SAT Math Questions

Discus: SAT/ACT Tests and Test Preparation: April 2004 Archive: SAT Math Questions
 By Ladybajan (Ladybajan) on Thursday, April 22, 2004 - 11:00 pm: Edit

I know you have to make them equations, but there are so many ways to solve them, that I got different answers each time. So please explain thoroughly.
1. In a game, all tokens of the same color are worth the same number of points. If one player won 2 red tokens and 4 blue tokens for a total score of 24 points, and another player won 3 red tokens and 2 blue tokens for a total score of 16 points, how many points is a blue token worth?

In this question, I know how to get the 3 and 7, I just don't know how to get the 5.
2. Points P, Q, and R lie in a plane. If the distance between P and Q is 5 and the distance between Q and R is 2, which of the following could be the distance between P and R?
I. 3
II. 5
III. 7
Answer? E) I, II, and III

This one I tried so many times and it is driving me crazy because it seems so easy, yet isn't.
3. A business is owned by 3 men and 1 woman, each of whom has an equal share. If one of the men sells 1/2 of his share to the woman, and another of the men keeps 2/3 of his share and sells the rest to the woman, what fraction of the business will the woman own?

This one is a grid-in question. I don't even understand why AD is 1.
4. Points A, B, C, and D lie on a line in that order. If AD/AC= 2/1 and AD/AB= 3/1, what is the value of AC/ BD?

Thanks in advance for all the help.

 By Brum (Brum) on Friday, April 23, 2004 - 07:54 am: Edit

can someone explain the law of sides of a triangle please?

ie is it: "the sum of the lengths of any 2 sides must be greater than (or greater than or equal to) the length of the 3rd side"?

 By Daromanian (Daromanian) on Friday, April 23, 2004 - 01:02 pm: Edit

2(value of red) + 4(value of blue) = 24 points
3(value of red) + 2(value of blue) = 16 points

double the second equation then subtract it from the first.
you get:
-4(value of red) = -8
(value of red) = 2
plug into first equation
4 + 4(value of blue) = 24
4(value of blue) = 20
(value of blue) = 5

for the second one you have a triangle with sides 5 and 2, so the last side can't be larger than 7, every size until zero works because the angles change.

ok beginning of 3rd one:
woman has 1/4 of shares
man1 has 1/4
man2 has 1/4
man3 has 1/4

then one man gives half of his (1/8) to the woman
woman has 3/8
man1 has 1/8
man2 has 1/4
man3 has 1/4

then another gives 1/3 of his shares (1/12)
woman has 11/24
man1 has 1/8
man2 has 2/12
man3 has 1/4

all it is is adding 1/4 + 1/8 + 1/12
= 11/24

last one, i dont reli understand it now but this might help?

2ac = 3ab
ab + bc + cd = ad

 By Xiggi (Xiggi) on Friday, April 23, 2004 - 01:15 pm: Edit

The third and fourth problems are solved easier by assigning simple values to the problems.

For the third one, sine you will deal with 1/2 and 1/3, pick a number like 6 and assign it to each owner. So, they start with 6 each. From there on, it is rather fast to add the original 6, 3 from the first man, and 2 from the second. Woman now owns 11 out of a total of 24 (4 times 6).

For the last problem, assign a value of the segment AD. Again based on the values offered (1/2 and 1/3) pick a number like 12. From there, simply claclulate the value of each segment. For instance is AD = 12, then AC = 12/2 or 6.

In this type of problem, it helps to draw a quick diagram and visualize the question at hand.

 By Fusiachi (Fusiachi) on Friday, April 23, 2004 - 04:00 pm: Edit

Relating to Daromanian's comment about the second problem about all values down to zero working, only every value from 3 to 7 is applicable. It cannot be less than 3 because then the side that is 5 would be larger than the other two. A straight line would leave 3 between the two, or the shortest distance.

Sorry, just being picky.

