Mathematics and Statistics (Commerce) Part 2 गणित व संख्याशास्त्र (वाणिज्य) भाग 2

Quick summary

Businesses use agents to buy and sell goods and pay them commission or brokerage. They also sell on credit using bills of exchange, which may be discounted with a bank. This chapter covers commission, brokerage, trade and cash discount, and the mathematics of bills: present worth, true discount, banker's discount and banker's gain.

Key points

• Agent: middleman who buys/sells for a principal. Commission is his reward, usually a percentage of sales/purchases.
• Commission agent – buys/sells for commission. Broker – brings buyer and seller together; earns brokerage. Factor – has possession of goods and sells in his own name. Auctioneer – sells goods by public auction to the highest bidder. Del credere agent – guarantees payment by buyers; gets extra commission (del credere commission).
• Trade discount: allowed on the list (catalogue) price to traders. Cash discount: allowed on the net price for prompt cash payment.
• Successive discounts are applied one after another, not added.
• Bill terms: date of bill, face value, period, nominal due date, legal due date = nominal due date + 3 days of grace.
• Present worth (P.W.): amount which, with interest for the unexpired period, becomes the face value.
• True discount (T.D.): interest on P.W. for the unexpired period.
• Banker's discount (B.D.): interest on face value for the unexpired period (what the bank actually deducts).
• Banker's gain (B.G.) = B.D. − T.D. = interest on T.D.

Important terms

• List price / Marked price – printed price before discount.
• Net price – price after trade discount.
• Discounted value – amount paid by bank = F.V. − B.D.

Formulas

• Commission = Rate × Sales/100
• Net price after trade discount d% = List price × (100 − d)/100
• T.D. = (P.W. × n × r)/100; F.V. = P.W. + T.D.
• B.D. = (F.V. × n × r)/100
• B.G. = B.D. − T.D. = (T.D. × n × r)/100
• Cash value (amount received from bank) = F.V. − B.D.

(n = unexpired period in years, r = rate % p.a.)

5 important questions

Q1. A salesman gets 8% commission on sales. Sales were ₹75,000. Find his commission.

Commission = 8/100 × 75,000 = ₹6,000.

Q2. A wholesaler allows 25% trade discount and 5% cash discount. Find the net amount for an article marked ₹1,600.

After trade discount: 1,600 × 75/100 = ₹1,200. After cash discount: 1,200 × 95/100 = ₹1,140.

Q3. Find the true discount on ₹2,040 due in 4 months at 6% p.a.

P.W. = F.V. ÷ (1 + nr/100) = 2,040 ÷ (1 + (1/3)(6)/100) = 2,040 ÷ 1.02 = ₹2,000. T.D. = 2,040 − 2,000 = ₹40.

Q4. A bill of ₹10,000 is discounted 73 days before the legal due date at 10% p.a. Find B.D. and cash value.

n = 73/365 = 1/5 year. B.D. = 10,000 × (1/5) × 10/100 = ₹200. Cash value = ₹9,800.

Q5. Explain del credere agent.

An agent who guarantees the principal that buyers will pay; for this extra risk he gets an additional del credere commission.

Exam tip

Count days carefully: exclude the date of discounting, include the legal due date, and use 365 days in a year.

Quick summary

Insurance protects people and businesses against financial loss, and annuities are regular equal payments made at fixed intervals. This chapter covers life, fire, marine and accident insurance, premium and claim calculations, and annuities — immediate annuity, annuity due, their present and future values, and sinking fund.

Key points

• Premium: amount paid by the insured to the insurer. Policy value / sum insured: amount for which the property or life is insured.
• Types: life insurance (pays on death or maturity), fire insurance, marine insurance (ship, cargo, freight), accident insurance.
• General insurance follows the principle of indemnity: compensation only up to the actual loss.
• If property is under-insured, the claim is proportionate (average clause).
• Annuity: a sequence of equal payments at equal intervals. Terms: annuitant, periodic payment C, rate per period i = r/100, number of periods n, present value P, accumulated (future) value A.
• Immediate annuity (ordinary annuity): payments at the end of each period. Annuity due: payments at the beginning of each period. Deferred annuity: payments start after some periods.
• For monthly payments, replace r by r/12 and n by 12n.
• Sinking fund: a fund created by setting aside equal amounts periodically to repay a debt or replace an asset.

