I kinda need help with some SAT math

[JeffreyToTheMax]

So, here's my situation. I took the SAT (http://en.wikipedia.org/wiki/SAT) when I was in high school, and I did pretty good on both of the parts that weren't math. Part of the reason for that was that I assumed calculators were not allowed, when in fact they were all but required. So, my math score wasn't as high as it could have been.

Now it's a few years later, and I have a 2-year transfer degree. The issue is, I don't really have money to afford finishing up my degree (computer science), so I was hoping if I re-took the SAT, this time armed with a TI-83, my score would miraculously be near-perfect and all the colleges in the world would be climbing on top of each other to get access to my juicy, pulsating brain. Or just one. I'm not picky.

So, I loaded up a website that had some practice tests, and realized with horror that I don't actually know how to do any of this stuff anymore. The test is in six days. From what I remember of the SAT, not only do I need to figure out how to do this kind of math, I need to get so good at it that I can complete the math portion within the incredibly strict time limit.

All the websites I have looked at are offering books as study aids, but not only can I not afford these books, I don't have the time to wait for one to ship. I need to learn quickly. Are there any resources out there for one such as I? Sites that will actually teach this stuff, in addition to providing practice tests?

Alternatively, I will also accept the assistance of anyone who wants to help me personally. That may seem a tall order, and honestly it really is, but this forum is home to some of the most helpful individuals I have come across. Especially when it comes to math.

So... I am at the mercy of the Playground. Only you can decide my fate!

[Player_Zero]

Do you have anything you need help with in particular?

There are any number of maths undergraduates on the forums.

[JeffreyToTheMax]

Here are some examples from a practice test I found.

The first two questions were trivial. 4/7 is greater than 1/2, and 5 feet per second equals 18,000 feet per hour. Elementary stuff so far.

Then comes the following:

3. What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

A. 90
B. 95
C. 100
D. 105
E. 110
I can figure that out, but not quickly. I pretty much added every one of those numbers together, and divided by how many there were to find the average. I'm assuming there's a faster way.

4. A cubical block of metal weighs 6 pounds. How much will another cube of the same metal weigh if its sides are twice as long?

A. 48
B. 32
C. 24
D. 18
E. 12
Again, with liberal and time-consuming use of my calculator, I can figure this out, but it needs to be quick.

5. In a class of 78 students 41 are taking French, 22 are taking German and 9 students are taking both French and German. How many students are not enrolled in either course?

A. 6
B. 15
C. 24
D. 33
E. 54

The next three questions are straightforward functions, which I've done recently enough to remember how to do them pretty quickly. But it starts to get interesting again:

(A diagram consisting of a circle, with three points labelled A, C, then B. Two curved lines link A to C, then C to B.)
9. Amy has to visit towns B and C in any order. The roads connecting these towns with her home are shown on the diagram. How many different routes can she take starting from A and returning to A, going through both B and C (but not more than once through each) and not travelling any road twice on the same trip?

A. 10
B. 8
C. 6
D. 4
E. 2
And the last one is geometry, which I'm okay at.

I know this is all extremely simple, which is why I think I can reasonably expect to master it in a few days. The last math I have taken is calculus, so this stuff is all way behind me. There are more tests, but this is the first one, and is probably a good place to start.

[Zanaril]

Here are some examples from a practice test I found.
All I can say is keep practising. It looks like you can do the maths, you just need to get better at recognising the logic behind the questions.

[JeffreyToTheMax]

All I can say is keep practising. It looks like you can do the maths, you just need to get better at recognising the logic behind the questions.

With some of them, I can figure it out given enough time. But it's not the way you're supposed to do it. Even though I know how to get the answer, eventually, that's a far cry from being able to work it out in less than a minute, and the process for doing so is likely entirely different.

[RS14]

Then comes the following:

3. What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

A. 90
B. 95
C. 100
D. 105
E. 110
I can figure that out, but not quickly. I pretty much added every one of those numbers together, and divided by how many there were to find the average. I'm assuming there's a faster way.

(10+20+...+190)=10(1+2+..19)

=10(20*19/2)

(Sum of consecutive integers 1...n is (n+1)n/2)

=1900

Edit: Then 19 numbers, so 1900/19=100, the average.



4. A cubical block of metal weighs 6 pounds. How much will another cube of the same metal weigh if its sides are twice as long?

A. 48
B. 32
C. 24
D. 18
E. 12
Again, with liberal and time-consuming use of my calculator, I can figure this out, but it needs to be quick.

