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IPMAT Indore 2022 — SA

All 15 SA previous year questions (PYQs) from the IPMAT Indore 2022 past year paper, with answers and full solutions.

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Q1:ipmat indore 2022QA › LogarithmsMediumSA · TITA
If log⁡(x2)y+log⁡(y2)x=1\log _{\left(x^{2}\right)} y+\log _{\left(y^{2}\right)} x=1 and y=x2−30y=x^{2}-30, then the value of x2+y2x^{2}+y^{2} is ___________.
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The Setup: We are given a logarithmic equation involving bases of x2x^2 and y2y^2, alongside a quadratic relation y=x2−30y=x^2-30. We must simplify the log equation to find xx and yy, then compute x2+y2x^2+y^2. Step 1: Simplify the logarithmic equation. Use the base power rule log⁡(bk)a=1klog⁡ba\log_{(b^k)} a = \frac{1}{k} \log_b a: 12log⁡xy+12log⁡yx=1\frac{1}{2}\log_x y + \frac{1}{2}\log_y x = 1 Multiply by 2: log⁡xy+log⁡yx=2\log_x y + \log_y x = 2 Step 2: Solve for the relationship between xx and yy. Let a=log⁡xya = \log_x y. By logarithmic inversion, log⁡yx=1a\log_y x = \frac{1}{a}. a+1a=2a + \frac{1}{a} = 2 a2−2a+1=0  ⟹  (a−1)2=0  ⟹  a=1a^2 - 2a + 1 = 0 \implies (a - 1)^2 = 0 \implies a = 1 Therefore, log⁡xy=1\log_x y = 1, which strictly implies y=xy = x. Step 3: Apply the quadratic relation to find the valid coordinates. Substitute y=xy = x into the second equation y=x2−30y = x^2 - 30: x=x2−30  ⟹  x2−x−30=0x = x^2 - 30 \implies x^2 - x - 30 = 0 Factor the quadratic: (x−6)(x+5)=0  ⟹  x=6 or x=−5(x - 6)(x + 5) = 0 \implies x = 6 \text{ or } x = -5 Logarithmic bases (x2x^2 and y2y^2) and arguments (xx and yy) must be strictly positive, discarding x=−5x = -5. Thus, x=6x = 6, and consequently y=6y = 6. Step 4: Calculate the final target expression. x2+y2=62+62=36+36=72x^2 + y^2 = 6^2 + 6^2 = 36 + 36 = 72 Final Answer: 72
Q2:ipmat indore 2022QA › ModulusMediumSA · TITA
The area enclosed by 2∣x∣+3∣y∣≤62|x|+3|y| \leq 6 is ____________ sq. units.
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The Setup: The inequality 2∣x∣+3∣y∣≤62|x| + 3|y| \le 6 forms a geometric shape (a rhombus) bounded by absolute value constraints. We must calculate its total enclosed area. Step 1: Find the intercepts of the boundary lines. Consider the boundary equation in the first quadrant where x≥0x \ge 0 and y≥0y \ge 0: 2x+3y=62x + 3y = 6 Find the xx-intercept (set y=0y = 0): 2x=6  ⟹  x=32x = 6 \implies x = 3. Find the yy-intercept (set x=0x = 0): 3y=6  ⟹  y=23y = 6 \implies y = 2. The vertices of the full rhombus across all four quadrants are (±3,0)(\pm 3, 0) and (0,±2)(0, \pm 2). Step 2: Calculate the area of the bounded shape. The area of a rhombus is given by 12×d1×d2\frac{1}{2} \times d_1 \times d_2, where d1d_1 and d2d_2 are the lengths of the diagonals. The horizontal diagonal connects (−3,0)(-3, 0) to (3,0)(3, 0), so d1=6d_1 = 6. The vertical diagonal connects (0,−2)(0, -2) to (0,2)(0, 2), so d2=4d_2 = 4. Area=12(6)(4)=12\text{Area} = \frac{1}{2} (6)(4) = 12 Final Answer: 12
Q3:ipmat indore 2022QA › Time, Speed & DistanceEasySA · TITA
When Geeta increases her speed from 1212 km/hr to 2020 km/hr, she takes one hour less than the usual time to cover the distance between her home and office. The distance between her home and office is ___________ km.
