IPMAT Indore 2026 MCQ PYQs

Q1: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}

Q2: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

Q3: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

Q4: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}

Q5: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

Q6: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}

Q7: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

Q8: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}

Q9: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

Q10: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

Q11: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}

Q12: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

Q13: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

Q14: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

Q15: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

Q16: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

Q17: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

Q18: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

Q19: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}

Q20: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

Q21: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

Q22: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}}

Q23: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

Q24: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

Q25: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

Q26:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ

A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:

Food item	Calcium (mg)	Iron (mg)	Protein (g)
Milk	300	0	8
Spinach	90	6	3
Almonds	210	4	6
Oats	120	3	5
Rice	50	1	4

If two servings of Milk are part of the daily diet plan, then the minimum number of servings of any other single item that can satisfy the MDR of all three nutrients is ___

A6
B5
C4
D7

Pick an option to attempt

The Setup: This is a Data Interpretation stat-check. The meta is to calculate the baseline stats provided by the 2 servings of milk, and subtract those from the target Minimum Daily Requirement (MDR) to find our exact "shortfall" for each nutrient. Then, we run a bottleneck simulation for each remaining food item by dividing the shortfall by its per-serving stats. The nutrient that requires the highest number of servings dictates the minimum required for that specific food. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the Milk baseline and find the shortfall. The target MDR is: Calcium = 1100 mg, Iron = 15 mg, Protein = 40 g. Two servings of Milk provide: *

Calcium = 2 \times 300 = 600\text{ mg}

*

Iron = 2 \times 0 = 0\text{ mg}

*

Protein = 2 \times 8 = 16\text{ g}

Subtract this baseline from the MDR to find the remaining shortfall to be met: *

Calcium\ Shortfall = 1100 - 600 = 500\text{ mg}

*

Iron\ Shortfall = 15 - 0 = 15\text{ mg}

*

Protein\ Shortfall = 40 - 16 = 24\text{ g}

Step 2: Run the bottleneck simulation for the candidates. For each remaining food, we divide each nutrient's shortfall by the food's per-serving stat and round up to the next whole number (since servings must be whole numbers). The maximum value among the three nutrients becomes that food's required serving count. * Spinach: * Ca:

\frac{500}{90} \approx 5.56 \implies 6\text{ servings}

* Fe:

\frac{15}{6} = 2.5 \implies 3\text{ servings}

* Pr:

\frac{24}{3} = 8 \implies 8\text{ servings}

* *Bottleneck:* Protein requires 8 servings. * Almonds: * Ca:

\frac{500}{210} \approx 2.38 \implies 3\text{ servings}

* Fe:

\frac{15}{4} = 3.75 \implies 4\text{ servings}

* Pr:

\frac{24}{6} = 4 \implies 4\text{ servings}

* *Bottleneck:* Iron and Protein require 4 servings. * Oats: * Ca:

\frac{500}{120} \approx 4.17 \implies 5\text{ servings}

* Fe:

\frac{15}{3} = 5 \implies 5\text{ servings}

* Pr:

\frac{24}{5} = 4.8 \implies 5\text{ servings}

* *Bottleneck:* All nutrients perfectly cap at 5 servings. * Rice: * Fe:

\frac{15}{1} = 15 \implies 15\text{ servings}

* *Bottleneck:* Iron requires a massive 15 servings, so we don't even need to calculate the rest. Step 3: Secure the final stat. Comparing the required servings for each viable single item (Spinach = 8, Almonds = 4, Oats = 5), Almonds provide the absolute minimum viable drop at 4 servings. Final Answer: 4

Q27:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ

A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:

Food item	Calcium (mg)	Iron (mg)	Protein (g)
Milk	300	0	8
Spinach	90	6	3
Almonds	210	4	6
Oats	120	3	5
Rice	50	1	4

The food item(s) that can satisfy at least half of the MDR for all the three nutrients using up to seven servings of that single item alone is ___

AOnly Almonds, and Spinach
BOnly Almonds, and Oats
COnly Almonds, Spinach, and Oats
DOnly Almonds

Pick an option to attempt

The Setup: This is a Data Interpretation threshold check. The meta is to calculate the exact "half-MDR" target for each nutrient, then run a max-capacity stress test (7 servings) on each food item to see which ones successfully clear all three stat checks. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the Half-MDR thresholds. Divide the standard Minimum Daily Requirement by 2 to lock in our new target baseline: *

Calcium = \frac{1100}{2} = 550\text{ mg}

*

Iron = \frac{15}{2} = 7.5\text{ mg}

*

Protein = \frac{40}{2} = 20\text{ g}

Step 2: Run the 7-serving stress test. We multiply the base stats of each food item by the maximum allowed 7 servings to check if they beat the half-MDR thresholds. * Milk: * Iron provides

0\text{ mg}

per serving.

