Free GMAT Prep Hour: Quant; Overlapping Sets

[@manhattanprepgmat6791]

Solution for problem 6 from Jeff Vollmer! Given Facts: 100 books 60 were illustrated 39 had fewer than 100 pages 43 had animals that talked 9 had none of the qualities 33 had two or three of the qualities. (This last fact is key; it says 33 books had at least two of the qualities!) First, get rid of the 9 books that have none of the qualities. That leaves 91 books that have at least one of these qualities. From here, we could solve this algebraically: Let A represent the # of books that have exactly one of these qualities, B represent the # of books that have exactly two of these qualities, and C represent the # of books that have all three. A + B + C = 91 -- total books 1A + 2B + 3C = 60 + 39 + 43 = 142 -- sum of all the qualities B + C = 33 Subtract the first equation from the second. That leaves B + 2C = 51. If B + C = 33, that means that C = 18. So the number of books that have all three qualities is 18.

[@retnodewati6906]

Moral of the story, skip this question! Lol. Jeff is amazing. My fav instructor by far! Thank you for this free prep hour!

[@manhattanprepgmat6791]

Solution for the altered version of problem 1 from Jeff Vollmer! Make sure you set up your matrix for this problem! The keys to maximizing the number of fed cows are to 1) maximize the total number of cows and fed animals, and 2) minimize the other components in those rows and columns; in other words, minimize the number of unfed cows and fed pigs. There are 240 animals on the farm, and less than a third are cows. So the number of cows is below 80. But we want to maximize the total number of cows, because we want to eventually max the number of fed cows. So set the number of total cows to 79, the maximum allowed by the problem. That leaves 161 pigs on the farm. If less than 1/3 of those pigs are unfed, that means more than 2/3 were fed. But we want to minimize the number of pigs that were fed, because that will allow us to max out the number of cows that were fed. Remember, the problem tells us that 180 of the animals on the farm were fed, and we want as many of them to be cows in this problem. 2/3 of 161 is just over 107, so the least possible value for unfed pigs is 108. If there are 108 pigs that were fed, that leaves room for 72 cows to have been fed.

[@Varmaakhil192]

This is by far the simplest explanation I have seen for overlapping sets problems!! Thank you so much!! Made my life much simpler :)

[@manhattanprepgmat6791]

You're very welcome!

[@ashishsinha9035]

Thanks Mr. Jeff.

[@ImadeIyamu]

lmao I love when he said flee, run away. Jeff is an amazingly engaging teacher and simplifies stuff so clearly.

[@hhaba1]

honestly jeff is my fav instructor , he is so clear !!

[@manhattanprepgmat6791]

Yay! We think Jeff is pretty great too 😁

[@leilarma7648]

I can t thank you enough for your help , You are such an amazing instructor

[@manhattanprepgmat6791]

Thank you!!!

[@josephzenteno8293]

Best explanation of how to solve these kind of problems

[@shubhrahalder25june]

Superb video sir.Very thankful sir!!!

[@imgauravrawat]

Question No. 5 : There is formula for that question total No. people= no. of people speaking different language(spanish+Italian+ french) - II ( No. of people speaking exactly 2 language) - 2III( No. of people speaking all language) 85=103(no. of people speaking different language)- II ( No. of people speaking exactly 2 language) - 2*3 ( No. of people speaking all languages is 3) 85= 97- II ( No. of people speaking exactly 2 language) 85-97= - II ( No. of people speaking exactly 2 language) -12 = - II ( No. of people speaking exactly 2 language) so the answer is 12( no. of people speaking exactly 2 language) hope this will work.

[@filippoiguera3224]

Thank you very much, man!!

[@CkDiegoo]

Problem 6: D 18. I assumed g was center. (d+e+f) = 33; so 91=33+g+a+b+c; solved a=60-d-e-g, b=39-d-g etc... plugged in 33 for (d+e+f) and solved for g. Please advice if wrong, and thank you in advance.

[@rickc9622]

Thanks, Jeffery, Great video!

[@manhattanprepgmat6791]

Glad you enjoyed it!

