Five friends A, B, C, D, and E go on a trip together. Each of them likes at least one of the following three activities: Countryside Sightseeing, Shopping, and Adventure Sports such that:
- No two friends like exactly the same set of activities.
- A, B and E like exactly two activities each.
- D likes only Countryside Sightseeing.
- C likes more number of activities compared to D.
The number of friends who like both Shopping and Adventure Sports is ___
Enter your answer to attempt
The Setup: This is a classic Logical Reasoning/Arrangements puzzle (IPMAT Indore vibes). Think of it as assigning unique loadouts to a 5-player squad from a pool of 3 perks (Countryside Sightseeing, Shopping, Adventure Sports). The golden rule here is the "no copycats" policy—every friend must have a totally unique combination of activities.
Step 1:Analyze the NPCs (D and C).
The prompt states D is a one-trick pony, only liking Countryside Sightseeing (CS). So, D's loadout size is exactly 1.
The rules also state C likes *more* activities than D. Since the max activities available is 3, C's loadout size must be either 2 or 3.
Step 2:Distribute the Duo Loadouts (A, B, E).
We are told A, B, and E each like *exactly* two activities. Let's calculate the total possible 2-activity combos from our pool of 3 (CS, Shopping=Sh, Adventure Sports=AS).
Using combinations:
3C2=3
The only possible duo sets are {CS, Sh}, {CS, AS}, and {Sh, AS}.
Step 3:Apply the "No Copycats" Rule.
Since no two friends can have the exact same set of activities, A, B, and E must each claim exactly one of these three unique duo combinations. All three 2-activity loadouts are now permanently locked and taken by this trio.
Step 4:Lock in C's Loadout.
Let's check back on C. We established C needs 2 or 3 activities. But wait—all the 2-activity slots are already hogged by A, B, and E! If C took a 2-activity set, they'd duplicate someone else's set, failing the vibe check (violating the main rule). Thus, C is forced to take the only remaining unique loadout size greater than 1: all 3 activities {CS, Sh, AS}. C is literally doing the 100% completionist run.
Step 5:Visualize the Final Roster.
Let's drop all this intel into a matrix to see exactly who is doing what.
Friend
CS
Sh
AS
D
✓
—
—
One of A/B/E
✓
✓
—
One of A/B/E
✓
—
✓
One of A/B/E
—
✓
✓
C
✓
✓
✓
From the table, look at the Sh and AS columns to see who is running the dual Shopping/Adventure Sports build. We count the rows where both show a ✓.
That's exactly the third A/B/E row and C. Total = 2.
Final Answer: 2
Ram purchased 3 oranges at Rs. 20 each and 5 mangoes at Rs. 38 each and then sold them by offering discounts of 25% and 20% on the fixed marked prices of oranges and mangoes, respectively. If he earned a profit of 50% on selling the oranges alone, and a total profit of 32% on selling all the fruits, then the marked price of each mango was ___
Enter your answer to attempt
The Setup: This is a classic Profit & Loss side quest. Ram is basically running a whole fruit cartel. We are given the final profit margins and need to reverse-engineer the original marked price (the bait) of the mangoes. We'll break this down by isolating the cost and revenue for each fruit type.
Step 1: Calculate the base stats (Total Cost Price).
Ram drops some cash upfront to acquire the inventory. We calculate the total cost price by multiplying the quantity by the price per unit.
TotalCP=3×20+5×38=60+190=250rupeesStep 2: Calculate the total loot (Total Selling Price).
The problem states he secures a massive 32% overall profit on the entire stash. We apply a 1.32 multiplier to the total cost.
TotalSP=250×1.32=330rupeesStep 3: Isolate the orange revenue.
The oranges alone gave a cracked 50% profit buff. The cost price for just the oranges was 60 rupees.
OrangesSP=60×1.5=90rupeesStep 4: Find the mango selling price.
By subtracting the orange revenue from the total revenue, we find exactly how much the mango bag sold for.
MangoesSP(for5)=330−90=240rupees
To find the selling price of just one mango, divide by the squad size of 5.