Important terms

• Insured – person who takes insurance. Insurer – the company.
• Claim – demand for compensation after a loss.
• Present value – today's value of future annuity payments.
• Future (accumulated) value – value of all payments at the end of the term.

Formulas

• Premium = Rate × Policy value / 100 (rate per ₹100)
• Claim (under-insurance) = (Policy value ÷ Property value) × Loss
• Immediate annuity: A = C[(1 + i)ⁿ − 1]/i; P = C[1 − (1 + i)⁻ⁿ]/i
• Annuity due: A′ = C(1 + i)[(1 + i)ⁿ − 1]/i; P′ = C(1 + i)[1 − (1 + i)⁻ⁿ]/i
• Relation: A = P(1 + i)ⁿ

5 important questions

Q1. A property worth ₹5,00,000 is insured for ₹4,00,000. Fire causes a loss of ₹1,00,000. Find the claim (average clause).

Claim = (4,00,000/5,00,000) × 1,00,000 = ₹80,000.

Q2. Find the premium on a policy of ₹2,40,000 at ₹2.50 per ₹100.

Premium = 2.50 × 2,40,000/100 = ₹6,000.

Q3. Find the accumulated value of an immediate annuity of ₹1,000 per year for 4 years at 10% p.a.

i = 0.1. A = 1000[(1.1)⁴ − 1]/0.1 = 1000(1.4641 − 1)/0.1 = ₹4,641.

Q4. Find the present value of an immediate annuity of ₹1,000 per year for 3 years at 10% p.a. [(1.1)⁻³ = 0.7513]

P = 1000(1 − 0.7513)/0.1 = 1000 × 2.487 = ₹2,487.

Q5. Distinguish between immediate annuity and annuity due.

Immediate annuity: payments at the end of each period. Annuity due: payments at the beginning of each period, so each payment earns interest for one extra period.

Exam tip

Write i as a decimal (10% → 0.1) and use the log or power table values given in the question carefully.

Quick summary

When two variables are correlated, regression helps us estimate (predict) the value of one variable from the other. This chapter explains the meaning and types of regression, fitting a simple linear regression line by the method of least squares, the two regression lines (Y on X and X on Y), regression coefficients and their properties.

Key points

• Regression means estimating an unknown value of one variable from a known value of another.
• Types: simple and multiple; linear and non-linear. This chapter studies simple linear regression.
• The line of best fit is found by the method of least squares (sum of squares of errors is minimum).
• Regression of Y on X: Y = a + bX, used to estimate Y when X is known. Regression coefficient bYX.
• Regression of X on Y: X = a′ + b′Y, used to estimate X when Y is known. Regression coefficient bXY.
• Both lines pass through the point of means (x̄, ȳ).
• Properties of regression coefficients: r² = bYX × bXY; r = ±√(bYX × bXY), with the same sign as the coefficients; both coefficients have the same sign; if one is greater than 1, the other is less than 1; they are independent of change of origin but not of scale.
• If r = ±1 the two lines coincide; if r = 0 they are perpendicular.

Important terms

• Dependent variable – the variable being estimated.
• Independent variable – the variable used to estimate.
• Regression coefficient – slope of the regression line.
• Least squares – method that minimises the sum of squared errors.

Formulas

• bYX = Cov(X, Y)/σx² = [Σxy − n x̄ ȳ] ÷ [Σx² − n x̄²]
• bXY = Cov(X, Y)/σy² = [Σxy − n x̄ ȳ] ÷ [Σy² − n ȳ²]
• bYX = r σy/σx; bXY = r σx/σy
• Line Y on X: (Y − ȳ) = bYX (X − x̄)
• Line X on Y: (X − x̄) = bXY (Y − ȳ)
• r = ±√(bYX × bXY)

5 important questions

Q1. Given x̄ = 10, ȳ = 20, bYX = 0.8. Find the regression line of Y on X and estimate Y when X = 15.

Y − 20 = 0.8(X − 10) ⇒ Y = 0.8X + 12. For X = 15, Y = 12 + 12 = 24.

