POST all difficult sat MATH problems!

Discus: SAT/ACT Tests and Test Preparation: June 2004 Archive: POST all difficult sat MATH problems!
 By Whtx (Whtx) on Wednesday, May 26, 2004 - 08:51 pm: Edit

Hi guys,

I'm sure we are all studying for the upcoming SAT and i think it'd be best to get a thread going that deals with all the difficult and tricky math problems. THe type of problems that is keeping you at 720 instead of 800.

Anyone got any problems or tricks or advice to share about these tricky problems? Thanks!

 By Mattlord (Mattlord) on Wednesday, May 26, 2004 - 11:51 pm: Edit

buhp! (bring up his post!)

 By Gameguy56 (Gameguy56) on Thursday, May 27, 2004 - 12:01 am: Edit

ok, this one bugs the h3ll out of me
"If three different circles are drawn on a peice of paper, at most how many points can be common to all three"

A.None
B 1
C 2
D 3
E 6

I can't see a mental picture of 3 circles in my head that have 2 comon points between all three

 By Eecs (Eecs) on Thursday, May 27, 2004 - 12:21 am: Edit

the most amount of points that can be common to two distinct circles is two. so if you add in a third circle, the common points between all three can't be more than 2. so just draw a third circle going thru those 2 common points from the first two circles

 By Gameguy56 (Gameguy56) on Thursday, May 27, 2004 - 03:27 pm: Edit

ahh, thanks Eecs, that one bugged me, but now I see how that works

 By Benzinspeicher (Benzinspeicher) on Friday, May 28, 2004 - 04:23 pm: Edit

Here are a few that I had fun doing:
A two digit number is flipped around, and the result is 45 greater than the original. What is the greatest possible value of the original number?

You stand in line. You notice that there are 9 more people in front than in behind. You also see that the total number of people in the line is 3 times the number of people behind. How many in the line?

An equilateral triangle of sides 12 is inscribed in a circle. Find the area of the circle.

Enjoy doing these!

 By Gastudent123 (Gastudent123) on Friday, May 28, 2004 - 04:41 pm: Edit

1) 94
2) F= B+9, F+B+1=3B
substitute to get 2B+10=3B, which results in B=10 and F=19. Then, add 10+19+1(for yourself) to get 30
3) 48pi

 By Benzinspeicher (Benzinspeicher) on Friday, May 28, 2004 - 10:53 pm: Edit

nailed them biatches!
Here's a few more that are fun:
1. A family counselor went to help a drugged up, and drunken family. One child said "I have as many brothers as sisters." Another said "I have twice as many brothers as sisters." How many children are there?

2. Montgomery's birthday took place on a Friday last year. Wayland's birthday took place 150 days later. How many Sundays passed between the birthdays?

 By Conker (Conker) on Friday, May 28, 2004 - 11:26 pm: Edit

Actually, #1 should 49. The flipped number must be greater than the original number. But good work with #2! I totally forgot to consider the person standing in line.

Anyway the new problems...

#1: 7
#2: 22

 By Optimizerdad (Optimizerdad) on Friday, May 28, 2004 - 11:27 pm: Edit

1. 3 girls + 4 boys = 7 children
2. 22

 By Tootall (Tootall) on Saturday, May 29, 2004 - 12:40 am: Edit

good job guys, i like math. It's so fun!!!!!!!!!!!1

 By Scion (Scion) on Saturday, May 29, 2004 - 02:59 pm: Edit

An equilateral triangle of sides 12 is inscribed in a circle. Find the area of the circle.

A family counselor went to help a drugged up, and drunken family. One child said "I have as many brothers as sisters." Another said "I have twice as many brothers as sisters." How many children are there?

