A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics and Chemistry. It was found that all students passed in atleast one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, atmost 11 students passed in both Mathematics and Physics, atmost 15 students passed in both Physics and Chemistry, atmost 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is \underline\hspace2cm .

Probability PYQ — JEE Main 2024 · 30 Jan · Shift 1 · Q51 with Solution

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Correct answer: 10

Let

$M$ = set of students who passed Mathematics,
$P$ = set of students who passed Physics,
$C$ = set of students who passed Chemistry.

We need the maximum possible value of $x=|M\cap P\cap C|.$

1. Use inclusion-exclusion

By the inclusion-exclusion principle, $|M\cup P\cup C|=|M|+|P|+|C|-|M\cap P|-|P\cap C|-|M\cap C|+|M\cap P\cap C|.$

Substitute the known values: $40=20+25+16-|M\cap P|-|P\cap C|-|M\cap C|+x.$

Since $20+25+16=61,$ we get $40=61-(|M\cap P|+|P\cap C|+|M\cap C|)+x.$ So, $|M\cap P|+|P\cap C|+|M\cap C|=21+x.$

2. Use the upper bounds on pairwise intersections

3. Pairwise intersections must each contain the triple intersection

In particular, $x\le |M\cap P|\le 11.$ So, $x\le 11.$

Now we check whether $x=11$ is possible.

4. Test $x=11$

Since each is at most $15$ , this is possible numerically. But we must also respect subject totals.

Let the numbers in exactly two subjects be:

$a$ = passed only $M\cap P$ ,
$b$ = passed only $P\cap C$ ,
$c$ = passed only $M\cap C$ ,
$x$ = passed all three.

If $x=11$ and $|M\cap P|=11$ , then $a=0.$

Now count students in $M$ : $|M|=(\text{only }M)+a+c+x=20.$ Since $a=0$ , $(\text{only }M)+c+11=20,$ so $(\text{only }M)+c=9.$ Hence, $c\le 9.$ Thus, $|M\cap C|=c+x\le 9+11=20,$ but more importantly from $c\le 9$ we get $|M\cap C|\le 20,$ which is still fine. Let us use Chemistry: $|C|=(\text{only }C)+b+c+x=16,$ so $(\text{only }C)+b+c+11=16,$ therefore $b+c\le 5.$ Hence, $|P\cap C|+|M\cap C|=(b+11)+(c+11)=b+c+22\le 27.$ But from above we need $|P\cap C|+|M\cap C|=21,$ which is possible. So we need a sharper check.

Let us use the exact-only-region formulation.

5. Venn diagram variables

Let

$m$ = only Mathematics,
$p$ = only Physics,
$c$ = only Chemistry,
$u$ = only $M\cap P$ ,
$v$ = only $P\cap C$ ,
$w$ = only $M\cap C$ ,
$x$ = all three.

Then $m+u+w+x=20 \quad (1)$ $p+u+v+x=25 \quad (2)$ $c+v+w+x=16 \quad (3)$ $m+p+c+u+v+w+x=40 \quad (4)$

Also,

\quad v+x\le 15, \quad w+x\le 15.$$ Now add (1), (2), (3): $$m+p+c+2(u+v+w)+3x=61.$$ Subtract (4): $$u+v+w+2x=21.$$ Thus, $$u+v+w=21-2x. \quad (5)$$ Since $u,v,w\ge 0$, we need $$21-2x\ge 0 \implies x\le 10.5.$$ Hence, $$x\le 10.$$ So the maximum possible value is at most $10$. --- ## 6. Check whether $x=10$ is achievable If $x=10$, then from (5), $$u+v+w=21-20=1.$$ We must also satisfy: $$u+10\le 11 \implies u\le 1,$$ $$v+10\le 15 \implies v\le 5,$$ $$w+10\le 15 \implies w\le 5.$$ So choosing, for example, $$u=1,\quad v=0,\quad w=0$$ is valid. Now from (1), (2), (3): $$m+1+0+10=20 \implies m=9,$$ $$p+1+0+10=25 \implies p=14,$$ $$c+0+0+10=16 \implies c=6.$$ All are non-negative, and total: $$9+14+6+1+0+0+10=40,$$ which works. Therefore, $x=10$ is achievable. --- ## 7. Final answer The maximum number of students who passed all three subjects is $$\boxed{10}.$$

1. Use inclusion-exclusion

By the inclusion-exclusion principle, $|M\cup P\cup C|=|M|+|P|+|C|-|M\cap P|-|P\cap C|-|M\cap C|+|M\cap P\cap C|.$

Substitute the known values: $40=20+25+16-|M\cap P|-|P\cap C|-|M\cap C|+x.$

Since $20+25+16=61,$ we get $40=61-(|M\cap P|+|P\cap C|+|M\cap C|)+x.$ So, $|M\cap P|+|P\cap C|+|M\cap C|=21+x.$

2. Use the upper bounds on pairwise intersections

3. Pairwise intersections must each contain the triple intersection

In particular, $x\le |M\cap P|\le 11.$ So, $x\le 11.$

Now we check whether $x=11$ is possible.

4. Test $x=11$

Since each is at most $15$ , this is possible numerically. But we must also respect subject totals.

Let the numbers in exactly two subjects be:

$a$ = passed only $M\cap P$ ,
$b$ = passed only $P\cap C$ ,
$c$ = passed only $M\cap C$ ,
$x$ = passed all three.

If $x=11$ and $|M\cap P|=11$ , then $a=0.$

Let us use the exact-only-region formulation.

5. Venn diagram variables

Let

$m$ = only Mathematics,
$p$ = only Physics,
$c$ = only Chemistry,
$u$ = only $M\cap P$ ,
$v$ = only $P\cap C$ ,
$w$ = only $M\cap C$ ,
$x$ = all three.

Then $m+u+w+x=20 \quad (1)$ $p+u+v+x=25 \quad (2)$ $c+v+w+x=16 \quad (3)$ $m+p+c+u+v+w+x=40 \quad (4)$

Also,

\quad v+x\le 15, \quad w+x\le 15.$$ Now add (1), (2), (3): $$m+p+c+2(u+v+w)+3x=61.$$ Subtract (4): $$u+v+w+2x=21.$$ Thus, $$u+v+w=21-2x. \quad (5)$$ Since $u,v,w\ge 0$, we need $$21-2x\ge 0 \implies x\le 10.5.$$ Hence, $$x\le 10.$$ So the maximum possible value is at most $10$. --- ## 6. Check whether $x=10$ is achievable If $x=10$, then from (5), $$u+v+w=21-20=1.$$ We must also satisfy: $$u+10\le 11 \implies u\le 1,$$ $$v+10\le 15 \implies v\le 5,$$ $$w+10\le 15 \implies w\le 5.$$ So choosing, for example, $$u=1,\quad v=0,\quad w=0$$ is valid. Now from (1), (2), (3): $$m+1+0+10=20 \implies m=9,$$ $$p+1+0+10=25 \implies p=14,$$ $$c+0+0+10=16 \implies c=6.$$ All are non-negative, and total: $$9+14+6+1+0+0+10=40,$$ which works. Therefore, $x=10$ is achievable. --- ## 7. Final answer The maximum number of students who passed all three subjects is $$\boxed{10}.$$

Probability question

Probability question

1. Use inclusion-exclusion

2. Use the upper bounds on pairwise intersections

3. Pairwise intersections must each contain the triple intersection

4. Test $x=11$

5. Venn diagram variables

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