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Sets and Relations question

2024 · 4 Apr · Shift 1 · Q57
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Sets and Relations question

2024 · 4 Apr · Shift 1 · Q57

JEE MainMathematicsSets and RelationsNumerical+4 / −1
In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let mmm and nnn respectively be the least and the most number of students who studied all the three subjects. Then m+n\mathrm{m}+\mathrm{n}m+n is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 45

Let

  • MMM = students studying Mathematics,
  • PPP = students studying Physics,
  • CCC = students studying Chemistry.

We are given:

  1. Total students =220=220=220
  2. 101010 studied none

So, ∣M∪P∪C∣=220−10=210|M\cup P\cup C|=220-10=210∣M∪P∪C∣=220−10=210

Also,

\quad 85\le |P|\le 95, \quad 75\le |C|\le 90$$ And $$|P\cap C|=30, \quad |C\cap M|=50, \quad |M\cap P|=40$$ Let $$x=|M\cap P\cap C|$$ We must find the least value $m$ and the greatest value $n$ of $x$. --- ## 1. Use inclusion-exclusion By the principle of inclusion-exclusion, $$|M\cup P\cup C|=|M|+|P|+|C|-|M\cap P|-|P\cap C|-|C\cap M|+|M\cap P\cap C|$$ Substitute the known values: $$210=|M|+|P|+|C|-40-30-50+x$$ $$210=|M|+|P|+|C|-120+x$$ Hence, $$x=330-(|M|+|P|+|C|)$$ So the value of $x$ depends only on the sum $$S=|M|+|P|+|C|$$ with $$x=330-S$$ --- ## 2. Find possible range of $S$ Given ranges: $$125\le |M|\le 130$$ $$85\le |P|\le 95$$ $$75\le |C|\le 90$$ Therefore, $$285=125+85+75\le S\le 130+95+90=315$$ So from $x=330-S$, - to make $x$ minimum, maximize $S$, - to make $x$ maximum, minimize $S$. This gives first: $$x_{\min}=330-315=15$$ $$x_{\max}=330-285=45$$ But we must also check feasibility with pairwise intersections. --- ## 3. Feasibility conditions Since $x=|M\cap P\cap C|$, it must satisfy $$x\le |M\cap P|=40, \quad x\le |P\cap C|=30, \quad x\le |C\cap M|=50$$ Thus, $$x\le 30$$ So although the formula gave a possible upper bound $45$, it is not feasible because triple intersection cannot exceed $|P\cap C|=30$. Hence, $$x\le 30$$ Now check whether $x=30$ is actually possible. If $x=30$, then $$|M|+|P|+|C|=330-30=300$$ This is possible because $300$ lies in the allowed range $[285,315]$. For example, $$|M|=130, \quad |P|=85, \quad |C|=85

which gives 130+85+85=300130+85+85=300130+85+85=300.

Also the Venn regions become:

  • only (P∩C)(P\cap C)(P∩C) but not MMM: 30−30=030-30=030−30=0
  • only (M∩P)(M\cap P)(M∩P) but not CCC: 40−30=1040-30=1040−30=10
  • only (C∩M)(C\cap M)(C∩M) but not PPP: 50−30=2050-30=2050−30=20

Now singles:

  • only M=130−(10+20+30)=70M =130-(10+20+30)=70M=130−(10+20+30)=70
  • only P=85−(10+0+30)=45P =85-(10+0+30)=45P=85−(10+0+30)=45
  • only C=85−(20+0+30)=35C =85-(20+0+30)=35C=85−(20+0+30)=35

All are non-negative, and total is 70+45+35+10+20+0+30=21070+45+35+10+20+0+30=21070+45+35+10+20+0+30=210 So x=30x=30x=30 is feasible.

Therefore, n=30n=30n=30

4. Find the least value of xxx

From above, x=330−Sx=330-Sx=330−S

To minimize xxx, maximize SSS. The largest possible sum is S=315S=315S=315 Rightarrow x=15$$

Check feasibility of x=15x=15x=15. Take

∣P∣=95,∣C∣=90\quad |P|=95, \quad |C|=90∣P∣=95,∣C∣=90

Then S=315S=315S=315.

Now the pair-only regions are:

  • (M∩P)(M\cap P)(M∩P) only =40−15=25=40-15=25=40−15=25
  • (P∩C)(P\cap C)(P∩C) only =30−15=15=30-15=15=30−15=15
  • (C∩M)(C\cap M)(C∩M) only =50−15=35=50-15=35=50−15=35

Singles:

  • only M=130−(25+35+15)=55M =130-(25+35+15)=55M=130−(25+35+15)=55
  • only P=95−(25+15+15)=40P =95-(25+15+15)=40P=95−(25+15+15)=40
  • only C=90−(35+15+15)=25C =90-(35+15+15)=25C=90−(35+15+15)=25

All are non-negative, and total is 55+40+25+25+15+35+15=21055+40+25+25+15+35+15=21055+40+25+25+15+35+15=210 So x=15x=15x=15 is feasible.

Thus, m=15m=15m=15

5. Compute m+nm+nm+n

m+n=15+30=45m+n=15+30=45m+n=15+30=45

Final Answer

45\boxed{45}45​

The derived answer matches the stored correct answer.

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