Top 30 Important Questions for Class 12 Applied Maths 2026

Top 30 Important Questions for Class 12 Applied Maths 2026: Applied Mathematics offers practical tools for various real-world scenarios. Students should focus on concepts is vital for competitive exams. They can get Applied Maths topics, focusing on conceptual clarity and effective solution strategies. The questions are designed to strengthen candidates' knowledge and understanding before facing the actual exam.

Top 30 Important Questions for Class 12 Applied Maths 2026

Candidates can get the Applied Maths practice questions. The problems have been designed as per the syllabus and exam pattern. Candidates can solve the questions by improving their test-taking abilities before facing the actual exam. 

Congruence Modulo

Problem: Find the value of 3¹⁶ modulo 4.

This problem uses the properties of congruence modulo. The approach involves finding a power of 3 that yields a simple remainder, like 1 or -1, when divided by 4.

Solution Steps:

1. Calculate 3² = 9.

2. Find the remainder when 9 is divided by 4: 9 ≡ 1 (mod 4), which simplifies to 3² ≡ 1 (mod 4).

3. Raise both sides of the congruence to the power of 8 to reach 3¹⁶: (3²)⁸ ≡ 1⁸ (mod 4).

4. This simplifies to 3¹⁶ ≡ 1 (mod 4).

5. Thus, the value of 3¹⁶ modulo 4 is 1.

Binomial Distribution

Problem: The mean and standard deviation of a Binomial Distribution are 12 and 2, respectively. Find the value of p (the probability of success).

Key Concepts & Formulas:

• Mean (μ) = np

• Variance (σ²) = npq

• Standard Deviation (σ) = √npq

• Relationship: p + q = 1

Pedagogical Emphasis: Avoid the common mistake of equating the standard deviation directly to npq. Remember that standard deviation is the square root of the variance (√npq).

Solution Steps:

1. Given Mean (np) = 12.

2. Given Standard Deviation (√npq) = 2.

3. Square the standard deviation to find the variance: (√npq)² = 2² ⇒ npq = 4.

4. Substitute np = 12 into the variance equation: 12 * q = 4.

5. Solve for q: q = 4 / 12 = 1/3.

6. Use p = 1 - q: p = 1 - 1/3 = 2/3.

Pipes and Cisterns

Problem: Pipe A can fill a tank in 12 minutes. Pipe B can empty it in 18 minutes. If both pipes are opened simultaneously, how long will it take to fill the tank?

Analysis:

• Pipe A (inlet pipe) fills at a rate of 1/12 tank per minute.

• Pipe B (outlet pipe) empties at a rate of 1/18 tank per minute.

Formula: When an inlet and an outlet pipe work together, their rates subtract: 1/Tₐₑ = 1/Tₐ - 1/Tₑ.

Solution Steps:

1. Substitute values: 1/Tₐₑ = 1/12 - 1/18.

2. Find a common denominator (36): 1/Tₐₑ = (3 - 2) / 36 = 1/36.

3. Therefore, Tₐₑ = 36 minutes.

Inferential Statistics (T-Test)

Problem: For a sample of 60 bulbs taken for a t-test of significance, what is the degree of freedom?

Key Concept: For a one-sample t-test, the formula for the degree of freedom (df) is: df = n - 1, where n is the sample size.

Solution:

1. Given sample size (n) = 60.

2. Calculate df = 60 - 1 = 59.

Linear Programming (LPP)

Problem: For an objective function Z = ax + by, the maximum value occurs at two corner points of the feasible region: (30, 30) and (0, 40). Find the condition on A and B.

Key Concept: If the optimal value of an objective function in an LPP occurs at two different corner points, there are infinite optimal solutions along the line segment connecting these points. This implies the objective function yields the same value at both points.