 By Benzinspeicher (Benzinspeicher) on Saturday, May 29, 2004 - 03:33 pm: Edit

Ok, the triangle problem:
From the center of the circle, draw a line to each vertex of the triangle, which is the radius. Since there are three lines, and therefore three sektors of the circle, each sektor should be 360/3=120 deg. in measure. Next, draw a line from the center of the circle to the midpoint of each side of the triange. Now you should have six 30-60-90 triangles (The 60 because it bisects the 120, the 90 because of the midpoint line to center is perpendicular to a triangle side). Take any of those triangles. The side opposite the 60 should be 6. Use the 30-60-90 triangle identity to solve for the hypotenuse, which is the radius. To do this, first solve for the side across the 30, which would be 6/(sqrt3). Next, for the hypotenuse, multiply this by 2, and get 12/(sqrt3). Now, you have radius. Use pi x r^2, which would be (144x pi)/3, or 48pi.

 By Benzinspeicher (Benzinspeicher) on Saturday, May 29, 2004 - 03:40 pm: Edit

Next, the case of the screwed up family. The first child says that he/she has just as many brothers as sisters. The next child says that he/she has twice as many brothers as sisters. From this, I can figure that there has to be more brothers. I can also see that the first statement is said by a boy, seeing as how he has less brothers because he excludes himself. Because he excludes himself, I can say that there is one more boy than girl, from the first statement. I write this into an equation: b=g+1. Next, the second statement is said by a girl, because a boy would repeat the first statement. So I use this statement, remembering that she excludes herself, to write a second equation: 2(g-1)=b, and I substitue for "b": 2(g-1)=g+1. I solve for "g" and I get g=3. Remember, b=g+1, so b=4. So 3 girls and 4 boys, so 7 children.

 By Ilovefood (Ilovefood) on Saturday, May 29, 2004 - 07:35 pm: Edit

I would say that there could be any number of children if the family is as drugged up as I imagine; the kids could be so f***ed up that they don't actually know who is real and who is a hallucination. Trippy...

 By Apocalypse_Now (Apocalypse_Now) on Saturday, May 29, 2004 - 08:36 pm: Edit

The decimal representation of m/n, where m and n are relatively prime positive integers and m<n, contains the digits 2, 5, and 1 consecutively, and in that order. Find the smallest value of n for which this is possible.

Tricky bugger...

 By Apocalypse_Now (Apocalypse_Now) on Sunday, May 30, 2004 - 10:35 am: Edit

buhp

 By Conker (Conker) on Sunday, May 30, 2004 - 10:32 pm: Edit

Here's a tough geometry problem (although it's probably only because I suck at geometry):

There is a triangle PQR. M and N are the midpoints of PQ and PR, respectively. A line is drawn from M to R, and from N to Q. The point where these two lines intersect is labeled S. Find the ratio of triangle MNS to triangle PQR.

 By Number9 (Number9) on Sunday, May 30, 2004 - 10:37 pm: Edit

PLease explain how I would go about this problem: 2. Montgomery's birthday took place on a Friday last year. Wayland's birthday took place 150 days later. How many Sundays passed between the birthdays?

Have I mentioned that I am horrid at math?

 By Benzinspeicher (Benzinspeicher) on Sunday, May 30, 2004 - 11:29 pm: Edit

That geometry problem: is it 1:8?
Number 9: That 150 day problem. Ok, I divide the number of days, by the number of weeks 150 by 7, and I take the quotient only, which is 21. That's the number of weeks that pass by. So if 21 weeks pass by, I can say that Sunday happens 21 times. But now, the remainder from 150/7 is 3. That means that after the 21 weeks, there's another 3 days until Wayland's birthday. So from Friday (21 weeks later), I count another 3 days, and Sunday passes again. 21+1=22 Sundays.
Apocalypse Now: I am working on that tough problem. No way that is an SAT problem, is it? That seem more AIME like. Anyway, good problem though.

 By Benzinspeicher (Benzinspeicher) on Sunday, May 30, 2004 - 11:29 pm: Edit

That geometry problem: is it 1:8?
Number 9: That 150 day problem. Ok, I divide the number of days, by the number of weeks: 150 by 7, and I take the quotient only, which is 21. That's the number of weeks that pass by. So if 21 weeks pass by, I can say that Sunday happens 21 times. But now, the remainder from 150/7 is 3. That means that after the 21 weeks, there's another 3 days until Wayland's birthday. So from Friday (21 weeks later), I count another 3 days, and Sunday passes again. 21+1=22 Sundays.
Apocalypse Now: I am working on that tough problem. No way that is an SAT problem, is it? That seem more AIME like. Anyway, good problem though.