Doubling the length in one dimension doubles the mass. We are doubling it in three dimensions, so

6*2*2*2=6*2^3=48




(A diagram consisting of a circle, with three points labelled A, C, then B. Two curved lines link A to C, then C to B.)
9. Amy has to visit towns B and C in any order. The roads connecting these towns with her home are shown on the diagram. How many different routes can she take starting from A and returning to A, going through both B and C (but not more than once through each) and not travelling any road twice on the same trip?

A. 10
B. 8
C. 6
D. 4
E. 2



Edit: I got that order wrong initially, now fixed.

A->C->B->A:
Two choices for first path, two for second, one for third. 2*2*1=4 total choices for this order.
A->B->C->A
One for first, two for second, two for third. 2*2*1=4 total choices for this order.

Thus 8 total choices.

[BizzaroStormy]

There is faster way to find the average of the numbers unfortunately, just add them all up, then divide by the amount of numbers.

Edit: Ignore that bit, didn't read the problem completely.


For #5 you have the 41 french, 22 german, and 9 for both.

first, subtract 9 from both german and french to account for the students who took botth, leaving you with 32 French, and 13 german. After that, subtract the 9, 13, and 32 from the total of 78. You're left with 24 student who didn't give a crap.

As for the second and fourth problems, I can't really help without a visual representation.

[JeffreyToTheMax]

(Extremely helpful stuff)

Thanks so much for this. I will meditate on these answers, by which I mean write them down and try them with different values until I memorize them.

More questions are likely coming, unless I have some kind of epiphany, but they'll all be pretty much at this level of simplicity.


first, subtract 9 from both german and french to account for the students who took botth, leaving you with 32 French, and 13 german. After that, subtract the 9, 13, and 32 from the total of 78. You're left with 24 student who didn't give a crap.

When you put it that way, it's incredibly simple. I have a feeling I'm going to feel pretty stupid when this is all said and done.


As for the second and fourth problems, I can't really help without a visual representation.
Yeah, sorry about that. If it becomes necessary, I'll figure out how to insert images.

[ForzaFiori]

Edit: I got that order wrong initially, now fixed.

A->C->B->A:
Two choices for first path, two for second, one for third. 2*2*1=4 total choices for this order.
A->B->C->A
One for first, two for second, two for third. 2*2*1=4 total choices for this order.

Thus 8 4 total choices.

Everything else you did was spot on, but I think you misread it here man. There weren't any paths from A->B, only A->C and C->B.

Therefor the answer is just 4, since you dont need all the striked out stuff.

[RS14]

A, B, C are on a circle.

I ran into the problem myself when looking back to see what SAT math problems were like.

Diagram:

http://www.majortests.com/sat/testpics/p001-9.gif

[JeffreyToTheMax]

Everything else you did was spot on, but I think you misread it here man. There weren't any paths from A->B, only A->C and C->B.

It no doubt would be much more clear if I wasn't lazy and included the diagram. Hopefully my hotlink can be forgiven:

http://www.majortests.com/sat/testpics/p001-9.gif

EDIT: Wow, that was quick work finding the image. I'm assuming we're looking at the same site.

[snoopy13a]

Remember to look out for tricky problems such as:

What is the square root of 16?

Usually, the first couple of problems are straightforward and not tricky while the last couple are. Don't spend too much time on a problem, if it looks like it'll tough, skip and come back later.

[Mando Knight]

(10+20+...+190)=10(1+2+..19)

=10(20*19/2)

(Sum of consecutive integers 1...n is (n+1)n/2)

=1900

Edit: Then 19 numbers, so 1900/19=100, the average.

*Bonk!*

It's a linear set of data! (y=10*x)
For such a set, take the first number and the last number, add them together, and divide by 2.
10+190=200. 200/2=100. Boom, done.

[FoeHammer]

The problem involving the cube is simple if thought about algebraically. The volume of a cube is x^3. If the sides are twice as long, replace x with 2x. we then have (2x)^3 which is 8(x^3). As volume scales linearly with weight, the weight is 48 lbs(6*8)

[JeffreyToTheMax]

Okay, I thought I could figure this last one out, but I can't seem to do it without resorting to time-burning trigonometry. I really hope that's not the only way to do it. Surely there's something I'm missing:

http://www.majortests.com/sat/testpics/p001-10.gif
10. In the figure above AD = 4, AB = 3 and CD = 9. What is the area of triangle AEC ?

A. 18
B. 13.5
C. 9
D. 4.5
E. 3

Thanks to my good friend Pythagoras, the length of AC is simple to get. I just need to get the length of AE somehow.

[snoopy13a]

Wrote too soon there...

