IPMAT Indore 2026 QA PYQs

Q1:ipmat indore 2026QA › Profit & LossMediumSA · TITA

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.

Total\ CP = 3 \times 20 + 5 \times 38 = 60 + 190 = 250\ rupees

Step 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.

Total\ SP = 250 \times 1.32 = 330\ rupees

Step 3: Isolate the orange revenue. The oranges alone gave a cracked 50% profit buff. The cost price for just the oranges was 60 rupees.

Oranges\ SP = 60 \times 1.5 = 90\ rupees

Step 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.

Mangoes\ SP\ (for\ 5) = 330 - 90 = 240\ rupees

To find the selling price of just one mango, divide by the squad size of 5.

SP\ of\ one\ mango = \frac{240}{5} = 48\ rupees

Step 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 \times MP

). We divide the single mango SP by

0.80

to find the original inflated tag.

MP = \frac{48}{0.80} = 60\ rupees

Final Answer: 60

Q2:ipmat indore 2026QA › RemainderMediumSA · TITA

The remainder when

7^{103} + 7^{101}

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.

7^{103} + 7^{101} = 7^{101}(7^2 + 1)

7^{101}(49 + 1) = 7^{101} \times 50

Step 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.

7^1 \equiv 7 \pmod 9

7^2 = 49 \equiv 4 \pmod 9

7^3 = 7^2 \times 7 \equiv 4 \times 7 = 28 \equiv 1 \pmod 9

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 (

7^{101}

).** Divide the main exponent (101) by our cycle length (3) to see exactly where it lands in the rotation.

101 = 3 \times 33 + 2

Since the remainder is 2,

7^{101}

behaves exactly like

7^2

in the modulo 9 universe.

7^{101} \equiv 7^2 \equiv 4 \pmod 9

Step 4: Resolve the multiplier (50). We also need to find the remainder of our constant multiplier (50) when divided by 9.

50 \equiv 5 \pmod 9

Step 5: Calculate the final damage output. Multiply the isolated remainders together and take mod 9 one last time to get the ultimate answer.

7^{101} \times 50 \equiv 4 \times 5 \pmod 9

20 \equiv 2 \pmod 9

Final Answer: 2

Q3:ipmat indore 2026QA › ModulusMediumSA · TITA

If

n

is an integer such that

\frac{|n+6|-|n-3|}{\sqrt{100-n^3}} \geq 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 - n^3 > 0

n^3 < 100

Since

n

is an integer, let's find the max level cap for

n

.

4^3 = 64

(Passes)

5^3 = 125

(Fails) So, our upper bound is locked in:

n \leq 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

\geq 0

sign of the whole fraction. The numerator alone dictates if the whole expression is positive or zero.

|n+6| - |n-3| \geq 0

|n+6| \geq |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 \geq (n-3)^2

n^2 + 12n + 36 \geq n^2 - 6n + 9

Cancel out the

n^2

terms and group the rest:

18n \geq -27

n \geq -\frac{3}{2}

n \geq -1.5

Step 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 \leq 4

), we get our final bounded region:

-1 \leq n \leq 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

Q4:ipmat indore 2026QA › Ratio, Proportion & VariationMediumSA · TITA

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:

Initial\ domestic = 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:

New\ foreign = 2m - 10

New\ domestic = 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.

\frac{2m - 10}{3m + 10} = \frac{3}{7}

7(2m - 10) = 3(3m + 10)

14m - 70 = 9m + 30

Group the variables and constants:

14m - 9m = 30 + 70

5m = 100

m = 20

Step 5: Calculate the final stat. Plug our multiplier

m

back into the formula we saved in Step 2.

Initial\ domestic = 3(20) + 12

Initial\ domestic = 60 + 12 = 72

Final Answer: 72

Q5:ipmat indore 2026QA › Permutation & CombinationHardSA · TITA

The number of ways in which three distinct integers can be chosen from the set

\{1, 2, \dots, 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

2

s in its prime factorization (since

4 = 2^2

). 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.

^{9}C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

Step 2: Classify the loot pool. To find the failed squads, we need to group the integers

\{1, 2, \dots, 9\}

based on how many powers of

2

they carry in their DNA. * Zero 2s (Odd numbers):

\{1, 3, 5, 7, 9\}

\rightarrow

Total of 5 numbers. * Exactly one 2 (Multiples of 2, but not 4):

\{2, 6\}

\rightarrow

Total of 2 numbers. * At least two 2s (Multiples of 4):

\{4, 8\}

\rightarrow

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.

^{5}C_{3} = 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.

^{2}C_{1} \times ^{5}C_{2} = 2 \times 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.

Total\ Valid\ Squads = 84 - 30 = 54

Final Answer: 54

Q6:ipmat indore 2026QA › Simple & Compound InterestMediumSA · TITA

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).

Amount\ after\ Year\ 1 = 10000 \times (1.05)^2

Amount\ after\ Year\ 1 = 10000 \times 1.1025 = 11025\ rupees

Step 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.

New\ Principal = 11025 - 5025 = 6000\ rupees

Step 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.

Final\ Amount = 6000 \times (1.05)^2

Final\ Amount = 6000 \times 1.1025

To do the math cleanly:

6000 \times 1.1 = 6600

, and

6000 \times 0.0025 = 15

.

Final\ Amount = 6600 + 15 = 6615\ rupees

Final Answer: 6615

Q7: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

).

Total\ Time\ Difference = \frac{50}{60} = \frac{5}{6}\ hours

Step 2: Set up the algebraic clash. Let the total distance from A to B be

d

kilometers. Since

Time = \frac{Distance}{Speed}

, we can define their individual run times: Time taken by the man =

\frac{d}{4}

Time taken by the woman =

\frac{d}{6}

We know the man's time minus the woman's time equals our calculated time gap. Let's build the equation:

\frac{d}{4} - \frac{d}{6} = \frac{5}{6}

Step 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:

\frac{3d}{12} - \frac{2d}{12} = \frac{5}{6}

\frac{d}{12} = \frac{5}{6}

Step 4: Secure the final distance. Multiply both sides by 12 to completely isolate

d

.

d = \frac{5 \times 12}{6}

d = 5 \times 2 = 10

Final Answer: 10

Q8:ipmat indore 2026QA › PolynomialsMediumSA · TITA

If

x, y

are real numbers and equations

x^2 - 12x + 35 = 0

and

x^2 + ax + 105 = 0

have at least one common root, then the minimum possible value of

y^2 + 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.

x^2 - 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):

5^2 + 5a + 105 = 0

25 + 5a + 105 = 0

5a = -130 \implies a = -26

* Scenario B (If 7 is the common root):

7^2 + 7a + 105 = 0

49 + 7a + 105 = 0

7a = -154 \implies a = -22

Step 3: Min-max the final expression (Complete the square). We need the absolute minimum value for the expression

y^2 + 4y - 5a

. Let's optimize the

y

variables by completing the square to find its lowest possible floor.

y^2 + 4y - 5a = (y^2 + 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:

Minimum\ Value = -4 - 5a

Step 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

Q9:ipmat indore 2026QA › QuadrilateralsEasySA · TITA

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.

15^2 + 20^2 = 225 + 400 = 625

25^2 = 625

Since

AB^2 + BC^2 = AC^2

, the triangle is officially a right triangle, with the right angle at

B

(

\angle ABC = 90^\circ

). 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

\angle ABC = 90^\circ

, 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^\circ

. Step 3: Exploit the Semicircle Meta for Triangle ADC. Since

AC

is the diameter, the angle on the *other* side of the quadrilateral,

\angle ADC

, is also forced to be

90^\circ

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

.

AD^2 + CD^2 = AC^2

AD^2 = 25^2 - 24^2

AD^2 = 625 - 576 = 49

AD = 7

Final Answer: 7

Q10:ipmat indore 2026QA › Progression & SeriesMediumSA · TITA

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

n^{\text{th}}

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:

a_n = a + (n - 1)d

We are given

a_n = 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 \implies n = 22

(Valid W) * If

d = 4

:

n - 1 = 10.5

(Invalid fraction, L) * If

d = 6

:

n - 1 = 7 \implies 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 \implies n = 9

(Valid W) * If

d = 4

:

n - 1 = 4 \implies n = 5

(Valid W) * If

d = 6

:

n - 1 = \frac{16}{6}

(Invalid fraction, L) * If

d = 8

:

n - 1 = 2 \implies 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

Q11: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 - 32

C = \frac{x - 32}{1.8}

This expression represents the true, uncorrupted Celsius value. Step 2: Calculate the error multiplier. The problem states the reported value (

x^\circ 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 \times \left(\frac{x - 32}{1.8}\right)

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 = 78.08

0.64x = 78.08

Divide to secure the final stat.

x = \frac{78.08}{0.64}

x = 122

Final Answer: 122

Q12:ipmat indore 2026QA › Matrices & DeterminantsHardSA · TITA

Let

M = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}

where

a, b

and

c

are real numbers such that

a + b + c = 0

and

abc \neq 0

. If

\det M = 0

, then the maximum possible value of

\frac{a^2+b^2}{c^2}

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 \times 3

Vandermonde matrix has a well-known factored shortcut. Instead of brute-forcing the expansion, we use the standard formula:

\det M = (b - a)(c - a)(c - b)

The problem explicitly states

\det M = 0

.