Solution Steps:

1. Equate Z at (30, 30) and (0, 40): Z(30, 30) = Z(0, 40).

2. Substitute into the objective function: a(30) + b(30) = a(0) + b(40).

3. Simplify: 30a + 30b = 40b.

4. Solve for the relationship: 30a = 10b ⇒ 3a = b or b - 3a = 0.

Alligation and Mixtures

Problem: In what ratio must water be mixed with milk to gain 16 2/3% by selling the mixture at the cost price of the milk?

Analysis:

• Water (cheaper, cost c = 0). Milk (dearer, cost d = x).

• Selling Price (SP) of mixture = Cost Price of milk = x.

• Profit = 16 2/3% = 50/3 % = 1/6.

Solution Steps:

1. Find the mean cost price (m) of the mixture: SP = m * (1 + Profit %).
   x = m * (1 + 1/6) = m * (7/6) ⇒ m = 6x/7.

2. Apply the Alligation Rule: (Ratio of Cheaper: Dearer) = (d - m) : (m - c).
   Ratio (Water : Milk) = (x - 6*x*/7) : (6*x*/7 - 0).
   Ratio = (x/7) : (6*x*/7).

3. Simplify the ratio: 1 : 6.

Pedagogical Emphasis: Always check the order requested for the ratio (e.g., Water: Milk or Milk: Water).

Perpetuity

Problem: Find the present value of a perpetuity of ₹900 payable at the beginning of each year, if money is worth 5% per annum.

Analysis:

• Regular Payment (R) = ₹900.

• "Payable at the beginning" indicates a perpetuity due (Type 2 Perpetuity).

• Interest Rate (i) = 5% = 0.05.

Formula for Perpetuity Due: PV = R + R/i.

Solution Steps:

1. Substitute values: PV = 900 + (900 / 0.05).

2. Calculate second term: 900 / 0.05 = 18,000.

3. Total present value: PV = 900 + 18,000 = ₹18,900.

Time Series Analysis

Problem: Identify the component of a time series represented by the phases: Prosperity, Recession, and Depression.

Analysis: These terms describe business cycle fluctuations over a period longer than one year.

Conclusion: This pattern represents the Cyclical Trend (or Cyclical Variation) component of a time series.

Area Under the Curve

Problem: Find the area bounded by the curve y = √x, the x-axis, and the lines x = 1 and x = 4.

Key Concept: The area under a curve y = f(x) from x = a to x = b is given by ∫[from a to b] y dx.

Pedagogical Emphasis: The function y = √x represents only the upper half of the parabola y² = x (where y ≥ 0). For subjective answers, include "square units" in the result.

Solution Steps:

1. Set up the integral: Area = ∫[from 1 to 4] √x dx = ∫[from 1 to 4] x¹ᐟ² dx.

2. Integrate: Area = [ (x³ᐟ²) / (3/2) ] from 1 to 4 = (2/3) * [x³ᐟ²] from 1 to 4.

3. Apply limits: Area = (2/3) * [4³ᐟ² - 1³ᐟ²] = (2/3) * [(√4)³ - 1] = (2/3) * [2³ - 1].

4. Simplify: Area = (2/3) * [8 - 1] = (2/3) * 7 = 14/3 square units.

Producer Surplus

Problem: A commodity's supply function is given by p = 4 + x. Find the producer surplus when 12 units are sold on the market.

Pedagogical Emphasis: Producer or consumer surplus questions are very likely in exams.

Key Concepts:

• Supply Function, S(x): p = 4 + x.

• Equilibrium Quantity (x₀) = 12.

• Producer Surplus (PS) Formula: PS = (p₀ * x₀) - ∫[from 0 to x₀] S(x) dx.

Solution Steps:

1. Find equilibrium price (p₀) for x₀ = 12: p₀ = 4 + 12 = 16.

2. Calculate total revenue (p₀ * x₀): 16 * 12 = 192.

3. Evaluate the integral of the supply curve:
   ∫[from 0 to 12] (4 + x) dx = [4x + x²/2] from 0 to 12.
   = (4*12 + 12²/2) - 0 = 48 + 72 = 120.