Q2. bYX = 0.6 and bXY = 0.9. Find r.

r = √(0.6 × 0.9) = √0.54 ≈ 0.735 (positive, since both coefficients are positive).

Q3. The two regression lines are x + 2y = 5 and 2x + 3y = 8. Find x̄ and ȳ.

Both lines pass through (x̄, ȳ). From the first, x = 5 − 2y. Substituting: 2(5 − 2y) + 3y = 8 ⇒ 10 − y = 8 ⇒ y = 2, x = 1. So x̄ = 1, ȳ = 2.

Q4. State any four properties of regression coefficients.

r² = bYX × bXY; both have the same sign; if one exceeds 1 the other is below 1; independent of change of origin, not of scale.

Q5. Distinguish between correlation and regression.

Correlation measures the degree of relationship; regression estimates one variable from another. Correlation is symmetric (rXY = rYX); regression coefficients are not (bYX ≠ bXY in general).

Exam tip

Find means first by solving the two regression equations simultaneously — they intersect at (x̄, ȳ).

Quick summary

A time series is a set of observations recorded at successive points of time, like yearly sales or monthly rainfall. This chapter explains uses of time series analysis, its four components, and three methods of measuring the secular trend: graphical (freehand) method, moving averages and least squares.

Key points

• Time series analysis helps understand past behaviour, forecast the future, plan business and compare data.
• Four components:
• Secular trend (T) – long-term general tendency to increase, decrease or remain steady (e.g., population growth).
• Seasonal variation (S) – regular changes within a year due to seasons, festivals, customs (e.g., sale of umbrellas in monsoon).
• Cyclical variation (C) – up-and-down movements over more than a year (business cycle: prosperity, recession, depression, recovery).
• Irregular variation (I) – random, unpredictable changes due to floods, wars, strikes.
• Graphical (freehand) method: plot data and draw a smooth line through the middle of the points. Simple but subjective.
• Moving averages: averages of successive groups (3-yearly, 4-yearly etc.). For an even period, centre the averages by taking a 2-item moving average again.
• Least squares: fit a straight line y = a + bx by minimising the sum of squared errors; gives the best-fit trend line and can be used for forecasting.
• To simplify least squares, shift origin: for an odd number of years take the middle year as origin (u = x − middle year).

Important terms

• Trend value – estimated value on the trend line.
• Period of moving average – number of items averaged together.
• Business cycle – recurring phases of prosperity and depression.

Formulas

• Trend line: y = a + bu
• Normal equations: Σy = na + bΣu; Σuy = aΣu + bΣu²
• When Σu = 0: a = Σy/n, b = Σuy/Σu²

5 important questions

Q1. State the components of a time series with one example each.

Secular trend (rising population), seasonal (more ice cream sales in summer), cyclical (business cycle), irregular (fall in sales due to a flood).

Q2. Compute 3-yearly moving averages for: 2015: 10, 2016: 12, 2017: 14, 2018: 16, 2019: 18.

(10 + 12 + 14)/3 = 12 (for 2016); (12 + 14 + 16)/3 = 14 (2017); (14 + 16 + 18)/3 = 16 (2018).

Q3. Fit a least squares trend line to the data in Q2.

Take u = year − 2017: u = −2, −1, 0, 1, 2; Σu = 0, Σu² = 10. Σy = 70, Σuy = −20 − 12 + 0 + 16 + 36 = 20. a = 70/5 = 14, b = 20/10 = 2. Trend: y = 14 + 2u.

Q4. Using the trend in Q3, estimate the value for 2021.

u = 2021 − 2017 = 4. y = 14 + 2 × 4 = 22.

Q5. State merits and demerits of the graphical method.

Merits: simple, quick, needs no calculation. Demerits: subjective (different people draw different lines), not suitable for forecasting.

Exam tip

Always show the table with columns year, y, u, u², uy — the method marks come from this table.

Quick summary

Index numbers measure relative changes in prices, quantities or values over time. This chapter covers definitions and types of index numbers, notation, construction by the simple aggregate method and weighted aggregate methods (Laspeyres, Paasche, Dorbish-Bowley, Fisher, Marshall-Edgeworth, Walsh), and the cost of living index number by the aggregative expenditure and family budget methods.

