Category Archives: ACT

10.02.14
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SAT – Probability

Q:  What is the probability of randomly guessing the last four digits of a telephone number correctly?

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A: 1/10,000

Explanation:  Each digit is chosen from the integers 0 through 9, which is ten possible choices for each digit.  There are four digits to choose independently, so there are:

10 x 10 x 10 x 10 = 10,000 ways to choose the four digits.

Then, the one correct sequence of 4 digits, is one out of 10,000.  The probability is 1/10,000.

Another way to arrive at the number 10,000 is as follows:

The 4-digit sequence may be anything from 0000 to 9999.  So, how many is that?  9999 – 0000 +1 = 10,000.  Why the ‘plus one’ on the end?  If you just subtract the numbers, it tells you how many spaces there are between the numbers.  Add one to include either the first number or the last number, whichever one has not been already counted.

 

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09.29.14
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SAT Question Of The Day – estimating for hard geometry

From the official SAT site, here’s the Question of the Day for 9/25/14:

http://sat.collegeboard.org/practice/sat-question-of-the-day?questionId=20140925&oq=1

sat925

The answer is C.  Choose it (on the site, using the link above), and you will see the explanation involving a 30-60-90 triangle.  This is definitely the way to go, if you ‘remember how’.

The problem is, so many students will look at this question and say ‘I don’t remember how to do this’ and then skip it.  DON’T SKIP IT!!!!

Just ‘measure’ it!  OK, it’s not really measuring.  But consider this:  The diagram does NOT say ‘not drawn to scale’.  Therefore, we can assume it IS reasonably drawn to scale.   Look at the diagram.  Triangle ABO does look equilateral, O does look like the center of the circle, and the circle does look like a circle.  So, take the given measurement – that the sides of triangle ABO are 6 – and run with it!  Look at segment AB below.  It is 6 units long.  Measure it with your fingers and then use the same distance to guess at the length of segment BC.  Below, I’ve shown that 6 units, placed twice end-to-end along segment BC, is just a little longer than segment BC.  So, segment BC is shorter than 12.  If you do it carefully, you’d probably guess segment BC is about 10-11 units long.

2014-09-27_104226ans

Look at the choices.  Calculate decimal approximations for A, B, and C by typing the answers into your calculator.  Which one is closest to our estimate? C! And it’s the right answer! 

 

 

2014-09-27_104256

You’d be surprised how often you can really do this and get the right answer.  Follow me, and you’ll see lots of these!

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09.18.14
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Precalculus – Domain of a function

Q: What is the domain of this function?

 

Answer:

   ‘Like’ if you got it!

Explanation:

The domain of a function is a description of the x-values that may be plugged into the function. For a square root function, the restriction is that the number under the radical must not be negative. Therefore,

 

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09.12.14
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ACT, Geometry – diagonal of a rectangular prism

Q: What is the (straight-line) distance between points A and B on the figure below?

rectangularprism2

Answer:  7

Explanation:  We are finding the length of the green segment below:

 

rectangularprism2wline

This requires the three-dimensional version of the distance formula:

  2014-09-12_225150

There is another way to do this, using a progression of two right triangles, with the green segment being the hypotenuse of the second triangle, but I find the above formula much simpler.  It is just an extension of the ‘usual’ (2-dimensional) distance formula we are so familiar with from Algebra and Geometry.

BUT if you want an even easier way, and I know you do, use my TI-84 Plus program ‘DIST3D’.

 

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09.12.14
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ACT , Algebra 1 – system of equations

MathPro Q&A Forum (ACT)

 Q: For what value of A will the following system of equations have infinitely many solutions?

3x – 2y = 4

Ax + 4y = -8

 

Answer: -6

Explanation: In order for the system to have infinitely many solutions, the two equations must describe the same line.  (Recall that the solution(s) of a system are the points where the graphs of the two lines meet.  For the two lines to meet in infinitely many points, they must be the same line.)

 To make the second equation the same as the first, multiply both sides of the first equation by -2:

 -2(3x – 2y) = -2(4)

-6x + 4y = -8

So, A = -6.

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09.10.14
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SAT – number properties

Q: The variables a, x, and b are positive integers. If ax = 7, xb = 28, and ab = 4, what is the value of a?

Answer: 1

Explanation:

a, x, and b are all positive integers, so the only possibilities for the variables are factors of 7, 28, and 4.

Since 7 is prime, a and x must be 7 and 1, but in which order? If x = 1, then b = 28, and then we can’t find an integer value for a from the last equation. So, x must instead be 7.

It follows that a = 1 and b = 4.

The answer is 1.

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09.10.14
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Geometry – polygons

Q: A hexagon has interior angles measuring (in degrees) 137, 100, 150, 114, 90, and x. What is the value of x?
 
 Answer: 129
Explanation:
The total measure of all the interior angles in a polygon is (n-2) * 180, where n is the number of sides of the polygon.
A hexagon has 6 sides, so n = 6. The total is (6-2) * 180 = 720.
Add up the five given angle measures: 137 + 100 + 150 + 114 + 90 = 591.
The measure of the remaining angle is 720 – 591 = 129.
My TI-84 Plus program POLYGON will find this total for you.
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09.10.14
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SAT – Counting and Ordering

Q: There are 11 basketball teams in a tournament. Each team plays each other team once. How many games are played?

 Answer: 55

Explanation 1: This question is really ‘How many ways are there to choose 2 teams from a total of 11 teams, if the order of the teams is not important?’ It’s a combination question. Find the nCr function on your calculator. On the TI-84 Plus it’s in MATH-PRB-3. Type in 11 nCr 2. It’s 55. (There is a longer way to do it by hand, using the formula for nCr.)

Explanation 2: Team A plays one game against each of the 10 other teams. That’s 10 games so far. Team B plays 9 other games (its game against Team A has already been counted). Team C plays 8 other games, … all the way down to the 1 last game between teams J and K. Add: 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55.

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09.10.14
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Algebra 2 – exponential functions

Q: In 2010, River City had a population of 500. If the population has increased by 6% each year since then, what is the population this year, in 2014?

Answer: 631

Explanation:

Alternatively,

Calcululate 6% of 500 by doing .06 * 500 = 30.

Increase 500 by 30, and you find that the population is 530 in 2011.

Now, increase 530 by 6% (530*.06, added to 530, or shortcut is to do 530*1.06).

The population is 561.8 in 2012. Don’t round this number yet – not until you get to the end of the problem.

Doing this calculation 2 more times gives you 595.508 for 2013 and 631.23… for 2014.

Now, round to the nearest whole number: 631. It’s not the most efficient way, but you get the right answer.

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09.10.14
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Precalculus – logarithms

Q: Solve the following equation involving logarithms:

Answer: x=4

Explanation:

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