4. Calculate PS = 192 - 120 = 72.

Differential Equations

Problem: Find the general solution of the differential equation: y log(y) dx - x dy = 0.

Method: Use the Variable Separable Method.

Solution Steps:

1. Rearrange: y log(y) dx = x dy.

2. Separate variables: (1/x) dx = 1 / (y log y) dy.

3. Integrate both sides: ∫ (1/x) dx = ∫ (1/y) / log(y) dy.

4. The right side is of the form ∫ f'(y)/f(y) dy, where f(y) = log(y) and f'(y) = 1/y.

5. Integrate: log|x| = log|log(y)| + C.

6. (Memory Tip: When all terms are logarithms, write the integration constant as log(C) for simplification.)
   log|x| = log|log(y)| + log|C|.

7. Use log property: log|x| = log|C * log(y)|.

8. Equate arguments: x = C * log(y), or log(y) = mx (where m = 1/C).

Boats and Streams

Problem: The speed of a boat in still water is 15 km/hr, and the rate of the current is 3 km/hr. Find the distance traveled by the boat downstream in 12 minutes.

Pedagogical Emphasis: Always ensure unit consistency. Convert minutes to hours.

Solution Steps:

1. Speed of boat (x) = 15 km/hr; Speed of current (y) = 3 km/hr.

2. Time (t) = 12 minutes = 12/60 hours = 1/5 hours.

3. Downstream Speed = x + y = 15 + 3 = 18 km/hr.

4. Distance = Speed × Time = 18 km/hr × (1/5) hr = 3.6 km.

Time Series (Moving Averages)

Problem: For a given set of five values [35, 70, 36, 59, …], calculate the 3-year moving averages.

Method: A 3-year moving average is the average of three consecutive data points.

Solution Steps:

1. First Moving Average: (35 + 70 + 36) / 3 = 141 / 3 = 47.

2. Second Moving Average: (70 + 36 + 59) / 3 = 165 / 3 = 55.

Hypothesis Testing Definitions

Problem: What is the Significance Level in hypothesis testing?

Here is a comparison of Significance Level and Confidence Level:

Integration by Substitution

Problem: Find the integral of 1 / (x + x log x).

(Memory Tip: If an integral is in the form ∫ f'(x) / f(x) dx, its solution is log|f(x)| + C.)

Solution Steps:

1. Factor out x from the denominator: ∫ 1 / (x(1 + log x)) dx.

2. Rewrite to match the shortcut form: ∫ (1/x) / (1 + log x) dx.

3. Identify f(x) = 1 + log x, and f'(x) = 1/x.

4. Apply the rule: The integral is log|1 + log x| + C.

Assertion & Reason (Increasing Functions)

• Assertion (A): The function f(x) = x³ - 12x is strictly increasing in the intervals (-∞, -2) U (2, ∞).

• Reason (R): For a strictly increasing function, its first derivative f'(x) must be greater than zero (f'(x) > 0).

Analysis:

1. Reason (R) is the correct definition of a strictly increasing function. So, R is true.

2. Evaluate Assertion (A):

• f'(x) = 3x² - 12.

• For strict increase, 3x² - 12 > 0 ⇒ 3(x² - 4) > 0 ⇒ (x - 2)(x + 2) > 0.

• This inequality holds for x < -2 or x > 2, which corresponds to intervals (-∞, -2) and (2, ∞).

• Therefore, Assertion (A) is also true.

3. Conclusion: Both A and R are true, and R provides the correct explanation for A.

Assertion & Reason (Rate of Interest)

• Assertion (A): If the nominal rate of interest is 12.5% and the inflation rate is 2%, then the effective rate of interest is 10.5%.

• Reason (R): If interest is calculated only at the end of a year (compounded annually), then the effective rate of interest is the same as the nominal rate of interest.

Analysis:

1. Assertion (A) is true. The "effective rate" here refers to the real rate of return after inflation: 12.5% - 2% = 10.5%.

2. Reason (R) is also true. If interest is compounded annually, the effective annual rate equals the nominal annual rate.