[ForzaFiori]

A, B, C are on a circle.

I ran into the problem myself when looking back to see what SAT math problems were like.

Diagram:

http://www.majortests.com/sat/testpics/p001-9.gif

My mistake. I apologize.

[Mando Knight]

Okay, I thought I could figure this last one out, but I can't seem to do it without resorting to time-burning trigonometry. I really hope that's not the only way to do it. Surely there's something I'm missing:

http://www.majortests.com/sat/testpics/p001-10.gif
10. In the figure above AD = 4, AB = 3 and CD = 9. What is the area of triangle AEC ?

A. 18
B. 13.5
C. 9
D. 4.5
E. 3

Thanks to my good friend Pythagoras, the length of AC is simple to get. I just need to get the length of AE somehow.
Similar triangles.

AE=AD*AB/(AB+CD)=1.

Thus, the answer is D.

[FoeHammer]

Find the area of ACD and subtract the area of CDE.

[Mando Knight]

Find the area of ACD and subtract the area of CDE.

When you don't know the area of CDE, finding the area of AEC directly is just as easy.

[JeffreyToTheMax]

Similar triangles.

AE=AD*AB/(AB+CD)=1.

This is almost definitely the key. Forgive my ignorance, but could you explain the concept of similar triangles to me? I'm having trouble seeing how you got this equation.

[RS14]

ABE and DCE have two of the same angles (this we see by noting that CD and AB are parallel lines, thus the inside angle made by CB with them is the same). Thus they have three of the same angles. Thus they are similar: each is a scalar factor of the other, rotated, transposed, etc.

Lines CD and AB correspond, in these triangles, because they fall between equal angles in similar triangles. Thus the ratio between them is equal to the ratio between any two corresponding sides in these triangles. The ratio is 3:1, so ED:AE is 3:1; because their sum is four, AE=1 and ED=3.

Then the area of AEC is 9*1/2=9/2-4.5 (9 altitude from AE to C; base 1 AE).

Sorry, forgot what the length AD was, and forgot to divide the area by half. Corrected above.

@V: Yeah, I realized.

[Mando Knight]

This is almost definitely the key. Forgive my ignorance, but could you explain the concept of similar triangles to me? I'm having trouble seeing how you got this equation.

The concept of similar triangles is this: when two triangles have equal angles, their sides must be directly proportional to the equivalent side on the other triangle.

The key to using this with the problem is to construct a triangle similar to AEB by placing a point opposite point D (I'll call it D', pronounced "D prime") so that the magnitudes of AD' and CD are equal, as are CD' and AD.

The new triangle, similar to AEB, has a base length equal to CD+AB, and a height length (proportional to AE) equal to AD. The only piece of information we're interested in here is the length of AE, so we solve the equation AE/AD=AB/(CD+AB) (similar triangles) for AE.

Now that we have AE, we can find the area of triangle AEC, which has a base of AE and a height of CD (or vice-versa). CD is 9, AE is 1, and the area of a triangle is half the product of the height and the base. Thus, the area is 4.5.

ABE and DCE have two of the same angles (this we see by noting that CD and AB are parallel lines, thus the inside angle made by CB with them is the same). Thus they have three of the same angles. Thus they are similar: each is a scalar factor of the other, rotated, transposed, etc.

Lines CD and AB correspond, in these triangles, because they fall between equal angles in similar triangles. Thus the ratio between them is equal to the ratio between any two corresponding sides in these triangles. The ratio is 3:1, so ED:AE is 3:1; because their sum is four, AE=1 and ED=3.

Then the area of AEC is 3*1=3 (3 altitude from AE to C; base 1 AE).
This is also true, except for the last line. Why you're multiplying ED and AE I don't know, but it's wrong.

[JeffreyToTheMax]

This is getting extremely frustrating. I may need to reconsider my major if I can't comprehend simple math. I guess I'll work on this more tomorrow. Hopefully someone will be online to help me then.

[RS14]

You might try this PDF; it may explain it better than we can.

http://www.artofproblemsolving.com/Books/IntroGeom/exc1.pdf

[drakir_nosslin]

On questions such as no.5
'In a class of 78 students 41 are taking French, 22 are taking German and 9 students are taking both French and German. How many students are not enrolled in either course?'

These kinds of questions are simplified a lot with the help of the following diagram:
http://i352.photobucket.com/albums/r352/drakir_nosslin/Fun%20pics/circles.jpg?t=1268003849
One circle is for the French group, one is for the German, the number in between is the number of students who are in both classes. From there you just need to subtract the number in the middle from the numbers in the big circles, subtract the sum of all three from the starting number and there you go! I find it a lot easier to solve the problem if I have more than just numbers, might work for you as well.

