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Each of 90 students participated at least 1 of the track try [#permalink]
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19 Feb 2013, 09:39
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Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts? (A) 25 (B) 20 (C) 15 (D) 10 (E) 5
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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Hi, So the formula that we can work from is: Group A + Group B + Group C + Neither  Doubles  2(Triples) = TotalSo 20 + 40 + 60 + 0  Doubles  2(5) = 90
110Doubles = 90
Doubles = 20The above formula is critical. Although it is simple, it is important to really understand it. Once you do, groups will be much easier. Happy Studies, HG.
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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19 Feb 2013, 11:08
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its B We are asked how much are in only two tryouts that means what is (H and L only) + (L and 100m only) + (H and 100m only) Total = H + L + 100m  both + All + None 90 = 20 + 40 + 60  both + 5 + 0 90 = 125  both > both = 35 35 = (H and L) + (L and 100m) + (H and 100m) H and L only = H and L  all > H and L  5 L and 100m only = L and 100m  all > L and 100m  5 H and 100m only = H and 100m  all > H and 100m  5 so (H and L only) + (L and 100m only) + (H and 100m only) = (H and L) + (L and 100m) + (H and 100m)  15 > 35  15 = 20. Regards, Abhijit.
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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mun23 wrote: Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25 (B) 20 (C) 15 (D) 10 (E) 5 A venn diagram can show you exactly what you are supposed to do. Attachment:
Ques4.jpg [ 18.31 KiB  Viewed 9800 times ]
The high jump circle has 20 students. It includes the pink region (only High jump), d, e and 5 (students who tried all 3) The long jump circle has 40 students. It includes the yellow region (only long jump), d, f and 5 (students who tried all 3) The 100 meter dash circle has 60 students. It includes the green region (only 100 meter), e, f and 5 (students who tried all 3) So when you add 20 + 40 + 60, you get 120 which is 30 more than 90. Why? Because in 20+40+60, you have counted, d, e and f twice and 5 thrice. You need to subtract two times 5 and d+e+f once to get 90. 120  2*5  (d+e+f) = 90 (d+e+f) = 20 This is how you get the formula: Group A + Group B + Group C + Neither  Doubles  2(Triples) = Total
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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25 Feb 2013, 03:08
VeritasPrepKarishma wrote: mun23 wrote: Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25 (B) 20 (C) 15 (D) 10 (E) 5 A venn diagram can show you exactly what you are supposed to do. Attachment: Ques4.jpg The high jump circle has 20 students. It includes the pink region (only High jump), d, e and 5 (students who tried all 3) The long jump circle has 40 students. It includes the yellow region (only long jump), d, f and 5 (students who tried all 3) The 100 meter dash circle has 60 students. It includes the green region (only 100 meter), e, f and 5 (students who tried all 3) So when you add 20 + 40 + 60, you get 120 which is 30 more than 90. Why? Because in 20+40+60, you have counted, d, e and f twice and 5 thrice. You need to subtract two times 5 and d+e+f once to get 90. 120  2*5  (d+e+f) = 90 (d+e+f) = 20 This is how you get the formula: Group A + Group B + Group C + Neither  Doubles  2(Triples) = Total I get your method here..however can you please explain the problem mentioned below as well...
The 38 movies in the video store fall into the following three categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action and drama, 3 are both action and comedy, and 4 are both drama and comedy. How many actiondramacomedies are there?
In this question, we are looking for those movies that are in all 3 categories.
The answer I am getting is 1 which is wrong
38=4812(i got this 12 by adding 5+3+4)2x therefore, 2=2x
can you please help me fix this?



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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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25 Feb 2013, 04:27
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Shini1 wrote:
I get your method here..however can you please explain the problem mentioned below as well...
The 38 movies in the video store fall into the following three categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action and drama, 3 are both action and comedy, and 4 are both drama and comedy. How many actiondramacomedies are there?
In this question, we are looking for those movies that are in all 3 categories.
The answer I am getting is 1 which is wrong
38=4812(i got this 12 by adding 5+3+4)2x therefore, 2=2x
can you please help me fix this?
I am glad you asked this question. This is the reason I tell people to not use formulas in sets questions and to rely on their understanding of venn diagrams. The formula discussed above is used in case you know d, e and f i.e. no of people participating in two and only two events. In this question, the given numbers 5, 3 and 4 are not d, e, and f. They are the number of movies which fall into two groups as well as three groups. 5 movies which are both action and drama include those movies which are comedy as well i.e. which fall in all three categories. In this case, the formula we use is P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C) 38 = 48  12 + x x = 2 Think about this formula. P(A) includes P(A n B) which includes P(A n B n C). To avoid double counting in case of P(A), P(B) and P(C), you subtract P(A n B), P(A n C) and P(B n C) once but now you have subtracted P(A n B n C) thrice which means you haven't accounted for it at all. Hence, you add it back.
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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25 Feb 2013, 07:33
Hi, I'm not a big fan of Venn diagrams and have found that for groups, these formulas, once you really practice them, are big time savers. In this case, the formula would be: Singles + Doubles + Triples = Total
24 + 12 + Triples = 38
Triples = 2This is just a derivative of the other formula. You have to play around with this to get the hang of it. OG 13 PS #178 is a good one to practice with. Let me know if you have any questions. Happy studies. HG.
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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25 Feb 2013, 08:23
I get Karishma's method .. Thanks @ HerrGrau how n here do you get your singles from tried the OG prob too thank u ..and do we have your debrief on the site as well ..your verbal is outstanding ..need to know how for that as well ..