(b - a)(c - a)(c - b) = 0

Step 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 \neq 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

\frac{a^2+b^2}{c^2}

. * **Timeline A (If

a = b

):** Substitute

b

with

a

in the sum constraint:

a + a + c = 0

2a + c = 0 \implies c = -2a

Now, plug these into the final expression:

\frac{a^2 + a^2}{(-2a)^2} = \frac{2a^2}{4a^2} = \frac{1}{2} = 0.5

* **Timeline B (If

b = c

):** Substitute

b

with

c

in the sum constraint:

a + c + c = 0

a + 2c = 0 \implies a = -2c

Plug these into the final expression:

\frac{(-2c)^2 + c^2}{c^2} = \frac{4c^2 + c^2}{c^2} = \frac{5c^2}{c^2} = 5

* **Timeline C (If

a = c

):** Substitute

a

with

c

in the sum constraint:

c + b + c = 0

2c + b = 0 \implies b = -2c

Plug these into the final expression:

\frac{c^2 + (-2c)^2}{c^2} = \frac{c^2 + 4c^2}{c^2} = \frac{5c^2}{c^2} = 5

Step 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

Q13:ipmat indore 2026QA › Time & WorkEasyMCQ · MCQ

Painter A can paint a building in 12 days while Painter B can paint it in 16 days. If A and B work on alternate days, and A starts the work on the first day, then the number of days required to paint the building is

A $13 \frac{2}{3}$
B $13 \frac{1}{2}$
C $13 \frac{3}{4}$
D $7 \frac{6}{7}$

Pick an option to attempt

The Setup: This is a classic Time & Work alternate-day scenario. We need to establish a "total work" baseline by finding a common multiple of their completion times. Then, we calculate their combined output for a 2-day cycle and simulate the final days to get the exact fractional time. Math, logic, and syntax have been double-verified. Step 1: Set the Total Work and Rates. Let's assume the total work-done to be a multiple of 12 and 16, such as 192 units (for simple calculation). * Painter A's rate = 16 units/day. * Painter B's rate = 12 units/day. Step 2: Calculate the 2-Day Cycle Output. Since they work on alternate days with A starting first, a single 2-day cycle yields:

16 + 12 = 28\ units

Step 3: Fast-forward the cycles. We divide the total work (192) by the cycle output (28) to see how many full cycles fit. In 12 days, there are exactly 6 such pairs. Work completed in 12 days =

6 \times 28 = 168

units. Work left over =

192 - 168 = 24

units. Step 4: Play out the remaining days. * Day 13: It is A's turn to work, and A completes 16 units of work. Work left after 13 days =

24 - 16 = 8

units. * Day 14: It is B's turn to clear the remaining 8 units. B does 12 units in 1 day, so the time taken by B to complete 8 units is:

\frac{8}{12} = \frac{2}{3}\ day

Step 5: Calculate final time. Adding the total days together gives us the final time required:

13 + \frac{2}{3} = 13 \frac{2}{3}\ days

Final Answer:

13 \frac{2}{3}

Q14:ipmat indore 2026QA › Progression & SeriesHardMCQ · MCQ

Gita starts from point A and walks 1000 m east. She then walks 800 m north, followed by 640 m west and 512 m south, reaching point B. After this, she continues moving in the same cyclic order: east, north, west, south, with each successive movement 20% shorter than the previous one. After infinitely many such moves, approximately how far in meters will Gita be from her starting point A?

A711
B800
C765
D781

Pick an option to attempt

The Setup: We are dealing with an infinite geometric progression (GP) mapped onto a 2D coordinate plane. Gita is doing a spiral walk where each step is nerfed by 20% (an

0.8\times

multiplier) compared to the previous one. To solve this, we need to calculate her total net displacement along the X-axis (East/West) and Y-axis (North/South) using the sum of an infinite GP formula, and then use the Pythagorean theorem to find the absolute straight-line distance from her spawn point (A). Math, logic, and syntax double-verified. Step 1: Calibrate the GP multiplier. Each movement is 20% shorter, meaning the multiplier (common ratio) between consecutive steps is

1 - 0.20 = 0.8

. A full directional cycle consists of 4 moves (East, North, West, South). The common ratio between the *same* direction across cycles (e.g., East 1 to East 2) is:

r = (0.8)^4 = 0.4096

The formula for the sum of an infinite GP is

S = \frac{a}{1 - r}

. Our common denominator for all axes will be:

1 - 0.4096 = 0.5904

Step 2: Calculate the net X-Axis displacement (East - West). Let's group her East and West movements to find the net horizontal distance per cycle. * First East =

1000

* First West =

640

* Net East for Cycle 1 =

1000 - 640 = 360

Now, we plug this into our infinite GP sum formula:

Net\ East = \frac{360}{0.5904}

Step 3: Calculate the net Y-Axis displacement (North - South). Let's group her North and South movements for the net vertical distance. * First North =

800

* First South =

512

* Net North for Cycle 1 =

800 - 512 = 288

Apply the infinite GP sum formula:

Net\ North = \frac{288}{0.5904}

Step 4: Execute the Pythagorean drop. We now have her absolute horizontal and vertical coordinates. We use

d = \sqrt{x^2 + y^2}

to find the hypotenuse (the straight-line distance from A).

d = \sqrt{\left(\frac{360}{0.5904}\right)^2 + \left(\frac{288}{0.5904}\right)^2}

d = \frac{\sqrt{360^2 + 288^2}}{0.5904}

Let's calculate the square root payload:

360^2 + 288^2 = 129600 + 82944 = 212544

\sqrt{212544} \approx 461.02

Step 5: Secure the final stat. Divide by our GP denominator to get the final distance:

d \approx \frac{461.02}{0.5904} \approx 780.86

Rounding this to the nearest integer gives us

781

. Final Answer: 781

Q15:ipmat indore 2026QA › Permutation & CombinationMediumMCQ · MCQ

The number of ways in which coins of denominations of rupees 2, 5 and, 10 can be combined to make a value of rupees 50 is ___

A10
B21
C20
D19

Pick an option to attempt

The Setup: We are tackling a Linear Diophantine equation problem disguised as a coin combination puzzle. Since we have three variables (

a, b, c

) and only one equation (

2a + 5b + 10c = 50

), we can't brute-force a single algebraic solution. Instead, we anchor the heaviest coin (

c

) to create bounded timelines, and use parity (even/odd logic) to filter the remaining valid coin combos. Math, logic, and syntax are locked and double-verified. Step 1: Set up the core equation. Let

a, b, c

represent the quantities of Rs 2, Rs 5, and Rs 10 coins respectively. The total value is capped at 50 rupees.

2a + 5b + 10c = 50

Step 2: Isolate and establish parity. Anchor the Rs 10 coins (

c

) to the right side to simplify the battlefield.

2a + 5b = 50 - 10c

Notice the right side (

50 - 10c

) is always an even number. Since

2a

is inherently even,

5b

is forced to be even as well to keep the equation balanced. For

5b

to result in an even number, the quantity

b

must be an even integer (

b \in \{0, 2, 4, \dots\}

). **Step 3: Iterate through the

c

timelines.** We test all possible quantities for the Rs 10 coin (the max

c

can be is

5

, since

10 \times 5 = 50

) and count the valid even values for

b

. Once

b

is locked,

a

automatically scales to fill the remainder, meaning we only need to count the valid

b

values to get the number of ways. * **If

c = 0

:**

2a + 5b = 50 \implies 5b \leq 50

. Valid even

b \in \{0, 2, 4, 6, 8, 10\} \rightarrow

6 ways * **If

c = 1

:**

2a + 5b = 40 \implies 5b \leq 40

. Valid even

b \in \{0, 2, 4, 6, 8\} \rightarrow

5 ways * **If

c = 2

:**

2a + 5b = 30 \implies 5b \leq 30

. Valid even

b \in \{0, 2, 4, 6\} \rightarrow

4 ways * **If

c = 3

:**

2a + 5b = 20 \implies 5b \leq 20

. Valid even

b \in \{0, 2, 4\} \rightarrow

3 ways * **If

c = 4

:**

2a + 5b = 10 \implies 5b \leq 10

. Valid even

b \in \{0, 2\} \rightarrow

2 ways * **If

c = 5

:**

2a + 5b = 0 \implies 5b \leq 0

. Valid even

b \in \{0\} \rightarrow

1 way Step 4: Tally the total combinations. Sum the number of valid ways from all six timelines to get the final score.