[snoopy13a]

This is getting extremely frustrating. I may need to reconsider my major if I can't comprehend simple math. I guess I'll work on this more tomorrow. Hopefully someone will be online to help me then.

Don't beat yourself up too much for not remembering obscure points from 10th grade math.

I really, really recommend going to the local Barnes and Noble and picking up a SAT review book and just do practice tests non-stop for five days. I realize that they cost about $30 or so but the alternative could be wasting the money you spent on the SAT.

[Ashtar]

The only answer to math is more math. I had to put in about 300 hours (30 days @ 10 hours a day) to score a 5.5 (out of 6) in my junior year at university.

I hope you can get the motivation high and remember just to put in those hours until it becomes: "Oh yes, I saw that problem before!".

My trick was to create a "recipe book" with the most common approaches to the problems along with 1-2 examples per page. That worked really well. Even if it wasn't allowed at the exam, just creating it and extracting the knowledge necessary for it was worth it.

Wish you all the best.

[Tirian]

With some of them, I can figure it out given enough time. But it's not the way you're supposed to do it. Even though I know how to get the answer, eventually, that's a far cry from being able to work it out in less than a minute, and the process for doing so is likely entirely different.

I don't want to get all Zen or Jedi wisdom on you, but it's the calculator that's killing your time here. I will go so far as to suspect that the questions you are quoting are specifically designed to punish someone who doesn't take a moment to see the math that's behind the numbers. Which, speaking as a math major, is a nice trick to play on you, because math becomes less and less about numbers the higher you go. (Maybe it stays about numbers for engineers and physicists, but I'm not sure I would bet on it.)

Take your time on the problems at first, and try to see behind that curtain. For instance, when I saw the metallic cube problem, my thought was "Okay, the bigger cube has doubled size, so the volume is 2^3 = 8 times the original cube. The weight of the original cube is 6 lbs, so the weight of the larger cube is 6x8=48 pounds." So, as long as you keep your head about you, you can do it without a calculator or scratch paper.

It takes time to develop that skill, so definitely dive into as many sample problems as you can. But it is a skill that will pay off every day as a math student and that you will get better at every day.

[toasty]

I don't want to get all Zen or Jedi wisdom on you, but it's the calculator that's killing your time here. I will go so far as to suspect that the questions you are quoting are specifically designed to punish someone who doesn't take a moment to see the math that's behind the numbers. Which, speaking as a math major, is a nice trick to play on you, because math becomes less and less about numbers the higher you go. (Maybe it stays about numbers for engineers and physicists, but I'm not sure I would bet on it.)

Take your time on the problems at first, and try to see behind that curtain. For instance, when I saw the metallic cube problem, my thought was "Okay, the bigger cube has doubled size, so the volume is 2^3 = 8 times the original cube. The weight of the original cube is 6 lbs, so the weight of the larger cube is 6x8=48 pounds." So, as long as you keep your head about you, you can do it without a calculator or scratch paper.

It takes time to develop that skill, so definitely dive into as many sample problems as you can. But it is a skill that will pay off every day as a math student and that you will get better at every day.

This. This this this a thousand times this. My math teacher has hammered into my head all year (my senior year in US Highschool). The numbers do NOT matter, the methods behind the numbers, the formulas, the functions, they matter. Learn the logic behind the functions and formulas so you can create your own or come up with your own and then you will be fine.

This is what the SAT math section picks up on. It doesn't test your skills at moving around numbers, it makes you look at a problem and instantly reocongize the logic behind the problem. The math IS NOT HARD. Its Sophmore/Junior Level Highschool Math (that is to say: if you have a 2 year degree all this should be familiar to you). However, they state questions in such a way that you need to look beyond the numbers and directly into logic.

However, seeing as I couldn't do that very well... good luck with that. :smallbiggrin: (Seriously, my math score was pretty weak, honestly).

[Mando Knight]

This. This this this a thousand times this. My math teacher has hammered into my head all year (my senior year in US Highschool). The numbers do NOT matter, the methods behind the numbers, the formulas, the functions, they matter. Learn the logic behind the functions and formulas so you can create your own or come up with your own and then you will be fine.

Even the formulas and functions do not matter, the principles behind them do. What trips students up the most in higher-level courses is looking for a single, simple equation they can plug their numbers into to find the answer rather than trying to understand what the formula is. Understand what something is, and you will know what to do with it.