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The Setup: A speed increase from 12 km/hr12\text{ km/hr} to 20 km/hr20\text{ km/hr} results in a 11-hour time reduction to cover a fixed distance. We must build a time-difference equation to find the distance. Step 1: Formulate the time equation. Let dd be the fixed distance between the home and office. Time at normal speed = d12\frac{d}{12}. Time at increased speed = d20\frac{d}{20}. The difference between these times is exactly 11 hour: d12−d20=1\frac{d}{12} - \frac{d}{20} = 1 Step 2: Solve for dd. Find a common denominator (6060) to combine the fractions: 5d60−3d60=1\frac{5d}{60} - \frac{3d}{60} = 1 2d60=1  ⟹  d30=1  ⟹  d=30\frac{2d}{60} = 1 \implies \frac{d}{30} = 1 \implies d = 30 Final Answer: 30
Q4:ipmat indore 2022QA › Permutation & CombinationEasySA · TITA
The number of triangles that can be formed by choosing points from 7 points on a line and 5 points on another parallel line is _________.
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The Setup: We are choosing points to form triangles from two parallel lines containing 77 and 55 points respectively. Step 1: Identify the combinatorial cases for triangle formation. A triangle requires exactly 33 non-collinear vertices. Since the points lie on two parallel lines, taking 33 points from the same line is invalid. Thus, we must select either: * Case 1: 11 point from the first line and 22 points from the second line. * Case 2: 22 points from the first line and 11 point from the second line. Step 2: Calculate configurations for Case 1. Choose 11 point from the line with 77, and 22 points from the line with 55: Ways1=(71)×(52)=7×(5×42)=7×10=70\text{Ways}_1 = \binom{7}{1} \times \binom{5}{2} = 7 \times \left(\frac{5 \times 4}{2}\right) = 7 \times 10 = 70 Step 3: Calculate configurations for Case 2. Choose 22 points from the line with 77, and 11 point from the line with 55: Ways2=(72)×(51)=(7×62)×5=21×5=105\text{Ways}_2 = \binom{7}{2} \times \binom{5}{1} = \left(\frac{7 \times 6}{2}\right) \times 5 = 21 \times 5 = 105 Step 4: Sum the valid triangle configurations. Total Triangles=70+105=175\text{Total Triangles} = 70 + 105 = 175 Final Answer: 175
Q5:ipmat indore 2022QA › Profit & LossMediumSA · TITA
Aruna purchases a certain number of apples for INR 20 each and a certain number of mangoes for INR 25 each. If she sells all the apples at 10%10 \% profit and all the mangoes at 20%20 \% loss, overall she makes neither profit nor loss. Instead, if she sells all the apples at 20%20 \% loss and all the mangoes at 10%10 \% profit, overall she makes a loss of INR 150. Then the number of apples purchased by Aruna is _________.
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The Setup: We have a system of profit/loss constraints across two different scenarios for an unknown quantity of apples and mangoes with fixed unit costs. We must solve the linear equations to find the exact number of apples. Step 1: Define variables and calculate per-item cost variations. Let AA be the number of apples (Cost = 2020 each). Let MM be the number of mangoes (Cost = 2525 each). * +10%+10\% on apples = +2+2 profit per apple. * −20%-20\% on mangoes = −5-5 loss per mango. * −20%-20\% on apples = −4-4 loss per apple. * +10%+10\% on mangoes = +2.5+2.5 profit per mango. Step 2: Formulate Scenario 1 to find the quantity ratio. In the first scenario, the net profit is exactly 00: 2A−5M=0  ⟹  2A=5M  ⟹  M=0.4A2A - 5M = 0 \implies 2A = 5M \implies M = 0.4A Step 3: Formulate Scenario 2 to solve for AA. In the second scenario, the net loss is 150150 (meaning profit is −150-150): −4A+2.5M=−150-4A + 2.5M = -150 Multiply by −1-1 for clarity: 4A−2.5M=1504A - 2.5M = 150 Step 4: Substitute the ratio and calculate AA. Substitute M=0.4AM = 0.4A into the second equation: 4A−2.5(0.4A)=1504A - 2.5(0.4A) = 150 4A−1A=1504A - 1A = 150 3A=150  ⟹  A=503A = 150 \implies A = 50 Final Answer: 50
Q6:ipmat indore 2022QA › ModulusMediumSA · TITA
Given that f(x)=∣x∣+2∣x−1∣+∣x−2∣+∣x−4∣+∣x−6∣+2∣x−10∣,x∈(−∞,∞)f(x)=|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|, x \in(-\infty, \infty) the minimum value of f(x)f(x) is _________.