Key points

• Index numbers are specialised averages expressed as percentages; base year index = 100.
• Types: price index, quantity index, value index.
• Notation: p0, q0 = base-year price and quantity; p1, q1 = current-year price and quantity; P01 = price index of current year w.r.t. base year.
• Simple aggregate method: all items given equal importance.
• Weighted aggregate methods use quantities as weights.
• Fisher's ideal index is the geometric mean of Laspeyres' and Paasche's indices; Dorbish-Bowley's is their arithmetic mean.
• Cost of living index (CLI): shows change in the cost of living of a group of people. Methods: aggregative expenditure method (same as Laspeyres) and family budget method (weighted average of price relatives, with weights W = p0q0).
• Uses of CLI: fixing wages and dearness allowance, measuring purchasing power, framing policies.

Important terms

• Base year – year of comparison (index 100).
• Price relative I = (p1/p0) × 100.
• Weight – relative importance of an item.

Formulas

• Simple aggregate: P01 = (Σp1/Σp0) × 100; Q01 = (Σq1/Σq0) × 100
• Value index: V01 = (Σp1q1/Σp0q0) × 100
• Laspeyres: P01(L) = (Σp1q0/Σp0q0) × 100
• Paasche: P01(P) = (Σp1q1/Σp0q1) × 100
• Dorbish-Bowley: P01(D-B) = ½ [Σp1q0/Σp0q0 + Σp1q1/Σp0q1] × 100
• Fisher: P01(F) = √[(Σp1q0/Σp0q0) × (Σp1q1/Σp0q1)] × 100
• Marshall-Edgeworth: P01(M-E) = [Σp1(q0 + q1)/Σp0(q0 + q1)] × 100
• Walsh: P01(W) = [Σp1√(q0q1)/Σp0√(q0q1)] × 100
• CLI (aggregative expenditure) = (Σp1q0/Σp0q0) × 100
• CLI (family budget) = ΣIW/ΣW, where I = (p1/p0) × 100 and W = p0q0

5 important questions

Q1. Σp0 = 50, Σp1 = 65. Find the simple aggregate price index.

P01 = 65/50 × 100 = 130.

Q2. Σp1q0 = 1056, Σp0q0 = 960, Σp1q1 = 1484, Σp0q1 = 1325. Find Laspeyres' and Paasche's indices.

L = 1056/960 × 100 = 110. P = 1484/1325 × 100 = 112.

Q3. Using Q2, find Dorbish-Bowley's and Fisher's indices.

D-B = (110 + 112)/2 = 111. Fisher = √(110 × 112) = √12320 ≈ 111.0.

Q4. Find CLI by the family budget method if ΣIW = 2,40,000 and ΣW = 2,000.

CLI = 2,40,000/2,000 = 120. Cost of living rose by 20%.

Q5. State any three uses of cost of living index numbers.

Fixing wages and dearness allowance; measuring purchasing power of money; helping government frame price and wage policies.

Exam tip

Prepare one big table with all products (p0q0, p1q0, p0q1, p1q1) at the start — then every index is just one division.

Quick summary

Linear programming is a mathematical technique for planning and decision-making: it finds the best (maximum or minimum) value of a linear objective function subject to linear constraints. This chapter explains the meaning of LPP, how to formulate a problem mathematically, and how to solve it by the graphical method.

Key points

• Objective function: Z = ax + by, to be maximised (profit) or minimised (cost).
• Decision variables: x, y — quantities to be decided.
• Constraints: linear inequalities from limited resources (time, material, money).
• Non-negativity constraints: x ≥ 0, y ≥ 0.
• Feasible region: common region satisfying all constraints. Every point in it is a feasible solution.
• Optimal solution: feasible solution that gives the best value of Z.
• Graphical method steps: (1) convert each inequality to an equation and draw the line using intercepts; (2) shade the region satisfying each inequality (test with origin); (3) identify the feasible region; (4) find its corner points; (5) evaluate Z at each corner point; (6) choose the maximum/minimum.
• The optimal value always occurs at a corner (vertex) of the feasible region (if it exists).
• A feasible region may be bounded or unbounded; there may be multiple optimal solutions or no solution.

