Math Question

Test Preparation SAT Preparation
1

<p>Okay BB v2 Section 6, question 15 about the triangle. page 597.</p>

<p>IN triangle, PQR above, QS/QV = 1/3 and PT/PR=3/4.</p>

<p>What is the value of the fraction (area tri PST)/ (area tri PQR).</p>

<p>The diagram is basically 1 triangle inside another. THe outermost is PQR from left to top to right. then inside that is PST where point S is on the line formed by Q and V which is on the base of the 2 triangles. </p>

<p>sorry i suck at explaining. i don't have a scanner so i can't draw this either but i would appreciate the help.</p>

2

<p><a href=“http://talk.collegeconfidential.com/sat-act-tests-test-preparation/100382-sat-question-right-place.html[/url]”>http://talk.collegeconfidential.com/sat-act-tests-test-preparation/100382-sat-question-right-place.html&lt;/a&gt;&lt;/p&gt;

<p>Let PT = 3y
Let TR = y</p>

<p>(follows logically from givens)</p>

<p>Let QS = x
Let SV = 2x</p>

<p>(also follows logically from the givens)</p>

<p>The area of the big thing is .5(4y)(3x) = 6xy
The area of the small thing is .5(3y)(2x) = 3xy</p>

<p>The big part is twice as big. The fraction is 1/2.</p>

<p>and <a href=“http://talk.collegeconfidential.com/sat-preparation/788933-bb2-math-problem.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/788933-bb2-math-problem.html&lt;/a&gt;&lt;/p&gt;

<p>QS/QV = 1/3</p>

<p>that means the length of QS is 1
and the entire length of QV is 3
(Obviously)
so using that we can determine that the length of SV by itself is 2</p>

<p>2nd part: PT /PR = 3/4</p>

<p>that means PT = 3 and PR = 4
(Again obviously)</p>

<p>so we need to find the value of the fraction area PST /area PQR</p>

<p>area of a triangle = bh 1/2</p>

<p>first lets find the area of PST.</p>

<p>the base is 3 as we determined line PT is 3. the height of SV is 2 as we determined above.
(3 x 2) x 1/2 = 3
thus the area of PST = 3</p>

<p>now let’s find the area of PQR.</p>

<p>base or length PR = 4 as stated in the original question
height or length QV = 3 as stated in the original question.
(4x3) x 1/2 = 6</p>

<p>so area PST / area PQR </p>

<p>= 3/6
= 1/2…or 0.5</p>

3

<p>thanks so much ^^</p>

4

<p>this is a easy one that i got the right answer to but the BB explanation is way too complex… is there a shortcut to BB v2 Test 4 Section 9 Q1? There is the same number of boys and girls on a school bus when it departs from school. At the first stop, 4 boys get off the bus and nobody gets on. AFter the first stop there are twice as many girls as boys on the bus. How many girls are on the bus?</p>

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<p>and this one. If p is an integer and 3 is the remainder when 2p+7 is divided by 5 then p could be? What is the correct equational way to solve it… i can do guess and check but i want to know the real math</p>

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<p>and question 12 section 9 BB v2 Test 4…</p>

7

<p>bump 10 char</p>

8

<p>"If p is an integer and 3 is the remainder when 2p+7 is divided by 5 then p could be? What is the correct equational way to solve it… i can do guess and check but i want to know the real math "</p>

<p>This should get you started:</p>

<p>In 2p + 7 = 5n + 3, n is an integer. </p>

<p>Thus
2p + 7 = 5n + 3
5n = 2p + 4
Now play with the value of p
n = (2p + 4) / 5</p>

<p>For instance, if p = 3, then 10/5 equals 2 and is an integerm but for p = 2, the answer is 8/5 and …</p>

9

<p>

</p>

<p>Let x be # girls
x – 4 = boys </p>

<p>There are twice as many girls as boys, so
x = 2(x – 4)
x = 2x – 8
x = 8</p>

10

<p>hows x=2x-8 suddenly become x=8? sorry im so slow sometimes…</p>

11

<p>x = 2x – 8</p>

<p>Flip -8 to other side and change sign
to get
x + 8 = 2x</p>

<p>Flip x to other side and change sign
to get
8 = 2x - x</p>

<p>and you get 8 = x</p>

<p>PS The usual way is taught in HS is to add or substract similar quantities on both sides</p>

<p>x = 2x – 8
add +8 on both sides</p>

<p>x**+8** = 2x - 8 **+8 **
for x + 8 = 2x </p>

<p>or another example: </p>

<p>x - 2x = 2x – 8 - 2x
is
-x = -8 and x = 8.</p>

12

<p>oh wow… i can’t believe i didn’t see that… wowowowowow thanks! and ps. im not dumb, if you thought i was ;p</p>

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<p>I would advise you to keep working through the problems and refresh your memory for the various mathematical operations. Make sure you review basic operations, exponents, simplifications, etc. It is important that the simpler operations become second nature. This WILL happen as you progress through the 10 tests in your blue book. </p>

<p>Test writers love to make the simplest things look complicated, and the more complicated problems very simple. Never feel bad about “not getting” a problem. It’s much worse to think you know it, and stumped the next time a similar problem shows up.</p>