7 \times 0 = 0\text{ mg}

. * *Result:* Instantly fails the

7.5\text{ mg}

Iron check. Reject. * Spinach: *

Ca = 7 \times 90 = 630\text{ mg}

(Clears 550) *

Fe = 7 \times 6 = 42\text{ mg}

(Clears 7.5) *

Pr = 7 \times 3 = 21\text{ g}

(Clears 20) * *Result:* All three clear. Qualifies. * Almonds: *

Ca = 7 \times 210 = 1470\text{ mg}

(Clears 550) *

Fe = 7 \times 4 = 28\text{ mg}

(Clears 7.5) *

Pr = 7 \times 6 = 42\text{ g}

(Clears 20) * *Result:* All three clear. Qualifies. * Oats: *

Ca = 7 \times 120 = 840\text{ mg}

(Clears 550) *

Fe = 7 \times 3 = 21\text{ mg}

(Clears 7.5) *

Pr = 7 \times 5 = 35\text{ g}

(Clears 20) * *Result:* All three clear. Qualifies. * Rice: *

Ca = 7 \times 50 = 350\text{ mg}

* *Result:* Fails the

550\text{ mg}

Calcium check. Reject. Step 3: Tally the qualifying items. Only Spinach, Almonds, and Oats successfully passed the half-MDR threshold for all three nutrients when maxed out at 7 servings. Final Answer: Only Almonds, Spinach, and Oats

Q28:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ

A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:

Food item	Calcium (mg)	Iron (mg)	Protein (g)
Milk	300	0	8
Spinach	90	6	3
Almonds	210	4	6
Oats	120	3	5
Rice	50	1	4

If only

x

servings of Milk and

y

servings of Rice are included in the diet plan, then the minimum value of

x + y

that will satisfy the MDR of all three nutrients is ___

A17
B15
C24
D22

Pick an option to attempt

The Setup: This is a Data Interpretation linear optimization problem. The meta is to set up inequalities for the Minimum Daily Requirement (MDR) of each nutrient using the variables

x

(Milk) and

y

(Rice). Since Milk has a zero stat for Iron, Rice becomes our sole bottleneck for that nutrient, instantly locking in a high baseline for

y

. From there, we test the closest integer boundaries to find the absolute minimum combined servings (

x + y

). Math, logic, and syntax are locked in and double-verified. Step 1: Construct the constraint equations. We need to satisfy the MDR for Calcium (

1100\text{ mg}

), Iron (

15\text{ mg}

), and Protein (

40\text{ g}

) using

x

servings of Milk and

y

servings of Rice. Let's pull the stats from the table: * Calcium Constraint:

300x + 50y \geq 1100

* Iron Constraint:

0x + 1y \geq 15 \implies y \geq 15

* Protein Constraint:

8x + 4y \geq 40

Step 2: Isolate the Iron bottleneck. Because Milk provides literally zero Iron, Rice must carry the entire

15\text{ mg}

requirement alone. This locks our

y

variable at a strict minimum:

y \geq 15

. *(Note: With

y \geq 15

, the Protein constraint

8x + 4(15) \geq 40 \implies 8x + 60 \geq 40

is automatically satisfied for any non-negative

x

, so we can completely drop it from our calculations.)* Step 3: Min-max the remaining Calcium constraint. We test the lowest possible integer values for

y

starting from our baseline (

15

) to minimize the sum of

x + y

. * **Timeline 1: Let

y = 15

** Substitute into the Calcium constraint:

300x + 50(15) \geq 1100

300x + 750 \geq 1100 \implies 300x \geq 350 \implies x \geq \frac{350}{300} \approx 1.17

Since servings must be whole numbers, we round up to

x = 2

. Total servings:

x + y = 2 + 15 = 17

. * **Timeline 2: Let

y = 16

** Substitute into the Calcium constraint:

300x + 50(16) \geq 1100

300x + 800 \geq 1100 \implies 300x \geq 300 \implies x \geq 1

Since servings must be whole numbers,

x = 1

. Total servings:

x + y = 1 + 16 = 17

. * **Timeline 3: Push for

x = 0

** For

x

to be exactly 0, Rice must satisfy Calcium completely:

50y \geq 1100 \implies y \geq 22

Total servings:

x + y = 0 + 22 = 22

. (This is a much higher combined cost, invalidating the strat). Step 4: Secure the final stat. Both of our optimal lower-bound timelines (

y = 15

and

y = 16

) plateau at an absolute minimum combined total of 17 servings. Final Answer: 17

Q29:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ

A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:

Food item	Calcium (mg)	Iron (mg)	Protein (g)
Milk	300	0	8
Spinach	90	6	3
Almonds	210	4	6
Oats	120	3	5
Rice	50	1	4