[Joran]

One suggestion I have, outside of doing lots of practice problems, is to go through all the questions and answer the ones you know you can do quickly first. Skip over the questions you know that will take you longer to do until you finish all the quick and easy ones first. Then go back and do the harder ones.

I find that I usually have time to brute force the hardest ones; worst case scenario, you end up with more questions answered than if you did them sequentially. If something is taking you too long, move on to another problem and let your subconscious work it over a little more.

[Griever]

This is getting extremely frustrating. I may need to reconsider my major if I can't comprehend simple math. I guess I'll work on this more tomorrow. Hopefully someone will be online to help me then.

Coming from an Electrical Engineering Major, Computer Science really never gets up there in the difficult maths; a lot of their stuff that I've studied and seen being things like Boolean Algebra and such, which has little to no correlation to plain ol' Algebra.

So, for my two cents- stick with it! Algebra and Trig are boring anyway, Calculus and Differential Equations are the fun courses!

[Joran]

Coming from an Electrical Engineering Major, Computer Science really never gets up there in the difficult maths; a lot of their stuff that I've studied and seen being things like Boolean Algebra and such, which has little to no correlation to plain ol' Algebra.

So, for my two cents- stick with it! Algebra and Trig are boring anyway, Calculus and Differential Equations are the fun courses!

Yeah, C.S. doesn't require too much math. At my school we were required to take Calculus, Statistics, and Boolean algebra. I can say, I've yet to use Calculus in my computer studies; most of the math has been of the Boolean and Statistical variety.

[Douglas]

Here are some examples from a practice test I found.

The first two questions were trivial. 4/7 is greater than 1/2, and 5 feet per second equals 18,000 feet per hour. Elementary stuff so far.

Then comes the following:

3. What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

A. 90
B. 95
C. 100
D. 105
E. 110
It's an arithmetic (that is, each number is the previous plus a constant) sequence of numbers. The average of the whole lot of them is the same as the average of the first and last two. 10+190 / 2 = 100.


4. A cubical block of metal weighs 6 pounds. How much will another cube of the same metal weigh if its sides are twice as long?

A. 48
B. 32
C. 24
D. 18
E. 12
You have an object, and the object gets doubled - three times, once for each dimension. 6 * 2*2*2 = 6 * 8 = 48.


5. In a class of 78 students 41 are taking French, 22 are taking German and 9 students are taking both French and German. How many students are not enrolled in either course?

A. 6
B. 15
C. 24
D. 33
E. 54
This one's been explained pretty well already.

(A diagram consisting of a circle, with three points labelled A, C, then B. Two curved lines link A to C, then C to B.)
9. Amy has to visit towns B and C in any order. The roads connecting these towns with her home are shown on the diagram. How many different routes can she take starting from A and returning to A, going through both B and C (but not more than once through each) and not travelling any road twice on the same trip?

A. 10
B. 8
C. 6
D. 4
E. 2
Ok, there are two possible orders here: A->B->C->A, and A->C->B->A. First, there is exactly 1 way to get directly from A to B. There are 2 ways to get directly from B to C, and 2 ways to get directly from C back to A. Thus, there are 1*2*2 = 4 possible routes for the first order. The second order is just the first one in reverse, and all roads can be traveled in both directions, so that also has 4 possible routes. 4+4 = 8.


Okay, I thought I could figure this last one out, but I can't seem to do it without resorting to time-burning trigonometry. I really hope that's not the only way to do it. Surely there's something I'm missing:

http://www.majortests.com/sat/testpics/p001-10.gif
10. In the figure above AD = 4, AB = 3 and CD = 9. What is the area of triangle AEC ?

A. 18
B. 13.5
C. 9
D. 4.5
E. 3

Thanks to my good friend Pythagoras, the length of AC is simple to get. I just need to get the length of AE somehow.
CB is a straight line. E is 9/(9+3) of the way along it horizontally from C, so it is that same portion along it vertically from C. AD is 4, so E is 3 from D and 1 from A. AEC is ADC - EDC, and figuring out the area from there should be simple. 4.5

[JeffreyToTheMax]

I don't want to get all Zen or Jedi wisdom on you, but it's the calculator that's killing your time here. I will go so far as to suspect that the questions you are quoting are specifically designed to punish someone who doesn't take a moment to see the math that's behind the numbers. Which, speaking as a math major, is a nice trick to play on you, because math becomes less and less about numbers the higher you go. (Maybe it stays about numbers for engineers and physicists, but I'm not sure I would bet on it.)