 By Billiam2 (Billiam2) on Sunday, May 30, 2004 - 11:40 pm: Edit

That's a real jackass thing to do. These kids are trying to study for the SAT math, and I don't know what you hope to accomplish by posting #14 from the 2003 AIME.

 By Optimizerdad (Optimizerdad) on Monday, May 31, 2004 - 12:12 am: Edit

OK, I cheated :-). I wrote a small 8_line program to search all (j/i) values, with j < i, upto i=1000. The smallest denominator I get is 127, from
j/i = 32/127 = 0.251969

It'll take a better mathematician than me to
prove that this is right, if it is. Any takers?

 By Conker (Conker) on Monday, May 31, 2004 - 04:42 am: Edit

"That geometry problem: is it 1:8?"

Nope. Try again.

 By Optimizerdad (Optimizerdad) on Monday, May 31, 2004 - 08:36 am: Edit

Conker:
You get two similar triangles (the inner triangle is an inverted version of the large outer triangle, with each of its sides having half the length of the corresponding side in the larger triangle). Since a triangle's area is
0.5 * base * height
the new area would be (1/4) the area of the original triangle.

 By Morgan (Morgan) on Monday, May 31, 2004 - 12:07 pm: Edit

some more tricky problems

1 How many non-congruent triangles with perimeter 7 have integer side length
A 1
B 2
c 3
d 4
e 5

2 r,s,t,v are different positive integers.
What is the value of r+s+t+v if:
(10-r)(10-s)(10-t)(10-v)=9
A 8
B 22
C 28
d 32
e 40

 By Qwert271 (Qwert271) on Monday, May 31, 2004 - 12:21 pm: Edit

1) B. 2,2,3; 1,3,3. No other combination will work because of the rule that the sum of any two side has to be greater than the third i.e. 5,1,1 won't work because 1+1 is less than 5.

2) E. Call each of the factorized terms i1, i2, i3, i4. The product of them equals 9. So, if the product of four distinct integers is 9, come up with the integers (you will see that there is only one combo). Call them i1 = -1, i2 = 1, i3 = -3, i4 = 3. Thus r = 11, s = 9, t =13, v = 7. Sum = 40 = E

 By Morgan (Morgan) on Monday, May 31, 2004 - 12:29 pm: Edit

good job, that's right.

 By Qwert271 (Qwert271) on Monday, May 31, 2004 - 01:14 pm: Edit

Not atall.

 By Benzinspeicher (Benzinspeicher) on Monday, May 31, 2004 - 03:36 pm: Edit

Yes, they are right. The first triangle question. Those are the only two that work. And that second one, you need two neg integers, and two pos integers: 3, -3, 1, -1. Then simply figure out how much from ten you would have to subtract to get those values. Then add them.

 By Qwert271 (Qwert271) on Monday, May 31, 2004 - 04:11 pm: Edit

I know. I did the problem. "Not atall" is an expression that means a combination of "you're welcome" and "glad to be of assistance."

 By Gameguy56 (Gameguy56) on Monday, May 31, 2004 - 04:44 pm: Edit

A flock of geese are on a pond and being observed continuosly. At 1:00 PM 1/5 of the geese flew away. at 2:00 PM 1/8 of the geese that remained flew away. At 3:00 PM, 3 times as many geese as had flown away at 1:00 PM flew away, leaving 28 geese. at no other times did geese arrive or fly away. How many geese were in the origional flock?

how do you do this one?

 By Optimizerdad (Optimizerdad) on Monday, May 31, 2004 - 06:23 pm: Edit

It's algebra, and not too complex. If the original #geese is x, we have

Time
1:00 #flying_away= 0.2x, and #remaining = x - 0.2x = 0.8x
2:00 #flying_away=(1/8)(0.8x)= 0.1x, and #remaining = 0.8x - 0.1x = 0.7x
3:00 #flying_away=(3)(0.2x)= 0.6x, and #remaining= 0.7x - 0.6x = 0.1x

Final #geese remaining = 0.1x = 28
so x = 280

 By Whtx (Whtx) on Wednesday, June 02, 2004 - 11:30 pm: Edit

Hey guys what the hell? Some of these are like brain teasers or from other math trivia questions.