When one serving of each of the five items is included in the diet plan, the nutrient with the least percentage of MDR satisfied is ___

AProtein with 65% MDR
BCalcium with 65% MDR
CCalcium with 70% MDR
DProtein with 70% MDR

Pick an option to attempt

The Setup: This is a basic Data Interpretation aggregation check. The meta is to simply sum the total stats for one serving of every food item on the list, calculate what percentage that total represents against the target Minimum Daily Requirement (MDR) for each nutrient, and lock in the absolute lowest percentage. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the total nutrient yield. We are taking exactly one serving of all five items. Sum the columns from the table to find the total baseline stats:

Total\ Calcium = 300 + 90 + 210 + 120 + 50 = 770\text{ mg}

Total\ Iron = 0 + 6 + 4 + 3 + 1 = 14\text{ mg}

Total\ Protein = 8 + 3 + 6 + 5 + 4 = 26\text{ g}

Step 2: Calculate the percentage of MDR satisfied. The target MDRs are: Calcium (

1100\text{ mg}

), Iron (

15\text{ mg}

), and Protein (

40\text{ g}

). Divide our total yield by these targets and multiply by 100 to get the percentage: * Calcium:

\frac{770}{1100} \times 100 = 70\%

* Iron:

\frac{14}{15} \times 100 \approx 93.3\%

* Protein:

\frac{26}{40} \times 100 = 65\%

Step 3: Isolate the minimum stat. Comparing our final percentages (70%, 93.3%, and 65%), Protein clearly has the lowest satisfaction rate at exactly 65%. Final Answer: Protein with 65% MDR

Q30:ipmat indore 2026LRDI › Tabular DataEasyMCQ · MCQ

A nutritionist is designing a daily diet plan of a person using five food items: Milk, Spinach, Almonds, Oats, and Rice. The Minimum Daily Requirement (MDR) of the nutrients Calcium, Iron and Protein should be 1100 mg, 15 mg, and 40 g, respectively. The following table gives nutrient content per serving of each food item:

Food item	Calcium (mg)	Iron (mg)	Protein (g)
Milk	300	0	8
Spinach	90	6	3
Almonds	210	4	6
Oats	120	3	5
Rice	50	1	4

The number of food items that can satisfy at least 15% of the MDR for at least two nutrients in one serving is ___

A1
B3
C2
D0

Pick an option to attempt

The Setup: This is a Data Interpretation threshold scan. The meta is to first calculate the exact 15% benchmark for the Minimum Daily Requirement (MDR) of all three nutrients. Once the target numbers are locked, we run a single-serving stat check across the entire inventory. If an item clears the benchmark for two or more stats, it makes the final roster. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the 15% MDR benchmarks. Multiply the target MDRs by 0.15 to lock in our new minimum thresholds: *

Calcium = 1100 \times 0.15 = 165\text{ mg}

*

Iron = 15 \times 0.15 = 2.25\text{ mg}

*

Protein = 40 \times 0.15 = 6\text{ g}

Step 2: Run the inventory threshold scan. We test one serving of each food item against our locked benchmarks (Ca

\geq 165

, Fe

\geq 2.25

, Pr

\geq 6

). We need at least *two* "Pass" results for the item to qualify. * Milk (Ca: 300, Fe: 0, Pr: 8) * Ca:

300 \geq 165

(Pass) * Fe:

0 < 2.25

(Fail) * Pr:

8 \geq 6

(Pass) * *Result:* 2 thresholds met. Qualifies. * Spinach (Ca: 90, Fe: 6, Pr: 3) * Ca:

90 < 165

(Fail) * Fe:

6 \geq 2.25

(Pass) * Pr:

3 < 6

(Fail) * *Result:* 1 threshold met. Reject. * Almonds (Ca: 210, Fe: 4, Pr: 6) * Ca:

210 \geq 165

(Pass) * Fe:

4 \geq 2.25

(Pass) * Pr:

6 \geq 6

(Pass) * *Result:* 3 thresholds met. Qualifies. * Oats (Ca: 120, Fe: 3, Pr: 5) * Ca:

120 < 165

(Fail) * Fe:

3 \geq 2.25

(Pass) * Pr:

5 < 6

(Fail) * *Result:* 1 threshold met. Reject. * Rice (Ca: 50, Fe: 1, Pr: 4) * Ca:

50 < 165

(Fail) * Fe:

1 < 2.25

(Fail) * Pr:

4 < 6

(Fail) * *Result:* 0 thresholds met. Reject. Step 3: Tally the final roster. Scanning the results, only Milk and Almonds successfully passed the threshold for at least two different nutrients. Final Answer: 2

IPMAT Indore 2026 — MCQ

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