[@filippoiguera3224]

Thank you for the wonderful video, Jeff! If I am targeting a 50 on Quant, should I perfectly know how to do these problems?

[@manhattanprepgmat6791]

Hey Filippo! We'd recommend checking out the GMAT Forums here: www.manhattanprep.com/gmat/forums/ That's the best place for questions like this one!

[@ashmapoojary]

Problem 5, Method...amazing.

[@manhattanprepgmat6791]

Fantastic!

[@smakshithalia5172]

Hi @jeff About Problem 6, do you think the following is a good approach of solving it? Total illustrated + total fewer than 100 pages + total with talking animals = 60+39+43= 142 142 - total books with qualities = 142 - 91 = 51 51 - books with at least two qualities = 51 - 33 = 18 Ans (D) If it's correct, could you explain why? Although I've reached the answer, I'm still a bit unclear on the logic. Thanks!

[@jeffreyvollmer8806]

This is correct! Yep, 91 of the books are responsible for 142 of the qualities (60 + 39 + 43). If each book had only 1 qualities, there should be 91 qualities. But there are 142 qualities... This implies that there are 51 extra qualities. So, where do those extra qualities come from? From the books that have more than 1 quality. Of the 91 books, 33 of them have more than 1 quality. So these 33 books are responsible for the 51 extra qualities. If each of these 33 books had exactly 2 qualities, then there should be 33 extra qualities. However, there are actually 51 extra qualities. This means that some of the books must have more than 2 qualities. And because 51 is 18 greater than 33, we need 18 of these books to have a third quality. (I like to think of the third qualities as an Extra extra quality) You can see this represented algebraically in the comment pinned up top!

[@filippoiguera3224]

Is the answer to problem 7 72?

[@harshitajain8151]

My mind is blown!

[@manhattanprepgmat6791]

🤯🤯🤯

[@somasundaram5744]

hello sir! In question 3...we assumed the intersection to be zero. The question says atleast one of the two session which means it can be more than 1. So in such a case if intersection is not zero we get diff value. Can you explain this please!

[@harshitajain8151]

A small doubt @jeff - if we do "extra counts - people who speak exactly two - people who speak exactly three" , we get "people who speak exactly one" - is that right?

[@jeffreyvollmer8806]

if we do "extra counts - people who speak exactly two - people who speak exactly three" we should actually get the number of people who speak exactly THREE languages. Extra counts = people who speak exactly two + 2(people who speak exactly three). The people who speak exactly one language don't contribute to the extra counts at all, so those people shouldn't be a part of the "extra counts" equation.

[@harshitajain8151]

@@jeffreyvollmer8806 thank you!

[@4tCa4mzUPqRZZo]

1:04:54, Sorry, I understand everything but why they were counted two extra times. I'm just so confused about the extra times?

[@jeffreyvollmer8806]

Hi Bill, I try to walk through this thinking process at 55:55, but here's a simpler example to get started: Imagine you are in a room with 10 people who own either a cat, a dog, or both. You ask all the cat owners to raise their right hands and 6 hands go up. Then you ask all the dog owners to raise their right hands and 5 hands go up. You've just counted 11 hands, but there can't be 11 pet owners in the room with you--there are only 10 other people in the room... So how is it possible that you counted 11 hands? Someone must have raised their hand more than once--you counted that person twice, or as I'm calling it here, ONE EXTRA time. The only reason you should count someone an extra time (twice instead of once) would be if they raised their hand twice. This would require them to be both a cat owner and a dog owner. Therefore ONE EXTRA count implies that someone is part of the cat-owner/dog-owner overlap. We can extend this thinking by considering a room full of people who own a cat, a dog, a rabbit, or more than one of these pet types. If someone owns two of these three pets, they will raise their hand twice during the counting phase. And if someone happens to own all three of these pets, they will raise their hand three times during the counting phase. If someone raises their hand three times, I am going to count that single person three total times. Now I'm expecting to count everyone once since they are all pet owners. But for the people who own more than one pet, I'm going to count that person more times than expected. I'm going to count them an EXTRA time or two. So, if someone owns all three pets, they will raise their hand three times, and I will count them three total times. I was expecting to count this pet owner once for sure, so the math I'm using here is this: 3 total hand-counts - 1 person responsible for these counts = 2 EXTRA COUNTS. Therefore, if a person owns all three pets (or speaks French, Italian and Spanish) I will have counted them 2 EXTRA TIMES. Hope that helps!

