SAT math question

<p>Hi, I'm doing practice problems and I need help with a certain problem:</p>

<p>At a high school basketball game, (3/5) of the students who attended were seniors, (1/3) of the students who attended were juniors, and the remaining 80 students who attended were all sophomores. How many seniors attended this game?</p>

<p>(a) 120
(b) 175
(c) 180
(d) 210
(e) 300</p>

<p>Can someone explain the working?</p>

<p>First we need to find the total number of students. To do this we will add the Senior and Junior fractions to find the fraction for sophomores.</p>

<p>3/5 is equivalent to 9/15 and 1/3 is equivalent to 5/15. We can now add, getting 14/15, the fraction of Seniors and Juniors. As the rest of the students are sophomores, we can subtract 14 from 15, getting 1, which above 15 is the fraction for sophomores (1/15).</p>

<p>80 times 15 gets us 1200, which means that (3/5)*(1200)=720, so 720 seniors were at the game, making all answers wrong.</p>

<p>Where did you get the problem?</p>

<p>See that’s the way that I did it, but as you said none of the answers correspond.</p>

<p>My SAT tutor gave it to me out of Barron’s SAT Math section book or something of the likes.
And I don’t have the answers with me</p>

<p>The one thing that I suppose the error could be is there being 8 sophomores instead of 80, making the answer 120.</p>

<p>Very strange question … I did it the same way as you two without looking at your posts</p>

<p>exactly how I did it as well. That should be the correct method. considernt 1/3 are juniors and 3/5 are seniors then the 80 sophomores represent 1/15 of the total crowd, leading to a grand total of 1200 students who attended so 720 seniors… Dafuq is wrong with this question(this is why I only use real SAT questions ). if you google this question, though someone else solved it like this:</p>

<p>After the 1/3 “of the others” left, 2/3 “of the others” remained. Therefore,
80 = 2/3 x
120 = x = the number after the seniors left. </p>

<p>The seniors constituted 3/5 of the origainal crowd, so x must be 2/5 of the original crowd:
120 = 2/5 y
300 = y = the original crowd
seniors = 3/5 (300) = 180</p>

<p>@observeraffect perhaps, but it’s so strange… </p>

<p>At least I’m glad that I’m not the only one that thinks this! </p>

<p>See another one of the problems my tutor gave me is of a similar nature, and again I can’t get the answer:</p>

<p>In a school election, Susan received (2/3) of the ballots cast, Mary received (1/5) of the remaining ballots, and Bill received all the other votes. If Bill received 48 votes, how many votes did Susan receive?</p>

<p>(a) 75
(b) 90
(c) 120
(d) 150
(e) 180</p>

<p>Help!</p>

<p>its c, 2/3x + 1/15x +48 = x(total number votes)
x=180 then 2/3 times 180 = 120</p>

<p>mary is 1/15 because mary received .2 of the remaining 1/3 ballots after subtracting susan’s 2/3. So you multiply .2 by 1/3 to get 1/15</p>

<p>@hagzzz</p>

<p>“The seniors constituted 3/5 of the origainal crowd, so x must be 2/5 of the original crowd:
120 = 2/5 y
300 = y = the original crowd
seniors = 3/5 (300) = 180”</p>

<p>If there were 180 seniors, then the total is 300 (The inverse of 3/5 times 180). If there were 300 students, and there was 1/3 of them as Juniors, then it follows there were 100 juniors. If there were 180 seniors, 100 juniors, and 80 sophomores, then 60 sophomores were also a junior or a senior!</p>

<p>Yeah I got that too. </p>

<p>Thanks guys!</p>

<br>

<br>

<p>See that’s the way that I did it, but as you said none of the answers correspond.</p>

<p>My SAT tutor gave it to me out of Barron’s SAT Math section book or something of the likes.
And I don’t have the answers with me<<<</p>

<p>Do yourself two favors:</p>

<ol>
<li>Do not waste your time on tests that are now written by ETS.</li>
<li>Most importantly, fire this clown who pretends to be a tutor. If this idiot does use Barron’s and cannot see how wrong those problems are, why would rely on him. </li>
</ol>

<p>Does not get simpler than that.</p>

<p>haha I will second xiggi’s words. there is no sense in solving incorrect problems and wasting time. it just leads to aggravation. i have seen this time and again with the problems written by test prep companies. stay away from those if you can.</p>