At a certain university, 1/4 of the applicants failed to meet the minimum standards and were rejected immediately. Of those who met the standards, 2/5 were accepted. If 1200 applicants were accepted, how many applied ?

please help me in this question, I'm confused ... thanks in advance :)

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    $\begingroup$ From the last two sentences, you can work out how many people met the standards. Knowing this, from the first sentence you can work out how many applicants there were. Now please don't just post a comment saying "I don't understand, pls explain". THINK ABOUT IT, for at least a couple of hours if necessary, before asking for more help. $\endgroup$ – TonyK Oct 2 '15 at 15:17
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    $\begingroup$ Here is a traditional way that avoids algebra. Let us "guess" $1000$ applied. Then $250$ were rejected immediately, leaving $750$. Of these, two-fifths got in, a total of $300$. Bad guess! But since $1200$ is four times $300$, there must have been $4000$ applicants. $\endgroup$ – André Nicolas Oct 2 '15 at 15:30

The total number of applicants are $N$.

$N/4$ were rejected, so we are left with $3N/4$.

$2/5$th of the $3N/4$ were accepted, that is:



$N = 4000$ applicants.

enter image description here

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    $\begingroup$ I am really impressed that you took your time to make a diagram of this ;) $\endgroup$ – BigbearZzz Oct 2 '15 at 16:14
  • $\begingroup$ I appreciate your comment very much. Thank you. $\endgroup$ – NoChance Oct 2 '15 at 16:20

Hint: Since $\frac{1}{4}$ of the applicants did not meet the minimum standards, there are $(1-\frac{1}{4})$ of the applicants who met the minimum standards.

$\frac{2}{5}$ of the applicants who met the minimum standards were accepted. This means it is $\frac{2}{5}$ of the $\frac{3}{4}$ of the applicants who were accepted.

Can you do the rest?

  • $\begingroup$ This is obviously wrong $-$ you can't accept more people than applied $-$ but please read my comment to the OP before correcting it. $\endgroup$ – TonyK Oct 2 '15 at 15:22
  • $\begingroup$ @TonyK Oops! I have made a typo error. Yes, I have just read your comment. Perhaps I will erase my workings and give hints instead. $\endgroup$ – ChrisJWelly Oct 2 '15 at 15:24

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