SPofonemango=5240=48rupeesStep 5: Reverse-engineer the Marked Price.
Ram offered a 20% discount on mangoes to pass the vibe check for his customers. This means the final selling price is exactly 80% of the marked price (SP=0.80×MP). We divide the single mango SP by 0.80 to find the original inflated tag.
MP=0.8048=60rupeesFinal Answer: 60
Q3:ipmat indore 2026QA › RemainderMediumSA · TITA
The remainder when 7103+7101 is divided by 9 is ____
Enter your answer to attempt
The Setup: We are diving straight into modular arithmetic. We need to find the remainder of a massive exponent expression. Instead of brute-forcing a number that would literally crash a calculator, we factor out the base and find the cyclicity (the repeating pattern) of its remainders. We are basically finding the modulo meta to speedrun the solution.
Step 1: Min-Max the expression (Factorization).
Don't fight the exponents directly. Pull out the highest common factor to simplify the battlefield.
7103+7101=7101(72+1)7101(49+1)=7101×50Step 2: Find the cyclicity of 7 mod 9.
We need to see how powers of 7 behave when divided by 9. We are looking for a remainder of 1, because that tells us when the pattern resets.
71≡7(mod9)72=49≡4(mod9)73=72×7≡4×7=28≡1(mod9)
Boom. A remainder of 1 means the cycle resets every 3 powers. The cycle length is exactly 3.
**Step 3: Analyze the main character's power level (7101).**
Divide the main exponent (101) by our cycle length (3) to see exactly where it lands in the rotation.
101=3×33+2
Since the remainder is 2, 7101 behaves exactly like 72 in the modulo 9 universe.
7101≡72≡4(mod9)Step 4: Resolve the multiplier (50).
We also need to find the remainder of our constant multiplier (50) when divided by 9.
50≡5(mod9)Step 5: Calculate the final damage output.
Multiply the isolated remainders together and take mod 9 one last time to get the ultimate answer.
7101×50≡4×5(mod9)20≡2(mod9)Final Answer: 2
Five friends A, B, C, D, and E go on a trip together. Each of them likes at least one of the following three activities: Countryside Sightseeing, Shopping, and Adventure Sports such that:
- No two friends like exactly the same set of activities.
- A, B and E like exactly two activities each.
- D likes only Countryside Sightseeing.
- C likes more number of activities compared to D.
The number of ways in which a pair of friends can be chosen for a trip from those who like Adventure Sports is ___
Enter your answer to attempt
The Setup: This is a direct sequel to the previous IPMAT arrangement lore. The base matrix rules are exactly the same, but the final quest objective has changed. We still have our 5-player squad, but now we need to calculate the combinatorics of forming a duo specifically from the 'Adventure Sports' mains. Math and logic have been double-verified as requested.
Step 1:Rebuild the Loadout Matrix.
Quick recap of the previous logic: D is locked to a 1-perk loadout (CS). A, B, and E each take one of the three unique 2-perk combos. C is forced to take the 100% completionist 3-perk build to pass the "no copycats" rule.
Step 2:Identify the Adventure Sports Mains.
We check the AS column in our roster. Who has the AS perk currently equipped?
* One of the A/B/E trio running {CS, AS}
* One of the A/B/E trio running {Sh, AS}
* C running the full {CS, Sh, AS}
That gives us exactly 3 friends who have Adventure Sports active in their rotation.
Step 3:Form the Duo.
The prompt asks for the number of ways to choose a *pair* (2 players) from this eligible pool of 3. This is a straight-up combinations formula since the order in which we pick them doesn't matter (a squad is a squad).
nCr=3C23C2=2×13×2=3Final Answer: 3
Q5:ipmat indore 2026QA › ModulusMediumSA · TITA
If n is an integer such that 100−n3∣n+6∣−∣n−3∣≥0, then the number of possible values of n is ___
Enter your answer to attempt
The Setup: We've got a rational inequality with a strict vibe check on the domain. The denominator has a square root, meaning it must be strictly positive (no imaginary numbers, and definitely no dividing by zero). Once we establish the domain boundaries, we let the numerator 1v1 itself to find the overlapping integer set. Math and logic double-verified.
