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# Among 400 students, 56% study sociology, 44% study mathematics and 40%

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Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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22 Jul 2010, 04:04
3
15
00:00

Difficulty:

55% (hard)

Question Stats:

73% (02:56) correct 27% (03:12) wrong based on 337 sessions

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Among 400 students, 56% study sociology, 44% study mathematics and 40% study biology. If 30% of students study both mathematics and sociology, what is the largest possible number of students who study biology but do not study either mathematics or sociology?

A. 30
B. 90
C. 120
D. 172
E. 188

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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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22 Jul 2010, 07:11
3
Okay, here's my take on this. This is mildly tricky. Let's first draw a Venn Diagram.

Attachment:
File comment: Venn Diagram

c73058.jpg [ 19.75 KiB | Viewed 5857 times ]

According to the diagram we need to find z.

So we are given the following:

$$x+y+z+a+b+c+p=400$$ (Totally there are 400 students)

$$a+c+x+p = \frac{56}{100}*400 = 224$$ (Total number of Sociology students)

$$a+b+y+p = \frac{44}{100}*400 = 176$$ (Total number of Math students)

$$b+c+z+p = \frac{40}{100}*400 = 160$$ (Total number of Biology students)

$$a+p = \frac{30}{100}*400 = 120$$ (Total number of students studying both Math and Sociology)

$$x+y+z+2(a+b+c)+3p=224+176+120 = 560$$

Substituting the first equation into this ($$x+y+z+a+b+c+p=400$$)

$$(a+b+c)+2p=560-400=160$$

Substituting $$a+p=120$$ in this equation, we get:

$$b+c+p=160-120=40$$

So, now we know that the total number of biology students = $$b+c+z+p = 160$$. So to get z, we just plug in $$b+c+p=40$$ to get $$z = 120$$

Hope this helps.

Try drawing a Venn Diagram and start simplifying it in terms of what's given and you'll hit the answer really soon most of the time.
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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22 Jul 2010, 08:15
whiplash2411 wrote:

$$a+c+x+p = \frac{56}{100}*400 = 224$$ (Total number of Sociology students)

$$a+b+y+p = \frac{44}{100}*400 = 176$$ (Total number of Math students)

whiplash2411 wrote:
I can't see anything in red, can you please let me know what's wrong? Thanks.

Think You mixed up X and Y there thats all! (colors apparently don't work in the [ m ] notation:

a+c+x+p = 224 (should be y)

a+b+y+p = 176 (should be x)
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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22 Jul 2010, 08:54
whiplash2411 wrote:
I can't see anything in red, can you please let me know what's wrong? Thanks.

Actually, the problem is in b & c when you wrote the equations of sociology & Maths

b & c should replace each other in these eqations.

a + c + x + p

There is no problem with x & y.
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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16 Oct 2010, 09:31
Lady's way is elegant but time-consuming.
Too many digits and variables... Can not we apply an overlapping sets formula to fix it more easy?
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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16 Oct 2010, 09:35
Financier wrote:
Lady's way is elegant but time-consuming.
Too many digits and variables... Can not we apply an overlapping sets formula to fix it more easy?

Check Bunuel's post in this thread formulae-for-3-overlapping-sets-69014.html
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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16 Oct 2010, 09:59
i kind of ended up guessing c with the following method:

3 sets theory formula

400 = 224 (S) +176 (M) +160 (B) - 2 * All 3 - SM - SB - MB

[Formula - Total = A + B + C - 2 ABC - AB - BC - CA + None]

2 All 3 + SB +MB = 160 - 120 [coz SM is 120]

2 All 3 + SB + MB = 40

Now 160 people do B

This includes people who do only B (what we need) + MB + SB + Do all 3

So only B, I did = 160 - 40 = 120

But according to the formula 2 All 3 + SB + MB = 40... so I am wondering why do I have "2 All 3" giving me the right answer?? Can someone explain this using the set theory formula? Thanks.
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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17 Oct 2010, 11:38
2
gosh... these 3 sets overlapping rules/equations driving me mad! I'd have just done this using simple logic only:

total no of students doing either maths or sociology = 400, but of which we know 120 r doing both, so there are a total 400-120 = 280 students doing maths or/and sociology. now, make this as one single entity in the venn diagram. (of course this only works as we don't really care how many r doing bio &/or maths/sociology, we are only concerned with the one doing bio only! so i didn't go for the complex rules and just focus on the bio group as the primary entity - i really hate 3 sets thing and always try to simplify it to two sets only... doesn't work all the time i know...)

then draw another one for your bio group, so u end up with two groups only!

group A = 280(maths or/and sociology), group B = 160 (bio)

obviously together they can't exceed 400 in number so u merge them together to form the overlap.

it hit 400 when the overlap = 40, ie minimum no of ppl doing both groups therefore max no. of students doing bio only = 160-40 = 120.

does this makes sense?