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The Setup: We must find the minimum value of a sum of absolute value functions f(x)=∣x∣+2∣x−1∣+∣x−2∣+∣x−4∣+∣x−6∣+2∣x−10∣f(x)=|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|. Step 1: Translate the function into a median problem. The minimum value of a sum of absolute deviations ∑ci∣x−ai∣\sum c_i |x - a_i| occurs at the median of the data points aia_i, weighted by their frequencies cic_i. List out the points based on their respective coefficients: * 00 (Frequency 1) * 11 (Frequency 2) * 22 (Frequency 1) * 44 (Frequency 1) * 66 (Frequency 1) * 1010 (Frequency 2) The complete ordered data set contains 88 points: {0,1,1,2,4,6,10,10}\{0, 1, 1, 2, 4, 6, 10, 10\}. Step 2: Identify the median of the set. Since there is an even number of data points (n=8n=8), the minimum is achieved anywhere along the interval between the 4th and 5th values. The 4th point is 22 and the 5th point is 44. Any x∈[2,4]x \in [2, 4] minimizes f(x)f(x). Step 3: Calculate the minimum value f(x)f(x). Substitute any value from the valid interval (e.g., x=2x = 2) into f(x)f(x): f(2)=∣2−0∣+2∣2−1∣+∣2−2∣+∣2−4∣+∣2−6∣+2∣2−10∣f(2) = |2-0| + 2|2-1| + |2-2| + |2-4| + |2-6| + 2|2-10| f(2)=2+2(1)+0+∣−2∣+∣−4∣+2∣−8∣f(2) = 2 + 2(1) + 0 + |-2| + |-4| + 2|-8| f(2)=2+2+0+2+4+16=26f(2) = 2 + 2 + 0 + 2 + 4 + 16 = 26 Final Answer: 26
Q7:ipmat indore 2022QA › Matrices & DeterminantsMediumSA · TITA
If A=[100001010]A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right], then the absolute value of the determinant of (A9+A6+A3+A)\left(A^{9}+A^{6}+A^{3}+A\right) is __________.
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The Setup: A specific permutation matrix AA is given. We must establish the cyclical property of its powers to evaluate the absolute determinant of a matrix polynomial (A9+A6+A3+A)(A^9+A^6+A^3+A). Step 1: Establish the power cycle of Matrix AA. A=[100001010]A = \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} Calculate A2A^2: A2=[100001010][100001010]=[100010001]=IA^2 = \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = I Since A2=IA^2 = I, the powers of AA alternate: odd powers yield AA, and even powers yield the Identity matrix II. Step 2: Simplify the matrix polynomial. The expression is A9+A6+A3+AA^9 + A^6 + A^3 + A. Apply the parity rules: * A9=AA^9 = A (odd) * A6=IA^6 = I (even) * A3=AA^3 = A (odd) * A=AA = A (odd) Sum Matrix=A+I+A+A=3A+I\text{Sum Matrix} = A + I + A + A = 3A + I Step 3: Construct the elements of the final matrix. 3A=[300003030]3A = \begin{bmatrix}3&0&0\\0&0&3\\0&3&0\end{bmatrix} Add the Identity matrix II: 3A+I=[400013031]3A + I = \begin{bmatrix}4&0&0\\0&1&3\\0&3&1\end{bmatrix} Step 4: Calculate the determinant. Expand along the top row: det⁡(3A+I)=4(1×1−3×3)=4(1−9)=4(−8)=−32\det(3A+I) = 4(1 \times 1 - 3 \times 3) = 4(1 - 9) = 4(-8) = -32 The absolute value of the determinant is ∣−32∣=32|-32| = 32. Final Answer: 32
Q8:ipmat indore 2022QA › Binomial TheoremEasySA · TITA
The sum of the coefficients of all the terms in the expansion of (5x−9)4(5 x-9)^{4} is __________.
