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Temka [501]
3 years ago
5

On a school spirit day 257 students wore gold or black, 34 wore both gold and black, and 178 wore black. How many students wore

only gold?
(Proof not needed, only LEGIT answers only please and thx)

Mathematics
2 answers:
barxatty [35]3 years ago
8 0
The answer is A hope this helps
pentagon [3]3 years ago
7 0

Answer:

2 possible answers

Step-by-step explanation:

Assumption: 178 wore black (that means some may wear gold + black aka as long as they wore abit of black, it's counted):

since there are 257 students and 34 wore both gold and black, and 178 wore black, the number of students who wore only black = 178 - 34 = 144

no of students who wore only gold = 257 - 144 - 34 = 79

Assumption: 178 students wore ONLY black:

no of students who wore only gold: 257 - 178 - 34 = 45

Topic: Set theory

If you like to venture further, feel free to check out my insta (learntionary). It would be best if you could give it a follow. I'll be constantly posting math tips and notes! Thanks!

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{\tt \leadsto \dfrac{(-5)}{6} \times \dfrac{9}{20} + \dfrac{(-5)}{6} + \dfrac{7}{25}}

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{\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{9}{20} + \dfrac{7}{25} \bigg)}

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{\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{9 \times 5}{20 \times 5} + \dfrac{7 \times 4}{25 \times 4} \bigg)}

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{\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{45 + 28}{100} \bigg)}

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{\tt \leadsto \dfrac{\cancel{(-5)} \times 73}{6 \times \cancel{100}} = \dfrac{(-1) \times 73}{6 \times 20}}

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{\tt \leadsto \dfrac{(-73)}{120}}

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{\red{\underline{\boxed{\bf So, \: the \: answer \: obtained \: is \: \: \dfrac{(-73)}{120}}}}}

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