 equal sections with 90 degree angles, like a pie. Each of these 4 sections has 100pi/4=25pi area. One of the these 4 sections can be sector E (the area BENEATH the UPPER arc) because sector ADC has both a 90 degree angle and a radius of 10, so sector ADC has area 25pi. </p>

<p>Because of deductions from the third paragraph, we find that area x has area 25pi(sector ADC with the upper arc)-50(triangle ADC). </p>

<p>You can do the exact same thing with the sector F (make a circle with center B, find its area, divide it into 90 degree sections, etc.) and you find that area y has area 25pi(sector ADC with the lower arc)-50(triangle BAC). x+y=25pi-50+25pi-50=50pi-100=50(pi-2)</p>

<p>The answer is B, 50(pi-2).</p>

5

<p>@tomatox1, my solution basically said that</p>

<p>[square ABCD] = [sector ABC] + [sector ADC] - [shaded region] by inclusion-exclusion. Therefore,</p>

<p>[shaded region] = [sector ABC] + [sector ADC] - [square ABCD], which is easy to compute.</p>