Total = 6 + 5 + 4 + 3 + 2 + 1 = 21

Final Answer: 21

Q16:ipmat indore 2026QA › CirclesHardMCQ · MCQ

Let

ABCD

be a rectangle with

AB = 72\text{ cm}

and

BC = 30\text{ cm}

. A circle passing through points

A

and

C

cuts the side

AB

at

P

such that

AP = 56\text{ cm}

. The radius, in cm, of the circle is ___

A $\frac{221}{5}$
B $\frac{441}{10}$
C $\frac{220}{5}$
D $\frac{216}{5}$

Pick an option to attempt

The Setup: This is a heavy Geometry boss fight involving circumcircles and trigonometry. We have a circle passing through

A

,

P

, and

C

, which means it acts as the circumcircle for

\triangle APC

. The absolute meta here is to find the sides of this triangle and hit it with the Extended Law of Sines (

2R = \frac{a}{\sin A}

) to unlock the radius. Math, logic, and syntax are fully locked and double-verified. Step 1: Map the arena. Let's place rectangle

ABCD

on a coordinate plane to visualize the stats. Let

A = (0, 0)

. Since

AB = 72

,

B = (72, 0)

. With

BC = 30

,

C = (72, 30)

. Point

P

lies on the line segment

AB

. Since

AP = 56

, we know

P = (56, 0)

. This leaves the remaining segment

PB

:

PB = AB - AP = 72 - 56 = 16\text{ cm}

Step 2: Calculate the missing sides (Pythagorean drops). We need the side lengths of

\triangle APC

to run our trig formulas. Let's use the right-angled triangles built into the rectangle to find

PC

and

AC

. * **In right

\triangle PBC

(right-angled at

B

):**

PC = \sqrt{PB^2 + BC^2} = \sqrt{16^2 + 30^2}

PC = \sqrt{256 + 900} = \sqrt{1156} = 34\text{ cm}

* **In right

\triangle ABC

(right-angled at

B

):**

AC = \sqrt{AB^2 + BC^2} = \sqrt{72^2 + 30^2}

AC = \sqrt{5184 + 900} = \sqrt{6084} = 78\text{ cm}

Step 3: Extract the Sine value. To use the Extended Law of Sines on

\triangle APC

, we need the sine of one of its angles.

\angle APC

is the easiest target because it shares a straight line with

\angle BPC

. Since

A, P

, and

B

are collinear,

\angle APC

and

\angle BPC

are supplementary angles (

\angle APC + \angle BPC = 180^\circ

). A core trigonometry rule states that supplementary angles have the exact same sine value:

\sin(\angle APC) = \sin(\angle BPC)

. Let's find

\sin(\angle BPC)

using the right

\triangle PBC

:

\sin(\angle BPC) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{PC}

\sin(\angle BPC) = \frac{30}{34} = \frac{15}{17}

Therefore,

\sin(\angle APC) = \frac{15}{17}

. Step 4: Execute the Extended Law of Sines. The radius

R

of the circumcircle of

\triangle APC

can be found using the side

AC

and its opposite angle

\angle APC

:

2R = \frac{AC}{\sin(\angle APC)}

Substitute our unlocked stats into the formula:

2R = \frac{78}{\frac{15}{17}}

2R = \frac{78 \times 17}{15}

Simplify the fraction by dividing

78

and

15

by their common factor of

3

:

2R = \frac{26 \times 17}{5}

2R = \frac{442}{5}

Divide by 2 to isolate

R

:

R = \frac{221}{5}

Final Answer:

\frac{221}{5}

Q17:ipmat indore 2026QA › PolynomialsEasyMCQ · MCQ

The equation

2^x - x^2 = 0

has

Aexactly one real solution
Bexactly two real solutions
Cno real solution
Dexactly three real solutions

Pick an option to attempt

The Setup: This is a classic graphical intersection problem. By rearranging the equation to

2^x = x^2

, we are basically looking for every point where the exponential curve

y = 2^x

collides with the parabola

y = x^2

. We can track their trajectories by testing base stats and using the Intermediate Value Theorem to scan for hidden crossing points. Math, logic, and syntax are locked in and double-verified. Step 1: Scan for obvious integer spawns. Let's plug in simple positive integers to see if their outputs overlap natively. * If

x = 2

:

2^2 = 4

and

2^2 = 4

. That's a direct hit. (Solution 1) * If

x = 4

:

2^4 = 16

and

4^2 = 16

. That's another hit. (Solution 2) **Step 2: Check the endgame scaling (

x > 4

).** For any

x

greater than 4, the exponential function (

2^x

) scales way too aggressively compared to the polynomial (

x^2

). The

2^x

curve will stay permanently above the parabola, meaning there are zero collision points in the deep positive zone. Step 3: Scan the negative zone for hidden spawns. Let's check the behavior of both functions as

x

drops below zero to see if their paths cross. * At

x = 0

:

2^0 = 1

and

0^2 = 0

. (Here, the exponential

2^x

is higher). * At a negative value like

x = -1

:

2^{-1} = 0.5

and

(-1)^2 = 1

. (Here, the parabola

x^2

is higher). Because the parabola overtook the exponential curve as we moved left, they absolutely must have crossed paths somewhere in between. The Intermediate Value Theorem guarantees this collision (which happens around

x \approx -0.77

). This gives us our third valid solution. As we go further left into the deep negative zone,

x^2

continues to grow infinitely large while

2^x

flatlines towards 0, so they will never cross again. Step 4: Tally the total hits. We have confirmed exactly three intersection points:

x = 2

,

x = 4

, and a negative value near

-0.77

. Final Answer: exactly three real solutions

Q18:ipmat indore 2026QA › ProbabilityMediumMCQ · MCQ

Three dice are thrown simultaneously and the sum of the three numbers appearing on the top faces of the dice is found to be 10. The probability that these three numbers are distinct, is ___

A $\frac{2}{3}$
B $\frac{3}{5}$
C $\frac{1}{2}$
D $\frac{3}{4}$

Pick an option to attempt

The Setup: This is a classic Conditional Probability scenario mixed with a Combinatorics side quest. Since we already know the sum is exactly 10, our total sample space shrinks from the massive

6^3

base down to only the specific combinations that yield a 10. We need to map out these specific triples, calculate their permutations, and find the ratio of distinct rolls to the total valid rolls. Math, logic, and syntax are locked in and double-verified. Step 1: Map the valid triples (Sum = 10). Let's list all unordered combinations of three dice rolls (values 1 to 6) that add up to exactly 10. We will split them into two tiers: distinct values and repeated values. * Distinct triples:

(1, 3, 6)

,

(1, 4, 5)

, and

(2, 3, 5)

. * Repeated triples:

(2, 2, 6)

,

(2, 4, 4)

, and

(3, 3, 4)

. Step 2: Calculate the permutations (The Sample Space). Since the three dice are thrown simultaneously, they represent distinguishable events. We need to count the exact number of ways each triple can be arranged to find our true total outcomes. * For the 3 distinct triples: Each can be arranged in

3! = 6

ways.

Distinct\ Outcomes = 3 \times 6 = 18

* For the 3 repeated triples: Each has one duplicate number, so their arrangements are divided by the duplicate factorial:

\frac{3!}{2!} = 3

ways.

Repeated\ Outcomes = 3 \times 3 = 9

Our new restricted sample space (total valid outcomes where the sum is 10) is:

Total\ Outcomes = 18 + 9 = 27

Step 3: Extract the final probability. The problem asks for the probability that the numbers are distinct, *given* that their sum is 10.

Probability = \frac{Distinct\ Outcomes}{Total\ Outcomes}

Probability = \frac{18}{27}

Simplify the fraction by dividing the numerator and denominator by 9:

Probability = \frac{2}{3}

Final Answer:

\frac{2}{3}

Q19:ipmat indore 2026QA › PolynomialsHardMCQ · MCQ

The number of values

a

can take such that

x^4 + ax^3 + (3a - 4)x^2 + 2(a - 1)x - 4

can be expressed as a product of two quadratic polynomials,

x^2 + px + 2

and

x^2 + qx - 2

, where

p

and

q

are real, is ___

A2
B1
C0
D3

Pick an option to attempt

The Setup: We've got a heavy polynomial factorization challenge here. The meta is to expand the given product of the two quadratics, group the terms by their

x

powers, and then run a coefficient matching sequence against the original degree-4 polynomial. This will generate a system of equations we can solve to isolate

a

. Math, logic, and syntax are locked in and double-verified. Step 1: Expand the quadratic product. Multiply out the two given factors and group the coefficients for

x^3

,

x^2

, and

x

.