Step 1: Check the denominator's HP (Domain Constraints).
For the expression to even exist in the real number universe, the stuff inside the square root must be strictly greater than 0.
100−n3>0n3<100
Since n is an integer, let's find the max level cap for n.
43=64 (Passes)
53=125 (Fails)
So, our upper bound is locked in: n≤4.
Step 2: Isolate the numerator.
Because a square root (when real and non-zero) always outputs a positive value, the denominator is essentially just a positive buff. It doesn't affect the ≥0 sign of the whole fraction. The numerator alone dictates if the whole expression is positive or zero.
∣n+6∣−∣n−3∣≥0∣n+6∣≥∣n−3∣Step 3: Square up.
Since both sides of the inequality are absolute values (meaning they are inherently non-negative), squaring both sides is perfectly safe and won't illegally flip the inequality sign. It's the cleanest way to strip the absolute value armor.
(n+6)2≥(n−3)2n2+12n+36≥n2−6n+9
Cancel out the n2 terms and group the rest:
18n≥−27n≥−23n≥−1.5Step 4: Find the overlapping integer zone.
Since n must be an integer, the lowest value it can take based on Step 3 is −1.
Combining this with our domain cap from Step 1 (n≤4), we get our final bounded region:
−1≤n≤4
The valid integer roster is {−1,0,1,2,3,4}.
Step 5: Count the squad.
Counting the integers in that set gives us 6 unique values.
Final Answer: 6
In a company, initially the ratio of the foreign and domestic workers was 5 : 8. When some foreign and 12 domestic workers left the company, this ratio became 2 : 3. Later, when 10 more foreign workers were replaced by 10 new domestic workers, this ratio became 3 : 7. The initial number of domestic workers in the company was ___
Enter your answer to attempt
The Setup: We've got a multi-stage ratio problem that plays out like a corporate battle royale. The roster size keeps getting nerfed and buffed across different phases. The rookie mistake here is starting at Phase 1 and creating a chaotic web of variables. The optimal strat is to anchor our equations to the *middle* phase, creating a clean bridge between the past and the future states. Math and logic double-verified.
Step 1: Anchor to the mid-game meta.
Let's look at the company after the first wave of layoffs/rage-quits. The ratio became 2:3.
Let the number of foreign workers at this point be 2m, and domestic workers be 3m.
Step 2: Trace back the domestic timeline.
We know exactly 12 domestic workers left to reach this 3m state. Therefore, the *initial* number of domestic workers was simply:
Initialdomestic=3m+12
Keep this formula safe in your inventory; it's our final win condition.
Step 3: Track the Phase 3 substitutions.
In the final phase, the company does a massive roster swap. 10 foreign workers leave and are instantly replaced by 10 new domestic workers.
Let's update the Phase 2 variables:
Newforeign=2m−10Newdomestic=3m+10
The prompt states this new team composition creates a ratio of 3:7.
Step 4: Execute the final algebra clash.
Set up the fraction and cross-multiply to secure the W.
3m+102m−10=737(2m−10)=3(3m+10)14m−70=9m+30
Group the variables and constants:
14m−9m=30+705m=100m=20Step 5: Calculate the final stat.
Plug our multiplier m back into the formula we saved in Step 2.
Initialdomestic=3(20)+12Initialdomestic=60+12=72Final Answer: 72
The number of ways in which three distinct integers can be chosen from the set {1,2,…,9} such that their product is divisible by 4, is ___
Enter your answer to attempt
The Setup: We need a 3-number squad whose product has at least two 2s in its prime factorization (since 4=22). Instead of brute-forcing all the winning combinations, it's mathematically way faster to use the "Complement Strat": calculate the absolute total number of possible combinations, then subtract the squads that fail the vibe check (products that are NOT divisible by 4). Math, logic, and syntax have been double-verified.
Step 1: Calculate the absolute total.
First, we find the total number of ways to draft 3 distinct integers from our pool of 9, with zero constraints.
