Problem Solving

kashefian wrote:I solved this problem like this:

P (A & B & C) = P(A) + P(B) + P(C) - P(A or B or C) or
(A & B & C) = (A) + (B) + (C) - (A or B or C)
x = 300 - 225 = 75

Therefore the probability is 75/300 = 25%

Note that A + B + C = 300 because we have a total of 300 people. It took me around 30 seconds to answer the question.

Stuart Kovinsky wrote:

agganitk wrote:According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?

A. 15%
B. 20%
C. 25%
D. 0%
E. 35%

Ans??

Let's say we have 100 people to make things simple. We want to minimize the triple group, so let's minimize how many people like each kind of fruit, giving us:

70 apple lovers, 75 banana lovers and 80 cherry lovers.

Now, 70 + 75 + 80 = 225, so we have 225 "fruit loves" spread out among 100 people.

Therefore, there are 125 more "fruit loves" than there are people.

Our job is to minimize the number of people who love all 3; to do so, we want to maximize the number of people who love exactly 2 of the 3 and minimize the number of people who love exactly 1 of the 3.

So, we can come up with two equations:

AB + AC + BC + 2ABC = 125

and

AB + AC + BC + ABC = 100

AB = number who like just apple/banana
AC = number who like just apple/cherry
BC = number who like just banana/cherry
ABC = number who like all 3

The first equation is derived from the triple-overlapping set equation:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups)

100 = 70 + 75 + 80 - AB - AC - BC - 2(ABC)

and when we rearrange to get all variables on one side:

AB + AC + BC + 2ABC = 125

The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC

So, back to our equations:

AB + AC + BC + 2ABC = 125
AB + AC + BC + ABC = 100

If we subtract the second from the first, we get:

ABC = 25... done!

Now, at this point you may be saying, "umm.. ok.. but I asked for a simple way to solve, that seemed super complicated and time consuming!"

However, if you understand the concepts behind triple-overlap (or double-overlap) questions, it's fairly intuitive; the complicated part is getting to the stage at which you have that deeper understanding.

Of course, this exact question won't appear on the GMAT. So, as always, after you do a question you ask yourself: "what did I learn from this question that's going to help me on future questions?"

Here's our takeaways:

1) there are multiple ways to solve overlapping sets questions. The more you familiarize yourself with the 3 major approaches (equations/venn diagrams/matrices(the last only works when there are 2 overlapping sets, unless you're really good at drawing a 3-dimensional matrix)), the more likely it is that the quickest approach will jump out at you on test day.

2) if you're shooting for a 600+, learn the two equations noted above.

3) whenever you're asked to minimize something, think "what do I need to maximize to achieve that result?"

Stuart Kovinsky wrote:

agganitk wrote:According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?

A. 15%
B. 20%
C. 25%
D. 0%
E. 35%

Ans??

Let's say we have 100 people to make things simple. We want to minimize the triple group, so let's minimize how many people like each kind of fruit, giving us:

70 apple lovers, 75 banana lovers and 80 cherry lovers.

Now, 70 + 75 + 80 = 225, so we have 225 "fruit loves" spread out among 100 people.

Therefore, there are 125 more "fruit loves" than there are people.

Our job is to minimize the number of people who love all 3; to do so, we want to maximize the number of people who love exactly 2 of the 3 and minimize the number of people who love exactly 1 of the 3.

So, we can come up with two equations:

AB + AC + BC + 2ABC = 125

and

AB + AC + BC + ABC = 100

AB = number who like just apple/banana
AC = number who like just apple/cherry
BC = number who like just banana/cherry
ABC = number who like all 3

The first equation is derived from the triple-overlapping set equation:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups)

100 = 70 + 75 + 80 - AB - AC - BC - 2(ABC)

and when we rearrange to get all variables on one side:

AB + AC + BC + 2ABC = 125

The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC

So, back to our equations:

AB + AC + BC + 2ABC = 125
AB + AC + BC + ABC = 100

If we subtract the second from the first, we get:

ABC = 25... done!

Now, at this point you may be saying, "umm.. ok.. but I asked for a simple way to solve, that seemed super complicated and time consuming!"