Please post more SAT math questions that are hard/tricky. Thanks

 By Haithman (Haithman) on Wednesday, June 02, 2004 - 11:42 pm: Edit

These are SAT questions...

 By Kewlkiwi102 (Kewlkiwi102) on Thursday, June 03, 2004 - 12:09 am: Edit

For the goose one, if you wanted to put it all into one equation, x being the original number of geese:

(x-x/5)-(1/8)(x-x/5)-3(x/5)=28
0.1x=28
x=280

 By Scion (Scion) on Thursday, June 03, 2004 - 12:10 am: Edit

can u please reexplain this one? thanks

r,s,t,v are different positive integers.
What is the value of r+s+t+v if:
(10-r)(10-s)(10-t)(10-v)=9
A 8
B 22
C 28
d 32
e 40

 By Xiggi (Xiggi) on Thursday, June 03, 2004 - 12:52 am: Edit

As always for the SAT1, using algebra yields the correct answer but does cost you a precious commodity time, and introduce a the risk of silly errors.

For a problem like the geese one, as soon as you see mentions of 1/5, 1/8 etc, switch to a 100 percent mode.

100% minus 20% leaves 80%
80% minus 10% leaves 70%
70% minus 50% leaves ... 10%

If 10% of the original group is now 28, the total was 28 X 10 or 280.

Time to solve? 20 seconds, no calculator, just you sharp pencil and a sharp mind, and best of all no mathematical error possible. That is what the SAT is all about.

Remember that the more convoluted a problem seems the SIMPLER the answer will be. Actually, you could have almost guessed that the answer would be 280 as soon as you saw 28. However, using 100 or 100% as beginning plug is so much simpler.

Simplicity is also key for the problem with the four brackets. The explanation given above is absolutely correct. However, the only thing you need to realize is that the numbers that compose the brackets can only be -1, +1, +3, and -3. Like stated above, that is the ONLY combination that could work. From there on, the best way to write it up is:

r = 10 + 3 = 13
s = 10 - 3 = 7
t = 10 - 1 = 9
v = 10 + 1 = 11

Add them up and that gives 40. However, an astute SAT test taker will have realized that the total of -1, +1, +3, -3 is simply ZERO, so the answer is really 10+10+10+10 or 40. Again, a very simple answer to a seemingly convoluted problem.

 By Morgan (Morgan) on Thursday, June 03, 2004 - 01:00 am: Edit

(1)*(3)*(-3)*(-1)=9
(10-r)(10-s)(10-t)(10-v)=9
just set 10-r=1 10-s=3 10-t=-3 10-v=-1
after you solve those equestions, you get r=9, s=7, v=11, t=13, then added them up.
r+s+t+v=11+7+13+9=40

 By Morgan (Morgan) on Thursday, June 03, 2004 - 01:19 am: Edit

loops, I didn't see xiggi alreday explained, but anyways, here are two more problems if your need more practice.

1) in a league with 6 teams, each team is scheduled to play each other team exactly twice. what is the total number of games that are scheduled in the league.

2)How many positive integers less than 1001 are divisible by either 2 or 5 or both?

 By Scion (Scion) on Thursday, June 03, 2004 - 02:40 am: Edit

for the first one about the teams, the answer's 30...right? but if it's right, what the fastest way to get it?

 By Oasis (Oasis) on Thursday, June 03, 2004 - 04:14 am: Edit

1. The first team would play 10 games (2 games with the each of the remaining 5 teams), the 2nd team would play 8 games (2 games with each of the remaining 4 teams) and so on. So the answer is 10+8+6+4+2 = 30.

2. There are 500 integers that could be divided by 2 from 1-1001. There are 200 integers that could be divided by 5 from 1-1001. Since this included some repeats, take out the ones we counted twice (divisible by 2 and 5, which would be multiples of 10). There are 100 such integers from 1-1001. So 500+200-100 = 600.