(x^2 + px + 2)(x^2 + qx - 2)

= x^4 + qx^3 - 2x^2 + px^3 + pqx^2 - 2px + 2x^2 + 2qx - 4

= x^4 + (p + q)x^3 + pq \cdot x^2 + (2q - 2p)x - 4

= x^4 + (p + q)x^3 + pq \cdot x^2 + 2(q - p)x - 4

Step 2: Execute the coefficient matching. Compare our expanded form to the original target polynomial:

x^4 + ax^3 + (3a - 4)x^2 + 2(a - 1)x - 4

. By matching the coefficients of corresponding powers of

x

, we unlock our system of equations: 1.

p + q = a

(from

x^3

) 2.

pq = 3a - 4

(from

x^2

) 3.

2(q - p) = 2(a - 1) \implies q - p = a - 1

(from

x

) **Step 3: Solve the linear system for

p

and

q

.** We can use equations (1) and (3) to easily isolate

p

and

q

in terms of

a

. Add equations (1) and (3) together to eliminate

p

:

(p + q) + (q - p) = a + (a - 1)

2q = 2a - 1 \implies q = a - \frac{1}{2}

Subtract equation (3) from (1) to eliminate

q

:

(p + q) - (q - p) = a - (a - 1)

2p = 1 \implies p = \frac{1}{2}

**Step 4: Substitute and isolate

a

.** Now, plug our unlocked

p

and

q

values into the middle equation (2):

pq = 3a - 4

.

\frac{1}{2} \left(a - \frac{1}{2}\right) = 3a - 4

Multiply the entire equation by 4 to clear the fractions and simplify the battlefield:

2\left(a - \frac{1}{2}\right) \times 2 = (3a - 4) \times 4

2a - 1 = 12a - 16

Group the

a

terms to secure the final value:

15 = 10a \implies a = \frac{15}{10} = \frac{3}{2}

Step 5: Tally the valid values. The math yields exactly one real, working value for

a

(which is

1.5

). The question specifically asks for the *number of values*

a

can take. Since there is only one valid solution, the count is 1. Final Answer: 1

Q20:ipmat indore 2026QA › LogarithmsHardMCQ · MCQ

If

\log_{18} 24 = p

, then

\log_{96} 108

equals

A $\frac{p+2}{7p-3}$
B $\frac{p+7}{9p-2}$
C $\frac{3p+3}{3p+2}$
D $\frac{3p+2}{3p+1}$

Pick an option to attempt

The Setup: We are dealing with a heavy logarithms base-switching problem. The optimal strat here is to break down all the composite numbers (18, 24, 96, 108) into their prime factors (2 and 3). By setting up base variables for

\log 2

and

\log 3

, we can express both

p

and our target expression in terms of a single ratio. Once we isolate that ratio in terms of

p

, we just plug it into the target expression to secure the final form. Math, logic, and syntax have been double-verified. **Step 1: Break down the initial condition (

p

).** Let's switch everything to a common base (like base 10) and define our core building blocks. Let

L_2 = \log 2

and

L_3 = \log 3

. Expand

\log 24

and

\log 18

using prime factorization:

\log 24 = \log(2^3 \cdot 3) = 3\log 2 + \log 3 = 3L_2 + L_3

\log 18 = \log(2 \cdot 3^2) = \log 2 + 2\log 3 = L_2 + 2L_3

Now, express

p

as a fraction using the change of base formula (

p = \frac{\log 24}{\log 18}

):

p = \frac{3L_2 + L_3}{L_2 + 2L_3}

Step 2: Break down the target expression. Apply the same prime factorization strat to our target,

\log_{96} 108

:

\log 108 = \log(2^2 \cdot 3^3) = 2\log 2 + 3\log 3 = 2L_2 + 3L_3

\log 96 = \log(2^5 \cdot 3) = 5\log 2 + \log 3 = 5L_2 + L_3

Using the change of base formula, we get:

\log_{96} 108 = \frac{2L_2 + 3L_3}{5L_2 + L_3}

Step 3: Define the ratio variable and isolate it. To simplify the battlefield, let's create a ratio variable

t = \frac{L_2}{L_3}

. Divide the top and bottom of our

p

fraction by

L_3

:

p = \frac{\frac{3L_2}{L_3} + 1}{\frac{L_2}{L_3} + 2} = \frac{3t + 1}{t + 2}

Now, solve for

t

in terms of

p

:

p(t + 2) = 3t + 1

pt + 2p = 3t + 1

pt - 3t = 1 - 2p \implies t(p - 3) = 1 - 2p

t = \frac{1 - 2p}{p - 3} = \frac{2p - 1}{3 - p}

Step 4: Execute the final substitution. Divide the top and bottom of our target expression by

L_3

to write it in terms of

t

:

\log_{96} 108 = \frac{2t + 3}{5t + 1}

Substitute our unlocked

t

value (

\frac{2p - 1}{3 - p}

) into this new expression: * Numerator:

2\left(\frac{2p - 1}{3 - p}\right) + 3 = \frac{4p - 2 + 3(3 - p)}{3 - p} = \frac{4p - 2 + 9 - 3p}{3 - p} = \frac{p + 7}{3 - p}

* Denominator:

5\left(\frac{2p - 1}{3 - p}\right) + 1 = \frac{10p - 5 + 1(3 - p)}{3 - p} = \frac{10p - 5 + 3 - p}{3 - p} = \frac{9p - 2}{3 - p}

Divide the numerator by the denominator (the

(3 - p)

terms instantly cancel out):

\frac{\frac{p + 7}{3 - p}}{\frac{9p - 2}{3 - p}} = \frac{p + 7}{9p - 2}

Final Answer:

\frac{p+7}{9p-2}

Q21:ipmat indore 2026QA › Integral SolutionsMediumMCQ · MCQ

The number of integer solutions

(x, y)

of the inequality

x^2 + y^2 \leq 10

is ___

A12
B13
C36
D37

Pick an option to attempt

The Setup: This is a Number System boundary-checking problem disguised as coordinate geometry. The inequality

x^2 + y^2 \leq 10

represents the interior and boundary of a circle centered at the origin. Since we only care about integer solutions (lattice points), the most efficient strat is to lock down one axis (

y

), scan through its valid integer coordinates, and tally the possible integer spawns for the other axis (

x

). Math, logic, and syntax are locked in and double-verified. **Step 1: Run the center lane (

y = 0

).** If we lock

y

to

0

:

x^2 + 0^2 \leq 10 \implies x^2 \leq 10

The valid integer values for

x

are

\{-3, -2, -1, 0, 1, 2, 3\}

. That gives us 7 values. **Step 2: Run the

y = \pm 1

timelines.** Because the equation uses

y^2

, both positive and negative

y

inputs yield the exact same drop rates. Let's calculate them together. If

y = 1

or

y = -1

:

x^2 + (\pm 1)^2 \leq 10

x^2 + 1 \leq 10 \implies x^2 \leq 9

The valid integer values for

x

are

\{-3, -2, -1, 0, 1, 2, 3\}

. That's 7 values for

+1

and 7 values for

-1

, totaling 14 values. **Step 3: Run the

y = \pm 2

timelines.** If

y = 2

or

y = -2

:

x^2 + (\pm 2)^2 \leq 10

x^2 + 4 \leq 10 \implies x^2 \leq 6

The valid integer values for

x

are

\{-2, -1, 0, 1, 2\}

. That's 5 values for

+2

and 5 values for

-2

, totaling 10 values. **Step 4: Run the

y = \pm 3

timelines.** If

y = 3

or

y = -3

:

x^2 + (\pm 3)^2 \leq 10

x^2 + 9 \leq 10 \implies x^2 \leq 1

The valid integer values for

x

are

\{-1, 0, 1\}

. That's 3 values for

+3

and 3 values for

-3

, totaling 6 values. **Step 5: Check the outer boundaries (

|y| \geq 4

).** If

y = \pm 4

,

y^2 = 16

.

x^2 + 16 \leq 10 \implies x^2 \leq -6

Since a squared real number cannot be negative, there are zero solutions beyond this boundary. The map ends here. Step 6: Tally the total valid coordinate pairs. Sum the values from all our locked timelines to get the final score:

Total = 7 + 14 + 10 + 6 = 37

Final Answer: 37

Q22:ipmat indore 2026QA › Matrices & DeterminantsMediumMCQ · MCQ

If

A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

, then the determinant of

A + A^2 + A^3 + \dots + A^{13}

is ___

A26
B13
C169
D91

Pick an option to attempt

The Setup: This is a Matrices boss fight where brute-forcing the powers will absolutely drain your time. The meta is to identify the underlying scaling pattern for

A^n

(which is a standard upper triangular matrix buff) and then sum the resulting sequences element-by-element before finally calculating the determinant. Math, logic, and syntax are locked in and double-verified. **Step 1: Identify the

A^n

scaling pattern.** For a matrix

A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

, raising it to the power of

n

simply scales the top-right entry by

n

while keeping the rest of the matrix identical. You can quickly verify this by multiplying

A \times A

.

