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by EbrahimHashem » Thu Aug 20, 2009 9:25 pm
In a symposium of 400 people, it was observed that 200 write with Reynolds, 70 with Parker and 50 with Paper mate pens. 30 people had both Reynolds and Parker. 20 had Reynolds and Paper mate and 8 had Parker and Paper mate. How many had Reynolds alone?

A)154
B)254
C)354
D)454
E)554
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Source: — Problem Solving |

by iamjakekim » Thu Aug 20, 2009 10:29 pm
must be lower than 200, so A!!
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by tom4lax » Fri Aug 21, 2009 4:59 am
Not sure I fully understand the question.

Reynolds = 200
Parker = 70
Papermate = 50
Total = 320

Rey & Par = 30
Rey & Pap = 20
Par & Pap = 8
Total = 58

Total = (# w/ one group) - (# w/ two groups) - 2(#w/ three groups)

400 = 320 - 58 - 2x
400 = 262 - 2x
Basically what I'm getting at is that total is less than 400, so there are no in the "3" group.

Looking only at Reynolds: 200 - 50 = 150.

I also would have guessed A here, but what am I missing?
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by real2008 » Fri Aug 21, 2009 8:35 am
150<=reynolds alone<=158
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