However, if you understand the concepts behind triple-overlap (or double-overlap) questions, it's fairly intuitive; the complicated part is getting to the stage at which you have that deeper understanding.

Of course, this exact question won't appear on the GMAT. So, as always, after you do a question you ask yourself: "what did I learn from this question that's going to help me on future questions?"

Here's our takeaways:

1) there are multiple ways to solve overlapping sets questions. The more you familiarize yourself with the 3 major approaches (equations/venn diagrams/matrices(the last only works when there are 2 overlapping sets, unless you're really good at drawing a 3-dimensional matrix)), the more likely it is that the quickest approach will jump out at you on test day.

2) if you're shooting for a 600+, learn the two equations noted above.

3) whenever you're asked to minimize something, think "what do I need to maximize to achieve that result?"

nitin9003 wrote:

Stuart Kovinsky wrote:
True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC

Hi Stuart,
Thanks for the vivid description. It was really helpful.
But I have some problems
I couldn't understand this equation and I have not seen it anywhere. Can you plz explain and also how come you assume the total in exactly 1 group to be 0.

One quarter of the cars on a parking lot have air conditioning and one third of the trucks on the same lot have air conditioning. If there is at least one car and at least one truck on the lot, no other vehicles on the lot, and a total of 72 vehicles on the lot, what's the maximum number of cars that could have air conditioning?

Stuart Kovinsky wrote:

agganitk wrote: than there are people.

Our job is to minimize the number of people who love all 3; to do so, we want to maximize the number of people who love exactly 2 of the 3 and minimize the number of people who love exactly 1 of the 3.

Hey Stuart thanks a lot for your elaborate explanation but,

I have a question regarding the second equation y we minimize number of ppl who love exactly one shud nt we also maximize them to get minimum love all three exactly types

And what if there are people that don't eat apples, bananas or cherries in the survey?

Stuart Kovinsky wrote:

agganitk wrote:According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?

A. 15%
B. 20%
C. 25%
D. 0%
E. 35%

Ans??

Let's say we have 100 people to make things simple. We want to minimize the triple group, so let's minimize how many people like each kind of fruit, giving us:

70 apple lovers, 75 banana lovers and 80 cherry lovers.

Now, 70 + 75 + 80 = 225, so we have 225 "fruit loves" spread out among 100 people.

Therefore, there are 125 more "fruit loves" than there are people.

Our job is to minimize the number of people who love all 3; to do so, we want to maximize the number of people who love exactly 2 of the 3 and minimize the number of people who love exactly 1 of the 3.

So, we can come up with two equations:

AB + AC + BC + 2ABC = 125

and

AB + AC + BC + ABC = 100

AB = number who like just apple/banana
AC = number who like just apple/cherry
BC = number who like just banana/cherry
ABC = number who like all 3

The first equation is derived from the triple-overlapping set equation:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups)

100 = 70 + 75 + 80 - AB - AC - BC - 2(ABC)

and when we rearrange to get all variables on one side:

AB + AC + BC + 2ABC = 125

The second equation is derived from another version of the triple-overlapping set equation:

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

We set the "total in exactly 1 group" to 0, so we get:

100 = 0 + AB + AC + BC + ABC

So, back to our equations:

AB + AC + BC + 2ABC = 125
AB + AC + BC + ABC = 100

If we subtract the second from the first, we get:

ABC = 25... done!

Now, at this point you may be saying, "umm.. ok.. but I asked for a simple way to solve, that seemed super complicated and time consuming!"

However, if you understand the concepts behind triple-overlap (or double-overlap) questions, it's fairly intuitive; the complicated part is getting to the stage at which you have that deeper understanding.

Of course, this exact question won't appear on the GMAT. So, as always, after you do a question you ask yourself: "what did I learn from this question that's going to help me on future questions?"

Here's our takeaways:

1) there are multiple ways to solve overlapping sets questions. The more you familiarize yourself with the 3 major approaches (equations/venn diagrams/matrices(the last only works when there are 2 overlapping sets, unless you're really good at drawing a 3-dimensional matrix)), the more likely it is that the quickest approach will jump out at you on test day.

2) if you're shooting for a 600+, learn the two equations noted above.

3) whenever you're asked to minimize something, think "what do I need to maximize to achieve that result?"