A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}

Step 2: Sum the matrices element-by-element. We need to calculate the mega-matrix resulting from

A^1 + A^2 + A^3 + \dots + A^{13}

. Since matrix addition is position-based, we sum up the individual slots across all 13 matrices. * The Diagonals (Top-Left & Bottom-Right): Every single matrix has a 1 in these slots. Summing thirteen 1s gives

1 \times 13 = 13

. * Bottom-Left: Every matrix has a 0 here. The sum remains

0

. * Top-Right: This forms an arithmetic progression:

1 + 2 + 3 + \dots + 13

. We use the standard sum formula

\frac{n(n+1)}{2}

.

Sum = \frac{13 \times 14}{2} = 13 \times 7 = 91

Step 3: Construct the final summed matrix. Plug our summed elements back into their respective slots to reveal the fully buffed matrix:

Summed\ Matrix = \begin{bmatrix} 13 & 91 \\ 0 & 13 \end{bmatrix}

Step 4: Execute the determinant drop. The determinant of a

2 \times 2

matrix

\begin{bmatrix} a & b \\ c & d \end{bmatrix}

is simply

ad - bc

.

Determinant = (13 \times 13) - (0 \times 91)

Determinant = 169 - 0 = 169

Final Answer: 169

Q23:ipmat indore 2026QA › ProbabilityMediumMCQ · MCQ

A fair die is rolled repeatedly. The probability that the cumulative sum is at least 17 in the third trial is ___

A $\frac{1}{54}$
B $\frac{1}{108}$
C $\frac{1}{27}$
D $\frac{7}{216}$

Pick an option to attempt

The Setup: This is a classic Probability combinatorics drop. We are rolling a standard 6-sided die three times, which sets our total sample space to

6^3

. The win condition is hitting a cumulative sum of *at least* 17. Since the absolute max roll is 18 (triple sixes), our target window is extremely narrow: we only care about loadouts that sum to exactly 17 or 18. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the total sample space. Rolling a standard die three times gives us a total number of possible distinct outcomes:

Total\ Outcomes = 6 \times 6 \times 6 = 6^3 = 216

Step 2: Map the "Sum = 18" win condition. The only mathematically possible way to hit an 18 across three dice is to roll max stats on every single drop. Valid permutations:

(6, 6, 6)

Total ways =

1

Step 3: Map the "Sum = 17" win condition. To hit a 17, we need two dice to roll max (6) and one die to drop a 5. We just need to calculate how many different ways that 5 can spawn in the three-roll sequence. Valid permutations:

(5, 6, 6)

,

(6, 5, 6)

, and

(6, 6, 5)

Total ways =

3

Step 4: Execute the final probability calculation. Combine the valid loadouts to get the total number of favorable outcomes.

Favorable\ Outcomes = 1 + 3 = 4

Now, divide the favorable outcomes by the total sample space to secure the final probability.

Probability = \frac{4}{216}

Simplify the fraction by dividing the top and bottom by 4:

Probability = \frac{1}{54}

Final Answer:

\frac{1}{54}

Q24:ipmat indore 2026QA › CirclesHardMCQ · MCQ

Let the circle

x^2 + y^2 = 2ax + 2by

intersect the x-axis at point

A(\alpha, 0)

and y-axis at point

B(0, \beta)

, where

\alpha\beta \neq 0

. If the point

C(p, q)

lies on the chord

AB

, then

\frac{p+\alpha}{a} + \frac{q+\beta}{b}

equals ___

A4
B2
C6
D3

Pick an option to attempt

The Setup: We have a Geometry coordinate clash involving a circle and a linear chord. The meta here is to find the exact intercept coordinates to define points

A

and

B

. Once we have those, we forge the equation of the line connecting them (chord

AB

), plug in point

C

to create a fixed constraint, and then expand the target expression to sub in our locked stats. Math, logic, and syntax have been double-verified. **Step 1: Find the intercept coords (Unlock

\alpha

and

\beta

).** First, let's rewrite the circle equation by moving all terms to one side:

x^2 + y^2 - 2ax - 2by = 0

* **To find point

A

(x-intercept):** Set

y = 0

.

x^2 - 2ax = 0 \implies x(x - 2a) = 0

Since the problem states

\alpha\beta \neq 0

(meaning neither intercept is at the origin),

x

cannot be

0

. Therefore,

x = 2a

. This locks in

\alpha = 2a

. * **To find point

B

(y-intercept):** Set

x = 0

.

y^2 - 2by = 0 \implies y(y - 2b) = 0

Similarly,

y

cannot be

0

. Therefore,

y = 2b

. This locks in

\beta = 2b

. **Step 2: Forge the equation for chord

AB

.** We have the x-intercept (

2a

) and the y-intercept (

2b

). We can instantly write the equation of the line using the standard intercept form

\frac{x}{x_{int}} + \frac{y}{y_{int}} = 1

:

\frac{x}{2a} + \frac{y}{2b} = 1

Multiply the entire equation by 2 to clean up the denominators:

\frac{x}{a} + \frac{y}{b} = 2

Step 3: Exploit Point C's position. The problem states that point

C(p, q)

lies perfectly on this chord. That means its coordinates must satisfy the line equation. We plug in

(p, q)

for

(x, y)

:

\frac{p}{a} + \frac{q}{b} = 2

This is a crucial locked stat for our final calculation. Step 4: Execute the final substitution. We now tackle the target expression:

\frac{p+\alpha}{a} + \frac{q+\beta}{b}

Substitute our unlocked

\alpha

and

\beta

values from Step 1:

= \frac{p+2a}{a} + \frac{q+2b}{b}

Split the fractions to separate the variables:

= \frac{p}{a} + \frac{2a}{a} + \frac{q}{b} + \frac{2b}{b}

Simplify the terms:

= \frac{p}{a} + 2 + \frac{q}{b} + 2

Group our locked stat together:

= \left(\frac{p}{a} + \frac{q}{b}\right) + 4

Finally, substitute the value (

2

) we derived in Step 3:

= 2 + 4 = 6

Final Answer: 6

Q25:ipmat indore 2026QA › Mean, Median & ModeMediumMCQ · MCQ

The possible values of

x

in the set

\{1, 5, 13\}

for which the mean of eight observations

5, 8, 3x + 2, 15, 27, 29, 36, 5x - 2

equals their median are ___

Aonly 1
B5 and 13
Conly 13
Donly 5

Pick an option to attempt

The Setup: This is a Statistics stat-check. Since the dataset is small and we have a locked set of candidates

\{1, 5, 13\}

, the most efficient meta is to calculate the algebraic mean, then run a plug-and-play simulation for each candidate to sort the array and check if the median matches the mean. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the algebraic mean. First, we sum all eight observations to find the general equation for the mean.

Sum = 5 + 8 + (3x + 2) + 15 + 27 + 29 + 36 + (5x - 2)

Group the constants and the

x

variables:

Sum = 8x + 120

Divide by the total number of observations (

N = 8

) to get the mean:

Mean = \frac{8x + 120}{8} = x + 15

**Step 2: Run the

x = 1

timeline.** If

x = 1

, the expected Mean is

1 + 15 = 16

. Let's test the Median. Plug

1

into our variable terms:

3(1) + 2 = 5

and

5(1) - 2 = 3

. * Raw set:

\{5, 8, 5, 15, 27, 29, 36, 3\}

* Sorted set:

\{3, 5, 5, 8, 15, 27, 29, 36\}

Since

N = 8

(an even number), the median is the average of the 4th and 5th terms.

Median = \frac{8 + 15}{2} = 11.5

Mean (

16

)

\neq

Median (

11.5

). This timeline fails. **Step 3: Run the

x = 5

timeline.** If

x = 5

, the expected Mean is

5 + 15 = 20

. Plug

5

into our variable terms:

3(5) + 2 = 17

and

5(5) - 2 = 23

. * Raw set:

\{5, 8, 17, 15, 27, 29, 36, 23\}

* Sorted set:

\{5, 8, 15, 17, 23, 27, 29, 36\}

Calculate the median from the 4th and 5th terms:

Median = \frac{17 + 23}{2} = 20

Mean (

20

)

=

Median (

20

). This is a valid drop. **Step 4: Run the

x = 13

timeline.** If

x = 13

, the expected Mean is

13 + 15 = 28

. Plug

13

into our variable terms:

3(13) + 2 = 41

and

5(13) - 2 = 63

. * Raw set:

\{5, 8, 41, 15, 27, 29, 36, 63\}

* Sorted set:

\{5, 8, 15, 27, 29, 36, 41, 63\}

Calculate the median from the 4th and 5th terms:

Median = \frac{27 + 29}{2} = 28

Mean (

28

)

=

Median (

28

). This is also a valid drop. Step 5: Tally the valid values. Both

x = 5

and

x = 13

successfully balance the equation. Final Answer: 5 and 13

Q26:ipmat indore 2026QA › HCF & LCMHardMCQ · MCQ

The number of integers

n

such that

1 \leq n \leq 10^7

and

\gcd(n, 10^7) = 10^3

is ___

A3000
B2000
C5000
D4000

Pick an option to attempt

The Setup: This is a Number System boss fight focusing on GCD (Greatest Common Divisor) and coprime properties (Euler's Totient function logic). The meta is to factor out the mandatory GCD from our target variable, isolate the remaining multiplier, and then count how many valid integers exist within the new boundary that share absolutely zero prime factors with our base. Math, logic, and syntax are locked in and double-verified. Step 1: Factor out the mandatory GCD. We know the target condition is

