Monthly Archives: September 2014

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 – Probability

Q: In a group of 10 randomly chosen people, what is the probability that at least two people share the same birthday? (same month and day)

Answer: 12% 

Explanation: (For simplicity, this explanation doesn’t take into account people with leap-day birthdays, but it wouldn’t change the answer by much!)

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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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09.10.14
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Geometry – volume

Q: If 20 solid gumballs with diameter 1 inch are placed inside a hollow sphere of diameter 6 inches, how much empty space is there in the larger sphere?

Answer:

Explanation:

 

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09.10.14
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Algebra 1 – Find the equation of a line

Q: If the slope of the line through (-3,6) and (1,y) is 1/2, what is the value of y?

Answer:  8

Explanation: Slope is computed by the formula (y2 – y1) / (x2 – x1) … the change in y divided by the change in x.

So,

(y – 6) / (1 – -3) = 1 / 2

(y – 6) / 4 = 1 / 2

Multiply both sides by 4 and you have:

y – 6 = 2

y = 8

BUT, if you’re taking a test and having trouble remembering the method above, try this instead: Plot the point (-3,6) on a graph and move from that point with a slope of 1/2. Move up 1, right 2, up1, right 2… and repeat this pattern until your x-value is 1. Then, read the y-value at the same point. This is not my favorite way, but it works!

And one more thing!  If this were a multiple choice question, you could use my TI-84 Plus program ‘DISTANCE’ (which finds distance, midpoint, and slope for any two ordered pairs).  Substitute one of the answer choices for x, and see if the program reports that slope is 1/2.

 

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09.10.14
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SAT, Geometry – triangles

Q: The lengths of two sides of a triangle are 4 and 6. If the third side is an integer, what is one possible value for the perimeter?

Answer:  The perimeter can be any of the following values: 13, 14, 15, 16, 17, 18, or 19.

Explanation: The length of the third side of any triangle must be longer than the difference of the other two sides, but shorter than the sum. (This is called the Triangle Inequality Theorem.) Here, the third side is longer than 6-4=2 and shorter than 6+4=10. It can be any integer between, but not including, 2 and 10. Recall that integers include only whole numbers and their negatives. The third side must then be either 3, 4, 5, 6, 7, 8, or 9. To get the possible perimeters, add the other two side lengths (4 and 6) to each of the possibilities for the third side. The result is that the perimeter can be 13, 14, 15, 16, 17, 18, or 19.

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09.10.14
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SAT – Average speed

Q:  Martin drives 120 miles from Oaktown to Mapleville. He drives for 60 minutes, stops for lunch for 30 minutes, then completes his journey in 60 more minutes. Including his stopping time, what is Martin’s average speed in miles per hour for this trip?

Answer:  48

Explanation:  Average speed = (total distance) / (total time).

The total distance is 120 miles.

The total time in minutes is 60 + 30 + 60 = 150 minutes.

Convert this to hours:  150/60 = 2.5 hours

Average speed = (120 miles) / (2.5 hours) = 48 miles per hour

Hint:  If you didn’t know the ‘formula’, just look at the units that the problem mentions:  miles per hour.  That’s miles divided by hours.  Do the total of each amount, and divide them.

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09.10.14
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Algebra 2 – proportions

Q:  If x is inversely proportional to the square of y, and x = 4 when y = 1, find x when y = 2.

Answer:  1

Explanation:  We are told that x and y² are inversely proportional, so x • y² = k, where k is a constant.

Let’s find out what k is, using the given info:

4 • 1² = k

k = 4

Now,  x • y² = 4.

Plug in the new value for y, to find the new value for x:

x • y² = 4

x • 2² = 4

x • 4 = 4

x = 1

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09.10.14
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Algebra 1, 2 – systems of equations

Q: Red pens cost twice as much as blue pens. Two blue pens and three red pens cost a total of $2.32. How much does a blue pen cost?

 Answer:  $0.29

 Explanation:  Let r be the cost of a red pen, and let b be the cost of a blue pen. 

Then, r = 2b and 2b + 3r = 2.32. Solve this system of equations. 

By substitution,

r + 3r = 2.32

4r = 2.32

r = 0.58

Plug r= 0.58 into the first equation (or second, but I like the first better):

0.58 = 2b

0.29 = b

 You can also use other algebraic methods such as elimination, Cramer’s Rule, etc.  Or use my TI-84 Plus program CRAMER once you rearrange the system to say: 2b – r = 0 and 2b + 3r = 2.32.

 

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