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The Setup: We are asked to find the arithmetic sum of the coefficients of all terms within the binomial expansion of (5x−9)4(5x-9)^4. Step 1: Apply the polynomial coefficient property. For any polynomial P(x)P(x), the sum of its fully expanded coefficients is mathematically obtained by evaluating the polynomial at x=1x = 1. Step 2: Evaluate the expression at x=1x = 1. P(1)=(5(1)−9)4P(1) = (5(1) - 9)^4 P(1)=(5−9)4P(1) = (5 - 9)^4 P(1)=(−4)4P(1) = (-4)^4 P(1)=256P(1) = 256 Final Answer: 256
Q9:ipmat indore 2022QA › Progression & SeriesMediumSA · TITA
A new sequence is obtained from the sequence of positive integers (1,2,3,…)(1,2,3, \ldots) by deleting all the perfect squares. Then the 2022nd 2022^{\text {nd }} term of the new sequence is ________.
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The Setup: A sequence of positive integers removes all perfect squares. We must determine the numerical value of the 2022nd2022^{\text{nd}} remaining integer in this sequence. Step 1: Establish the formula for sequence terms. Let NN be the actual value of the term. The sequence loses one position for every perfect square it passes. The number of perfect squares less than or equal to NN is ⌊N⌋\lfloor\sqrt{N}\rfloor. Thus, the position of NN in the modified sequence is given by: N−⌊N⌋=2022N - \lfloor\sqrt{N}\rfloor = 2022. Step 2: Estimate NN to find bounds. Because NN will be slightly larger than 20222022, we estimate N≈2022\sqrt{N} \approx \sqrt{2022}. 442=193644^2 = 1936 and 452=202545^2 = 2025. Therefore, the square root will be bounded by approximately 4545. Step 3: Test the logical value for NN. Add our 4545 estimated skipped squares to the 2022nd2022^{\text{nd}} position: Let's test N=2022+45=2067N = 2022 + 45 = 2067. Check how many perfect squares exist up to 20672067: The largest integer square less than 20672067 is 452=202545^2 = 2025. Thus, exactly 4545 squares are skipped. Subtract the skipped squares to verify the position: Position=2067−45=2022\text{Position} = 2067 - 45 = 2022 The number 20672067 precisely lands on the 2022nd2022^{\text{nd}} position, and it is not itself a perfect square, so it genuinely survives into the new sequence. Step 4: Confirm the answer is unique. The counting function N−⌊N⌋N - \lfloor\sqrt{N}\rfloor never decreases as NN grows, so it takes the value 20222022 on at most one non-square NN. Hence 20672067 is the only possibility. Final Answer: 2067
Q10:ipmat indore 2022QA › TrigonometryHardSA · TITA
If sin⁡α+sin⁡β=23\sin \alpha+\sin \beta=\frac{\sqrt{2}}{\sqrt{3}} and cos⁡α+cos⁡β=13\cos \alpha+\cos \beta=\frac{1}{\sqrt{3}}, then the value of (20cos⁡(α−β2))2\left(20 \cos \left(\frac{\alpha-\beta}{2}\right)\right)^{2} is _________.
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The Setup: We are given constraint equations sin⁡α+sin⁡β=23\sin\alpha+\sin\beta = \frac{\sqrt{2}}{\sqrt{3}} and cos⁡α+cos⁡β=13\cos\alpha+\cos\beta = \frac{1}{\sqrt{3}}. We need to evaluate the compound trigonometric expression (20cos⁡(α−β2))2\left(20\cos\left(\frac{\alpha-\beta}{2}\right)\right)^2. Step 1: Square and add the two primary equations. Equation 1: (sin⁡α+sin⁡β)2=(23)2  ⟹  sin⁡2α+sin⁡2β+2sin⁡αsin⁡β=23(\sin\alpha+\sin\beta)^2 = \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2 \implies \sin^2\alpha + \sin^2\beta + 2\sin\alpha\sin\beta = \frac{2}{3} Equation 2: (cos⁡α+cos⁡β)2=(13)2  ⟹  cos⁡2α+cos⁡2β+2cos⁡αcos⁡β=13(\cos\alpha+\cos\beta)^2 = \left(\frac{1}{\sqrt{3}}\right)^2 \implies \cos^2\alpha + \cos^2\beta + 2\cos\alpha\cos\beta = \frac{1}{3} Summing them together: (sin⁡2α+cos⁡2α)+(sin⁡2β+cos⁡2β)+2(cos⁡αcos⁡β+sin⁡αsin⁡β)=23+13(\sin^2\alpha + \cos^2\alpha) + (\sin^2\beta + \cos^2\beta) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = \frac{2}{3} + \frac{1}{3} Using Pythagorean and subtraction identities: 1+1+2cos⁡(α−β)=1  ⟹  2+2cos⁡(α−β)=11 + 1 + 2\cos(\alpha-\beta) = 1 \implies 2 + 2\cos(\alpha-\beta) = 1 Step 2: Relate the expression to the half-angle formula. Factor out the 22: 2(1+cos⁡(α−β))=12(1 + \cos(\alpha-\beta)) = 1 Apply the power-reduction identity 1+cos⁡θ=2cos⁡2(θ2)1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right): 2(2cos⁡2(α−β2))=12 \left(2\cos^2\left(\frac{\alpha-\beta}{2}\right)\right) = 1 4cos⁡2(α−β2)=14\cos^2\left(\frac{\alpha-\beta}{2}\right) = 1 Step 3: Evaluate the target expression. The target is (20cos⁡(α−β2))2\left(20\cos\left(\frac{\alpha-\beta}{2}\right)\right)^2. Expand the target: 400cos⁡2(α−β2)400\cos^2\left(\frac{\alpha-\beta}{2}\right) Substitute the derived identity block: 100×[4cos⁡2(α−β2)]=100×1=100100 \times \left[4\cos^2\left(\frac{\alpha-\beta}{2}\right)\right] = 100 \times 1 = 100 Final Answer: 100
Q11:ipmat indore 2022QA › Progression & SeriesMediumSA · TITA
The 3rd ,14th 3^{\text {rd }}, 14^{\text {th }} and 69th 69^{\text {th }} terms of an arithmetic progression form three distinct and consecutive terms of a geometric progression. If the next term of the geometric progression is the nth n^{\text {th }} term of the arithmetic progression, then nn equals ________.
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The Setup: Three selected terms (3rd,14th,69th3^{\text{rd}}, 14^{\text{th}}, 69^{\text{th}}) of an Arithmetic Progression (AP) form a separate Geometric Progression (GP). The 4th term of this GP corresponds to the nthn^{\text{th}} term of the AP. We must find index nn. Step 1: Establish the GP relationship. Let the AP have a first term aa and common difference dd. T3=a+2dT_3 = a + 2d T14=a+13dT_{14} = a + 13d T69=a+68dT_{69} = a + 68d Because these form a GP, the square of the middle term equals the product of the outer terms: (a+13d)2=(a+2d)(a+68d)(a + 13d)^2 = (a + 2d)(a + 68d) a2+26ad+169d2=a2+70ad+136d2a^2 + 26ad + 169d^2 = a^2 + 70ad + 136d^2 Step 2: Relate the starting term to the difference. 33d2=44ad33d^2 = 44ad Assuming d≠0d \neq 0 (otherwise all terms are identical and limits break), divide by 11d11d: 3d=4a  ⟹  a=34d3d = 4a \implies a = \frac{3}{4}d Step 3: Calculate the GP common ratio. Substitute a=0.75da = 0.75d into the first two terms of the GP: T3=0.75d+2d=2.75d=114dT_3 = 0.75d + 2d = 2.75d = \frac{11}{4}d T14=0.75d+13d=13.75d=554dT_{14} = 0.75d + 13d = 13.75d = \frac{55}{4}d The common ratio r=T14T3=55/411/4=5r = \frac{T_{14}}{T_3} = \frac{55/4}{11/4} = 5. The third term is T69=0.75d+68d=68.75d=2754dT_{69} = 0.75d + 68d = 68.75d = \frac{275}{4}d. (Verified: 554d×5=2754d\frac{55}{4}d \times 5 = \frac{275}{4}d). Step 4: Calculate the next GP term and map it to the AP index. The 4th GP term is T69×rT_{69} \times r: Next GP Term=2754d×5=13754d\text{Next GP Term} = \frac{275}{4}d \times 5 = \frac{1375}{4}d Set this equal to the generic nthn^{\text{th}} term of the AP (a+(n−1)da + (n-1)d): 34d+(n−1)d=13754d\frac{3}{4}d + (n-1)d = \frac{1375}{4}d (n−1)d=13724d  ⟹  n−1=343  ⟹  n=344(n-1)d = \frac{1372}{4}d \implies n-1 = 343 \implies n = 344 Final Answer: 344
Q12:ipmat indore 2022QA › Set TheoryEasySA · TITA
Let P(X)P(X) denote power set of a set XX. If AA is the null set, then the number of elements in P(P(P(P(A))))P(P(P(P(A)))) is _________.