[@4tCa4mzUPqRZZo]

@@jeffreyvollmer8806 Wow, thank you for this detailed explanation! I really appreciate it!

[@jeffreyvollmer8806]

@@4tCa4mzUPqRZZo Glad it helps!

[@swarajkumardas2988]

Aren't der any other process to tackle overlapping sets other den matrix method?

[@jeffreyvollmer8806]

Hi Swaraj, Yes, there are algebraic methods you can use to solve these problems as well. (You'll often find that there exist more than one way to solve problems on the GMAT!) For a double-set problems, you can use the formula A + B + neither - both = Total. In this equation, A stands for the number of items that have quality A, and B stands for the number of items that have quality B. (Note, in this example, it is possible for something to have both qualities A and B!) However, if we use this equation, it can be difficult to divide those categories up. For example, in that equation, A here would include the items that have both qualities A and B, and A would also include the items that have quality A but not quality B. The matrix helps me to better categorize those divisions. Here's how you might apply this method to the first problem in the video: Cows + Fed Animals + Animals that are neither cows nor fed - Animals that are both cows and fed = Total. Cows = 80, Fed Animals = 180, Animals that aren't cows (and therefore must be pigs) and haven't been fed = 45, Total = 240 80 + 180 + 45 - both cows and fed = 240 305 - both cows and fed = 240. Both cows and fed = 65 (You can check this as around the 20:30 mark in the video.) We can do something analogous for triple sets. The equation for three potentially overlapping qualities A, B, and C, looks like this: A + B + C - number of items with at least 2 qualities - number of items with all three qualities + items with none of the qualities = Total. Again, this has the potential to be confusing because some groups here are subsets of other groups. For example, an item that has all three qualities is part of A, B, C, the items with at least 2 qualities, and the items with all three qualities. Of course, we only want to count this item once for our total, so when we count it three times as part of A, B, and C up front, we must un-count it twice... Additionally, an item counted in A, might also have quality C, but it might not have quality B. It can be very difficult to keep track of all of this, so if we are going to use a formula such as this one, we'll just want to be super careful about what each group actually means, and what subsets, if any, it might contain!

[@filippoiguera3224]

Answer Problem 6 : x = book with all the three qualities 100 = 60 + 3 +43 - (33 + x) + 9 91 = 142 - 33 - x 91 = 109 - x x = 18 Answer: D

[@olivieryeung398]

Problem 3 , wordings very ambiguous, which could easily lead to double counting. Very unclear to those who didn't' write/think up the question

[@jeffreyvollmer8806]

Hi! Which part of Problem 3 do you find ambiguous? Maybe I can help out here? I didn't write the problem--this one was an official problem from their free resources--but I still might be able to help out.

[@saifulisfree]

The trick that you did with "taking attendance" only worked out nicely because you have students taking either 1,2, or 3 classes. It wouldn't work if you had 4 classes correct? Because if you knew the "excess" votes in attendance and knew how many took all 4 classes, the remaining "excess" vote couldn't be parsed because you don't know how many took 3 or 2 classes. It's just by design the problem allows for this trick - right?

[@jeffreyvollmer8806]

Hi WW! If there were 4 classes, a fourth "variable" would be introduced, so we would need more information to solve a similar problem. If you want to create a four-class problem and post it here, I'd be interested in thinking through it. (But keep in mind that the frequency of encountering quadruple overlapping set questions is somewhere between ultra rare and non-existent on the GMAT!) The design of most triple overlapping set problems allows us to solve them with the "taking attendance" method--I haven't done anything particularly special to allow for that method's success here. You can apply the same method to #238 and #261 in the 2021 Official Guide.

[@shreyashgupta3903]

😂

[@tikiortaka]

April Fooled.