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HarveyS
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Of the 100 athletes at a soccer club, 40 play defense and 70 23 Mar 2014, 01:40
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C
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35% (medium)

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71% (01:47) correct 29% (01:47) wrong based on 605 sessions

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Of the 100 athletes at a soccer club, 40 play defense and 70 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between

A. 10 to 20
B. 10 to 40
C. 30 to 40
D. 30 to 70
E. 40 to 70
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C
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Bunuel
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Of the 100 athletes at a soccer club, 40 play defense and 70 23 Mar 2014, 05:33
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Mountain14
Of the 100 athletes at a soccer club, 40 play defense and 70 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between

A. 10 to 20
B. 10 to 40
C. 30 to 40
D. 30 to 70
E. 40 to 70
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First of all notice that since only 40 athletes play defense, then the number of athletes that play both midfield and defense cannot possibly be greater than 40. Eliminate D and E.

{Total} = {defense} + {midfield} - {both} + {neither}

100 = 40 + 70 - {both} + {neither}

{both} = {neither} + 10.

Since the least value of {neither} is given to be 20, then the least value of {both} is 20+10=30. Eliminate A and B.

Answer: C.
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Of the 100 athletes at a soccer club, 40 play defense and 70 25 Jul 2014, 02:49
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Refer diagram below

We require to find the shaded region

Setting up the equation

40 + 70-x + 20 = 100

x = 30

For defence = 40; the maximum value for which x can reach is also 40

Range = 30 to 40

Answer = C
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foot.png [ 4.77 KiB | Viewed 9769 times ]

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Of the 100 athletes at a soccer club, 40 play defense and 70 12 Sep 2015, 00:34
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Mountain14
Of the 100 athletes at a soccer club, 40 play defense and 70 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between

A. 10 to 20
B. 10 to 40
C. 30 to 40
D. 30 to 70
E. 40 to 70
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Fill the attached table and ull get the answer in less than a min

Since max number of athletes who can play defense s only 40, after filling the table u can conclude that the (c) 30 to 40 is the answer
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Screenshot_2015-09-12-13-01-11.png
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Of the 100 athletes at a soccer club, 40 play defense and 70 09 Aug 2017, 04:46
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HarveyS
Of the 100 athletes at a soccer club, 40 play defense and 70 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between

A. 10 to 20
B. 10 to 40
C. 30 to 40
D. 30 to 70
E. 40 to 70
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Here is how you can think about it:

We are given the minimum value of "Neither" and we need to find the minimum and maximum value of "Both".

Total = A + B - Both + Neither

Both = 40 + 70 - 100 + Neither

Both = 10 + Neither

Minimum value of Neither is 20.

Maximum value of Neither will be obtained when "defense" circle in inside the "midfield" circle. So 100 - 70 = 30 would be Neither.

So minimum value of Both is 10+20 = 30 and maximum value is 10+30 = 40.

Answer (C)
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Of the 100 athletes at a soccer club, 40 play defense and 70 03 Apr 2019, 18:08
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HarveyS
Of the 100 athletes at a soccer club, 40 play defense and 70 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between

A. 10 to 20
B. 10 to 40
C. 30 to 40
D. 30 to 70
E. 40 to 70
Show more

We can use the formula for overlapping sets:

Total = Midfield + Defense - Both + Neither

Now, using the least number of athletes who play neither position (20 players), we have:

100 = 70 + 40 - x + 20

100 = 130 - x

x = 30

So 30 is the least number of athletes who play both positions. However, the number of athletes who play both positions can’t exceed the number of athletes who play defense. Therefore, the greatest number of athletes who play both positions is 40.

Answer: C
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Of the 100 athletes at a soccer club, 40 play defense and 70 09 Jan 2024, 08:20
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https://gmatclub.com/forum/of-the-400-m ... 23692.html
Same technique as Bunuel's Table Matrix here.
Just be aware of the constraint that the max value can be 40 instead of 70
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