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The Setup: We need to compute the total number of elements generated through four successive iterations of the power set operation P(X)P(X), starting with a null set AA. Step 1: Track the set cardinality through each power set iteration. The cardinality property dictates that if a set XX has kk elements, its power set P(X)P(X) will have 2k2^k elements. * Iteration 0: The set AA is a null set. ∣A∣=0|A| = 0. * Iteration 1: ∣P(A)∣=20=1|P(A)| = 2^0 = 1. * Iteration 2: ∣P(P(A))∣=21=2|P(P(A))| = 2^1 = 2. * Iteration 3: ∣P(P(P(A)))∣=22=4|P(P(P(A)))| = 2^2 = 4. * Iteration 4: ∣P(P(P(P(A))))∣=24=16|P(P(P(P(A))))| = 2^4 = 16. Final Answer: 16
Q13:ipmat indore 2022QA › Progression & SeriesMediumSA · TITA
The numbers −16,2x+3−22x−1−16,22x−1+16-16,2^{x+3}-2^{2 x-1}-16,2^{2 x-1}+16 are in an arithmetic progression. Then xx equals ________.
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The Setup: Three expressions involving exponential powers of 22 and constants are given as consecutive terms of an Arithmetic Progression. We must solve for the unknown exponent xx. Step 1: Establish the arithmetic mean equation. For three terms A,B,CA, B, C in an AP, 2B=A+C2B = A + C. Given terms: A=−16A = -16, B=2x+3−22x−1−16B = 2^{x+3} - 2^{2x-1} - 16, and C=22x−1+16C = 2^{2x-1} + 16. 2(2x+3−22x−1−16)=−16+(22x−1+16)2(2^{x+3} - 2^{2x-1} - 16) = -16 + (2^{2x-1} + 16) Step 2: Simplify the algebraic constraints. 2⋅2x+3−2⋅22x−1−32=22x−12 \cdot 2^{x+3} - 2 \cdot 2^{2x-1} - 32 = 2^{2x-1} Combine the exponential terms: 2x+4−22x−32=22x−12^{x+4} - 2^{2x} - 32 = 2^{2x-1} Group the terms involving 2x2x on the right: 2x+4−32=22x−1+22x2^{x+4} - 32 = 2^{2x-1} + 2^{2x} Factor out 22x2^{2x}: 2x+4−32=22x(12+1)=32⋅22x2^{x+4} - 32 = 2^{2x} \left(\frac{1}{2} + 1\right) = \frac{3}{2} \cdot 2^{2x} Step 3: Use substitution to solve the resulting quadratic. Let y=2xy = 2^x. Then 2x+4=16y2^{x+4} = 16y, and 22x=y22^{2x} = y^2. 16y−32=32y216y - 32 = \frac{3}{2}y^2 Multiply by 22 and rearrange into standard quadratic form: 3y2−32y+64=03y^2 - 32y + 64 = 0 Factor the quadratic equation: (3y−8)(y−8)=0(3y - 8)(y - 8) = 0 Step 4: Evaluate valid roots for integer xx. This yields y=8/3y = 8/3 or y=8y = 8. Since y=2xy = 2^x, resolving for y=8y = 8 yields 2x=8  ⟹  x=32^x = 8 \implies x = 3. *(The fractional root y=8/3y=8/3 yields an irrational logarithmic decimal out of scope for standard numeric data entry).* Final Answer: 3
Q14:ipmat indore 2022QA › Permutation & CombinationMediumSA · TITA
Mrs and Mr Sharma, and Mrs and Mr Ahuja along with four other persons are to be seated at a round table for dinner. If Mrs and Mr Sharma are to be seated next to each other, and Mrs and Mr Ahuja are not to be seated next to each other, then the total number of seating arrangements is _________.