I wish someone had told me about this years ago, although to be honest, I probably wouldn't have listened. Back then I was more concerned with all the awesome stuff my graphing calculator could do, rather than actually learning math. I guess that's why I'm in the tough spot I am now.

My earlier venting aside, I really appreciate all the help everyone has been giving me. This is getting a little easier as I work at it. I'm still not 100% sure regarding the triangle issue, but rather than get hung up on it, I feel I should blaze ahead and learn as much as I can before the test.

Right now, if the SAT version I take looks almost exactly like that first practice test, I peg myself at a solid 90%. Unfortunately, when I loaded another practice test (#12 for some reason), I noticed that they had the gall to use questions that were different. So, my journey is not yet complete. I could figure out most of them, but here are the ones I'm not sure about:

5. The total weight of a tin and the cookies it contains is 2 pounds. After ¾ of the cookies are eaten, the tin and the remaining cookies weigh 0.8 pounds. What is the weight of the empty tin in pounds?

A. 0.2
B. 0.3
C. 0.4
D. 0.5
E. 0.6

http://www.majortests.com/sat/testpics/p012-6.gif
6. Which of the following pairs of angles must be equal?

A. a and e only
B. a and e, and c and d only
C. c and d only
D. d and e only
E. c and d and d and e only

http://www.majortests.com/sat/testpics/p012-7.gif
8. Refer to the chart from the previous question.
If median bonus amount = m, mean bonus amount = n, and modal bonus amount = p, which of the following represents the correct ordering of m, n and p?

A. m < n < p
B. m < n = p
C. m = p
D. p < m < n
E. p = n < m

9. Which of the following is the equation of a line passing through the origin and parallel to the line 2x – y = 5?

A. 5x – y = 0
B. 2x – y = 0
C. 2x + y = 5
D. 2x + y = 0
E. x + 2y = 0

http://www.majortests.com/sat/testpics/p012-10.gif
10. The graph shows a quadratic function with a minimum at (-1,1)
If f(a) = f(1), which if the following could be the value of a?

A. 2
B. 0
C. -1
D. -2
E. -3

I really hope this helps. I'm trying to use my calculator as little as possible, so as long as the actual SAT doesn't use entirely different and unrelated questions, I should be okay.

[snoopy13a]

[QUOTE=JeffreyToTheMax;8037389]

5. The total weight of a tin and the cookies it contains is 2 pounds. After ¾ of the cookies are eaten, the tin and the remaining cookies weigh 0.8 pounds. What is the weight of the empty tin in pounds?

A. 0.2
B. 0.3
C. 0.4
D. 0.5
E. 0.6

3/4 of the cookies are 1.2 pounds. So all of the cookies are 1.6 pounds and the tin must be 0.4 pounds.

[
http://www.majortests.com/sat/testpics/p012-7.gif
8. Refer to the chart from the previous question.
If median bonus amount = m, mean bonus amount = n, and modal bonus amount = p, which of the following represents the correct ordering of m, n and p?

A. m < n < p
B. m < n = p
C. m = p
D. p < m < n
E. p = n < m

Mode is 100
Median is 100 (there's 37 at 100, 7. below and 6 above so the median one must be 100

You could figure out the mean via calculator but as a short cut, I'll say that the sum is going to be subtracted by 350 (7*50) and added to by 400 (4*50 + 2*100). So the mean is going be something like $101.

Actually you don't even need to figure out the mean because C, (median = mode) is a correct answer.

9. Which of the following is the equation of a line passing through the origin and parallel to the line 2x – y = 5?

A. 5x – y = 0
B. 2x – y = 0
C. 2x + y = 5
D. 2x + y = 0
E. x + 2y = 0

For lines, write them as y = mx +b
So, your equation becomes:
y= 2x - 5

So, a parallel line will need a slope of 2 and pass through 0,0
Thus, answer should be y= 2X + 0 which is equivalent to B

[Douglas]

5. The total weight of a tin and the cookies it contains is 2 pounds. After ¾ of the cookies are eaten, the tin and the remaining cookies weigh 0.8 pounds. What is the weight of the empty tin in pounds?

A. 0.2
B. 0.3
C. 0.4
D. 0.5
E. 0.6
This is basic algebra in word form. Let's call the weight of the tin T, and the weight of the cookies C. The first sentence tells you T + C = 2. The second sentence tells you T + C/4 = .8. There are several ways you can proceed from here. The most generally applicable route is 1) pick an equation and solve for one of the variables in that equation, so something like T + C = 2 changes into C = 2 - T; 2) Substitute the result of step one into the other equation. With this example, take "T + C/4 = .8" and replace C with "2 - T". This gets you "T + (2 - T)/4 = .8"; 3) Solve the new equation for the remaining variable. For this example, it looks like this:
T + (2-T)/4 = .8 Get rid of the division on the left by multiplying by 4
4T + 2 - T = 3.2 Combine the T terms
3T + 2 = 3.2 Remove the non-T part of the left side by subtracting 2
3T = 1.2 Get rid of that 3 by dividing by 3
T = .4


http://www.majortests.com/sat/testpics/p012-6.gif
6. Which of the following pairs of angles must be equal?

