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In an examination, at least 70% of the students failed in physics, at least 72% failed in chemistry, at least 80% failed in mathematics and at least 85% failed in english. How many at least must have failed in all the four subjects.(all information is sufficient to calculate the answer)

Well, I don't understand what does that 'at least failed' means. And how would you know the exact number of students failing/passing in each subject and then calculating the students failing in all four from venn diagram which would be theoretically very difficult. Please help me!

Asaf Karagila
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Piyush Chaudhary
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  • See a simple example. If there were three students, Alice, Bob and Cecil; and at least two of them failed in maths (suppose they were AC), and at least two of them failed in chemistry (say, AC), then at least one of them (Cecil) must have failed in both, right...? – CiaPan Aug 14 '18 at 14:51

4 Answers4

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Given

Number of students failed in Physics $=$ at least $70$%

Number of students failed in Chemistry $=$ at least $72$%

Number of students failed in Mathematics $=$ at least $80$%

Number of students failed in English $=$ at least $85$%

Number of students passed in Physics $=$ at most $30$%

Number of students failed in Chemistry $=$ at most $28$%

Number of students failed in Mathematics $=$ at most $20$%

Number of students failed in English $=$ at most $15$%

Therefore, the number of students passed all the given subjects $=$ at most $(30+28+20+15)$% $=$ at most $93$%

Therefore, the number of students failed in all the four subjects $=$ at least $100$%$-93$% $=$ at least $7$%

Key Flex
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Hint : Take the worst case (minimum case) and try to solve it by Venn Diagram.

Anik Bhowmick
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Hint

  • If $70\%$ fails Physics and $72\%$ fails Chemistry then, at least, $42\%$ fails both. Don't they?
  • If $42\%$ fails Physic and Chemistry and $80\%$ fails Maths then, at least, $22\%$ fails the tree.

Can you get the answer now?

mfl
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Given: $$\begin{cases}P(Ph)\ge 0.7,\\ P(Ch)\ge 0.72,\\P(M)\ge 0.8,\\ P(E)\ge 0.85 \end{cases}$$ we get: $$P(Ph\cap Ch)=P(Ph)+P(Ch)-P(Ph\cup Ch)\ge P(Ph)+P(Ch)-1\ge 0.42;\\ P(M\cap E)=P(M)+P(E)-P(M\cup E)\ge P(M)+P(E)-1\ge 0.65;\\ P((Ph\cap Ch)\cap(M\cap E))=P(Ph\cap Ch)+P(M\cap E)-P(Ph\cup Ch\cup M\cup E)=\\ P(Ph\cap Ch)+P(M\cap E)-1\ge 0.42+0.65-1=0.07. $$ Similar post.

farruhota
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