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The Setup: A combinatorics problem where 8 individuals are arranged at a round table. We apply circular permutation blocks and complementary counting to satisfy simultaneous grouping constraints (one pair together, one pair separated). Step 1: Calculate permutations satisfying the first constraint. Total people = 8. Constraint 1: The Sharma pair must sit together. Bind the Sharmas into a single block. We now have 7 distinct entities to arrange in a circle (the Sharma block + 6 others). Circular arrangements=(n−1)!=(7−1)!=6!\text{Circular arrangements} = (n-1)! = (7-1)! = 6! The Sharmas can also swap seats internally within their block (2!2!). Total ways (Sharmas together)=6!×2=720×2=1440\text{Total ways (Sharmas together)} = 6! \times 2 = 720 \times 2 = 1440 Step 2: Calculate overlapping violations (both pairs together). Constraint 2: The Ahuja pair must NOT sit together. We find the restricted subset where they *do* sit together to subtract it from our total. Bind the Sharmas into one block and the Ahujas into another block. We now have 6 distinct entities to arrange. Circular arrangements=(6−1)!=5!\text{Circular arrangements} = (6-1)! = 5! Both the Sharma block and the Ahuja block have internal swaps (2!×2!2! \times 2!). Ways (Both pairs together)=5!×2×2=120×4=480\text{Ways (Both pairs together)} = 5! \times 2 \times 2 = 120 \times 4 = 480 Step 3: Apply complementary counting. Subtract the violating cases from the initial grouping constraints: Valid Seating=(Sharmas together)−(Both pairs together)\text{Valid Seating} = (\text{Sharmas together}) - (\text{Both pairs together}) Valid Seating=1440−480=960\text{Valid Seating} = 1440 - 480 = 960 Final Answer: 960
Q15:ipmat indore 2022QA › Mean, Median & ModeMediumSA · TITA
Let 50 distinct positive integers be chosen such that the highest among them is 100, and the average of the largest 25 integers among them exceeds the average of the remaining integers by 50. Then the maximum possible value of the sum of all the 50 integers is _________.
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The Setup: We are asked to maximize the total sum of 5050 distinct positive integers bounded by 100100. The average of the top 2525 integers is rigidly offset 5050 units above the average of the bottom 2525 integers. Step 1: Translate the average constraint into sum parameters. Let SLS_L be the sum of the largest 2525 integers, and SSS_S be the sum of the remaining (smallest) 2525 integers. Average(SL)−Average(SS)=50\text{Average}(S_L) - \text{Average}(S_S) = 50 SL25−SS25=50  ⟹  SL−SS=1250\frac{S_L}{25} - \frac{S_S}{25} = 50 \implies S_L - S_S = 1250 The goal is to maximize the grand total sum: SL+SSS_L + S_S. Step 2: Express the objective strictly in terms of SLS_L. Rearranging the constraint yields SS=SL−1250S_S = S_L - 1250. Total Sum=SL+(SL−1250)=2SL−1250\text{Total Sum} = S_L + (S_L - 1250) = 2S_L - 1250 To maximize the Total Sum, we strictly need to maximize the component SLS_L. Step 3: Maximize SLS_L using number properties. The set consists of strictly distinct positive integers with a ceiling of 100100. To maximize SLS_L (the sum of the 25 largest items), we pick the absolute highest available discrete integers descending from 100100: {100,99,98,…,76}\{100, 99, 98, \dots, 76\}. Using the arithmetic series sum formula: SL=n2(a1+an)=252(100+76)=25×88=2200S_L = \frac{n}{2}(a_1 + a_n) = \frac{25}{2}(100 + 76) = 25 \times 88 = 2200 Step 4: Verify the lower boundary SSS_S is physically viable. With SL=2200S_L = 2200, the forced offset calculates SS=2200−1250=950S_S = 2200 - 1250 = 950. We must verify if 2525 distinct integers smaller than 7676 can legitimately sum to 950950. The absolute minimum sum of 2525 distinct positive integers is 1+2+⋯+25=3251 + 2 + \dots + 25 = 325. The largest such sum is 51+52+⋯+75=157551 + 52 + \dots + 75 = 1575. Because 325≤950≤1575325 \le 950 \le 1575, the required SSS_S sits inside the achievable range. An explicit set works: {26,27,…,50}\{26, 27, \dots, 50\} consists of 2525 distinct integers, all below 7676, and sums to 252(26+50)=25×38=950\frac{25}{2}(26 + 50) = 25 \times 38 = 950. So the configuration is genuinely realisable, not merely arithmetically consistent. Step 5: Compute the final maximum total. Max Total Sum=SL+SS=2200+950=3150\text{Max Total Sum} = S_L + S_S = 2200 + 950 = 3150 Final Answer: 3150

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