9C3=3×2×19×8×7=84Step 2: Classify the loot pool.
To find the failed squads, we need to group the integers {1,2,…,9} based on how many powers of 2 they carry in their DNA.
* Zero 2s (Odd numbers):{1,3,5,7,9}→ Total of 5 numbers.
* Exactly one 2 (Multiples of 2, but not 4):{2,6}→ Total of 2 numbers.
* At least two 2s (Multiples of 4):{4,8}→ Total of 2 numbers.
Step 3: Count the failed squads (The Complement).
For a product to *fail* being divisible by 4, its prime factorization can only contain a maximum of one 2. This means the total power of 2 in the product must be exactly 0 or exactly 1.
* Case 0 (Zero powers of 2): All three numbers must be drafted from the odd pool.
5C3=10
* Case 1 (Exactly one power of 2): We need exactly one number from the {2,6} pool, and the remaining two must be purely odd.
2C1×5C2=2×10=20
Total failed squads = 10+20=30.
Step 4: Calculate the final W.
We take the absolute total and subtract our calculated fail state to find the exact number of successful combinations.
TotalValidSquads=84−30=54Final Answer: 54
Savitri borrowed 10000 rupees from a bank for a period of two years at a fixed interest rate of 10% per annum, compounded semi-annually, and paid back 5025 rupees at the end of first year. Then, the amount, in rupees, to be paid at the end of second year is ___
Enter your answer to attempt
The Setup: We are dealing with compound interest, but the bank is running a semi-annual meta. This means they apply the interest buff every 6 months instead of waiting for the full year. Savitri drops a mid-game payment to lower her debt aggro, so we must calculate this run in two distinct phases. Math, logic, and syntax have been double-verified.
Step 1: Calibrate the interest stats.
The annual rate is 10%, but since it compounds semi-annually, the bank splits it into two hits per year.
Half-year rate = 5%.
This translates to a growth multiplier of 1.05 every 6 months.
Step 2: Calculate the Year 1 damage.
The initial principal is 10000. In one year, there are two compounding cycles (two half-years).
AmountafterYear1=10000×(1.05)2AmountafterYear1=10000×1.1025=11025rupeesStep 3: Process the mid-game transaction.
Savitri pays back 5025 rupees at the exact 1-year mark to clear some of the accumulated debt. We subtract this from the total to find our new baseline for Phase 2.
NewPrincipal=11025−5025=6000rupeesStep 4: Calculate the final Year 2 boss phase.
This new 6000 rupee balance now has to survive the second year, which means taking two more hits of the 1.05 multiplier.
FinalAmount=6000×(1.05)2FinalAmount=6000×1.1025
To do the math cleanly: 6000×1.1=6600, and 6000×0.0025=15.
FinalAmount=6600+15=6615rupeesFinal Answer: 6615
Q9:ipmat indore 2026QA › Time, Speed & DistanceMediumSA · TITA
A man starts walking from point A to point B at 4 km/hr. After 30 minutes, a woman starts from point A at 6 km/hr. If the woman reaches point B 20 minutes earlier than the man, then the distance between A and B in km is ___
Enter your answer to attempt
The Setup: This is a classic Time, Speed, and Distance (TSD) chase sequence. The man is walking on default settings, but the woman has a major speed buff. Because they are traversing the exact same map (Point A to Point B), we can use their total time difference to reverse-engineer the map's distance. Math, logic, and syntax have been double-verified for your database.
Step 1: Calculate the true time gap.
The woman spawns in 30 minutes late, but still manages to beat the man to the finish line by 20 minutes.
This means she spent significantly less time actually traveling.
Total time saved = 30+20=50 minutes.
We absolutely must convert this to hours to match our speed stats (km/hr).
TotalTimeDifference=6050=65hoursStep 2: Set up the algebraic clash.
Let the total distance from A to B be d kilometers.
Since Time=SpeedDistance, we can define their individual run times:
Time taken by the man = 4d
Time taken by the woman = 6d
We know the man's time minus the woman's time equals our calculated time gap. Let's build the equation:
4d−6d=65Step 3: Execute the fraction math.
