Approaching Problems involving matrices/ Venn diagrams

[pappup5]

Of the 300 students who participated in athletics , 45 percent experienced fatigue, 35 percent experienced weakness, and 70 percent experienced convulsions. If all of the students experienced at least one of these effects and 25 percent of the students experienced exactly two of these effects, how
many students experienced only one of these effects?

Hi Ron,

There is a similar problem in OG13. So basically (if we do not use any formula), this boils down to 5 equations and 7 unknowns and with mathematical manipulation we can solve for all the effects (e.g x+y+z).

However, this seemed a little time consuming (I mean "way greater" than the average time for solving a standard GMAT quant problem). I checked online and found multiple instances where people had used formulae to shorten time.

Since memorizing multiple formulae is a little difficult for me, and in general I face problems with these "Venn Diagram" situations, can you please give me a simple insight on how to solve such problems in a time efficient manner? I tried looking at Thursday Video archives and could not locate any video that deals with such problems. If not already present, can you consider this for a future "Thursday's with Ron" session?

Thanks in advance.

[RonPurewal]

well, i don't know any formulas for those things (and i'm certainly not capable of remembering them, even if they do exist).

don't forget what is basically THE WHOLE POINT OF THIS ENTIRE EXAM, which is flexible thinking.
if using all those variables is obviously going to take until the sun goes down and comes back up ... well, then, don't use all those variables!
(:

[RonPurewal]

i assume this is a multiple-choice problem, and that the choices are numbers.
if that's the case, then you can immediately make the following realizations:
• "exactly two of these effects" corresponds to 3 different regions of the venn diagram.
• it doesn't matter how the 25 percent is distributed among those regions. (if it mattered, they'd have to give you enough information to solve for that distribution.)
• so... you can just distribute that 25 percent into those three regions however you want. literally, just make up a distribution. 10 here, 10 there, 5 in the other one. or, 25 in one of them and 0 in the others... doesn't matter.

once you do that, you can put "x" in the middle zone (corresponding to having all three problems). then you can just subtract a bunch of times to fill in the rest of the diagram in terms of x, and then everything has to add to 100.

or, you could try backsolving from the answer choices, once you've gotten to that point.

the point is -- as usual -- if you're not making progress with some approach, QUIT and try something else!

[pappup5]

i assume this is a multiple-choice problem, and that the choices are numbers.
if that's the case, then you can immediately make the following realizations:
• "exactly two of these effects" corresponds to 3 different regions of the venn diagram.
• it doesn't matter how the 25 percent is distributed among those regions. (if it mattered, they'd have to give you enough information to solve for that distribution.)
• so... you can just distribute that 25 percent into those three regions however you want. literally, just make up a distribution. 10 here, 10 there, 5 in the other one. or, 25 in one of them and 0 in the others... doesn't matter.

once you do that, you can put "x" in the middle zone (corresponding to having all three problems). then you can just subtract a bunch of times to fill in the rest of the diagram in terms of x, and then everything has to add to 100.

or, you could try backsolving from the answer choices, once you've gotten to that point.

the point is -- as usual -- if you're not making progress with some approach, QUIT and try something else!

Hi Ron,

Yes it is a multiple choice problem solving question. Thanks for your detailed explanation. I have submitted a request for a Thursday session for this topic also .

Actually within this question there are certain variations involving "double counting" which I was unable to understand clearly i.e. situations when the 2 intersects have elements that are counted in the "x" as well.

[RonPurewal]

Actually within this question there are certain variations involving "double counting" which I was unable to understand clearly i.e. situations when the 2 intersects have elements that are counted in the "x" as well.

those ^^ are probably taken care of, if you pick your own numbers as a way of distributing the 25 percent (as discussed above).

[pappup5]

Actually within this question there are certain variations involving "double counting" which I was unable to understand clearly i.e. situations when the 2 intersects have elements that are counted in the "x" as well.

those ^^ are probably taken care of, if you pick your own numbers as a way of distributing the 25 percent (as discussed above).

Sure thing Ron - I get it now somewhat(I will practice more with your valuable inputs) - back solving , reducing percentage situations to n=100 etc. are a smart way out. Please do consider my Thursday Session Request.

Thanks for your time.

[RonPurewal]

well... it might be the case that there just aren't enough problems like that to justify the creation of a study hall. if there's only one or two such problems in the entire canon of GMAC quant problems, then, it wouldn't make sense to try to create a study hall around them.

i can try to see what i find, though.

[pappup5]

well... it might be the case that there just aren't enough problems like that to justify the creation of a study hall. if there's only one or two such problems in the entire canon of GMAC quant problems, then, it wouldn't make sense to try to create a study hall around them.

i can try to see what i find, though.

Thanks Ron !!!

[RonPurewal]

sure.