A. a and e only
B. a and e, and c and d only
C. c and d only
D. d and e only
E. c and d and d and e only
This one I can only solve by eliminating the obviously wrong answers. Just looking at the diagram, d and e are obviously not the same. That eliminates options D and E. Now, if you look at the a/e and c/d combinations, you should notice that they are analogous to each other. Without numbers to go by, any reasoning that applies to one must clearly also apply to the other. That eliminates options A and C. The answer must be B.


http://www.majortests.com/sat/testpics/p012-7.gif
8. Refer to the chart from the previous question.
If median bonus amount = m, mean bonus amount = n, and modal bonus amount = p, which of the following represents the correct ordering of m, n and p?

A. m < n < p
B. m < n = p
C. m = p
D. p < m < n
E. p = n < m
This seems primarily a test of whether you know the definitions of those terms and how to evaluate them. Median is the one where if you line up all the numbers in order, the median would be exactly in the center of the line. With a problem like this, you can find it by removing numbers from both ends until you get to the middle. Take off the 2 of the top category and 2 from the bottom. That leaves 5 in the bottom. Take off the 4 in the 150 category and 4 from the bottom, leaving 1 in the bottom. Take off that last 1 and 1 from the 100 category, and you're down to all 100's. The median is 100. Alternatively, you could just recognize that the 100 category is so much bigger than the rest that it has to contain the median.

The mean is the most common definition of "average". Take all the numbers, add them up, and divide by how many there are. 50*7 + 100*37 + 150*4 + 200*2 = 5050. 7+37+4+2 = 50. 5050/50 = 101.

The mode is the easiest of all to find, it's just which number is most common. There are more at 100 than anywhere else, so the mode is 100.

So, median and mode are both 100 and equal, while mean is 101 and greater than the rest. The option that describes that is E.


9. Which of the following is the equation of a line passing through the origin and parallel to the line 2x – y = 5?

A. 5x – y = 0
B. 2x – y = 0
C. 2x + y = 5
D. 2x + y = 0
E. x + 2y = 0
"Passing through the origin" means the equation is correct when x and y are both 0. That knocks out C. Parallel happens when the ratio of coefficients of x and y are the same in both equations. They made this one easy by keeping the coefficients exactly the same. It's B.


http://www.majortests.com/sat/testpics/p012-10.gif
10. The graph shows a quadratic function with a minimum at (-1,1)
If f(a) = f(1), which if the following could be the value of a?

A. 2
B. 0
C. -1
D. -2
E. -3
The graph is showing f(a) on the vertical axis and a on the horizontal axis. They're telling you that f(a), how far up or down the point is, is the same as where a = 1. Conceptually, find the point on the line where a = 1. Draw a horizontal line through this point. They are asking for the value of a at the other point where your horizontal line crosses their line. Now, they told you it's a quadratic function, and it has a minimum at a = -1. Quadratic functions, also called parabolas, are completely symmetric. In this case, it is completely symmetric around a vertical line at a = -1. So, the point you started with is at a=1, which is 2 to the right of the line of symmetry. The other point where your horizontal line crosses the parabola is 2 to the left of the line of symmetry. -1 - 2 = -3. The answer is E.

[RS14]

5. The total weight of a tin and the cookies it contains is 2 pounds. After ¾ of the cookies are eaten, the tin and the remaining cookies weigh 0.8 pounds. What is the weight of the empty tin in pounds?

A. 0.2
B. 0.3
C. 0.4
D. 0.5
E. 0.6

Weight of tin: t
Weight of cookies: c

e1: c+t=2
e2: c(1/4)+t=0.8

Gaussian Elimination:

Multiply e2 by 4.

e1: c+t=2
e2: c+4t=3.2

Subtract e1 from e2.

e1: c+t=2
e2: 3t=1.2

Thus t=0.4, from e2. (Plugging this into e1, c=1.6, but this is not needed).



http://www.majortests.com/sat/testpics/p012-6.gif
6. Which of the following pairs of angles must be equal?

