Monthly Archives: September 2014

09.10.14
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SAT – Counting and Ordering

Q:  20 students are to choose a president and vice president from among themselves. How many possible ways can they be chosen?

Answer:  380

Explanation 1:  Using the Counting Principle, set up two blanks to represent the president and vice president, from left to right:

____    ____

There are 20 choices for the president.  Put 20 in the first blank:

_20_    ____

The same person cannot be chosen for both positions, so there are 19 left to choose from for the vice president.  Put 19 in the second blank:

_20_    _19_

Now, multiply these numbers.  20 x 19 = 380.

Explanation 2:  This is a permutation problem since you are choosing 2 people from a total of 20 people, and the order is important.  The order is important because the two people chosen must be given specific positions, and switching those positions results in a different scenario.  Do  20 nPr 2 , or use the formula for nPr.  On the TI-84 Plus, find nPr in MATH-PRB-2.  The result is 380.

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09.09.14
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SAT, Algebra 1 – percent change

Q: Last year, Emmy was 40 inches tall.  This year, she is 42 inches tall.  By what percent has her height increased?

 Answer: 5%

 Explanation:  Percent change is calculated by the following formula:

 (New – Original)/Original  x 100%

 (42 – 40)/40 = 2/40 = .05

 .05 x 100% = 5%

 Since it is positive, this is a percent increase.

My TI-84 Plus program PRCNTCHG will calculate this, too.

 

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09.09.14
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Geometry – Areas

Q: What is the area of this figure?

area

 Answer: 57

 Explanation:  Cut the figure into two rectangles as shown.  Find the missing vertical length by noticing that the two shorter vertical lengths must add up to the long vertical length.  The missing vertical length is 3, because 4 + 3 = 7.  Do the same for the missing horizontal length.  It is 5, because 6 + 5 = 11.

areaanswer

 Now, find the area of each rectangle (A = length x width).  The one on the left is 7×6 = 42.  The one  on the right is 3×5 = 15.  Add these areas to find the area of the original figure:  42 + 15 = 57.

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09.07.14
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SAT – number properties

Q: What is the sum of all of the integers from 1 to 100?

Answer: 5050

Explanation: Don’t add 1 + 2 + 3 + 4 + … all the way up to 100! While it will work, it is a giant waste of time. Look for a pattern of numbers to pair up:

 1 + 99 = 100

2 + 98 = 100

3 + 97 = 100

…

49 + 51 = 100

 That’s 49 one hundreds. 49×100 = 4900. Now, add the numbers we have not counted yet: 100 and 50.

4900 + 100 + 50 = 5050.

 My TI-84 Plus program ‘SUMATOB’ will find this sum, too.

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09.07.14
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Algebra 2 – linear functions

Q:  f(x) is a linear function with a y-intercept of 6. If f(-2) = 9, what is f(2)?

Answer:   3

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Explanation:

f(x) = mx + b              (because it is linear)

m is the slope of the line passing through (0,6) and (-2,9).

m = (9-6) / (-2-0)

m = – 3/2

b is the y-intercept.  b = 6.

So, f(x) = – 3/2 x + 6

Now, find f(2):

f(2) = (- 3/2)(2) + 6 = 3

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09.06.14
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Algebra 2 – Sequences

Q: What is the tenth term of the sequence: 4, 12, 36, 108, …?

Answer:  78732

Explanation:  It’s a geometric sequence because each term is multiplied by 3 to get the next term. The common ratio (r) is 3. You *could* just keep multiplying by 3 until you get to the tenth term. It’s not the most efficient way, but it works.

The ‘best’ way, if you know it, is to use the formula for the nth term of a geometric sequence:

2014-09-06_202336

Oh, and my TI-84 Plus program ‘NTHTERM’ will find this answer for you.

 

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09.06.14
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Geometry – Lines and segments

Q: Points D, E, and F are collinear with DE < EF and DF < EF.  If DE = 2x+6, DF = 3x-4, and EF = 22, what is DE?

 Answer: 14

 Explanation:

 First, recall that ‘DE’ refers to the length of the segment with endpoints D and E.  It is the distance between D and E.  Next, DON’T ASSUME that E is between the D and F!  Since DE and DF are each less than EF, and all three points lie on the same line (collinear), D must be between E and F.

 Draw the diagram:

__________________________

E                         D                   F

 Now label the distances as they are given in the problem:

  <————–22——————->

__________________________

E        2x+6         D     3x-4      F

 Use the concept: Part + Part = Whole (also called the Segment Addition Postulate) to write the equation:

 2x+6 + 3x-4 = 22

 Solve:

5x + 2 = 22

5x = 20

x = 4

 Plug x=4 back in to the ‘DE’ part:

DE = 2x+6 = 2(4) + 6 = 14.

 Tip:  To be sure your answer is correct, plug x=4 into the ‘DF’ part, too:

DF = 3x-4 = 3(4)-4 = 8.  Now, see that the sum of the two ‘parts’ does equal the ‘whole’:

14 + 8 = 22.

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09.04.14
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Algebra 1, 2 – solving quadratic equations

Q: The square of a positive number is equal to 5 more than the product of 4 and the same number. What is the number?

Answer: 5

Explanation:

2014-09-04_231259

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09.04.14
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SAT – Counting and Ordering

Q: In how many ways can the letters of the word ‘FIVE’ be arranged?

Answer:   24

Explanation 1:  SInce there are 4 different letters, do 4!  (that’s ‘4 factorial’).   4! = 4 x 3x 2x 1 = 24.  Your calculator might have a factorial (!) function.  On the TI-84 Plus, find it in MATH – PRB.

Explanation 2:  SInce you are choosing 4 items from a total of 4 items, and the order is important, you can do 4 nPr 4, or use the formula for nPr.

 

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09.03.14
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SAT – number properties

Q: How many positive integers less than 500 are multiples of 3, 7, and 11?

Answer:  2

Explanation:  3, 7, and 11 do not have any common factors except 1.  So, their least common multiple is 3 x 7 x 11 = 231.  The next two common multiples are 231 x 2 = 462 and 231 x 3 = 693.  693 is greater than 500.  There are only two that are less than 500.  They are 231 and 462. 

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