Explanations of SAT Question of the Day?

<p>My brain is dying!
I can't reason this out -- help anybody?
It's the SAT Question of the Day from the CollegeBoard.
...do I feel a bit dense! heh. Thanks. x</p>

<p>I didn't understand CB's sense of logic on this one, either...</p>

<p>Here's how I did it:</p>

<p>Since a/b = 3, then I can assume that a=21 and b=7.
21/7 = 3, so it works.</p>

<p>Since b/c = 7, then I can assume that b=7 and c=1. (Notice that the value of "b" from above has to equal the value of "b" here... b=7).</p>

<p>Plug in the information above into this equation:
(a+b)/(b+c)</p>

<p>(21+7)/(7+1) = 28/8 = 7/2</p>

<p>Or, without playing the numbers game and risking there isn't an easy integral solution</p>

<p>a/b = 3 -- multiply both sides by "b"
a = 3b</p>

<p>b/c = 7 --- multiply both sides by "c"
b = 7c ---- divide both sides by 7
c = b/7</p>

<p>So, when you plug in (a+b)/(b+c)...</p>

<p>3b + b / b/7 + b
7b/7 + b/7
4b / 8b/7
(4b)/(8b/7) ---- multiply each part of the fraction by 7
28b/8b ------ cancel out b's and factor out 4 for answer.....7/2</p>

<p>I prefer the "numbers game", but here's a straight forward algebra (unlike a convoluted CB solution).
a=3b,
b=7c therefore
a=21c,
(a+b)/(b+c) =
(21c+7c)/(7c+c) =
(28c)/(8c) =
7/2.</p>

<p>I did it the same way as gcf101. In problems like this, you can try to get everything in terms of one variable, so that the variable drops out.</p>

<p>I don't get how collegeboard can expect people to follow their reasoning. It took me awhile but this is what they did:</p>

<p>a/b = 3
a/b + 1 = 4
a/b + b/b = 4
(a+b)/b = 4</p>

<p>b/c = 7
c/b = 1/7
c/b + 1 = 1/7 + 1
(c+b)/b = 8/7
b/(b+c) = 7/8</p>

<p>And then multiply.</p>

<p>But if you came up with collegeboard's solution first, you're probably thinking too hard. =x</p>

<p>very strange question.I will be happy if there isn`t a question like this one on 24th</p>

<p>First off, I used proportions to find out that a=3b and b=7c. Then, I did some algebra playing and found out that:</p>

<p>a=3b=21c
b=1/3a=7c
c=1/21a=1/7b</p>

<p>If you plug the a values into the equation, you'll wind up with (a+1/3a)/(1/3a+1/21a)=(4/3a)/(8/21a). If you cancel out the a's, you're left with the complex fraction (4/3)/(8/21), which simplifies to 7/2.</p>

<p>Okay, now that we are into alternate ways, here's one more:
(a+b)/(b+c)=
((a+b)/b) / ((b+c)/b)=
(a/b+1)/(1+c/b)=
(3+1)/(1+1/7)=
7/2</p>

<p>Here's a different way</p>

<p>Manipulate the fractions to get
a = 3b and b = 7c.</p>

<p>Therefore, (a + b)/(b + c) = (3b + b)/(7c + c) = (b/c)/2.</p>

<p>We know b/c = 7, so plug that in to get 7/2.</p>

<p>Just for fun.</p>

<p>Multiply
a/b=3 and b/c=7:
(a/b)(b/c) = (3)(7)
a/c = 21</p>

<p>(a+b)/(b+c) =
a/(b+c) + b/(b+c) =
1/((b+c)/a) + 1/((b+c)/b) =
1/(b/a+c/a) + 1/(b/b+c/b) =
1/(1/3+1/21) + 1/(1+1/7) =
21/8 + 7/8 =
7/2.</p>

<p>Yeah azukit has a good explanation of CB's method</p>

<p>But that is a @#$% way to do it</p>

<p>Many other ways, most of them already been said</p>

<p>I personally did it ldftalk's way</p>

<p>i just plugged in numbers.. and after doing a+b/(b+c) i ended up getting an answer of 2.5
then i looked and saw 7/2=2.5 so the answer was A.</p>

<p>Fun doesn't end.
a:b = 3:1 = 21:7
b:c = 7:1
a:b:c = 21:7:1
(a+b) : (b+c) = 28:8 = 7: 2</p>

<p>Okay! Guys, I appreciate the help, but I now have a total...20 ways to do this problem, I get it, I get it! :P :P</p>