# Techniques in Answering Common MMC Oral Questions-Part 1

Hello guys! We are now in November, and all that means is that the country’s most widely participated mathematics contest, the Metrobank-MTAP-DepEd Math Challenge or MMC is fast approaching. In barely 2 months, the battle will begin. So, for my first article here, I want to share some techniques that MAY be useful to those who will be competing in the oral rounds of the MMC, especially in the team competitions.

As a little bit of a disclaimer, what I will share for now are techniques on how to solve some problems that I think most of you find VERY EASY. But, in the oral competitions, several distractions like nervousness come into play that kind of disrupts your solving. So, it is important not only that you know the concepts but also be able to solve the problem correctly in the fastest possible time. Hence, I hope that these techniques would help those having problems in solving these kinds of questions especially when under pressure. I know most of you may already know all of these, but for those who still don’t, I hope these help. So without further ado, let’s get started.

PROBLEM SET A: Heads – Feet Problems

This type of problems very frequently appears on MMC questions, especially in the Grade 7-8 levels.

Ex: Johnny, a poultry farmer, raises some pigs and chickens. As a regular routine, he does a daily inventory of his animals. One day, he was so groggy that all he remembers is that he was able to count 75 heads and 264 feet, but he doesn’t know exactly how many of each type of animal he has. How many pigs does he have?

Solution: This can be solved easily by basic algebra. But of course, when you are in an oral competition, and say, in the easy round, papers are not allowed. So, you can solve this Mentally by the formula:

In this case, we have the pigs as the most number of feet. Also, since in the group, the least possible no. of feet is 2 (since chickens have two feet),

$Pigs=\displaystyle\frac{264-2(75)}{2}$

$Pigs=\displaystyle\frac{264-150}{2}$

$Pigs=\displaystyle\frac{114}{2}$

$Pigs=57$

Hence, he has 57 pigs.

Note: This formula can be used in similar problems involving bicycles and tricycles, etc. However, it is still limited to 2 individuals (in this case, pigs and chickens). In the following articles, we’ll try to find the shortest way to solve those involving 3 or more individuals.

PROBLEM SET B: Working Together/Work Problems

This is also a very common problem that actually reaches up to the 4th year level of MMC oral competitions. I know a number of people getting confused in this type of problems, so I hope this helps.

Ex: Chuck and a woodchuck would chuck wood to earn a living. Chuck can chuck 100 logs of wood in 3 hours, while the woodchuck can chuck the same number in 4 hours. How many hours will it take if Chuck and the woodchuck would chuck wood together?

Solution: Again, this can be solved by simple algebra, but to save ink and paper (and trees. Lol!), we could solve the time t using the formula:

$t=\displaystyle\frac{ab}{a+b}$

where a and b are the rates of the 2 workers. Hence, for this problem we have

$t=\displaystyle\frac{3\cdot 4}{3+4}$

$t=\displaystyle\frac{12}{7}$ hrs

Now, what if, say, we have three workers A, B, and C with rates a, b, and c, respectively? We could derive the formula by using simple algebra. We know for a fact that letting t be the time it would take them working together, we have

$t\big (\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}\big )=1$

$t\big (\displaystyle\frac{ab+bc+ac}{abc}\big )=1$

$t=\displaystyle\frac{abc}{ab+bc+ac}$

By looking at this result we could derive a formula for t, that is, for any number of workers n,

For the next articles, I’ll try to unfold other tips/techniques in answering other common MMC oral competition questions.