 By Shadow_Wolf (Shadow_Wolf) on Friday, April 23, 2004 - 05:47 pm: Edit

1. You just make 2 equations based on the 2 people...and the variables are R=Red and B=Blue. Soo..

Person 1: 2R + 4B = 24
Person 2: 3R + 2B = 16

Now, we have 2 equations and 2 unknowns...so we'll make the Rs drop out (because we want the value of B) by multiplying the 1st equation by 3 and the second equation by 2 giving us the result:

Person 1: 6R + 12B = 72
Person 2: 6R + 4B = 32

Now in order to get the Rs to drop out you will have to multiply one of the equations by -1. I chose to multiply the 2nd equation giving me the result:

Person 1: 6R + 12B = 72
Person 2: -6R -4B = -32

Now we add the equations by adding the terms that are on top of each other together. Like this:

(6R + -6R) = 0
(12B + -4B) = 8B
(72 + -32) = 40

So our new equation looks like this:

0 + 8B = 40

Now solve for B by dividing both sides by 8 and you are left with:

B=5

2. P, Q, and R could be the vertices of an isosceles triangle (PQR) with PQ=5, QR=2, PR=5.

3. The Woman starts with ¼ of the total company then the guy gives her 1/2 of his 1/4 or 1/8 (You get this by multiplying 1/2 * 1/4 in essence taking half of the 1/4 that he started out with). Making her total 1/4 + 1/8 = 3/8. Then the other guy gives her 1/3 of his 1/4 which is 1/12 (1/3 * ¼). Add this to the 3/8 that the woman already has and you get 11/24.

4. Well..I don't know if this is right but if you're talking ratios (AD/AC = 2 to 1) then I would just choose an arbitrary number. Ok, so for AD/AC = 2/1 I would choose a number for AD...lets say 6. Now AC would have to be 1/2 of that which would give you AC=3. Verify: AD/AC = 6/3=2/1 (So far so good).
OK, so now you do the same thing with AD/AB. Because AD is already equal to 6, then AB would have to be 1/3 that or 2. Verify: AD/AB = 6/2 = 3/1 (Good).
Our line is starting to look something like this
A----------D = 6
A-----C = 3
A--B = 2
Now we need to find AC/BD...judging by our representation (AC-AB = BC : 3-2 = 1 : BC=1) and (AD-AC = CD : 6-3 = 3 : CD=3). Therefore (BC+CD = BD : 1 + 3 = 4) BD = 4. And you already have AC = 3. So now your ratio of AC/BD = 3/4. I think that is the answer.

 By Gastudent123 (Gastudent123) on Friday, April 23, 2004 - 07:17 pm: Edit

I also got 3/4 for the last problem.

 By Ladybajan (Ladybajan) on Friday, April 23, 2004 - 08:08 pm: Edit

Oh yes, I am sorry, you're both right, the answer is 3/4. My father was kicking me off the computer so I was in a hurry and forgot the answer. Thanks everyone for all the helpful answers. I actually get all these tricky problems now.

 By Jooolia (Jooolia) on Monday, April 26, 2004 - 06:29 pm: Edit

wait a minute.... you're saying that these questions are the ones that 17-18 year olds spend so much time stressing over? this is what american universities base their admissions on? holy crap, your school system is •••••• up. the canadian system isnt that great, and still these questions are at an elementary school level, or simple grade 9 math at the most.
no offense to americans, it doesn't mean you're dumb, it just surprises me that this is supposedly a challenge. i don't know how anyone could think that the american school system is ahead of canada.

Trolls should watch their language. Have fun posting in Canada, Joolia!
Moderator Trinity

 By Justperfect (Justperfect) on Monday, April 26, 2004 - 10:11 pm: Edit

for the token question, a good idea for you since you couldnt solve it was to plug in the answers for the blue and check it with the others.this is pry the easiest way to get past mc questions. but this method only works on some mc. pluggin in the answers is basically trial and error. good luck "lady"