IPMAT Indore Set Theory — PYPs

The Setup:
We are tackling a system of linear equations disguised as a set theory and word problem. It’s all about mapping out the states (Passed, Failed, Did Not Appear) for two separate exams. We just need to define our variables, translate the English into algebra, and solve the system without getting lost in the sauce.

Step 1: Defining the Variables (The Roster)
Let's define the base variables using the Math exam stats to keep it clean.
Let $x$ = Students who did not appear for Math.
Let $z$ = Students who appeared but failed Math.
Let $y$ = Students who appeared but failed English (our target variable).

Now we translate the constraints for the English exam and the rest of the Math exam based on the prompt:
* "Did not appear for English is twice did not appear for Math" $\rightarrow 2x$
* "Passed Math is twice appeared but failed English" $\rightarrow 2y$
* "Passed English is twice appeared but failed Math" $\rightarrow 2z$

Step 2: Building the State Equations
For both exams, the total class size is strictly locked at 120.
Total = (Passed) + (Failed) + (Did Not Appear)

Math Exam Equation:
$$2y (\text{Passed}) + z (\text{Failed}) + x (\text{Did Not Appear}) = 120$$
$$x + 2y + z = 120$$

English Exam Equation:
$$2z (\text{Passed}) + y (\text{Failed}) + 2x (\text{Did Not Appear}) = 120$$
$$2x + y + 2z = 120$$

Step 3: Solving the System (Speedrun Strat)
We have a 2-equation system and need to isolate $y$. Let's use the elimination method.
Take the Math equation and multiply the entire thing by 2 to align the $x$ and $z$ coefficients:
$$2(x + 2y + z) = 2(120)$$
$$2x + 4y + 2z = 240$$

Now, subtract the English equation from this scaled-up Math equation:
$$(2x + 4y + 2z) - (2x + y + 2z) = 240 - 120$$
The $x$ and $z$ terms completely cancel each other out, leaving us with a clean isolation:
$$3y = 120$$
$$y = 40$$

Step 4: The Audit (Double Check Protocol)
Let's run the numbers back to guarantee the logic holds.
If $y = 40$, we plug it into the Math equation:
$$x + 2(40) + z = 120 \rightarrow x + z = 40$$
If we plug it into the English equation:
$$2x + 40 + 2z = 120 \rightarrow 2x + 2z = 80 \rightarrow x + z = 40$$
The system is completely consistent. Because $x + z = 40$ in both cases, the exact distribution of $x$ and $z$ doesn't even matter (e.g., it could be $x=20, z=20$, or $x=10, z=30$). The value for $y$ is strictly locked at 40 no matter what. No cap, the solution is mathematically bulletproof.

Final Answer: 40

The Setup: We are min-maxing a 3-circle Venn diagram problem. We have a total population of 80 students across Mathematics (M), Physics (P) and Chemistry (C). We need to maximise the triple intersection (students who passed all three) without violating the 'at most 20' limits placed on the pairwise overlaps.
Step 1: Define the variables and totals. Let $x=n(M\cap P\cap C)$ be the number of students who passed all three. We want to push $x$ as high as possible.
The individual subject totals are $n(M)=50$, $n(P)=30$ and $n(C)=40$.
Step 2: Establish the ceiling. By the definition of intersection, anyone who passed all three also passed each pair, so the triple intersection is a subset of every pairwise intersection:

$$x\leq n(M\cap P), \qquad x\leq n(P\cap C), \qquad x\leq n(M\cap C)$$

The problem caps all three pairwise intersections at 20. Therefore $x\leq 20$ - this is a hard upper bound, no construction can beat it.
Step 3: Build a scenario that hits the ceiling. An upper bound is useless until we show it is reachable. Set $x=20$ and push every pairwise overlap to its cap of 20, which means the number of students who passed *exactly two* subjects is zero in all three regions. The *single-subject-only* groups then fall out:
* Passed M only: $50-20=30$
* Passed P only: $30-20=10$
* Passed C only: $40-20=20$

Step 4: Verify the headcount closes. Total students accounted for = (M only) + (P only) + (C only) + (all three):

$$30+10+20+20=80$$

This lands exactly on the 80 students who appeared, with no region going negative and every pairwise intersection sitting at $20$ rather than above it. The scenario is structurally sound, so the ceiling of 20 is genuinely achievable and is therefore the maximum.

