[Pri20150306RCU] A Very Crowded Class

Question

Solution

     Let us write down the given information in a Ratio diagram.

     As you can see, we have ratios with different units, which seems difficult to solve.  However, notice that the number of boys stayed the same throughout.  We know that the LCM of 6 and 7 is 42.  So let us use another type of unit, say “heart” units, with the number of boys corresponding to 42 of these units.  This can be done by multiplying the first column by 7 and multiplying the second column by 6.  This is what we get

     With the “heart” units, now it is very obvious that one “heart” is equivalent to 2.  From here we easily deduce that the number of boys is 84.

Commentary

     This is a type of “problem” where one quantity (the number of boys) is kept constant while another (the number of girls) changes, giving rise to different ratios.  It is similar to the "Boys, Girls and Party" problem, and you can certainly solve this problem using the bridging method shown there.  However, here we exploit the fact that the number of boys stayed the same, and we use the LCM to create a common type of unit (“heart” unit).  Once this is done, we can easily compare the number of girls using this common unit, and then the problem unravels.  Don’t you © hearts?

     Anyway, talking about authenticity in mathematics problems ... the number of boys is already 84, if you work out the total i.e. including the girls, you get ... (Do This Yourself).  Won’t you find this class a little too crowded?

     The person who set this question should probably have moved the pupils to the auditorium, yes?