Mike's Math Corner: Another SAT Example!

Back again with another difficult SAT math problem. This one involves some tough algebra that many students might trip up on. If you're an algebra ace, this problem won't take too much of your time. If you're not, then luckily we have tricks to solve multiple choice questions without doing the algebra at all. In fact, all we'll need to know is PEMDAS. Using the trick, in this case, wont even waste much time.

If x=5+4t^2 and y=3+2t, what is x in terms of y?

a) y^2-6y+14
b) y^2+6y+14
c) 4y^2-24y+36
d) 4y^2-24y+41
e) 4y^2+24y+41

Yet again, I'll start out using tricks that involve less advanced algebra. Pick a number for 't'. I usually stay away from 0 or 1 when doing this, as multiplying by 1 or 0 can make multiple answers look possible. I'm going with 2. So x=5+4(2)^2-->x=21. And y=3+2(2)-->y=7. Now we just need to plug y=7 into the answer choices and see which one gives us
21.

a)7^2-6(7)+14-->49-42+14=21.... there's our answer already.
b)7^2+6(7)+14-->49+42+14=63
c)4*7^2-24*7+36 -->196-168+36=64
d)4*7^2-24*7+41 -->196-168+41=69
e)4*7^2+24*7+41 -->196+168+41=405

*notice that by choosing 2 for t gave us different numbers for every answer choice. This makes it easy to see the right answer right away.

And if you prefer an algebraic approach to the question, you could solve t in terms of y, and substitute into the equation for x:
y=3+2t-->(y-3)/2=t
Now substitute this value of t into the x equation:
x=5+4t^2--> x=5+4[((y-3)/2)^2] *be very careful in squaring the y-3 here, FOIL* x=5+4(y^2-6y+9)/4 --> x=5+y^2-6y+9 --> x=y^2-6y+14 ...

This is answer a.

Always keep in mind that there is (almost) always more than one way to solve a math problem. On multiple choice questions, working backwards can sometimes be easier and faster than going forwards. Testing your answer choices to match the original question is a great tactic for any question you might get stuck on!

Mike's Math Corner

          Getting ready to take the MCAS or SAT? Do some of those math questions leave you scratching your head wondering where to even begin? Each week I'll be solving a sample MCAS or SAT question. I will start out with ways to eliminate wrong answer choices, or if possible, solve the problem outright without using complicated mathematics. I will then solve the problem 'the hard way', using the tricks from those higher-level math classes.

         This week's problem is a SAT math question involving geometry that I found on analyzemath.com:
 
AC is a diameter of the circle shown below and B is a point on the circle such that triangle ABC is isosceles. If the circle has a circumference of 8 Pi, what is the area of the shaded region?

.

A) 16 - 8Pi
B) 64
C) 8Pi - 2
D) 4Pi -8
E) 4(2 - Pi) 


           This is definitely a 'hard' question. Circles, areas, and not even a radius to work off of. Lets say for a moment you forget how to work with circles completely. Start with the answer choices and USE YOUR CALCULATOR! Answers A and E are both negative numbers.  Areas are never negative, so these are out. Also, if you remember that the formula for the area of a circle has pi in it, any piece of that area has to involve pi as well. Answer B is wrong because it does not have pi, and because its WAY too large. Get this far, and you have a 50-50 shot, and we havn't even done anything yet. Answer C is around 23, and D is around 5. Looks like a pretty small portion of the circle, so I'd guess D if I had to.
        Now for some math. Luckily, they tell you the circumference, and the formula for that is provided on your reference sheet: C=2*pi*r-->8*pi=2*pi*r--->4=r At this point, since the answer choices vary by a large ammount, I would approximate BC as being a bit longer than the radius (say 5), and make a rectangle with height of ~1 (about a quarter of the radius) to approximate the shaded region. This gives you an area of ~5 square units which, yet again, points to answer D. 

        If you're still curious on how to get an exact answer: Because angle B lies on the arc of a circle, and intercepts a diameter, it must measure 90* (this is a geometry theorem). Therefore, the isosceles triangle is a 45-45-90.  Drawing a radius from the center of the circle (call it point O) to point B will make 45-45-90 triangle BOC (altitudes of isosceles triangles drawn from the vertex angle are also angle bisectors). Angle BOC, being 90*, represents a quarter of the whole circle. We can find that area with the formula (1/4)(pi*r^2)=(1/4)(pi*4^2)=4*pi. We can then subtract the area of triangle BOC, to leave us with the exact area of the shaded region. For triangle BOC: A=1/2*B*H=(1/2)(4*4)=8. The shaded region is then: A=4*pi-8, or roughly 5 square units. Answer D

           OK, so in review, on this hard SAT question, it was possible to make an educated guess after doing VERY little math and come up with the correct answer. Using only the formula for circumference, and approximating the shaded region as a simple rectangle also gave us the correct answer. Actually doing the math out was far from easy and, I imagine, time consuming (although congratulations are in order if you did manage to solve it!). Remember that the SAT is a timed test! It is designed around logic and challenges you to come up with simple solutions to complex problems. As such, solving this completely would be considered a waste of valuable time on the real test, and could leave you struggling to finish the section in time.