Do this SAT problem the EASY way!

So you ‘don’t know how’ to do this SAT Question of the Day? http://sat.collegeboard.org/practice/sat-question-of-the-day?questionId=20141121&oq=1 The explanation given on the SAT site (look please!) is a good one, but ONLY if you would have thought of it! That’s the way your Algebra teacher wants you to do the problem. But DON’T SKIP IT! Do it MY WAY!
                                        

Q: A woman drove to work at an average speed of 40 miles per hour and returned along the same route at 30 miles per hour. If her total traveling time was 1 hour, what was the total number of miles in the round trip?


                          

 
There’s always a way to get rid of a choice or two. Let’s look…

 

(A) 30

(B) 30 1/7

(C) 34 2/7

(D) 35

(E) 40

Is it 30? No. Driving a distance of 30 miles in a total of 1 hour would give an overall average speed of 30 miles per hour. 30 mi/hr is the average speed for the second part of the trip, but the speed for the first part is 40 mi/hr. So, the overall average speed would be somewhere between 30 and 40, and therefore the number of miles would be between 30 and 40. (We’re talking about 1 hour, so the speeds and the distances will be the same numbers.) Cross out choice (A). Also cross out choice (E) for the same reason.

You’re down to three choices. At this point if you just guessed, you stand to gain more than you would lose. But you can do even better. The (wrong) answer they want you to fall for is (D). It’s too easy. It’s not just the average of 30 and 40, since the times spent at 30 vs 40 miles per hour are different. Cross out choice (D).

Two choices left. You were convinced above that 30 was the wrong answer. 30 1/7 is too close to 30 to make that big of a difference. The extra 1/7 of a mile wouldn’t be enough to raise the average speed to 40 for the first part of the trip. Cross out choice (B).

Done. The answer is C.

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How to use the TI-84 Plus on the SAT for graphing!

SAT Official Question of the Day for 11/18, but do it MY WAY!

2014-11-21_000219
See the question here: http://sat.collegeboard.org/practice/sat-question-of-the-day?questionId=20141118&oq=1 The solution given on the SAT site (look at it, please) is soooo much harder than doing THIS! Go get your graphing calculator and read on…
 Look at the choices:
a) -8
b) -4
c) 4
d) 8
e) 10

Try one. How about -8? So, we assume b = -8. Press ‘Y=’ on your TI-84 Plus (top left). Type into Y1: -2x^2-8x+5. That’s the function you get if you substitute -8 for b. Now press GRAPH (top right). (Do ‘ZOOM-6’ if you need a better picture.) What is the x-value of the highest point? It looks like about -2. The question says it should be 2. So, it’s not choice (a).

Try the next one: -4. Change Y1 (just type 4 over the 8), and press GRAPH again. This time it looks like the highest point is at x=-1. Still not right. It’s not choice (b).

Try 8. Change Y1 (just type +8 over the -4), and press GRAPH again. This time it looks like the highest point is at x=2. Choice (d) is correct.

And *THIS* is why I tell all of you to get the TI-84 Plus Graphing Calculator! Get one now: http://amzn.to/1tXyOTH

 

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Distance, Midpoint, and Slope

This week’s topic has been Distance, Midpoint, and Slope.

Let’s go through that blue SAT book everyone has, http://amzn.to/1wADKLS, or http://amzn.to/1wtPZuD, and this ACT book for problems about distance, midpoint, and slope that can be solved using my TI-84 Plus programs DISTANCE. Sorry, I won’t post the actual problems here.  But go dig out your books and calculator (buy them here: SAT, ACT , TI-84 Plus) and follow along.

SAT book

ISBN-13: 9780874478525.  The page numbers included here are from that book, but if you have the other one (9780874479799), please refer to the Test number and Section number instead.

DISTANCE:

Test 1, Section 3, p. 397, problem # 6

Test 1, Section 3, p. 400, problem #15

Test 3, Section 8, p. 546, problem #10

Test 4, Section 6, p. 596, problem #10

Test 5, Section 8, p. 669, problem #8

Test 7, Section 3, p. 769, problem #4

Test 8, Section 3, p. 830, problem #2

Test 9, Section 5, p. 905, problem #8

Test 10, Section 2, p. 953, problem #17  

ACT book

I’m working out of an older edition that only has 3 tests.  I believe they are the same as the first 3 tests of the 2nd and 3rd editions.  Once I have the newer book, I’ll add to this list.

DISTANCE:

Test 1, Section 2, problems #35, 45

Test 2, Section 2, problem #33

Test 3, Section 2, problems #38, 48  

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College students – sign up for this!

 

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Slope

This week’s topic is:  Distance, Midpoint, and Slope.

Q:  The slope of the line passing through (-1,-3) and (7,y) is -1/2.  What is the value of y?

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Answer: -7

Algebraic solution by hand:

Use the slope formula:

 

2014-11-07_153414

Solution using my TI-84 Plus program DISTANCE:

Run DISTANCEHow?

Use your multiple choice answer choices.  One of them is -7.  (If you don’t have choices, draw a picture.  Plot the point (-1,-3) and do a slope of -1/2 from there – down 1, right 2, down 1, right 2, … until you get to a point where x is 7.  Read the y-value of that point – it will be -7 and check it using this program.)

x1=-1

y1=-3

x2=7

y2=-7

Ignore the distance information – we don’t need it for this problem.  Press ENTER for the midpoint information, which we don’t need either.  Press ENTER once more to find that the slope is -1/2.