(the beauty of this is u don't need to even work out what 56% or 44% is, all u need is to focus is the overlap between the first 2 groups which is nice round no 120 and the totla no of students which is 400!)
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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30 Nov 2010, 21:03
pls assess my approach and correct me if i am wrong

s=56 and m = 44 , s+m-(s n m ) = 56+44-30= 70%

Now 70% are in either S or M ( not in both ), so 30% of the total students will be in B , which is 120

Pls correct me if my approach is wrong

thanks
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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30 Nov 2010, 23:21
Hi Dhiren,

I would like to think that since there is a 30% overlap between S and M...S would be 26% and M would be 14% that makes it total of 40% for S or M but not both.

Coming to the question: (S + M + SM(overlap)) ==> 26+14+30 = 70.
Now the question mentions 40% study biology, and the largest number of students who can study biology would be 30%(since the rest 10% would be overlap with Sociology and Mathematics..to put it another way 10% would be the least overlap with Sociaology and Mathematics)

==> 30%(400) = 120

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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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02 Dec 2010, 11:34
hirendhanak wrote:
pls assess my approach and correct me if i am wrong

s=56 and m = 44 , s+m-(s n m ) = 56+44-30= 70%

Now 70% are in either S or M ( not in both ), so 30% of the total students will be in B , which is 120

Pls correct me if my approach is wrong

thanks

Your approach is fine except for one tiny thing:
Now 70% are in either S or M ( not in both ), - instead this is 70% are in S or M or both. (When you subtract 30 above, you are doing it to remove double counting )
So we only have 30% left for only B.
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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02 Dec 2010, 16:02
My approach is similar to hirendhanak's

Div 400 / 4 to get a easy 100 to work with. Therefore,

Total S = 56
Total M = 44
Total B = 40
[S & M] = 30

Number of Only S or Only M or Only B or [S&M&B] or [B&S] or [B&M] = S + M - (S & M) = 70

Since we want max Only B, we set [S&M&B], [B&S] and [B&M] to 0, so

Only S or Only M or Only B = 70
Only B = 70 - Only S - Only M
Only B = 70 - (56 - 30) - (44 - 30) = 30

4 * Only B = 120
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Among 400 students, 56% study sociology, 44% study mathematics a  [#permalink]

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03 Apr 2015, 23:06
1
2
arjtryarjtry wrote:
Among 400 students, 56% study sociology, 44% study mathematics and 40% study biology. If 30% of students study both mathematics and sociology, what is the largest possible number of students who study biology but do not study either mathematics or sociology?

A. 30
B. 90
C. 120
D. 172
E. 188

This is how i solved this questions:
First calculate # of students studying M and S, since the question has given us information about % of students who study both of these. Therefore,
# of students studying M: 44% of 400 = 176
# of students studying S: 56% of 400 = 224
# of students studying both M and S: 30% of 400 = 120
Now we can calculate total number of students studying both M and S : 176+224-120= 280. Therefore largest number of students studying B would be
400(Total)-280 (Studying both M and S)=120
Since the question asked for largest possible number for B and not B only so it was pretty straightforward.
Hope this Helps!
Thanks
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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18 Apr 2015, 05:40
2
Dear Neha

Your solution was quite elegant! This was a good way of solving the question methodically, without having to dabble with too many variables in one go (which, as someone pointed out above, can get tedious).