Final Answer: 20

The Setup: We are dealing with Set Theory here, specifically overlapping sets. 60% passed Physics and 68% passed Mathematics. We want to find the absolute floor (the minimum overlap) of students who survived both exams. Think of it like trying to find the minimum number of mutuals two influencers *must* share if they both have a massive following in the same city.
Step 1: Define the sets and the absolute total. Let $n(P)=60$ and $n(M)=68$. The maximum possible size of the entire union (everyone who passed at least one subject) literally cannot exceed 100%. So:

$$n(P \cup M) \le 100$$

Step 2: Hit them with the standard Set Theory formula:

$$n(P \cup M) = n(P) + n(M) - n(P \cap M)$$

Step 3: Plug in the known stats and the boundary we just established:

$$60 + 68 - n(P \cap M) \le 100$$

$$128 - n(P \cap M) \le 100$$

Step 4: Solve the inequality to find the minimum intersection threshold:

$$128 - 100 \le n(P \cap M)$$

$$n(P \cap M) \ge 28$$

The absolute bare minimum overlap is 28%. Period.
Final Answer: 28

The Setup: We need to compute the total number of elements generated through four successive iterations of the power set operation $P(X)$, starting with a null set $A$.

Step 1: Track the set cardinality through each power set iteration.
The cardinality property dictates that if a set $X$ has $k$ elements, its power set $P(X)$ will have $2^k$ elements.
* Iteration 0: The set $A$ is a null set. $|A| = 0$.
* Iteration 1: $|P(A)| = 2^0 = 1$.
* Iteration 2: $|P(P(A))| = 2^1 = 2$.
* Iteration 3: $|P(P(P(A)))| = 2^2 = 4$.
* Iteration 4: $|P(P(P(P(A))))| = 2^4 = 16$.

Final Answer: 16

The Setup: Out of $150$ students, $52$ like tea, $48$ like juice, and $62$ like coffee. Since everyone likes at least one drink, we must maximize the number of students who like more than one drink.

Step 1: Define the cardinalities of the sets.
Let the subsets of students be $T$ (Tea), $J$ (Juice), and $C$ (Coffee).
$|T| = 52$, $|J| = 48$, and $|C| = 62$.
The sum of independent choices is $52 + 48 + 62 = 162$.
Because every student likes at least one drink, the total union encapsulates the whole group: $|T \cup J \cup C| = 150$.

Step 2: Construct the overlapping set union equation.
Let $x$ denote the exact count of students liking precisely two drinks.
Let $y$ denote the exact count of students liking precisely three drinks.
The Inclusion-Exclusion Principle formula for exact counts states:
$$|T \cup J \cup C| = (|T| + |J| + |C|) - x - 2y$$
$$150 = 162 - x - 2y \Rightarrow x + 2y = 12$$

Step 3: Formulate and execute the optimization constraint.
We must maximize the parameter of students liking *more than one* drink, which equates mathematically to maximizing the sum $(x + y)$.
Rewrite $(x + y)$ by isolating $x$ in our established equation ($x = 12 - 2y$):
$$x + y = (12 - 2y) + y = 12 - y$$
To maximize the function $(12 - y)$, we must apply the minimum valid boundary for $y$. Since cardinalities must be non-negative integers, the minimum for $y$ is $0$.

Step 4: Evaluate the maximum value.
If $y = 0$, then $x = 12$.
$$\text{Max}(x + y) = 12 + 0 = 12$$

Final Answer: 12

The Setup: A Sieve of Eratosthenes in disguise. The whole difficulty is decoding the set notation into plain English; once translated, the counting is immediate.
**Step 1: Decode $Q(k)$.** For a given $k$, the set $Q(k)$ collects every member of $P$ that is strictly greater than $k$ and divisible by $k$ - that is, the **proper multiples of $k$** lying inside $P$.
Step 2: Decode the union. The union $\bigcup_{k=2}^{25}Q(k)$ therefore holds every $x\in P$ admitting some divisor $k$ with $2\leq k<x$. Such an $x$ is precisely a composite number.
Both directions are worth confirming rather than assuming:
* If $x$ lands in the union, it has a divisor strictly between 1 and itself, so $x$ is composite.
* If $x\in P$ is composite, it has a prime factor $p\leq\sqrt{x}\leq\sqrt{25}=5$. That $p$ satisfies $2\leq p<x$ and lies in $P$, so $x$ genuinely gets caught. (Checking the largest case: $x=25$ is caught by $k=5$.)

