Is there a way to do these problems without resorting to plugging in?

Test Preparation SAT Preparation
1

<p>A right triangle has its height increased by 2 centimeters, if the resulting area is 5 square centimeters greater, then what was the base in centimeters of the original triangle?
A. 25
B. 20
C. 10
D. 8
E. 5</p>

<p><a href="http://img384.imageshack.us/img384/7104/460345cq2345041cj3.gif%5B/url%5D"&gt;http://img384.imageshack.us/img384/7104/460345cq2345041cj3.gif&lt;/a>
In rectangle PQRS above, what is the sum of a+b in terms of x?
A. 90+x
B. 180-x
C. 180+x
D. 270-x
E. 360-x</p>

<p>I have no problem with plugging in, but it's just that I've been so used to finding algebraic ways to solve things, that the thought of plugging in your own number doesn't occur to me. Then, when I can't solve it algebraically, I still don't think to plug in, and I get screwed.</p>

2

<ol>
<li>A=0.5bh
(A+5)=0.5b(h+2)
Substitute: (0.5bh)+5=0.5b(h+2)
Simplify and you get:b=5
The answer is E</li>
</ol>

<p>2.The 2 angles next to x are 90-a and 90-b, and all 3 add up to 90, so you could say
90=x+(90-a)+(90-b)
Simplify for a+b and you get 90+x</p>

<p>But in all honesty it’s probably easier to plug in.</p>

3

<p>I never plug in…I always try to find the proper solution to these problems</p>

4

<p>I’m glad the old New York math curriculum is being abolished. This is the type of stuff Math A/B doesn’t teach.</p>

5

<p>These problems aren’t “taught”. You just figure them out :P</p>

6

<p>What about this problem? I can find it through plugging in, but I want to find it the correct way:</p>

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In the figure above, AB = AC and AB||CD. What is the value of y in terms of x?
A. x
B. (180-x)/2
C. (360-x)/2
D. x/2
E. 90-x</p>

7

<p>Angle BCD=90-y
Angle ABC=(180-x)/2
Because AB//CD, Angle BCD=Angle ABC
or,90-y=(180-x)/2
Solving u get y=x/2
Therefore the answer is D</p>

<p>However I would like to know how plugging values will work here.</p>

8

<p>agreed with Arachnotron.There is nothing to be taught here.It’s all about figuring them out.</p>

9

<p>Unfortunately, not everyone has the time or thought speed to think of solutions to some of these problems during a SAT. Also, the reason I bring up the ridiculous soon-to-be defunct Math A/B system was because it didn’t have certain topics that should be known to students in New York. The more a test-taker knows, the better; but anyhow, I’m not going to go into the faults of the SAT here…</p>

10

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<p>If everyone did, the test would have no power to discriminate ability :P</p>

11

<p>Over the years I became a proponent of plugging in when it guarantees higher speed and/or accuracy on the test. Learning to solve analytically is still very important for better understanding.</p>

12

<p>Revisting post #6.
We are free assume that that perpendicular from the point D passes through the point A (it won’t change the answer); that makes ABDC a rhombus, therefore y=x/2.</p>

<p>=============
When plugging numbers, all the manipulations are the same as in strictly math method, but are easier to follow.
Let x=80, then
<ABC = 50,
<BCD = <ABC = 50 as alternate interior angles,
y = 90 - 50
y=40.
Only answer D satisfies x=80 and y=40.</p>

13

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<p>Parallel to ajwchin (post #2):
let a=b=80 (then a+b=160)
x=90-10-10=70
Answer A is the only working choice:
90+70=160.</p>

14

<p>What`s wrong with plugging ?You actually save time and no one cares if you can solve the problem algebraically</p>

15

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<p>In the image, there is a quadrilateral within the rectangle that has x as one of its angles. To find a+b in terms of x, an equation can be created that involves the sum of the angles of the quadrilateral equaling to 360. The northeast angle of the figure is obviously 90, but the two obtuse angles are equal to 180-a and 180-b because they’re on straight lines. Since all of the angles of the quadrilateral are now known, the equation 90+180-a+180-b+x=360 can be used to find a+b, which is equal to 90+x.</p>