Find the least common multiple (LCM) for the denominators 4 and 6, which is 12. Multiply the fractions to sync up the denominators:
123d−122d=6512d=65Step 4: Secure the final distance.
Multiply both sides by 12 to completely isolate d.
d=65×12d=5×2=10Final Answer: 10
If x,y are real numbers and equations x2−12x+35=0 and x2+ax+105=0 have at least one common root, then the minimum possible value of y2+4y−5a is ___
Enter your answer to attempt
The Setup: We've got a dual-equation boss fight with a "common root" mechanic, followed by a quadratic optimization side quest. We need to find the roots of the first equation, test them as keys for the second equation to unlock the possible values of a, and then min-max the final expression. Math, logic, and syntax are locked in and double-verified.
Step 1: Solve the base equation (Find the keys).
The first quadratic is fully revealed, so we factor it to find its roots.
x2−12x+35=0(x−5)(x−7)=0
The roots are x=5 and x=7.
**Step 2: Test the common root mechanic (Unlock a).**
Since the two equations share at least one root, we plug our keys from Step 1 into the second equation to see what values of a drop.
* Scenario A (If 5 is the common root):52+5a+105=025+5a+105=05a=−130⟹a=−26
* Scenario B (If 7 is the common root):72+7a+105=049+7a+105=07a=−154⟹a=−22Step 3: Min-max the final expression (Complete the square).
We need the absolute minimum value for the expression y2+4y−5a. Let's optimize the y variables by completing the square to find its lowest possible floor.
y2+4y−5a=(y2+4y+4)−4−5a(y+2)2−4−5a
Because any real number squared is non-negative, the lowest (y+2)2 can ever go is 0 (which happens when y=−2). Thus, the minimum value of the entire expression is hard-capped at:
MinimumValue=−4−5aStep 4: Calculate the final damage.
To make −4−5a as small as mathematically possible, we just plug in our two possible a values and see which one gives the lower score.
* **Using a=−22:**
−4−5(−22)=−4+110=106
* **Using a=−26:**
−4−5(−26)=−4+130=126
The absolute minimum possible value across all timelines is 106.
Final Answer: 106
Five friends A, B, C, D, and E go on a trip together. Each of them likes at least one of the following three activities: Countryside Sightseeing, Shopping, and Adventure Sports such that:
- No two friends like exactly the same set of activities.
- A, B and E like exactly two activities each.
- D likes only Countryside Sightseeing.
- C likes more number of activities compared to D.
The number of friends that like Countryside Sightseeing is ___
Enter your answer to attempt
The Setup: We are back in the IPMAT arrangement lore for part 3 of this puzzle sequence. The base matrix logic remains exactly the same as the previous drops, but our final objective has shifted. Instead of querying pairs or Adventure Sports, we just need to scan the entire roster for anyone who has the 'Countryside Sightseeing' (CS) perk equipped. Math and logic have been double-verified.
Step 1: Rebuild the Loadout Matrix.
Let's quickly reconstruct the squad's loadouts based on the established rules:
* D: Locked to a 1-perk loadout, which is exclusively {CS}.
* A, B, and E: They take the only three unique 2-perk combinations available: {CS, Sh}, {CS, AS}, and {Sh, AS}.
* C: Must have more activities than D (so 2 or 3). Since the 2-perk slots are all taken, C is the 100% completionist running the full 3-perk build: {CS, Sh, AS}.
Step 2: Query the CS Column.
We check each unique loadout to see if Countryside Sightseeing (CS) is active.
* Friend D: {CS} →Active (1)
* A/B/E Duo 1: {CS, Sh} →Active (2)
* A/B/E Duo 2: {CS, AS} →Active (3)
* A/B/E Duo 3: {Sh, AS} → Inactive
* Friend C: {CS, Sh, AS} →Active (4)
Step 3: Tally the squad.
Counting the "Active" statuses, exactly 4 friends have Countryside Sightseeing in their activity rotation.