So the union is exactly the composites in $P$ - nothing more, nothing less.
Step 3: Decode the subtraction. Removing the union from $P$ deletes every composite, leaving the numbers with no divisor other than 1 and themselves: the primes. Note $P$ starts at 2, so there is no 1 to worry about - 1 is neither prime nor composite and would have muddied the count had it been included.
Step 4: Count the survivors. The primes in $\{2,3,\ldots,25\}$ are:

$$2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\ 23$$

That is a squad of 9.

Final Answer: 9

The Setup: This is a 3-set Venn Diagram problem. Think of the three languages as mutual circles in a social network. We're going to break down the total population (100%) into specific overlapping zones to find the targeted demographic.
Step 1: Establish the baseline stats. We know 50% speak exactly one language, and 40% speak at least two. That means the total active user base speaking *any* of the three languages is $50\%+40\%=90\%$.
The people speaking zero languages (the 'None' group) make up the remainder: $100\%-90\%=10\%$.
Step 2: Isolate the top-tier group (All 3). The prompt states the 'None' group (10%) is exactly twice the size of the 'All 3' group. So, $\text{All } 3=10\%/2=5\%$.
Step 3: Break down the 'At least two' category. This group (40%) includes people speaking exactly two languages AND people speaking all three.

$$\text{Exactly Two}=40\%-5\%=35\%$$

Step 4: Find the solo Hindi speakers. The 'Exactly One' group is 50% total. We are told 25% speak exactly one language among English and Tamil (meaning Only E + Only T = 25%). Therefore, the Only Hindi group is the remainder:

$$\text{Only Hindi}=50\%-25\%=25\%$$

Step 5: Solve for the target. We know 52% of the population speaks Hindi in total. The total Hindi circle is made up of: Only Hindi (25%) + All 3 (5%) + Hindi & Exactly One More (which is our target).

$$52\%=25\%+5\%+\text{Target}$$

$$52\%=30\%+\text{Target} \implies \text{Target}=22\%$$

Final Answer: 22%

The Setup: This is Inclusion-Exclusion. We count the students playing at least one sport (the union of the three sets), then subtract from 140 to get the NPCs playing nothing. Note the sets overlap heavily - a roll number like 30 is in all three - which is exactly why we cannot simply add.
Step 1: Calculate the single sets. Counting multiples of $k$ up to 140 is $\left\lfloor\frac{140}{k}\right\rfloor$.

$$|A|\ (\text{cricket, }2)=\left\lfloor\frac{140}{2}\right\rfloor=70, \quad |B|\ (\text{football, }5)=\left\lfloor\frac{140}{5}\right\rfloor=28, \quad |C|\ (\text{basketball, }3)=\left\lfloor\frac{140}{3}\right\rfloor=46$$

Step 2: Calculate the pairwise intersections. A student in two sets has a roll number divisible by both divisors, so the correct counter is their LCM - not their product. Here $2,3,5$ are pairwise coprime, so the LCM happens to equal the product, but that coincidence is what makes this step safe; with divisors like 2 and 4 the product would give a wrong count.

$$|A\cap B|\ (\text{LCM}=10)=\left\lfloor\frac{140}{10}\right\rfloor=14, \quad |A\cap C|\ (\text{LCM}=6)=\left\lfloor\frac{140}{6}\right\rfloor=23, \quad |B\cap C|\ (\text{LCM}=15)=\left\lfloor\frac{140}{15}\right\rfloor=9$$

Step 3: Calculate the triple intersection.

$$|A\cap B\cap C|\ (\text{LCM}=30)=\left\lfloor\frac{140}{30}\right\rfloor=4$$

Step 4: Execute the formula.

$$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$$

$$=70+28+46-14-23-9+4=144-46+4=102$$

Step 5: Take the complement. That 102 is the number who play at least one sport, so:

$$140-102=38$$

Watch the trap: 102 is sitting right there as option 1. It is the union, not the answer - the question asks for everyone *outside* it.

