व्यावसायिक गणित व सांख्यिकी Business Mathematics and Statistics

थोडक्यात सारांश

Ratio, proportion and percentage are the basic tools of business arithmetic. They are used in sharing profits among partners, mixing goods, calculating profit and loss, discounts, commission and changes in prices. This topic covers the types of ratios, the properties of proportion, and the main percentage applications asked in university exams.

महत्त्वाचे मुद्दे

• Ratio (गुणोत्तर) compares two quantities of the same kind: a : b = a/b. It has no unit.
• A ratio stays the same if both terms are multiplied or divided by the same non-zero number.
• Types: duplicate (a² : b²), triplicate (a³ : b³), sub-duplicate (√a : √b), inverse (b : a), compound (product of ratios).
• Proportion (प्रमाण): equality of two ratios, a : b = c : d, i.e. a × d = b × c (product of extremes = product of means).
• Continued proportion: a : b = b : c, so b² = ac; b is the mean proportional.
• Direct variation: one quantity rises in the same ratio as the other; inverse variation: one rises as the other falls (more workers, fewer days).
• Percentage means "per hundred"; x% = x/100.
• Successive changes are multiplied, not added: +20% then −20% gives a net 4% fall.
• Profit and loss are always calculated on cost price (CP) unless stated otherwise.
• Trade discount is on list (marked) price; cash discount is on the price after trade discount.

महत्त्वाच्या संज्ञा

• Antecedent / consequent – first and second terms of a ratio.
• Fourth proportional – x in a : b = c : x.
• Third proportional – c in a : b = b : c.
• Commission – percentage of sales paid to an agent.
• Brokerage – commission paid to a broker on deals.

नमुने व सूत्रे

• Dividing an amount S in ratio a : b : c → shares S·a/(a+b+c), S·b/(a+b+c), S·c/(a+b+c)
• Fourth proportional of a, b, c: x = bc/a
• Mean proportional of a and c: b = √(ac)
• Percentage change = (Change ÷ Original) × 100
• Profit % = (Profit ÷ CP) × 100; SP = CP × (100 + profit%)/100
• Net effect of two successive changes x% and y% = x + y + xy/100

Worked examples:

५ महत्त्वाचे प्रश्न

Q1. Find the fourth proportional to 3, 5 and 12.

3 : 5 = 12 : x, so 3x = 5 × 12 = 60 and x = 20.

Q2. A salesman gets 5% commission on sales up to ₹50,000 and 8% on sales above that. Find his commission on sales of ₹80,000.

On ₹50,000: 5% = ₹2,500. On the remaining ₹30,000: 8% = ₹2,400. Total commission = ₹4,900.

Q3. Divide ₹1,20,000 among P, Q and R so that P : Q = 2 : 3 and Q : R = 4 : 5.

Make Q common: P : Q = 8 : 12 and Q : R = 12 : 15, so P : Q : R = 8 : 12 : 15 (35 parts). Each part = 1,20,000 ÷ 35 ≈ ₹3,428.57. Shares: P ≈ ₹27,428.57, Q ≈ ₹41,142.86, R ≈ ₹51,428.57 (total ₹1,20,000).

Q4. If 12 workers finish a job in 15 days, how many days will 20 workers take?

Inverse variation: workers × days is constant. 12 × 15 = 180 worker-days. Days = 180 ÷ 20 = 9 days.

Q5. A trader marks goods 25% above cost and allows a 10% discount. Find his profit percent.

Let CP = ₹100. Marked price = ₹125. SP = 125 × 0.90 = ₹112.50. Profit = ₹12.50 on ₹100, so profit = 12.5%.

परीक्षेसाठी टीप

Show the formula, the substitution and the final answer with a unit (₹, days, %). Most marks are lost by skipping the working.

थोडक्यात सारांश

Interest is the price paid for using borrowed money. Under simple interest it is charged only on the original principal; under compound interest it is charged on principal plus the interest already added. This topic covers both methods, conversion periods, the effective rate of interest and compound depreciation, with worked numericals.

