a set problem...

jamesk486 · Posted: Tue Aug 19, 2008 7:25 pm Post subject: a set problem...

There are 60 people, 48 invested in A, 35 invested in B, and 27 invested in C. among 19 of people who invested in both A and C, 12 also invested in B. All people who invested in B also invested in A. How many people did not invest in any of the three investment?
A. 2 B. 3 C. 4 D. 5 E. 6

OA is C

Cant figure it out

madeline · Posted: Tue Aug 19, 2008 8:25 pm Post subject:

Now I really need a whiteboard. I will attempt to explain this without one and see how it goes!

Draw a rectangle on your paper. This is your sample space of 60 people.
Draw a circle inside, label it A. This circle represents 48 people.
Draw a circle inside of A, label it B. This represents those who invested in B, all of whom also invested in A.
Draw a third circle that intersects both A and B, label it C. This represents the 27 who invested in C.
Your job is to find the number of people who are outside any of the circles but inside the rectangle.

As you can see, since B is entirely a subset of A, any number to do with B is in fact irrelevant here, since we want those that are OUTSIDE of both A and C.

A intersect C = 19
A union C = A + C - (A intersect C) = 48 + 27 - 19 = 56
Answer = S - 56 = 60 - 56 = 4

Stuart Kovinsky · Posted: Tue Aug 19, 2008 8:40 pm Post subject: Re: a set problem...

ritula · Posted: Tue Aug 19, 2008 8:47 pm Post subject:

madeline, that was an excellent explanation. The questions on sets really baffle me. can u pls explain how to solve such question if B was not totally inside A?

warlock · Posted: Tue Aug 19, 2008 8:50 pm Post subject:

n(a or b or c) = n(a) + n(b) + n(c) - n(a&b) - n(b&c) - n(c&a) +(a&b&c)

given: There are 60 people. So
n(a or b or c) = 60.

48 invested in A, 35 invested in B, and 27 invested in C. So..
n(a) = 48
n(b) = 35
n(c) = 27.

among 19 of people who invested in both A and C, 12 also invested in B. So..
n(a&c) = 19 but n(a&b&c) = 12. therefore only n(a&c) = 7.

All people who invested in B also invested in A. So..
n(a&b) = 35. but n(a&b&c) = 12..so only n(a&b) = 23. also only n(b) = 0, only n(b&c) = 0. and

finally only n(c) = 8
[n(c) = only n(c) + only n(a&c) + only n(b&c) + n(a&b&c)]

likewise only n(b) =0
[n(b) = only n(b) + only n(a&b) + only n(b&c) + n(a&b&c)]

likewise only n(a) =6
[n(a) = only n(a) + only n(a&b) + only n(a&c) + n(a&b&c)]

Therefore...Finally..phew....

n(a or b or c) = only n(a) + only n(b) + only n(c) + only n(a&b)+ only n(b&c)+ only n(a&c)+ n(a&b&c)

6+0+8+23+0+7+12 = 56.

this means that people who invest in atleast one plan is 56.
therefore ppl who dont invest in any is 60-54 = 4.

i know that this is too long an explanation..and i hope that u get it...
if not i have an image uploaded in ven diagram that will help u understand better..

jamesk486 · Posted: Tue Aug 19, 2008 10:55 pm Post subject:

wow thanks!

madeline · Posted: Wed Aug 20, 2008 8:18 am Post subject:

ritula, Stuart and warlock offer some nice explanation for if B wasn't totally inside of A, and warlock's Venn diagram (and formula) is the way to go! This would work for ANY problem with 3 sets intersecting.

I usually like to find shortcuts, such as ignoring B in this problem, which made my calculations easier than using the universal formula. However, sticking to the formula is the best way to go if you're easily confused or not very comfortable with these questions in general.

bradley281 · Posted: Wed Aug 20, 2008 10:01 am Post subject:

You can also answer this one logically.

So if you have 60 people and 48 invested in A you are left with 12 people that are un-identified.

As mentioned earlier B is irrelevant b/c all of those people are accounted for under A.

You are left with 27 in C of which 19 invested in A giving you 27-19=8 people that weren't accounted for under A

Therefore,

60-48=12-8=4

bradley281 · Posted: Wed Aug 20, 2008 10:03 am Post subject:

You can also answer this one logically.

So if you have 60 people and 48 invested in A you are left with 12 people that are un-identified.

As mentioned earlier B is irrelevant b/c all of those people are accounted for under A.

You are left with 27 in C of which 19 invested in A giving you 27-19=8 people that weren't accounted for under A

Therefore,

60-48=12-8=4

nervesofsteel · Posted: Wed Aug 20, 2008 5:55 pm Post subject:

C for me too..