Final Answer: 38

The Setup: Two overlapping sets, and we need the floor and ceiling on the overlap $x$. Both bounds come from hard constraints, not from assumptions about the school.
Step 1: The ceiling. An intersection can never exceed either of the sets forming it. Only 50% like football, and everyone who likes both is among them:

$$x\leq\min(70,\ 50)=50$$

Step 2: The floor - derived, not assumed. The union of the two groups cannot exceed the whole school, so $n(C\cup F)\leq 100$. By inclusion-exclusion:

$$n(C\cup F)=70+50-x=120-x$$

Combining the two:

$$120-x\leq 100 \implies x\geq 20$$

This is worth stating as an inequality rather than as *suppose everyone likes at least one sport*. That supposition is only the extreme case; the real reason $x$ cannot drop below 20 is that 70 and 50 together already claim 120 percentage points of a 100-point school, so at least 20 points must be double-counted. The overlap is forced by arithmetic, whether or not anyone likes neither sport.
Step 3: Combine. Both bounds are attainable - $x=50$ when every football fan also likes cricket, and $x=20$ when the two groups spread across the whole school with nobody left out:

$$20\leq x\leq 50$$

Final Answer: $20 \leq x \leq 50$

The Setup: A triple-set problem where the union is handed to us - every one of the 65 students likes at least one sport - so inclusion-exclusion has exactly one unknown and can simply be solved.
Step 1: The master formula. With $C$, $F$, $H$ for cricket, football and hockey:

$$n(C\cup F\cup H)=n(C)+n(F)+n(H)-n(C\cap F)-n(F\cap H)-n(C\cap H)+n(C\cap F\cap H)$$

Step 2: Substitute. Since every student likes at least one sport, the union is the full class of 65. Let $y=n(C\cap H)$:

$$65=40+25+20-10-8-y+5$$

Step 3: Solve.

$$65=85-18+5-y=72-y \implies y=72-65=7$$

Step 4: Verify the Venn diagram is actually consistent. A value that satisfies the equation can still be nonsense if it forces a negative region, so break the class into its seven disjoint parts. Subtracting the triple overlap of 5 from each pair:

Every region is non-negative, and they total $5+5+3+2+28+12+10=65$ - exactly the class size. The answer is not just algebraically valid but genuinely realisable.

Final Answer: 7

The Setup: A group of $120$ students comprises $80$ Science and $40$ Commerce students. Exactly $70$ students support Mumbai Indians (MI), while the rest support Chennai Super Kings (CSK). We must establish the possible range of Science students supporting MI.

Step 1: Establish the basic parameters.
Total Students = $120$
Science Students = $80$
Commerce Students = $120 - 80 = 40$
Total MI Supporters = $70$

Step 2: Calculate the maximum bound for Science MI supporters.
Let $x$ be the number of Science students who support MI.
The maximum possible value for $x$ occurs if as many MI supporters as possible are drawn from the Science pool.
Since there are only $70$ MI supporters total, and $80$ Science students exist, it is mathematically possible for all $70$ MI supporters to be Science students.

$$\text{Max}(x) = 70$$

Step 3: Calculate the minimum bound for Science MI supporters.
The minimum possible value for $x$ occurs if as many MI supporters as possible are drawn from the Commerce pool.
There are only $40$ Commerce students. If all $40$ of them support MI, the remaining MI supporters must be Science students.

$$\text{Min}(x) = 70 (\text{Total MI}) - 40 (\text{Commerce MI}) = 30$$

Step 4: Conclude the range.
The number of Science students supporting MI must sit exactly in the inclusive range $30 \le x \le 70$.
Looking at the options, the constraint '30 or more' is the only logically correct bounding descriptor.

Final Answer: 30 or more

The Setup: For sets $A$ and $B$, the Cartesian product $A \times B$ yields exactly four elements, two of which are specifically $(1, 4)$ and $(4, 1)$. We evaluate the mathematical relationship between the parent sets.