I would just like to add a bit of explanation after the step where you calculate that the number of students studying both M and S = 120

First calculate # of students studying M and S, since the question has given us information about % of students who study both of these. Therefore,
# of students studying M: 44% of 400 = 176
# of students studying S: 56% of 400 = 224
# of students studying both M and S: 30% of 400 = 120

We see that the total number of students who study either Maths or Sociology = 176 + 224 - 120 = 280

So, in the image we know that the number of students in the zone with the black boundary = 280

Let's assume the number of students who study only biology to be b (this is the number that we have to maximize)

And, let's assume the number of students who study none of the three subjects, that is the number of students in the white space = w

Since the total number of students = 400, we can write:
280 + b + w = 400

Or, b + w = 400 - 280 = 120

That is, b = 120 - w

So, the maximum value of b will happen for w = 0

This is how we get, the maximum value of b = 120

I wanted to explicitly draw out the attention of the students to w. Because, a few of the older solutions above have not even taken w into account. They could still get the right answer because the question here was asking about the maximum value of b (for which, as we saw, w = 0). But in another question, this non-consideration of w (the number of students who study none of the 3 subjects) could lead to a wrong answer.

Hope this helped.

- Japinder
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Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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09 May 2016, 19:05
Hussain15 wrote:
Among 400 students, 56% study sociology, 44% study mathematics and 40% study biology. If 30% of students study both mathematics and sociology, what is the largest possible number of students who study biology but do not study either mathematics or sociology?

A. 30
B. 90
C. 120
D. 172
E. 188

I think the question is more of a logic than equation based only...
we are told from the beginning that 40% students study biology. that's 160 students.
we have 30% (120 students) who study both math and sociology.
if 120 students study all 3 subjects, then 40 of the students study biology only.
if 120 students study only math and sociology, then the maximum # of students to study only biology is 160.
since we do not have 160 here, let's narrow down to what we have.
we know that 40<=# we are looking for<=160

A, D, and E are right away eliminated.
between B and D.

suppose 40 students study all 3 subjects. then 80 study only sociology and math.
therefore, we might have 0 who study sociology and biology, and 0 who study math and biology.
in this case, sociology only = 104, and math only = 56.
we are then left with biology only = 120.

since C is possible, and is definitely greater than B, then C must be the answer.
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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26 Oct 2016, 07:10
Solved it, after few months, in a different way...though it takes longer I believe...
3 venn diagram...
A is in the center
B is Sociology and Biology
C is Sociology and Math
D is Math and Biology
E is biology only
F is sociology only
G is math only

we are then told:
(1) A+B+C+F = 224
(2) A+C+D+G=176
(3) A+B+D+E=160
(4) A+C=120

(5) A+B+C+D+E+F+G=400
from (5), subtract (1)
we are left with:
D+E+G=176
but this is equal to A+C+D+G
D+E+G=A+C+D+G -> E=A+C
but we know for sure that A+C = 120!
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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23 Jun 2017, 03:20
total is 100 (100%)

s=56 m=44 and b=40
s & m = 30

s+m+b= 56+44+40 =140

total - (s+m+b) = 140 - 100 = 40
40% is the total that will be common among s,m,b

40 - s&m = 40 -30 = 10 %

so 10% from b will be shared with either s or m or both.

so b-10 = 40-10 =30 %

30% 400 = 120

ans C

Kudos if it helps
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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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26 Jun 2017, 18:00
Hussain15 wrote:
Among 400 students, 56% study sociology, 44% study mathematics and 40% study biology. If 30% of students study both mathematics and sociology, what is the largest possible number of students who study biology but do not study either mathematics or sociology?

A. 30
B. 90
C. 120
D. 172
E. 188

We can let S denote sociology, M denote mathematics, and B denote biology.

Since %(S or M) = %(S) + %(M) - %(S and M), we have:

%(S or M) = 56 + 44 - 30 = 70

Thus, 70% of the students study sociology or mathematics or both. If we add this percentage to the 40% who study biology, we have 110%, which is not possible since it’s over 100%. So, we must have (at least) 10% of the students who study biology and also study either mathematics or sociology or both. However, we can exclude that 10% of the students and say that the remaining 40% - 10% = 30% of the students study biology only. Thus, the largest possible number of students who study biology but do not study either mathematics or sociology is 0.3 x 400 = 120.

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Re: Among 400 students, 56% study sociology, 44% study mathematics and 40%  [#permalink]

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21 Sep 2018, 05:05
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# Among 400 students, 56% study sociology, 44% study mathematics and 40%

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