3. Conclusion: Both A and R are individually true, but R (about compounding frequency) does not explain A (about inflation's impact).

EMI Calculation (Flat Rate Method)

Problem: Sonia borrowed ₹1,00,000 from a society at 10% per annum for 2 years. Find the EMI using the Flat Rate method.

Analysis:

• Principal (P) = ₹1,00,000.

• Time (T) = 2 years.

• Number of installments (n) = 2 years * 12 months/year = 24.

• Monthly interest rate (i) = 10% per annum / 12 = 0.10/12 = 1/120.

Formula for EMI (Flat Rate Method): EMI = (P + Total Interest) / n. Total Interest = P * n * i.

Solution Steps:

1. Calculate total interest: Total Interest = 1,00,000 * 24 * (1/120) = ₹20,000.

2. Total amount to repay: 1,00,000 + 20,000 = ₹1,20,000.

3. Calculate EMI = 1,20,000 / 24 = ₹5,000.

Skew-Symmetric Matrix

Problem: Given that matrix A is skew-symmetric, find the values of a, b, and c and then evaluate the expression 2a + 3b - c.

Matrix A =

| 0 | -1 | 28 |

| a-8 | 0 | 3b |

| -c+2| 2 | 0 |

Key Property: For a skew-symmetric matrix, aᵢⱼ = -aⱼᵢ for all i ≠ j (and diagonal elements are zero).

Solution Steps:

1. Using a₁₂ = -a₂₁: -1 = -(a - 8) ⇒ 1 = a - 8 ⇒ a = 9.

2. Using a₁₃ = -a₃₁: 28 = -(-c + 2) ⇒ 28 = c - 2 ⇒ c = 30.

3. Using a₂₃ = -a₃₂: 3b = -(2) ⇒ b = -2/3.

4. Evaluate 2a + 3b - c: 2(9) + 3(-2/3) - 30 = 18 - 2 - 30 = -14.

LPP Formulation

Problem: Two tailors, A and B, earn ₹300 and ₹400 per day. Production capabilities and requirements are given. Formulate this as an LPP to minimise labour cost while meeting production targets.

Information Summary Table:

LPP Formulation:

1. Decision Variables: Let x = number of days Tailor A works, y = number of days Tailor B works.

2. Objective Function (to minimise labour cost): Minimise Z = 300x + 400y.

3. Constraints:

• Shirts: 6x + 10y ≥ 80 (simplifies to 3x + 5y ≥ 40).

• Trousers: 4x + 4y ≥ 32 (simplifies to x + y ≥ 8).

4. Non-Negativity Constraint: x ≥ 0, y ≥ 0.

Races and Games

Problem: In a 500-meter race, A defeats B by 60 meters or 12 seconds. Find the time taken by A to complete the race.

Logical Interpretation: When A finishes, B is 60 meters behind and takes 12 more seconds to cover that 60 meters.

Solution Steps:

1. B covers 60 meters in 12 seconds.

2. Speed of B = 60 m / 12 s = 5 m/s.

3. Time for B to complete 500m = 500 m / 5 m/s = 100 seconds.

4. A defeated B by 12 seconds: Time for A = 100 s - 12 s = 88 seconds.

Applications of Derivatives (Rate of Change)

Problem: The volume of a spherical balloon is increasing at a rate of 3 cm³/s. Find the rate of change of its surface area when its radius is 2 cm.

Setup:

• Given: dV/dt = 3 cm³/s.

• To Find: dA/dt when r = 2 cm.

• Formulas: V = (4/3)πr³, A = 4πr².

Solution Steps:

1. Find dr/dt: dV/dt = d/dt [(4/3)πr³] = 4πr²(dr/dt).

• 3 = 4πr²(dr/dt) ⇒ dr/dt = 3 / (4πr²).

2. Find dA/dt: dA/dt = d/dt [4πr²] = 8πr(dr/dt).

• Substitute dr/dt: dA/dt = 8πr * [3 / (4πr²)] = 6/r.

