Math Competition Problems - Problem of the Week - 18

In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar? - AMC 8 2012

Solution:

This is an easy problem once you interpret the question right. So let's see how we can write the equations from the given info. Since the question talks about red, green and blue marbles, let

g - be the number of green marbles

b - be the number of blue marbles

r - be the number of red marbles

Since "all but 6 are red marbles" - the number of green and blue marbles must be 6.

So g+b=6 ---> eqn 1

"all but 8 are green" - which means the number of red and blue marbles must be 8.

So r+b=8 ---> eqn 2

Similarly, all but 4 are blue, means the number of red and blue marbles must be 4

So, r+g=4 ---> eqn 3

Now that we have our 3 equations we can solve them. While there are a variety of ways to solve them the below method is the easiest.

Adding all 3 equations together we get

2r+2b+2g=18

Dividing the whole equation by 2 we get

r+b+g=9

So the total number of marbles in the jar should be 9. From here you can find the individual number of marbles, but that's not necessary for this question :)

Want to prepare for AMC 8? Follow my hashtag #PrepareForAMC8 for problems and tips to Prepare for the exam.

Did you like the problem? Would you like to have these problems and solutions delivered to your inbox as I publish them? Click here to subscribe

---

#Competition, #Math #Problems #OnlineMathTutoring, #MiddleSchoolMathEnrichment, #OnlineMathEnrichmentPrograms #AMC8 #AMC8Tutor #OnlineMathTutor #ProblemOfTheWeek

---

Previous Week's Problem

Problem of the Week 17

Problem of the week 17

Problem of the Week - 19

Next Week's Problem

Problem of the Week 19