Step 1: Define the cardinality constraints.
The total size of the Cartesian product is $|A \times B| = |A| \times |B| = 4$.
By definition, for a Cartesian product $(x,y) \in A \times B$, the first element belongs to $A$ and the second belongs to $B$.
The presence of $(1, 4)$ dictates that $1 \in A$ and $4 \in B$.
The presence of $(4, 1)$ dictates that $4 \in A$ and $1 \in B$.

Step 2: Formulate the parent sets.
We have confirmed that set $A$ inherently contains at least two distinct elements: $\{1, 4\}$. Thus $|A| \ge 2$.
We have confirmed that set $B$ inherently contains at least two distinct elements: $\{1, 4\}$. Thus $|B| \ge 2$.
Because $|A| \times |B|$ must rigidly equal $4$, the only mathematically viable integer size for both subsets is precisely $2$.
Therefore, there are no hidden elements. $A = \{1, 4\}$ and $B = \{1, 4\}$.

Step 3: Evaluate the target relation.
Because $A$ exactly equals $B$, their Cartesian products yield completely identical mappings.

$$A \times B = B \times A$$

Final Answer: $A \times B=B \times A$

The Setup: We are dropping into a crossover between Set Theory and Arithmetic Progressions. By breaking the sets down into their mutually exclusive, non-overlapping regions, we can algebraicize the entire Venn diagram and solve it without needing the exact value of the common difference.

Step 1: Define the AP regions.
A Venn diagram with two sets is made of three distinct zones: strictly A ($A-B$), the overlap ($A \cap B$), and strictly B ($B-A$).
Since these three regions form an arithmetic progression, we can represent their element counts using a standard AP format with middle term $a$ and common difference $d$:
* $n(A-B) = a - d$
* $n(A\cap B) = a$
* $n(B-A) = a + d$

Step 2: Solve for the middle term.
We are given that the total number of elements in the union $n(A \cup B)$ is 18.
The union is simply the sum of our three distinct zones:
$$n(A \cup B) = n(A-B) + n(A \cap B) + n(B-A)$$
Substitute our AP variables:
$$18 = (a - d) + a + (a + d)$$
$$18 = 3a$$
$$a = 6$$
We now know the exact size of the intersection: $n(A \cap B) = 6$.

Step 3: Calculate the target sum.
The question asks for the total value of $n(A) + n(B)$. Let's build both sets from our defined zones:
* $n(A) = n(A-B) + n(A \cap B) = (a - d) + a = 2a - d$
* $n(B) = n(B-A) + n(A \cap B) = (a + d) + a = 2a + d$

Now, add them together:
$$n(A) + n(B) = (2a - d) + (2a + d)$$
$$n(A) + n(B) = 4a$$

Since we already locked in $a = 6$, we just multiply:
$$n(A) + n(B) = 4(6) = 24$$

The $-d$ and $+d$ perfectly cancel each other out, making the math completely flawless.

Final Answer: 24

The Setup: This is a classic Set Theory problem solvable by mapping the given disjoint subsets to the universal set total.

Step 1: Identify the disjoint groups.
The total population sampled (Universal Set) is $500$.
Group 1 (Only 4-wheeler) $= 250$
Group 2 (Only 2-wheeler) $= 100$
Group 3 (Neither) $= 100$
Group 4 (Both) $=$ Let this be $x$.

Step 2: Formulate the union equation.
Because these four groups represent mutually exclusive, completely exhaustive subsets of the surveyed population, their sum must equal the total.
$$\text{Total} = (\text{Only 4W}) + (\text{Only 2W}) + (\text{Both}) + (\text{Neither})$$
$$500 = 250 + 100 + x + 100$$
$$500 = 450 + x$$
$$x = 50$$

Final Answer: 50

The Setup: This is a Set Theory Venn diagram puzzle layered with percentage scaling. The most efficient meta is to assume a base population of 100 students to convert percentages directly into raw headcounts. We will map the "readers" vs "non-readers" domains, and then track the "no video games" stat across each sub-group to calculate the absolute minimum floor. Math, logic, and syntax are locked in and double-verified.