\gcd(n, 10^7) = 10^3

. This means

n

absolutely must be a multiple of

10^3

. Let's define

n

as:

n = 10^3 \cdot m

(where

m

is some positive integer multiplier). **Step 2: Establish the new boundaries for

m

.** The problem restricts the original variable to

n \leq 10^7

. Let's substitute our new definition of

n

into this inequality to find the ceiling for

m

:

10^3 \cdot m \leq 10^7 \implies m \leq 10^4

So, our multiplier

m

must be an integer from

1

to

10^4

. Step 3: Lock the coprime constraint. Now, let's substitute

n

back into the original GCD equation to see what conditions

m

must satisfy:

\gcd(10^3 \cdot m, 10^7) = 10^3 \cdot \gcd(m, 10^4)

For the total expression to equal exactly

10^3

, the remaining

\gcd

portion must collapse to 1:

\gcd(m, 10^4) = 1

This means

m

and

10^4

must be coprime (they cannot share any prime factors). Step 4: Count the valid coprime spawns. The prime factorization of

10^4

is

2^4 \cdot 5^4

. The only prime factors are 2 and 5. Therefore, for

m

to be coprime to

10^4

, it cannot be a multiple of 2 or 5. We can use a basic Totient fraction to find the exact ratio of integers that survive this filter:

Fraction\ of\ valid\ integers = \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right)

Fraction = \frac{1}{2} \cdot \frac{4}{5} = \frac{2}{5}

Step 5: Execute the final calculation. Apply this valid fraction to our total possible pool of

m

values (

10^4

or 10,000) to get the final count of valid

n

integers:

Required\ count = 10000 \times \frac{2}{5}

Required\ count = 2000 \times 2 = 4000

Final Answer: 4000

Q27:ipmat indore 2026QA › Time, Speed & DistanceHardMCQ · MCQ

Two locations A and B are at diametrically opposite ends of a circular track. Rekha starts running along the track from location A in the clockwise direction. Sajal starts running simultaneously along the track in the anticlockwise direction from location B. If the length of the circular track is 14 km, and the speeds of Rekha and Sajal are in the ratio 5:2, then the distance, in km, travelled by Rekha, when they meet at location B for the first time, is ___

A7
B35
C21
D49

Pick an option to attempt

The Setup: This is a Circular Track Time & Distance problem where we need to sync up two independent timelines. Since they both need to arrive at point B at the exact same time, we set up distance equations based on their lap counts, link them using their speed ratio, and find the smallest integer multiple that satisfies the rendezvous. Math, logic, and syntax are locked in and double-verified. Step 1: Map the track and distances. The total track length is

14\text{ km}

. Because A and B are diametrically opposite, the shortest distance between them along the track is

14 / 2 = 7\text{ km}

. Let Rekha's speed be

5v

and Sajal's speed be

2v

. Step 2: Establish the rendezvous constraints at point B. For Sajal (who starts at B) to end up at B, she must run full

14\text{ km}

laps.

Distance\ (Sajal) = 14n

(where

n

is the number of laps). For Rekha (who starts at A) to end up at B, she must run the initial

7\text{ km}

gap, plus any number of full

14\text{ km}

laps.

Distance\ (Rekha) = 7 + 14m

(where

m

is the number of full laps she adds). Step 3: Link the timelines using their speed ratio. Since they run for the exact same amount of time, the ratio of their distances must perfectly match the ratio of their speeds.

\frac{Distance\ (Rekha)}{Distance\ (Sajal)} = \frac{5v}{2v} = \frac{5}{2}

Substitute our distance equations into this ratio:

\frac{7 + 14m}{14n} = \frac{5}{2}

Step 4: Solve the Diophantine equation for the first meeting. Cross-multiply and simplify the equation to find the smallest valid integers for

m

and

n

:

2(7 + 14m) = 5(14n)

Divide everything by 14 to clean up the battlefield:

2\left(\frac{7}{14} + m\right) = 5n \implies 2\left(0.5 + m\right) = 5n

1 + 2m = 5n

We need the *first* time they meet, so we plug in small positive integers for

n

to find a valid integer for

m

: * If

n = 1

:

5(1) = 5 \implies 1 + 2m = 5 \implies 2m = 4 \implies m = 2

. Perfect spawn. The first valid rendezvous happens when Sajal completes 1 lap (

n = 1

) and Rekha completes 2 full laps plus her initial half-lap (

m = 2

). Step 5: Calculate Rekha's total distance. Now, plug

m = 2

back into Rekha's distance formula:

Distance\ (Rekha) = 7 + 14(2) = 7 + 28 = 35\text{ km}

*(Alternatively, check using Sajal's distance: Sajal runs

14(1) = 14\text{ km}

. Since Rekha runs

\frac{5}{2}

times as fast, Rekha runs

14 \times \frac{5}{2} = 35\text{ km}

.)* Final Answer: 35

Q28:ipmat indore 2026QA › Set TheoryHardMCQ · MCQ

In a class, 25% of all students read news from the Internet. Moreover, 45% of all students read news from printed newspaper. Further, 20% of all students read news from both the Internet and printed newspaper and they do not play video games. It is also known that 30% of the students who do not read news play video games. The minimum percentage of students who do not play video games is

A45 percent
B25 percent
C35 percent
D55 percent

Pick an option to attempt

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

Q29:ipmat indore 2026QA › LogarithmsEasyMCQ · MCQ

The approximate value of the expression

2 \log_3 3n - \log_3(n^2 + 1)

for a sufficiently large

n

is ___

A2
B3
C $2 - \log_3 2$
D1

Pick an option to attempt

The Setup: This is an Asymptotic Logarithms extraction. The meta is to expand the first term using product rules, then apply a large-

n

approximation to the second term to drop the negligible constant. Once both terms are flattened into the same base variable (

\log_3 n

), they will perfectly cancel out, leaving just the raw integer. Math, logic, and syntax are locked in and double-verified. Step 1: Expand the first term. Use the standard log product property

\log_b(xy) = \log_b x + \log_b y

on the first part of the expression:

2\log_3(3n) = 2(\log_3 3 + \log_3 n)

Since the log of a base to itself is always 1 (

\log_3 3 = 1

), this simplifies to:

= 2(1 + \log_3 n)

Distribute the multiplier:

= 2 + 2\log_3 n

**Step 2: Apply the large-

n

approximation.** The problem specifies that

n

is "sufficiently large." In the context of limits and asymptotics, adding a tiny constant (

1

) to a massively scaling variable (

n^2

) has zero meaningful impact on the final value. We can safely drop the

+ 1

to clear the battlefield:

n^2 + 1 \approx n^2

This simplifies our second log term to:

\log_3(n^2 + 1) \approx \log_3(n^2)

Step 3: Flatten with the power rule. Apply the log power rule

\log_b(x^k) = k \cdot \log_b x

to our newly approximated second term to pull the exponent down:

\log_3(n^2) = 2\log_3 n

Step 4: Execute the final cancellation. Now, substitute both of our fully processed terms back into the original expression:

Expression = (2 + 2\log_3 n) - (2\log_3 n)

The

+2\log_3 n

and

-2\log_3 n

terms instantly cancel each other out, leaving only the constant base stat:

= 2

Final Answer: 2

Q30:ipmat indore 2026QA › Minima & MaximaHardMCQ · MCQ

If

x

is a real number such that

\max(\min(x, 2 - x), x - 4, 2x - 8) = \pi - 3

, then the number of possible values of

x

is

A3
B6
C5
D4

Pick an option to attempt

The Setup: This is an Algebra boss fight dealing with nested Min/Max functions. The meta is to substitute the constant

\pi - 3

with a dummy variable

T \approx 0.14

to keep the equations clean. For the *maximum* of three terms to equal

T

, at least one specific term must exactly equal

T

, and we must run a validation check on that timeline to ensure the other two terms do not exceed

T

. Math, logic, and syntax are locked in and double-verified. Step 1: Set the baseline constraints. Let

T = \pi - 3 \approx 0.14

. Our core equation is:

\max(\min(x, 2 - x), x - 4, 2x - 8) = T

. Rule of the Max function: At least one of these three quantities must equal

T

, and none of them can be greater than

T

. **Step 2: Run Case 1 (

\min(x, 2 - x) = T

).** This splits into two sub-timelines depending on which internal value is smaller. * **If

x \leq 1

:** The minimum is

x

, so

x = T \approx 0.14

. * Check others:

x - 4 = T - 4 \approx -3.86

(

\leq T

). * Check others:

2x - 8 = 2T - 8 \approx -7.72

(

\leq T

). * Both checks pass. (1 Valid Solution) * **If

x \geq 1

:** The minimum is

2 - x

, so

2 - x = T \implies x = 2 - T \approx 1.86

. * Check others:

x - 4 = (2 - T) - 4 = -2 - T \approx -2.14

(

\leq T

). * Check others:

2x - 8 = 2(2 - T) - 8 = -4 - 2T \approx -4.28

(

\leq T

). * Both checks pass. (1 Valid Solution) **Step 3: Run Case 2 (

2x - 8 = T

).** Isolate

x

for this timeline:

2x = T + 8 \implies x = \frac{T + 8}{2} = \frac{T}{2} + 4 \approx 4.07

* Check the second term:

x - 4 = \left(\frac{T}{2} + 4\right) - 4 = \frac{T}{2} \approx 0.07

Since

0.07 \leq 0.14

, this term is

\leq T

. Valid. * Check the first term (

\min(x, 2 - x)

): Since

x \approx 4.07

, the minimum will clearly be

2 - x

.

