In a school, half of the boys and one-fourth of the girls enrolled for a course. If $\frac{7}{8}$th of the boys and $\frac{3}{4}$th of the girls actually joined the course, what is the total number of students that did not join the course?

My solution :

$\frac{7}{8} \times \frac{1}{2} = \frac{7}{16}$ boys joined. Number of boys that did not join $= 1 - \frac{7}{16} = \frac{9}{16}$. Similarly, number of girls who did not join $= \frac{13}{16}$. Hence, number of students that did not join $= \frac{22}{16}$. But that is not possible.

$\boxed{\text{The answer is} \frac{1}{16}}$

Where am I wrong, and how to continue? Thank You. Update:- Answer given in my book is :- enter image description here

  • $\begingroup$ Check the Edit in my answer below. It's more an issue with the wording of the problem for which you were not exactly getting it. $\endgroup$ – Mathejunior Nov 4 '17 at 13:10

Yeah, basically $\frac{7}{8}\text{ th of }\frac{1}{2}$ of the total number of boys joined, which means, $\left[1-\left(\frac{7}{8} \times \frac{1}{2}\right)\right]=\frac{9}{16}$ of the total number of boys didn't join.

Similarly, $\frac{13}{16}$ of the total number of girls didn't join the course.

You were correct till now. But mind the words: *The fractions are upon the total number of boys or girls and not on an individual

Let suppose the class has $b$ number of boys and $g$ number of girls. So, the total number of students is $b+g$. And, the number of candib+dates who didn't join: $\frac{9}{16}b+\frac{13}{16}g$.

Note that the answer can't be $\frac{1}{16}$ as per the given question. Please look into the question carefully.

The ratio of students who didn't take part is $\frac{\frac{9}{16}b+\frac{13}{16}g}{b+g}\neq \frac{1}{16}$. As a matter of fact, if it were equal to $\frac{1}{16}$, then $$~~~~~~ ~~~\frac{\frac{9}{16}b+\frac{13}{16}g}{b+g} = \frac{1}{16} \\\implies 9b+13g=b+g~ \\\implies 8b+12g = 0~~~~~~~~ \\\implies b=g=0 \text{ student}$$

Hope you get the insight to the problem!

Let's take an example to make things clearer:

Example -

Total number of boys be $16$ and total number of girls be $32$. Then the total number of boys who didn't participate comes out to be $9$ and the number of girls who didn't participate is $26$ which means a total of $35$ out of $48$ candidates didn't participate. And, $\frac {35}{48}\neq \frac{1}{16}$. Note that $\frac {9b+13g}{16(b+g)}=\frac {9\times 16 + 13\times 32}{16(16+32)}=\frac {35}{48}$ as claimed.

Also, a major mistake in the problem is that boys or girls have to be a multiple of $16$.


The wording is a bit weird: As per the question, I also considered the students who didn't enroll for the course as a part of the non-joiners. The way you began isn't wrong. You did consider the way I considered.

But as per your book's answer, it turns out that they are asking for the students who enrolled for the course but didn't join. In that case, I hope you shouldn't have any problem with the provided solution.

It's a trouble with the wording of the question.

  • $\begingroup$ sir so the answer is 9/16 b + 13/16 g? no more steps? $\endgroup$ – Ram Keswani Nov 4 '17 at 12:13

Where are you wrong??

My solution is: 7/8 * 1/2 = 7/16 boys joined.

How can we have 7/16th of a boy? Assume there are $x $ boys. Then, $\frac {7x}{16}$ boys joined.

Similarly no. of girls that did not join is 13/16.

How can we have 13/16th part of a girl? Assume there are $y $ girls and then analyse the problem.

Then add the two parts to get the answer in $x $ and $y$.

  • $\begingroup$ 9/16 x + 13/16 y = ? How to continue sir, I still dont get it. $\endgroup$ – Ram Keswani Nov 4 '17 at 9:00
  • $\begingroup$ @RamKeswani That's all. That is the answer. $\endgroup$ – Rohan Nov 4 '17 at 9:01
  • $\begingroup$ @RamKeswani Note that you can accept an answer by ticking the green tick mark in the answer. $\endgroup$ – Rohan Nov 4 '17 at 9:02
  • $\begingroup$ answer is 1/16. How can 9/16x + 13/16 y be the answer? $\endgroup$ – Ram Keswani Nov 4 '17 at 12:12

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