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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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25 Feb 2013, 13:39
Hi, You can find the singles by subtracting all of the overlaps from the fake total (I call it the fake total because it is the total that counts all of the movies that are in different categories). In this case we are told that there are 12 overlaps. You have to multiply that by 2 because each overlap is in two groups (for instance, there are 5 doubles in action and in drama so those 5 doubles take up 10 slots). So the singles are 4824 = 24. There is no doubt that the "formula" approach to groups takes practice but it does work very well once you are comfortable with the formulas and how the different groups interact. Maybe a combined approach with Venn Diagrams would work for you. Let me know if you need more advice on this. HG. PS: I am planning on writing a debrief but haven't yet found the time:(
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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mun23 wrote: Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25 (B) 20 (C) 15 (D) 10 (E) 5 Refer diagram below We require to find value of shaded region = a + b + c (Values in Blue are calculated)Setting up the equation 20 + 55  (b+c) + c + 35  (a+c) = 90 a+b+c = 20 Answer = B
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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I get your method here..however can you please explain the problem mentioned below as well...
The 38 movies in the video store fall into the following three categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action and drama, 3 are both action and comedy, and 4 are both drama and comedy. How many actiondramacomedies are there?
In this question, we are looking for those movies that are in all 3 categories.
The answer I am getting is 1 which is wrong
38=4812(i got this 12 by adding 5+3+4)2x therefore, 2=2x
can you please help me fix this?
Refer diagram below We require to find x 10 + 11 + x + 4  x + 11 + x = 38 x = 2
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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22 Nov 2014, 10:06
Hope this helps! Attachment:
Capture.PNG [ 7.47 KiB  Viewed 6975 times ]
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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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07 Nov 2017, 08:11
PareshGmat wrote: I get your method here..however can you please explain the problem mentioned below as well...
The 38 movies in the video store fall into the following three categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action and drama, 3 are both action and comedy, and 4 are both drama and comedy. How many actiondramacomedies are there?
In this question, we are looking for those movies that are in all 3 categories.
The answer I am getting is 1 which is wrong
38=4812(i got this 12 by adding 5+3+4)2x therefore, 2=2x
can you please help me fix this?
Refer diagram below
We require to find x
10 + 11 + x + 4  x + 11 + x = 38
x = 2 Liked your diagram a lot... Just one thing  do we really need to fill all areas in circles? or it is just for illustration? I mean once I have this 5x and 3x and 4x for doubles and X is for triples I can go ahead and resolve it for 10 + 20 + 18  (5x)  (3x)  (4x)  2x = 38 without a need to deduct what the remaming area of circile action is = 10 (5x)  (3x)  x = 2+x.



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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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09 Nov 2017, 09:30
VeritasPrepKarishma wrote: Shini1 wrote: I am glad you asked this question. This is the reason I tell people to not use formulas in sets questions and to rely on their understanding of venn diagrams. The formula discussed above is used in case you know d, e and f i.e. no of people participating in two and only two events. In this question, the given numbers 5, 3 and 4 are not d, e, and f. They are the number of movies which fall into two groups as well as three groups. 5 movies which are both action and drama include those movies which are comedy as well i.e. which fall in all three categories.
In this case, the formula we use is P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C) 38 = 48  12 + x x = 2
Think about this formula. P(A) includes P(A n B) which includes P(A n B n C). To avoid double counting in case of P(A), P(B) and P(C), you subtract P(A n B), P(A n C) and P(B n C) once but now you have subtracted P(A n B n C) thrice which means you haven't accounted for it at all. Hence, you add it back.
For practical reasons, I wonder there could be more than a problem with more than 3 populations (diagrams)?



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Re: Each of 90 students participated at least 1 of the track try [#permalink]
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12 Nov 2017, 07:27
mun23 wrote: Each of 90 students participated at least 1 of the track tryouts: High jump, long jump, 100 meter dash. If 20 students participated in high jump tryout, 40 students participated in the long jump tryout and 60 students participated in the 100 meter dash tryout, and if 5 students participated in all 3 tryouts, how many students participated in only two of these tryouts?
(A) 25 (B) 20 (C) 15 (D) 10 (E) 5 We can use the following formula: Total = number of high jump + number of long jump + number of 100meter dash  number who did two events  2(number who did all 3 events) + number who did zero events Since we see that each student participated in at least 1 event, the number who did zero events is zero. Filling in the rest of the equation, we have: 90 = 20 + 40 + 60  D  2(5) + 0 90 = 110  D D = 20 Answer: B
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