महत्त्वाचे मुद्दे

• Principal (P) is the sum borrowed or invested; amount (A) = principal + interest.
• Simple interest (सरळ व्याज) is the same every year.
• Compound interest (चक्रवाढ व्याज) grows every period because interest earns interest.
• Conversion period: the interval after which interest is added to principal – yearly, half-yearly, quarterly, monthly.
• For half-yearly compounding: rate is halved and periods are doubled; for quarterly: rate ÷ 4, periods × 4.
• Nominal rate is the stated annual rate; effective rate is the actual annual rate after compounding.
• For the same P, R and time over 1 year, CI is greater than SI; for exactly 1 year with yearly compounding they are equal.
• Depreciation on reducing balance follows the compound formula with a minus sign.

महत्त्वाच्या संज्ञा

• Rate of interest (R) – interest on ₹100 for one year.
• Period (N or n) – time in years (or number of conversion periods).
• Effective rate – rate that gives the same interest in one year with yearly compounding.
• Compound depreciation – fall in value at a fixed percentage of the reduced value every year.

नमुने व सूत्रे

• SI = (P × N × R) ÷ 100; A = P + SI
• CI amount: A = P(1 + r/100)ⁿ; CI = A − P
• Compounded k times a year: A = P(1 + r/(100k))^(nk)
• Effective rate = [(1 + r/(100k))^k − 1] × 100
• CI − SI for 2 years = P(R/100)²
• Depreciated value: V = P(1 − r/100)ⁿ

Worked examples (P = ₹10,000):

५ महत्त्वाचे प्रश्न

Q1. Find the SI on ₹15,000 at 9% p.a. for 2 years 6 months.

N = 2.5 years. SI = 15,000 × 2.5 × 9 ÷ 100 = ₹3,375. Amount = ₹18,375.

Q2. At what rate will a sum double itself in 8 years at simple interest?

Interest = P when it doubles. P = P × 8 × R ÷ 100, so R = 100 ÷ 8 = 12.5% p.a.

Q3. Find the compound interest on ₹20,000 at 12% p.a. for 1 year, compounded quarterly.

Rate per quarter = 3%, n = 4. A = 20,000 × (1.03)⁴ = 20,000 × 1.12550881 = ₹22,510.18. CI = ₹2,510.18.

Q4. A machine costs ₹50,000 and depreciates at 10% p.a. on the reducing balance. Find its value after 2 years and the total depreciation.

V = 50,000 × (0.90)² = 50,000 × 0.81 = ₹40,500. Total depreciation = 50,000 − 40,500 = ₹9,500.

Q5. The difference between CI and SI on a sum for 2 years at 5% p.a. is ₹50. Find the sum.

CI − SI = P(R/100)² → 50 = P × (0.05)² = 0.0025P, so P = 50 ÷ 0.0025 = ₹20,000.

परीक्षेसाठी टीप

Before using the CI formula, convert the rate and the number of periods for the compounding frequency. Write (1.05)⁴, not 1.05 × 4.

थोडक्यात सारांश

An annuity is a series of equal payments made at equal intervals, like monthly rent, insurance premiums, SIP instalments or loan EMIs. This topic explains the types of annuities and how to find their future (accumulated) value and present value. It ends with the EMI formula used by banks for home, vehicle and personal loans.

महत्त्वाचे मुद्दे

• Annuity (वार्षिकी/हप्ता): equal periodic payments for a fixed number of periods.
• Ordinary annuity (immediate): payment at the end of each period (most loans, EMIs).
• Annuity due: payment at the beginning of each period (rent, insurance premium, SIP on the 1st).
• Annuity certain: fixed number of payments; contingent annuity: depends on an event (life insurance pension); perpetuity: payments go on forever.
• Future value (amount) answers "how much will my instalments grow to?"
• Present value answers "what lump sum today equals all future instalments?"
• An annuity due earns one more period of interest than an ordinary annuity, so its value = ordinary value × (1 + i).
• Sinking fund: equal amounts set aside regularly to repay a debt or replace an asset later; it uses the future-value formula.
• EMI (Equated Monthly Instalment) is an ordinary annuity whose present value equals the loan amount; each EMI has an interest part (large at first) and a principal part (grows over time).