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Distance

This week’s topic is:  Distance, Midpoint, and Slope.

Q:  For what value(s) of x is the point (x,4) exactly 5 units away from the point (6,8)?

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Answer: 3 or 9

Solution by hand:

Use the distance formula:

distance

Solution using my TI-84 Plus program DISTANCE:

Draw a picture:

distgraph

Guess that x=2, maybe.  It’s probably an integer if this is an SAT problem.  Use the answer choices if available.

Run DISTANCEHow?

x1=2

y1=4

x2=6

y2=8

The distance is 5.66.  It’s too big – you wanted it to be 5 – so move the point a little closer to the middle.

Maybe x=3:

x1=3

y1=4

x2=6

y2=8

The distance is exactly 5.  Perfect.  (There’s also another possible answer, x=9.  If you have multiple choice answers, you will see 3 and 9 as a choice, so try both and see that they both work.  For a grid-in SAT problem, you would only have to find one answer anyway.)           :

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Ratios and Proportions

This week’s topic has been Ratios and Proportions.

Let’s go through that blue SAT book everyone has, http://amzn.to/1wADKLS, or http://amzn.to/1wtPZuD, and this ACT book for problems about ratios that can be solved using my TI-84 Plus programs RATIO and PROPORTN.  (See these posts for examples: 10/27, 10/28, 10/29, 10/30)

Sorry, I won’t post the actual problems here.  But go dig out your books and calculator (buy them here: SAT, ACT , TI-84 Plus) and follow along.

SAT book

ISBN-13: 9780874478525.  The page numbers included here are from that book, but if you have the other one (9780874479799), please refer to the Test number and Section number instead.

RATIO:

Test 1, Section 7, p. 417, problem # 13

Test 4, Section 6, p. 594, problem #6

Test 10, Section 2, p. 953, problem #16

Test 10, Section 8, p. 980, problem #12

PROPORTN:

Math Multiple-Choice Sample Questions, p. 307, problem #6

Math Student-Produced Response Sample Questions, p. 347, problem #1

Test 1, Section 7, p. 417, problems # 11, 13

Test 2, Section 2, p. 452, problem #2

Test 3, Section 5, p. 528, problem #9

Test 3, Section 8, p. 543, problem #1

Test 4, Section 3, p. 583, problem #6

Test 5, Section 2, p. 639, problem #5

Test 6, Section 4, p. 715, problem #10

Test 7, Section 3, p. 768, problem #1

Test 9, Section 8, p. 916, problem #4

Test 9, Section 8, p. 918, problem #11

Test 10, Section 2, p. 950, problem #7

ACT book

I’m working out of an older edition that only has 3 tests.  I believe they are the same as the first 3 tests of the 2nd and 3rd editions.  Once I have the newer book, I’ll add to this list.

RATIO:

Test 2, Section 2, problem #15

PROPORTN:

Test 1, Section 2, problems #19, 39, 44

Test 2, Section 2, problems #3

 

 

 

 

 

 

 

 

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Ratios

Q: Last month, the ratio of sunny days to rainy days was 3:2.  How many rainy days  were there last month, if the month had 30 days?  (Assume all days were either sunny or rainy.)

 

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Answer: 12

Solution by hand:

Number of sunny days = 3x

Number of rainy days = 2x

3x + 2x = 30

5x=30

x = 6

There were 2x rainy days, so there were 12 rainy days.

 

Solution using my TI-84 Plus Program RATIO:

Run RATIOHow?

Total number of items (days) = 30

Number of categories of items = 2  (sunny and rainy)

Ratio part for category 1 (sunny) = 3

Ratio part for category 2 (rainy) = 2

Distribution is: sunny = 18 and rainy = 12.

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Proportions

Q:  x is inversely proportional to w.  If x = 6 when w = 8, what is x when w = 12?

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Answer: 4

Solution by hand:

For inverse proportions, (x1)(w1) = (x2)(w2)

(6)(8) = (x)(12)

48 = 12x

x=4

Solution using my TI-84 Plus Program PROPORTN:

Run PROPORTN.  How?

x1 = 6

y1 = 8  (here, we have renamed w1 as y1)

Have y2 (really w2)

y2 = 12

Choose ‘Inverse’ proportion.

Then, x2 = 4.

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Proportions

Let’s change yesterday’s question around a bit:

Q:  Wandering around blindly in a parking lot (don’t try it, please), the probability that the first car you bump into is blue is 2/5.  If there are 500 blue cars, how many cars are in the parking lot?

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Answer: 1250

Solution by hand:

Set up the proportion using the concept ‘blue / total = blue / total’:

2/5 = 500/x

2x = 2500

x = 1250

Solution using my TI-84 Plus Program PROPORTN:

Run PROPORTN.  How?

x1 = 2  (blue)

y1 = 5  (total)

Have x2 (blue)

x2 = 500

Choose ‘Direct’ proportion (because it doesn’t say ‘inverse’ in the problem).

Then, y2 (the total) = 1250.

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