A. a and e only
B. a and e, and c and d only
C. c and d only
D. d and e only
E. c and d and d and e only

Interesting. :smallconfused: (this took me substantially more than a minute, though I could have guessed correctly fairly quickly.)

Use a process of elimination: eliminate four incorrect answers by thinking of counterexamples. Symmetry helps. You can eliminate answers that are not symmetrical immediately, (i.e. A, C, because if a=e, d=c by symmetry).

Name the points D,E,B clockwise from the top left.

Answers D and E is eliminated by letting E be very close to B, so D and E are obviously very different; e.g. E~=90, D~=0.

Not a proof that B is correct, but the best way to do it on the SAT.




http://www.majortests.com/sat/testpics/p012-7.gif
8. Refer to the chart from the previous question.
If median bonus amount = m, mean bonus amount = n, and modal bonus amount = p, which of the following represents the correct ordering of m, n and p?

A. m < n < p
B. m < n = p
C. m = p
D. p < m < n
E. p = n < m

Mode: most common. p=100, because that company has the most employees.
Mean: (50*7+100*37+150*4+200*2)/(7+37+4+2). Use a calculator.
n=101
Median: Order the bonuses, e.g. {50,50,50...100,100,100...,150,150,150...,200,200}
Eliminate the equal numbers of extreme values from the ends. E.g. elimiante (50,200), (50,200), (50,150)...

We can simply note that the seven highest and seven lowest, when eliminated, are everything but the $100 elements. Thus m=100.




9. Which of the following is the equation of a line passing through the origin and parallel to the line 2x – y = 5?

A. 5x – y = 0
B. 2x – y = 0
C. 2x + y = 5
D. 2x + y = 0
E. x + 2y = 0

To be parallel, the ratio of the coefficients of x and y must be the same. Thus 2x-y=c is parallel, as is 4x-2y=c.

To pass through the origin, (0,0), when plugged in, must satisfy the equation. This happens only if 2x-y=0.



http://www.majortests.com/sat/testpics/p012-10.gif
10. The graph shows a quadratic function with a minimum at (-1,1)
If f(a) = f(1), which if the following could be the value of a?

A. 2
B. 0
C. -1
D. -2
E. -3

In a quadratic, the two prongs rise/sink equally about the minimum/maximum. Draw a horizontal line across the parabola from one arm to the other. The midpoint will be directly above the minimum. Thus the distance from -1 to 1 is the same as the distance from a to -1. Thus a=-3.

[this is at best a proof sketch; it's good enough for the SAT because doing it thoroughly takes more time.]

[JeffreyToTheMax]

It may be time to give up on this. I barely understand the things that everyone has been kind enough to guide me through step by step, and there are a huge number of other possible questions that I have no idea where to even begin with. On my first attempt, I got 37 questions right, missed 12, and omitted 5. Only in the best possible of circumstances could I see myself getting anywhere near that this time.

I wish I could buy the official SAT prep book, but I have literally no money. This whole thing was a last-ditch effort to make it to college, and now I wonder if I even belong there. I have a two year transfer degree from a community college, and that is worth exactly nothing. Even minimum wage jobs are beyond my ability since I have the people skills of a brick wall.

I know I'm kind of hijacking my own thread here, but even if a miracle happens and I do well enough on the SAT that others take notice, that doesn't do anything to resolve my immediate situation. I guess the only thing I can do right now is keep plugging away at this. I'll take another run at it tomorrow.

[snoopy13a]

It may be time to give up on this. I barely understand the things that everyone has been kind enough to guide me through step by step, and there are a huge number of other possible questions that I have no idea where to even begin with. On my first attempt, I got 37 questions right, missed 12, and omitted 5. Only in the best possible of circumstances could I see myself getting anywhere near that this time.

I wish I could buy the official SAT prep book, but I have literally no money. This whole thing was a last-ditch effort to make it to college, and now I wonder if I even belong there. I have a two year transfer degree from a community college, and that is worth exactly nothing. Even minimum wage jobs are beyond my ability since I have the people skills of a brick wall.

I know I'm kind of hijacking my own thread here, but even if a miracle happens and I do well enough on the SAT that others take notice, that doesn't do anything to resolve my immediate situation. I guess the only thing I can do right now is keep plugging away at this. I'll take another run at it tomorrow.

You almost certainly can transfer to a college with your community college grades. I'd look at local state colleges and talk to their admission people to see what they look for in community college grades, SAT scores, and GPA. After you find a school that's a good match, you canresearch federal loans to help pay for it. It may be too late now to get financial aid for the fall so you might have to work for a few months and apply for January.