Important terms

• LPP – Linear Programming Problem.
• Corner point – vertex of the feasible region.
• Bounded region – feasible region enclosed on all sides.
• Infeasible problem – no point satisfies all constraints.

Formulas

• Maximise / Minimise Z = c1x + c2y
• Subject to: a1x + b1y ≤ (or ≥) d1; a2x + b2y ≤ (or ≥) d2; x ≥ 0, y ≥ 0

5 important questions

Q1. Define objective function, constraints and feasible region.

Objective function is the linear function to be optimised; constraints are the linear restrictions; feasible region is the common region of all constraints.

Q2. Maximise Z = 3x + 5y subject to x + y ≤ 4, x ≥ 0, y ≥ 0.

Corner points: (0, 0), (4, 0), (0, 4). Z = 0, 12, 20. Maximum Z = 20 at (0, 4).

Q3. Minimise Z = 4x + 6y subject to x + y ≥ 5, x ≤ 4, y ≤ 4, x, y ≥ 0.

Corner points of the feasible region: (1, 4), (4, 1), (4, 4). Z = 4 + 24 = 28, 16 + 6 = 22, 16 + 24 = 40. Minimum Z = 22 at (4, 1).

Q4. Formulate: A firm makes chairs (profit ₹20) and tables (profit ₹30). Each chair needs 2 hours and each table 3 hours of carpentry; 120 hours are available. Maximise profit.

Let x chairs, y tables. Maximise Z = 20x + 30y subject to 2x + 3y ≤ 120, x ≥ 0, y ≥ 0.

Q5. Why does the optimal solution lie at a corner point?

Because Z is linear, its value increases steadily in one direction across the feasible region, so the extreme value is reached at a boundary vertex.

Exam tip

Draw lines with correct intercepts on graph paper and label all corner points with coordinates before calculating Z.

Quick summary

An assignment problem allocates n jobs to n workers (or machines) so that the total cost or time is minimum (or profit maximum). A sequencing problem finds the order of processing jobs on machines that minimises total elapsed time. This chapter covers the Hungarian method for assignment and Johnson's rule for sequencing n jobs on two and three machines.

Key points

• Assignment conditions: number of jobs = number of workers; each worker gets exactly one job; each job goes to exactly one worker.
• Hungarian method (minimisation):

1. Row reduction: subtract the smallest element of each row from every element of that row.
2. Column reduction: subtract the smallest element of each column from every element of that column.
3. Cover all zeros with the minimum number of horizontal/vertical lines.
4. If the number of lines = n, an optimal assignment can be made; otherwise subtract the smallest uncovered element from all uncovered elements, add it at intersections, and repeat.
5. Make assignments on zeros — one in each row and column.

• Special cases: unbalanced problem (add a dummy row/column of zeros), maximisation (subtract every element from the largest element, then minimise), restricted assignment (put a very large cost M where assignment is not allowed).
• Sequencing: elapsed time is the time from start of the first job to completion of the last job; idle time is time a machine waits.
• Johnson's rule (n jobs, 2 machines A and B, order A → B): find the smallest time in the table; if it is on machine A, place that job first; if on B, place it last; remove the job and repeat. Ties may be broken arbitrarily.
• n jobs, 3 machines (A, B, C): reduce to two fictitious machines G = A + B and H = B + C if min A ≥ max B or min C ≥ max B; then apply Johnson's rule.

Important terms

• Dummy – row or column added with zero costs to balance the problem.
• Total elapsed time (T) – total time to finish all jobs.
• Idle time – time a machine is not working.
• Optimal sequence – order giving minimum elapsed time.

Formulas

• Idle time of machine B = T − (sum of processing times on B)
• Three-machine reduction: Gi = Ai + Bi, Hi = Bi + Ci (valid if min Ai ≥ max Bi or min Ci ≥ max Bi)

5 important questions

Q1. Solve the assignment problem (costs): Worker 1: (10, 12) ; Worker 2: (14, 11) for Jobs P, Q.