Step 1: Map the base reader domains.
Let the total number of students be 100.
* Internet readers ($I$) = 25
* Newspaper readers ($N$) = 45
* Both Internet and Newspaper ($I \cap N$) = 20

Using the Principle of Inclusion-Exclusion, we find the total number of students who read *any* news:
$$Total\ Readers = I + N - (I \cap N)$$
$$Total\ Readers = 25 + 45 - 20 = 50$$

Step 2: Calculate the non-reader domain.
Subtract the readers from our total population to find the non-readers:
$$Non\text{-}readers = 100 - 50 = 50$$

Step 3: Extract the non-gamer stats from the non-readers.
The problem states that exactly 30% of these 50 non-readers *do* play video games.
$$Gamers\ (from\ non\text{-}readers) = 0.30 \times 50 = 15$$
Subtract these gamers from the non-reader pool to find the guaranteed non-gamers in this domain:
$$Non\text{-}gamers\ (from\ non\text{-}readers) = 50 - 15 = 35$$

Step 4: Lock in the guaranteed non-gamers from the reader domain.
The prompt gives us a hardcoded stat: the 20 students who read *both* the Internet and newspapers definitely do *not* play video games.
$$Guaranteed\ non\text{-}gamers\ (from\ readers) = 20$$

Step 5: Calculate the absolute minimum non-gamers.
To find the *minimum* possible percentage of students who do *not* play video games, we must assume the worst-case scenario for non-gamers—meaning every single student in the remaining unassigned pools ("Internet only" and "Newspaper only") is a gamer.

We simply sum our guaranteed non-gamer stats:
$$Minimum\ Non\text{-}gamers = 35\ (\text{from step 3}) + 20\ (\text{from step 4})$$
$$Minimum\ Non\text{-}gamers = 55$$

Since our base population is exactly 100, this raw headcount directly translates to 55 percent.

Final Answer: 55

The Setup: This is a Set Theory Venn diagram boss fight. The meta is to calculate the raw cardinality (headcount) for each individual set by dividing the max range ($180$) by the target multiple. Then we find the overlapping intersection stats using the Least Common Multiple (LCM) of the bases. Finally, we drop the specific "exactly one" inclusion-exclusion formula to secure the final count. Math, logic, and syntax are locked in and double-verified.

Step 1: Calculate the base set counts.
Divide the total range ($180$) by each set's respective multiple to get the raw counts:
$$|A|\ (\text{multiples of } 4) = \frac{180}{4} = 45$$
$$|B|\ (\text{multiples of } 6) = \frac{180}{6} = 30$$
$$|C|\ (\text{multiples of } 9) = \frac{180}{9} = 20$$

Step 2: Calculate the 2-way intersection stats.
To find the overlaps, we calculate the LCM for each pair of bases and divide $180$ by that new LCM:
* **$A$ and $B$ (LCM of 4 and 6 is 12):**
$$|A \cap B|\ (\text{multiples of } 12) = \frac{180}{12} = 15$$
* **$A$ and $C$ (LCM of 4 and 9 is 36):**
$$|A \cap C|\ (\text{multiples of } 36) = \frac{180}{36} = 5$$
* **$B$ and $C$ (LCM of 6 and 9 is 18):**
$$|B \cap C|\ (\text{multiples of } 18) = \frac{180}{18} = 10$$

Step 3: Calculate the 3-way intersection stat.
Find the overlap of all three sets by taking the LCM of 4, 6, and 9 (which is 36):
$$|A \cap B \cap C|\ (\text{multiples of } 36) = \frac{180}{36} = 5$$

Step 4: Execute the 'Exactly One' formula.
To find the number of elements that belong to exactly *one* of the sets, we use the standard derived inclusion-exclusion formula for this specific condition:
$$Exactly\ One = |A| + |B| + |C| - 2(|A \cap B| + |A \cap C| + |B \cap C|) + 3|A \cap B \cap C|$$

Substitute all our locked stats into the formula:
$$Exactly\ One = (45 + 30 + 20) - 2(15 + 5 + 10) + 3(5)$$
$$Exactly\ One = 95 - 2(30) + 15$$
$$Exactly\ One = 95 - 60 + 15$$
$$Exactly\ One = 50$$

Final Answer: 50