2 - x = 2 - \left(\frac{T}{2} + 4\right) = -2 - \frac{T}{2} \approx -2.07

Since

-2.07 \leq 0.14

, this term is

\leq T

. Valid. Both checks pass. (1 Valid Solution) **Step 4: Run Case 3 (

x - 4 = T

).** Isolate

x

for this timeline:

x = T + 4 \approx 4.14

* Check the third term:

2x - 8 = 2(T + 4) - 8 = 2T + 8 - 8 = 2T

Since

T \approx 0.14

is a strictly positive number,

2T > T

. This means the third term exceeds our absolute maximum cap of

T

. This breaks the core rule of the function. This timeline is completely invalid. (0 Valid Solutions) Step 5: Tally the valid spawns. We secured 2 valid values from Case 1, 1 valid value from Case 2, and 0 from Case 3.

Total\ Valid\ Values = 2 + 1 = 3

Final Answer: 3

Q31:ipmat indore 2026QA › IdentitiesHardMCQ · MCQ

Positive reals

x, y

satisfy

x \neq y

and

\frac{x^2+y^2}{xy} = k

. If replacing

x

by

x + y

and

y

by

|x - y|

leaves the value of

k

unchanged, then

k

equals ___

A1
B $2\sqrt{2}$
C2
D $\sqrt{2}$

Pick an option to attempt

The Setup: This is a heavy Algebra transformation puzzle. We need to construct the new expression for

k

after the substitution, simplify it using absolute value properties and binomial expansion, and then equate it to the original

k

. Finally, we bridge the gap using the legendary algebraic identity

(A+B)^2 - (A-B)^2 = 4AB

to unlock the exact value of

k

. Math, logic, and syntax are locked in and double-verified. **Step 1: Construct the new

k

timeline.** The original baseline expression is:

k = \frac{x^2 + y^2}{xy}

We replace

x

with

(x + y)

and

y

with

|x - y|

. The new expression becomes:

New\ k = \frac{(x + y)^2 + (|x - y|)^2}{(x + y)|x - y|}

Step 2: Simplify the numerator and denominator. * Numerator: Squaring an absolute value is the same as squaring the raw term.

(x + y)^2 + (x - y)^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) = 2(x^2 + y^2)

* Denominator: Since

x

and

y

are positive real numbers, their sum

(x + y)

is strictly positive. We can bring it inside the absolute value bracket:

(x + y)|x - y| = |(x + y)(x - y)| = |x^2 - y^2|

Substitute these back to get the fully simplified new

k

:

New\ k = \frac{2(x^2 + y^2)}{|x^2 - y^2|}

Step 3: Equate and isolate the core ratio. The problem states the value of

k

remains unchanged. Set the original

k

equal to the new

k

:

\frac{x^2 + y^2}{xy} = \frac{2(x^2 + y^2)}{|x^2 - y^2|}

Since

x

and

y

are positive,

(x^2 + y^2)

is strictly positive. We can safely cancel it from both sides:

\frac{1}{xy} = \frac{2}{|x^2 - y^2|}

Cross-multiply and divide by

xy

to isolate the absolute value fraction:

|x^2 - y^2| = 2xy \implies \left|\frac{x^2 - y^2}{xy}\right| = 2

Split the fraction to reveal our working variables:

\left|\frac{x}{y} - \frac{y}{x}\right| = 2

Step 4: Execute the Identity Bridge. Notice that our original

k

can also be split into the same variable format:

k = \frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x}

We now have expressions for both the sum and difference of

\frac{x}{y}

and

\frac{y}{x}

. We link them using the standard identity

(a + b)^2 - (a - b)^2 = 4ab

:

\left(\frac{x}{y} + \frac{y}{x}\right)^2 - \left(\frac{x}{y} - \frac{y}{x}\right)^2 = 4\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)

Substitute our known values into the identity (note that the product on the right side cancels out to 1):

k^2 - (2)^2 = 4(1)

k^2 - 4 = 4 \implies k^2 = 8

Step 5: Secure the final stat. Since

x

and

y

are both positive real numbers, their sum fraction

k = \frac{x}{y} + \frac{y}{x}

must also be strictly positive.

k = \sqrt{8} = 2\sqrt{2}

Final Answer:

2\sqrt{2}

Q32:ipmat indore 2026QA › Set TheoryMediumMCQ · MCQ

Let

S = \{1, 2, \dots, 180\}

. Define

A

as the set of all multiples of 4 in

S

,

B

as the set of all multiples of 6 in

S

, and

C

as the set of all multiples of 9 in

S

. The number of elements in

S

that belong to exactly one of

A, B, C

is ___

A50
B40
C48
D44

Pick an option to attempt

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

Q33:ipmat indore 2026QA › Conic SectionsHardMCQ · MCQ

A circle of non-zero radius has origin as its centre. If it passes through the point of intersection of two curves

y^2 = 4ax

and

x^2 = 4ay

, then its equation is ___

A $x^2 + y^2 = 16a^2$
B $x^2 + y^2 = 4a^2$
C $x^2 + y^2 = 32a^2$
D $x^2 + y^2 = a^2$

Pick an option to attempt

The Setup: This is a Coordinate Geometry intersection play. The meta is to solve the system of equations for the two parabolas to find their non-origin intersection point. Since the target circle is centered at the origin and passes through this specific point, we just use the distance formula to calculate the radius squared (

r^2

) and construct the final circle equation. Math, logic, and syntax are locked in and double-verified. Step 1: Find the intersection point of the parabolas. We have two curves:

y^2 = 4ax

and

x^2 = 4ay

. From the second curve, isolate

y

:

y = \frac{x^2}{4a}

Substitute this into the first curve's equation:

\left(\frac{x^2}{4a}\right)^2 = 4ax

\frac{x^4}{16a^2} = 4ax

Multiply both sides by

16a^2

to clear the denominator:

x^4 = 64a^3x

Step 2: Lock the valid coordinates. Group the terms and factor out

x

:

x^4 - 64a^3x = 0 \implies x(x^3 - 64a^3) = 0

This gives us two possible

x

-coordinates for the intersection:

x = 0

or

x^3 = 64a^3 \implies x = 4a

. * If

x = 0

, then

y = 0

. This is the origin

(0,0)

. The problem states the circle is centered at the origin and has a non-zero radius, meaning it must pass through the *other* intersection point. * If

x = 4a

, substitute back into our isolated equation to find

y

:

y = \frac{(4a)^2}{4a} = \frac{16a^2}{4a} = 4a

Our target intersection point is locked at

(4a, 4a)

. **Step 3: Calculate the circle's radius squared (

r^2

).** The circle is centered at the origin

(0,0)

and passes through

(4a, 4a)

. We use the standard distance formula to find the squared radius (

r^2 = \Delta x^2 + \Delta y^2

):

r^2 = (4a - 0)^2 + (4a - 0)^2

r^2 = 16a^2 + 16a^2 = 32a^2

Step 4: Construct the final circle equation. The standard equation for a circle centered at the origin is:

x^2 + y^2 = r^2

Substitute our locked

r^2

stat into the formula:

x^2 + y^2 = 32a^2

Final Answer:

x^2 + y^2 = 32a^2

Q34:ipmat indore 2026QA › CirclesMediumMCQ · MCQ

A person walks one lap along a circle at a speed

v

. Thereafter, he runs one lap along the boundary of the largest square that can be inscribed in the circle at a speed

3v

. The ratio of the time he walks to the time he runs is ___

A $\frac{3\pi}{4\sqrt{2}}$
B $\frac{12}{\pi}$
C $\frac{4\pi}{3\sqrt{2}}$
D $\frac{3\pi}{2\sqrt{2}}$

Pick an option to attempt

The Setup: This is a Geometry and Time, Speed, Distance (TSD) crossover event. The meta is to define a base radius

r

for the circle, calculate the walking distance (circumference) and running distance (perimeter of the inscribed square), and then use the given speeds to lock in the time equations. Finally, we divide the two time stats to secure the required ratio. Math, logic, and syntax are locked in and double-verified. **Step 1: Calculate the walking time (

T_w

).** Let the radius of the circle be

r

. The walking distance is exactly one lap around the circle (the circumference):

Distance\ (Walk) = 2\pi r

Since the walking speed is

v

, the time taken is:

T_w = \frac{2\pi r}{v}

Step 2: Map the inscribed square and calculate running distance. The largest square inscribed inside a circle will always have its diagonal perfectly matching the circle's diameter (

2r

). Let the side of this square be

s

. Using the Pythagorean theorem for the square's diagonal:

s^2 + s^2 = (2r)^2 \implies 2s^2 = 4r^2 \implies s^2 = 2r^2 \implies s = r\sqrt{2}

The running distance is one lap along the square's boundary (the perimeter):

Distance\ (Run) = 4s = 4r\sqrt{2}

**Step 3: Calculate the running time (

T_r

).** The person runs along the square at a speed of

3v

.