3. Evaluate dA/dt at r = 2 cm: dA/dt = 6 / 2 = 3 cm²/s.

Consumer Surplus with Monopolist Equilibrium

Problem: The demand function for a profit-maximising monopolist is P = 274 - x², and the marginal cost is MC = 4 + 3x. Find the consumer surplus.

Key Concept: For a profit-maximising monopolist, equilibrium occurs where Marginal Revenue (MR) equals Marginal Cost (MC).

Solution Steps:

1. Find MR: Revenue R(x) = P * x = (274 - x²)x = 274x - x³.

• MR = dR/dx = 274 - 3x².

2. Find equilibrium quantity (x₀) by setting MR = MC:

• 274 - 3x² = 4 + 3x ⇒ 3x² + 3x - 270 = 0 ⇒ x² + x - 90 = 0.

• (x + 10)(x - 9) = 0. Since x > 0, x₀ = 9.

3. Find equilibrium price (p₀): p₀ = 274 - (9)² = 274 - 81 = 193.

4. Calculate Consumer Surplus (CS): CS = ∫[from 0 to x₀] D(x) dx - (p₀ * x₀).

• ∫[from 0 to 9] (274 - x²) dx = [274x - x³/3] from 0 to 9.

• = (274*9 - 9³/3) - 0 = 2466 - 243 = 2223.

• p₀ * x₀ = 193 * 9 = 1737.

• CS = 2223 - 1737 = 486.

Solving Systems of Linear Equations (Cramer's Rule)

Problem: Solve the following system of linear equations using Cramer's Rule:

2x + 3y = 10

x + 6y = 4

Method: Cramer's rule uses determinants: x = Dₓ/D and y = Dᵧ/D.

Solution Steps:

1. Main determinant (D):
   D = | 2 3 |
   | 1 6 |
   D = (2 * 6) - (3 * 1) = 12 - 3 = 9. (Unique solution exists as D ≠ 0).

2. Determinant Dₓ: (Replace x-coefficients with constants)
   Dₓ = | 10 3 |
   | 4 6 |
   Dₓ = (10 * 6) - (3 * 4) = 60 - 12 = 48.

3. Determinant Dᵧ: (Replace y-coefficients with constants)
   Dᵧ = | 2 10 |
   | 1 4 |
   Dᵧ = (2 * 4) - (10 * 1) = 8 - 10 = -2.

4. Solve for x and y:

• x = Dₓ / D = 48 / 9 = 16/3.

• y = Dᵧ / D = -2 / 9 = -2/9.

5. Verification: Substitute x=16/3 and y=-2/9 into the original equations.

• Equation 1: 2(16/3) + 3(-2/9) = 32/3 - 6/9 = 32/3 - 2/3 = 30/3 = 10. (Holds true)

• Equation 2: 16/3 + 6(-2/9) = 16/3 - 12/9 = 16/3 - 4/3 = 12/3 = 4. (Holds true)

Probability Distributions

It involves finding an unknown constant in a probability distribution and then calculating specific probabilities.

Let X be a random variable representing the number of hours a person watches TV. The probability distribution P(x) is:

• P(x) = 0.2 if x = 0

• P(x) = kx if x = 1 or x = 2

• P(x) = k(5 - x) if x = 3

• P(x) = 0 otherwise

1. Probability Distribution Table:

2. Find the value of k: The sum of all probabilities must be 1 (ΣP(x) = 1).

• 0.2 + k + 2k + 2k = 1

• 0.2 + 5k = 1

• 5k = 0.8

• k = 0.8 / 5 = **4/25**.

3. Calculate Required Probabilities:

• P(x = 2): P(x=2) = 2k = 2 * (4/25) = **8/25**.

• P(x ≥ 2): P(x ≥ 2) = P(x=2) + P(x=3) = 2k + 2k = 4k.

• P(x ≥ 2) = 4 * (4/25) = **16/25**