extremefob

i dont understand #5 on pg 408. does anyone know how to do it??!?!

AND is p 409 #8 a mistake? how can it possibly be B? (-.5)(.7ish) is nowhere close to -.8ish its more like -.35ish.

zach447

For #5, they're basically asking what's the probability that a > b. There's probably an easier way to do this using combinations or something, but I can't think of it:

So, if a = 6, then there are 5 possible numbers you can land on for b for a > b. If a = 5, there are 4 possibilities; if a = 4, there are 3 possibilities, and so on...

So your total probability that a > b is going to be a fraction (situations which would make the statement true over total possible situations). Since these are the possibilities that are going to make it true, you add them together and put them in the numerator:

5+4+3+2+1 = 15

And your denominator is total possible combinations (not just a > b, but also a < b and a = b), which is 6 x 6.

So the answer is A) 15/36

Sorry for the crappy explanation... hopefully someone can give a better one.

As for #8, you're looking at Q wrong. It's between 1 and 2, not 0 and 1.

extremefob

thanks so so much

Yojo

To make an orange dye, 3 parts of red dye are mixed with 2 parts of yellow dye. To make a green dye, 2 parts of blue dye are mixed with 1 part of yellow dye. If equal amounts of green and orange are mixed, what is the proportion of yellow dye in the new mixture?

setzwxman

Yojo, the "official" solution to that problem is on CB's website. The answer is (C), 11/30, to see what they wrote. Their algebraic methods are very easy to follow/understand, but I tend to flub w/ the time crunch on the real SAT I when it comes to possible algebraic reasoning, unfortunately. I'm working on improving that pronto, lol. Anyway...

What I did was this...

So we know that for orange, you have a total of 5 parts. 3 of them are red, and 2 of them are yellow. IOW, 2/5 of the parts are yellow.

In the green sample, we have a total of 3 parts. 2 of them are blue, and 1 of them is yellow. IOW, 1/3 of the parts is yellow.

Now let's tack on some units to make it more realistic/less abstract/easier to think about.

Equal amounts of green and orange are mixed, so why not 1 L of orange and 1 L of green? In this case, we'd have 2/5 L of yellow from orange, and 1/3 L yellow from green. (2/5) + (1/3) = 11/15 L yellow total in the orange + green mixture. Now, we have to find the total volume to be able to find the PROPORTION of yellow in the combined sample. 1 L + 1 L = 2 L total.

[(11/15) L]/[2 L] = 11/30
*Note how the units cancel out to leave you with just the proportion.

gcf101

I agree with setzwxman - if you can do without equations (and I do like tackling them!), then do it. In many cases it's faster and more error proof, because you see everything clearly, not just some abstract constructions.
I would use setzwxman's approach with one change. It's easier to work with whole numbers, so in questions on mixes/parts choose LCD of total parts, then ingredients will be integers too.

Orange mix - 5 parts (R3+Y2)
Green mix - 3 parts (B2+Y1).
IMPORTANT: please note that parts in Orange and Green mixes are not assumed to be the same, even though they can.
For example, if we had 10L of Orange and 6L of Green, each part in both would be 2L.
Anyway, LCD(5,3)=15. If we have 15L of each mix, then
== in Orange one part is (15/5)L = 3L.
and there are 3L * 2 = 6L of Yellow stuff in it;
== in Green one part is (15/3)L = 5L
and that's exactly how much of Yellow dye is in it.
Joining equal amounts 15L of each Orange and Green we get 30L of new mix.
There are 6L + 5L = 11L of Yellow dye in it.
11 out of 30 is 11/30 - the answer.
++++++
Just out of curiosity:
from Orange come (15/5)L * 3 = 9L of Red
from Green come (15/3)L * 2 = 10L of Blue.
So in 30L of a final mix there'll be
11L of Yellow
9L of Red
10L of Blue.
Their ratio is 11:9:10.
We are ready to answer any possible question on this mix.
What's the ratio Yellow to Blue, or (Yellow and Blue) to Red, etc.
==============
Question to artists:
is this new dye of any particular color?

gcf101

Don't send in the Feds!

fbc001

>the "official" solution to that problem is on CB's website.

Could you point out where is the link/location to the official soln referred to above?

Perle

As for #5... I don't understand why we must add up and not multiply the probabilities together.

Can anyone help me please ?