[tex]\large\underline{\sf{Solution-}}[/tex]
Let assume that total number of students be 100.
According to statement,
[tex]\sf \: n(C) = 60\% \: = \: \frac{60}{100} \times 100 = 60 \\ \\ [/tex]
[tex]\sf \: n(F) = 30\% \: = \: \frac{30}{100} \times 100 = 30 \\ \\ [/tex]
[tex]\sf \: n(B) = 40\% \: = \: \frac{40}{100} \times 100 = 40 \\ \\ [/tex]
[tex]\sf \: n(C\cap F) = 10\% \: = \: \frac{10}{100} \times 100 = 10 \\ \\ [/tex]
[tex]\sf \: n(F\cap B) = 10\% \: = \: \frac{10}{100} \times 100 = 10 \\ \\ [/tex]
[tex]\sf \: n(C\cap B) = 15\% \: = \: \frac{15}{100} \times 100 = 15 \\ \\ [/tex]
[tex]\sf \: n(F\cup C\cup B) = 100\% \: = \: \frac{100}{100} \times 100 = 100 \\ \\ \\ [/tex]
We know,
[tex]\bf \: n(F\cup C\cup B) \\ \\ [/tex]
[tex]\sf \: = n(F) + n(C) + n(B) - n(F\cap C) - n(C\cap B) - n(B\cap F) + n(F\cap B\cap C) \\ \\ [/tex]
[tex]\sf \: 100 = 60 + 30 + 40 - 10 - 10 - 15 + n(F\cap C\cap B) \\ \\ [/tex]
[tex]\sf \: 100 = 130 - 35 + n(F\cap C\cap B) \\ \\ [/tex]
[tex]\sf \: 100 = 95 + n(F\cap C\cap B) \\ \\ [/tex]
[tex]\sf \: 100 - 95 = n(F\cap C\cap B) \\ \\ [/tex]
[tex]\sf \: \bf\implies \: n(F\cap C\cap B) = 5 \\ \\ [/tex]
[tex]\sf \: \bf\implies \: n(F\cap C\cap B) = \frac{5}{100} \times 100 = 5\% \\ \\ [/tex]
Hence, 5 % students play all the three games.
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
[tex]\large\underline{\sf{Solution-}}[/tex]
Let assume that total number of students be 100.
According to statement,
[tex]\sf \: n(C) = 60\% \: = \: \frac{60}{100} \times 100 = 60 \\ \\ [/tex]
[tex]\sf \: n(F) = 30\% \: = \: \frac{30}{100} \times 100 = 30 \\ \\ [/tex]
[tex]\sf \: n(B) = 40\% \: = \: \frac{40}{100} \times 100 = 40 \\ \\ [/tex]
[tex]\sf \: n(C\cap F) = 10\% \: = \: \frac{10}{100} \times 100 = 10 \\ \\ [/tex]
[tex]\sf \: n(F\cap B) = 10\% \: = \: \frac{10}{100} \times 100 = 10 \\ \\ [/tex]
[tex]\sf \: n(C\cap B) = 15\% \: = \: \frac{15}{100} \times 100 = 15 \\ \\ [/tex]
[tex]\sf \: n(F\cup C\cup B) = 100\% \: = \: \frac{100}{100} \times 100 = 100 \\ \\ \\ [/tex]
We know,
[tex]\bf \: n(F\cup C\cup B) \\ \\ [/tex]
[tex]\sf \: = n(F) + n(C) + n(B) - n(F\cap C) - n(C\cap B) - n(B\cap F) + n(F\cap B\cap C) \\ \\ [/tex]
[tex]\sf \: 100 = 60 + 30 + 40 - 10 - 10 - 15 + n(F\cap C\cap B) \\ \\ [/tex]
[tex]\sf \: 100 = 130 - 35 + n(F\cap C\cap B) \\ \\ [/tex]
[tex]\sf \: 100 = 95 + n(F\cap C\cap B) \\ \\ [/tex]
[tex]\sf \: 100 - 95 = n(F\cap C\cap B) \\ \\ [/tex]
[tex]\sf \: \bf\implies \: n(F\cap C\cap B) = 5 \\ \\ [/tex]
[tex]\sf \: \bf\implies \: n(F\cap C\cap B) = \frac{5}{100} \times 100 = 5\% \\ \\ [/tex]
Hence, 5 % students play all the three games.