Row reduction: (0, 2) and (3, 0). Two zeros in different rows and columns → Worker 1 → P, Worker 2 → Q. Minimum cost = 10 + 11 = 21.

Q2. How do you solve a maximisation assignment problem?

Subtract all elements from the largest element in the table to get an 'opportunity loss' matrix, then apply the Hungarian method for minimisation.

Q3. What is an unbalanced assignment problem? How is it solved?

When the number of rows ≠ number of columns. Add a dummy row or column with zero costs to make it square.

Q4. Find the optimal sequence for jobs (A, B times): I (3, 6), II (3, 4), III (7, 2), IV (4, 5), V (2, 5).

Smallest = 2 for III on B (last) and V on A (first). Then 3 for I and II on A → next positions. Remaining IV. Sequence: V, I, II, IV, III (or V, II, I, IV, III).

Q5. State Johnson's rule for n jobs on two machines.

Pick the smallest processing time; if it is on the first machine place the job as early as possible, if on the second machine as late as possible; strike it off and repeat.

Exam tip

After finding the sequence, always draw the time table (in-out for each machine) to compute total elapsed time and idle times.

Quick summary

A random variable assigns a number to each outcome of a random experiment. This chapter covers discrete and continuous random variables, probability mass function (p.m.f.), probability density function (p.d.f.), cumulative distribution function (c.d.f.), expected value and variance, Bernoulli trials, the binomial distribution, and the Poisson distribution.

Key points

• Random variable (r.v.): a real-valued function on the sample space, written X, Y, Z.
• Discrete r.v.: takes countable values (number of heads). Continuous r.v.: takes any value in an interval (height, time).
• p.m.f.: P(X = xi) = pi with pi ≥ 0 and Σpi = 1.
• c.d.f.: F(x) = P(X ≤ x).
• p.d.f. f(x) for a continuous r.v.: f(x) ≥ 0 and the total area under f(x) is 1; P(a < X < b) = ∫[a to b] f(x) dx.
• Expected value E(X) = Σ xi pi (mean). Variance Var(X) = E(X²) − [E(X)]²; standard deviation = √Var(X).
• Bernoulli trial: only two outcomes (success/failure), probability of success p constant, trials independent.
• Binomial distribution B(n, p): number of successes in n Bernoulli trials.
• Poisson distribution: for rare events in a large number of trials (e.g., printing mistakes per page), with parameter m (mean).

Important terms

• Success probability p, failure probability q = 1 − p.
• Parameter – constant that defines a distribution (n, p for binomial; m for Poisson).
• Mean and variance – centre and spread of a distribution.

Formulas

• E(X) = Σ xi pi; E(X²) = Σ xi² pi; Var(X) = E(X²) − [E(X)]²
• Binomial: P(X = x) = ⁿCx pˣ qⁿ⁻ˣ, x = 0, 1, …, n
• Binomial mean = np; variance = npq
• Poisson: P(X = x) = e⁻ᵐ mˣ / x!, x = 0, 1, 2, …; mean = variance = m
• For large n and small p, binomial ≈ Poisson with m = np

5 important questions

Q1. Two fair coins are tossed. X = number of heads. Find the p.m.f. and E(X).

P(X = 0) = 1/4, P(X = 1) = 1/2, P(X = 2) = 1/4. E(X) = 0 + 1/2 + 2/4 = 1.

Q2. Find the variance for the distribution in Q1.

E(X²) = 0 + 1 × 1/2 + 4 × 1/4 = 1.5. Var(X) = 1.5 − 1² = 0.5.

Q3. X ~ B(n = 5, p = 1/2). Find P(X = 2).

⁵C₂ (1/2)² (1/2)³ = 10 × 1/32 = 5/16.

Q4. For a binomial distribution, mean = 6 and variance = 4. Find n and p.

npq = 4, np = 6 ⇒ q = 4/6 = 2/3, p = 1/3, n = 6 ÷ (1/3) = 18.

Q5. If X has a Poisson distribution with m = 2, find P(X = 0). (e⁻² = 0.1353)

P(X = 0) = e⁻² × 2⁰/0! = 0.1353.

Exam tip

Check that probabilities add to 1 before finding the mean; in binomial questions write n, p and q clearly first.