Final Answer: 4
Let ABCD be a cyclic quadrilateral with AB=15, BC=20, CD=24 and AC=25. Then AD equals ___
Enter your answer to attempt
The Setup: This is a geometry boss fight dealing with a cyclic quadrilateral. The meta here is recognizing Pythagorean triples to unlock hidden right angles. If we can prove a triangle is a right triangle inside a circle, its hypotenuse instantly becomes the circle's diameter, which breaks the rest of the problem wide open. Math and logic are locked in and double-verified.
Step 1:Stat-check Triangle ABC.
We check the given side lengths for triangle ABC: AB=15, BC=20, and AC=25. Let's run the Pythagorean theorem to see if it passes the vibe check for a right-angled triangle.
152+202=225+400=625252=625
Since AB2+BC2=AC2, the triangle is officially a right triangle, with the right angle at B (∠ABC=90∘).
Step 2:Unlock the Circle's Diameter.
Because ABCD is a cyclic quadrilateral (meaning all its vertices are trapped on the edge of a single circle) and ∠ABC=90∘, the chord AC must be the diameter of that circumcircle. This is a fundamental geometry rule: any angle inscribed in a semicircle is exactly 90∘.
Step 3:Exploit the Semicircle Meta for Triangle ADC.
Since AC is the diameter, the angle on the *other* side of the quadrilateral, ∠ADC, is also forced to be 90∘ for the exact same reason. This turns triangle ADC into another right-angled triangle, sharing the exact same hypotenuse (AC=25).
Step 4:Execute the final Pythagorean drop.
Now we just plug the known stats into triangle ADC to find the missing side AD. We know AC=25 and CD=24.
AD2+CD2=AC2AD2=252−242AD2=625−576=49AD=7Final Answer: 7
Let S denote an arithmetic progression whose first term is either 132 or 158, and the common difference is an even integer less than 10. If the nth term of S is 174, then the number of possible distinct values of n is ___
Enter your answer to attempt
The Setup: We are dealing with an Arithmetic Progression (AP) that has branched storylines. We have two possible starting stats (a=132 or a=158) and a restricted loot pool for our common difference (d). The ultimate goal is to hit a final term of 174. We need to plug these into the standard AP formula and filter out any timelines that result in fractional n values, because a term's position index (n) must pass the vibe check of being a positive integer. Math and logic are locked and double-verified.
Step 1: Define the core equation.
The standard formula for the n-th term of an AP is:
an=a+(n−1)d
We are given an=174. Let's rearrange this to isolate the variable combo:
(n−1)d=174−a
**Step 2: Identify the valid d pool.**
The problem states d is an "even integer less than 10". Since our target (174) is strictly greater than both possible starting points (132 and 158), the sequence must be climbing. This means d has to be a positive integer.
Valid d roster: {2,4,6,8}.
**Step 3: Run the a=132 timeline.**
Substitute a=132 into our core equation.
(n−1)d=174−132=42
Now we test our d roster to see which ones divide 42 cleanly to give an integer n.
* If d=2: n−1=21⟹n=22 (Valid W)
* If d=4: n−1=10.5 (Invalid fraction, L)
* If d=6: n−1=7⟹n=8 (Valid W)
* If d=8: n−1=5.25 (Invalid fraction, L)
**Step 4: Run the a=158 timeline.**
Substitute a=158 into the core equation.
(n−1)d=174−158=16
Test the d roster against 16.
* If d=2: n−1=8⟹n=9 (Valid W)
* If d=4: n−1=4⟹n=5 (Valid W)
* If d=6: n−1=616 (Invalid fraction, L)
* If d=8: n−1=2⟹n=3 (Valid W)
Step 5: Tally the unique loadouts.
From both timelines, the valid integer values for n are {22,8,9,5,3}.
Since none of these values duplicate, we count the total number of distinct values in this set.