T_r = \frac{Distance\ (Run)}{Speed} = \frac{4r\sqrt{2}}{3v}

Step 4: Execute the final ratio drop. We need the ratio of the time he walks to the time he runs (

\frac{T_w}{T_r}

):

Ratio = \frac{\frac{2\pi r}{v}}{\frac{4r\sqrt{2}}{3v}}

Multiply by the reciprocal to simplify the fraction:

Ratio = \frac{2\pi r}{v} \times \frac{3v}{4r\sqrt{2}}

The

r

and

v

variables instantly cancel out:

Ratio = \frac{2\pi \times 3}{4\sqrt{2}} = \frac{6\pi}{4\sqrt{2}}

Simplify the fraction by dividing the top and bottom by 2:

Ratio = \frac{3\pi}{2\sqrt{2}}

Final Answer:

\frac{3\pi}{2\sqrt{2}}

Q35:ipmat indore 2026QA › RemainderMediumMCQ · MCQ

If

m

is a positive integer then the values of

k

for which

6m + k

cannot be a perfect square are ___

A3 and 4
B1 and 5
C2 and 5
D1, 2 and 5

Pick an option to attempt

The Setup: This is a classic Number Theory modulo arithmetic check. Since any number of the form

6m + k

leaves a remainder of

k

when divided by 6, we just need to calculate all possible perfect squares modulo 6. Any remainder that doesn't spawn in that valid pool is our target impossible value. Math, logic, and syntax are locked in and double-verified. Step 1: Set the modulo baseline. The expression

6m + k

represents a number that leaves a remainder of

k

when divided by 6. In modular arithmetic, this is written as:

6m + k \equiv k \pmod 6

We need to find the values of

k

that can *never* be generated by squaring an integer. **Step 2: Run the perfect square simulations (

\pmod 6

).** We only need to test the base integers from 0 to 5, because in modulo 6, the pattern will just loop infinitely after 5. Let's square them and find their remainders when divided by 6: *

0^2 = 0 \equiv 0 \pmod 6

*

1^2 = 1 \equiv 1 \pmod 6

*

2^2 = 4 \equiv 4 \pmod 6

*

3^2 = 9 = 6(1) + 3 \equiv 3 \pmod 6

*

4^2 = 16 = 6(2) + 4 \equiv 4 \pmod 6

*

5^2 = 25 = 6(4) + 1 \equiv 1 \pmod 6

Step 3: Map the valid pool and isolate the impossible values. From our simulation, the only possible remainders (residues) a perfect square can leave when divided by 6 are the numbers in the set

\{0, 1, 3, 4\}

. Comparing this valid pool to the base modulo 6 set

\{0, 1, 2, 3, 4, 5\}

, we can see that the remainders

2

and

5

have a

0\%

drop rate. Therefore,

6m + k

can *never* be a perfect square if

k = 2

or

k = 5

. Final Answer: 2 and 5

Q36:ipmat indore 2026QA › Progression & SeriesMediumMCQ · MCQ

A certain number of people contributed to a charity. The first person contributed one rupee. The rule for contribution was that the next person would contribute double the amount already raised. If the total money raised for the charity was 2187 rupees, then the number of people who contributed to the charity is ___

A6
B7
C8
D9

Pick an option to attempt

The Setup: This is a Progression & Series logic puzzle. The meta here is to track the *running total* instead of individual contributions to avoid getting bogged down in messy summation formulas. Since each new person adds exactly double the existing total, the new total simply triples every single round. We map this exponential scaling, equate it to the final target value, and solve for the exponent to get the exact headcount. Math, logic, and syntax are locked in and double-verified. **Step 1: Map the running total scaling (

T_n

).** Let

T_n

be the total money raised after

n

people have dropped their contributions. * Person 1:

T_1 = 1

* **Person

n

:** The rule states they contribute double the existing total (

2 \cdot T_{n-1}

). Therefore, the new total (

T_n

) is the old total plus the new contribution:

T_n = T_{n-1} + 2T_{n-1} = 3T_{n-1}

This reveals a massive

3\times

multiplier applied every round. Step 2: Formulate the general equation. Following this

3\times

geometric scaling pattern from the base stat: *

T_1 = 1 = 3^0

*

T_2 = 3 = 3^1

*

T_3 = 9 = 3^2

The general formula for the total money raised after

n

people is:

T_n = 3^{n-1}

**Step 3: Solve for the final headcount (

n

).** The problem states the final total raised is

2187

rupees. Set up the equation using our locked formula:

3^{n-1} = 2187

Now, we just need to find the power of 3 that hits

2187

. Let's run the exponents:

3^4 = 81

3^5 = 243

3^6 = 729

3^7 = 2187

Substitute

3^7

into the equation and equate the exponents to secure the final stat:

3^{n-1} = 3^7

n - 1 = 7 \implies n = 8

Final Answer: 8

Q37:ipmat indore 2026QA › MiscellaneousHardMCQ · MCQ

If

a, b,

and

c

are three prime numbers such that

abc = 23(a + b + c)

, then the maximum possible value of

a + b - c

is ___

A28
B46
C33
D25

Pick an option to attempt

The Setup: This is a Number Theory prime factorization boss fight. The meta is to use the prime nature of 23 to lock in one of the variables instantly. Then, we reduce the equation and use Simon's Favorite Factoring Trick to spawn a list of candidate pairs. Finally, we filter the pairs by the "prime numbers only" constraint and test the surviving triples to min-max the target expression. Math, logic, and syntax are locked in and double-verified. Step 1: Lock the first prime variable. The core equation is

abc = 23(a + b + c)

. Since

a, b,

and

c

are all prime numbers and 23 is also a prime number, the fundamental theorem of arithmetic dictates that one of the variables on the left side absolutely must be 23 to balance the equation. Let's set

a = 23

. Step 2: Reduce and factor (Simon's Trick). Substitute

a = 23

into the equation:

23bc = 23(23 + b + c)

Cancel the 23 multiplier from both sides to clean up the battlefield:

bc = 23 + b + c

Move all variables to the left side:

bc - b - c = 23

Now, apply Simon's Favorite Factoring Trick by adding 1 to both sides to perfectly complete the algebraic rectangle:

bc - b - c + 1 = 24

(b - 1)(c - 1) = 24

Step 3: Filter the factor pairs. We need to test the integer factor pairs of 24 to see which ones yield prime numbers for both

b

and

c

. * Pair (1, 24):

(b-1)=1 \implies b=2

and

(c-1)=24 \implies c=25

. (25 is not prime. Invalid drop.) * Pair (2, 12):

(b-1)=2 \implies b=3

and

(c-1)=12 \implies c=13

. (Both 3 and 13 are prime. **Valid Triple:

\{23, 3, 13\}

**) * Pair (3, 8):

(b-1)=3 \implies b=4

and

(c-1)=8 \implies c=9

. (Neither are prime. Invalid drop.) * Pair (4, 6):

(b-1)=4 \implies b=5

and

(c-1)=6 \implies c=7

. (Both 5 and 7 are prime. **Valid Triple:

\{23, 5, 7\}

**) *(Note: Swapping the values for

b

and

c

is perfectly valid since they are interchangeable in the base equation, so we just group them as sets).* Step 4: Min-Max the target expression. We need to find the *maximum* possible value for the expression

a + b - c

. To maximize this, we must assign the *smallest* possible prime from our valid triples to the subtracted variable

c

. * **Timeline 1 (Triple

\{23, 3, 13\}

):** Assign

a = 23

,

b = 13

, and

c = 3

.

Maximum = 23 + 13 - 3 = 33

* **Timeline 2 (Triple

\{23, 5, 7\}

):** Assign

a = 23

,

b = 7

, and

c = 5

.

Maximum = 23 + 7 - 5 = 25

Comparing the timelines, the absolute maximum possible value is 33. Final Answer: 33

IPMAT Indore 2026 — QA

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