<p>Hello, I'm preparing for the June SAT and was going through the CB blue book (first edition) and don't understand a math problem. The problem:
After the first term, each term in a sequence is 3 greater than 1/3 of the preceding term. If t is the first term of the sequence and t≠0, what is the ratio of the second term to the first term?
A)t+9/3
B)t+3/3
C)t+9/3t
D) t+3/3t
E) 9-2t/3
I answered A but the answer was C, this is tagged as a hard question. Why C? Please help.</p>
<p>how did u get A? i got B and C…
i picked t=3 so the next term would be 4. there has to be another way though…</p>
<p>t2/t = (t/3+3)/t = (t/3+9/3)/t = (t+9)/(3t)</p>
<p>After the first term, each term in a sequence is 3 greater than 1/3 of the preceding term. If t is the first term of the sequence and t≠0, what is the ratio of the second term to the first term?
A)t+9/3
B)t+3/3
C)t+9/3t
D) t+3/3t
E) 9-2t/3</p>
<p>Okay. The first term is t.</p>
<p>The second term must be t/3 + 3.</p>
<p>Make 3 a fraction
t/3 + 3/1</p>
<p>Get the same denominator
t/3 + 9/3</p>
<p>Add numerators
(t+9)/3</p>
<p>That is the value of the second term.</p>
<p>Ratio of the SECOND to the FIRST would be (t+9)/3 : t</p>
<p>Or,
(t+9)/3) / t</p>
<p>Which is equivalent to
(t+9)/3 ÷ t/1</p>
<p>Which is
(t+9)/3 * 1/t</p>
<p>Which is
(t+9)/3t</p>
<p>Matches answer C. It’s not answer A. You were probably thinking of what the second term would be, and didn’t find the RATIO of the second term to the first term. That’s very important.</p>
<p>Thanks Bassir. I didn’t find the ratio of the two. Would it be different if it said “what is the ratio of the FIRST term to the SECOND term?” Again thank you.</p>
<p>Yes, it would.</p>
<p>That would be the first term OVER the second term. Big difference.</p>
<p>ok this is actually easy and dosent waste time if you choose a number, choose a number that is divisible by 3 for t, i choose 6, so t would be 6, and 6/3 + 3 = 5, so 6 is the first term and 5 is the second term, so you want 5/6 as your ratio. plug in 6 as your t value untill you can reduce one of the answers to 5/6.
i did this in my head in under a minute, so this was is way easier than actaully creating equations, for this particular problem…</p>
<p>and the answer is C btw</p>
<p>Creating equations is much faster for me.</p>
<p>Thank you all who has answered my question.</p>
<p>Another problem
If 2r+5t=50, and r and t are positive integers, what is one possible value of r?</p>
<p>it could be 5, 15, or 20</p>
<p>Yes that’s right! How did you solve it? Please explain, thank you!</p>
<p>"If 2r+5t=50, and r and t are positive integers, what is one possible value of r? "
2r+5t=50
=> 2r=50-5t
=> r = (50-5t)/2</p>
<p>t is some positive integer, so substitute for t=1,2,3, etc., until r becomes negative
t=1, r=45/2, not an integer, similarly with all odd values of t
t=2, r=40/2 = 20
t=4, r=30/2 = 15
t=6, r=20/2 = 10
t=8, r=10/2 = 5
t=10, r=0, but r is positive
So r is 5, 10, 15 or 20.</p>
<p>i strongly recommend u to read the chinese secondary school books</p>
<p>@Jehron Thank you!
Nikang, why chinese?</p>
<p>What are you talking about?</p>