Final Answer: 5
Q14:ipmat indore 2026QA › Linear EquationsEasySA · TITA
The Celsius scale of temperature can be converted to Fahrenheit scale using the relationship F=1.8C+32, where F and C are temperature recorded in degrees Fahrenheit and Celsius, respectively. Ajay noted down the temperature of a chemical reaction as x degrees Fahrenheit but reported it as x degrees Celsius by mistake. If this led to a recording of the temperature that was 144% higher than the actual temperature in Celsius scale, then x equals ___
Enter your answer to attempt
The Setup: Ajay totally fat-fingered the data entry, confusing his Fahrenheit and Celsius stats. This typo created a massive 144% inflation buff to the recorded temperature. We need to set up an algebraic equation linking his fake reported stat to the actual true stat to reverse-engineer the exact value of x. Math, logic, and syntax are double-verified.
Step 1: Translate the true base stat.
The actual temperature was accurately read as x degrees Fahrenheit. Let's convert this to its true Celsius equivalent by rearranging the given formula for C.
1.8C=F−32C=1.8x−32
This expression represents the true, uncorrupted Celsius value.
Step 2: Calculate the error multiplier.
The problem states the reported value (x∘C) was 144% *higher* than the actual value. This means the reported value is the original 100% base value plus the extra 144%.
Total multiplier = 100%+144%=244%.
In pure decimal form, our multiplier is 2.44.
Step 3: Set up the algebraic clash.
We equate Ajay's glitched reported value (x) to 2.44 times the true Celsius expression.
x=2.44×(1.8x−32)Step 4: Execute the math to isolate the variable.
Multiply both sides by 1.8 to clear the denominator and clean up the equation.
1.8x=2.44(x−32)
Distribute the 2.44 into the parenthesis.
1.8x=2.44x−78.08
Group the x variables together to isolate the unknown.
2.44x−1.8x=78.080.64x=78.08
Divide to secure the final stat.
x=0.6478.08x=122Final Answer: 122
Let M=1aa21bb21cc2 where a,b and c are real numbers such that a+b+c=0 and abc=0. If detM=0, then the maximum possible value of c2a2+b2 is ____
Enter your answer to attempt
The Setup: This is a Matrices & Determinants boss fight. The matrix provided is a classic Vandermonde matrix. Because its determinant is locked to zero, we can exploit its factored form to force a relationship between the variables. From there, it's a branched timeline problem where we test different variable "dupes" against the a+b+c=0 constraint to find the absolute max stat for our final expression. Math, logic, and syntax are double-verified.
Step 1: Exploit the Vandermonde Meta.
The determinant of a 3×3 Vandermonde matrix has a well-known factored shortcut. Instead of brute-forcing the expansion, we use the standard formula:
detM=(b−a)(c−a)(c−b)
The problem explicitly states detM=0.
(b−a)(c−a)(c−b)=0Step 2: Establish the zero condition.
For the product of three factors to be exactly zero, at least one of the brackets must equal zero. This means at least two of the variables must be exact clones of each other.
Possible timelines: a=b, or b=c, or a=c.
*(Note: They cannot all be equal. If a=b=c, then a+b+c=0 forces them all to be 0, which violates the abc=0 anti-cheat rule).*
Step 3: Run the branched timelines (Case Analysis).
We must test each clone scenario using the constraint a+b+c=0 to unlock the final expression c2a2+b2.
* **Timeline A (If a=b):**
Substitute b with a in the sum constraint:
a+a+c=02a+c=0⟹c=−2a
Now, plug these into the final expression:
(−2a)2a2+a2=4a22a2=21=0.5
* **Timeline B (If b=c):**
Substitute b with c in the sum constraint:
a+c+c=0a+2c=0⟹a=−2c
Plug these into the final expression:
c2(−2c)2+c2=c24c2+c2=c25c2=5
* **Timeline C (If a=c):**
Substitute a with c in the sum constraint:
c+b+c=02c+b=0⟹b=−2c
Plug these into the final expression:
c2c2+(−2c)2=c2c2+4c2=c25c2=5Step 4: Max-out the final stat.
We review the damage outputs from our three timelines: 0.5, 5, and 5. The question asks for the absolute maximum possible value.
The highest yield is 5.
Final Answer: 5