2,000 people take part in a survey to determine the preferences of the public in terms of pizza toppings. 632 say that they like pepperoni on their pizza, 339 like bacon, and 477 like sausage. 223 like both pepperoni and sausage, 112 like pepperoni and bacon, and 66 like bacon and sausage. If 22 like all 3 toppings, how many people do not like any of the toppings on their pizza?
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Very good trick! I liked it. Now I do not need more to make the confused diagram.
Done the same way Finn. This type of problem is coming in my exam this month. Thanks for making me practise!!!😃😊😆 LOL.
exactly the same , it is P.I.E
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It is PIE, however I've simplified it a bit. Do you see why?
nice.................................
I usually use Venn diagram to solve this type of problems but have a feeling your method is quicker.
Venn diagrams are confusing and impractical (in my opinion). I like to look at numbers, not circles.
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i agree, venn diagram < P.I.E
Same here.
2000 - ((632 + 339 + 477) - ((223 + 112 + 66) - 22)) = 2000 - (1448 - (401 - 22)) = 2000 - (1448 - 379) = 2000 - 1069 = 931
x be the number of people who do not like any of the toppings. Then,
Letx = 2 0 0 0 − ( 3 1 9 + 2 0 1 + 2 2 + 9 0 + 4 4 + 1 8 3 + 2 1 0 ) = 2 0 0 0 − 1 0 6 9 = 9 3 1
Using n( AUBUC) formula we can find the number of people who liked either of the 3 toppings so subtracting it from 2000 we get 931
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2 , 0 0 0 − ( 6 3 2 + 3 3 9 + 4 7 7 − 2 2 3 − 1 1 2 − 6 6 + 2 2 ) = 9 3 1 .