महत्त्वाच्या संज्ञा

• i – rate of interest per period (annual rate ÷ periods per year, as a decimal).
• n – total number of payments.
• P (or a) – periodic payment.
• Amortisation – repaying a loan through instalments; the schedule shows interest and principal in each EMI.
• Perpetuity – annuity with no end.

नमुने व सूत्रे

• Future value of ordinary annuity: A = P × [(1 + i)ⁿ − 1] ÷ i
• Present value of ordinary annuity: V = P × [1 − (1 + i)⁻ⁿ] ÷ i
• Annuity due: multiply the ordinary result by (1 + i)
• Present value of perpetuity: V = P ÷ i
• EMI = L × r × (1 + r)ⁿ ÷ [(1 + r)ⁿ − 1], where L = loan, r = monthly rate, n = number of months
• Sinking fund instalment: P = A × i ÷ [(1 + i)ⁿ − 1]

Worked examples:

EMI example: loan ₹1,00,000 at 12% p.a. for 12 months. r = 12 ÷ 12 = 1% = 0.01, n = 12, (1.01)¹² = 1.126825.
EMI = 1,00,000 × 0.01 × 1.126825 ÷ 0.126825 = ₹8,884.88 (approx).
Total paid ≈ 12 × 8,884.88 = ₹1,06,618.56, so total interest ≈ ₹6,618.56.
First month: interest = 1% of 1,00,000 = ₹1,000; principal repaid = 8,884.88 − 1,000 = ₹7,884.88.

५ महत्त्वाचे प्रश्न

Q1. Distinguish between ordinary annuity and annuity due.

In an ordinary annuity each payment is made at the end of the period (loan EMI). In an annuity due it is made at the beginning (rent paid in advance). For the same P, i and n, an annuity due has a higher value by a factor (1 + i).

Q2. What is a sinking fund?

A fund built by setting aside equal amounts at regular intervals, invested at compound interest, to collect a known sum at a future date, e.g. to repay debentures or replace a machine.

Q3. Find the amount of an annuity of ₹2,000 paid at the end of every 6 months for 2 years at 8% p.a. compounded half-yearly.

i = 4% = 0.04, n = 4. A = 2,000 × [(1.04)⁴ − 1] ÷ 0.04 = 2,000 × (1.16985856 − 1) ÷ 0.04 = 2,000 × 4.246464 = ₹8,492.93.

Q4. Find the EMI for a loan of ₹1,00,000 at 12% p.a. repayable in 12 monthly instalments. Also find the total interest.

Steps as in the worked example: r = 0.01, n = 12 → EMI ≈ ₹8,884.88 → total paid ≈ ₹1,06,618.56 → interest ≈ ₹6,618.56. Mention that interest falls and principal rises every month.

Q5. A company must repay ₹5,00,000 after 5 years. How much should it set aside each year at 10% p.a. compound interest?

Sinking fund: P = 5,00,000 × 0.1 ÷ [(1.1)⁵ − 1] = 50,000 ÷ 0.61051 = ₹81,898.74 (approx) at the end of each year.

परीक्षेसाठी टीप

First decide: end-of-period or beginning, future value or present value. Then pick the formula. Write i per period, not per year.

थोडक्यात सारांश

A matrix is a rectangular arrangement of numbers in rows and columns. Businesses use matrices to store and combine data such as quantities sold, prices and costs across branches. A determinant is a single number linked with a square matrix; it is used to find the inverse of a matrix and to solve simultaneous equations by Cramer's rule.

महत्त्वाचे मुद्दे

• A matrix with m rows and n columns has order m × n.
• Types: row, column, square, zero (null), diagonal, scalar, identity (unit) matrix I, upper/lower triangular, symmetric (A' = A).
• Addition and subtraction are possible only for matrices of the same order (add corresponding elements).
• Multiplication AB is possible only if columns of A = rows of B; the result has order (rows of A) × (columns of B).
• In general AB ≠ BA (matrix multiplication is not commutative).
• Transpose A' (or Aᵀ): rows become columns.
• A determinant (निर्धारक) exists only for a square matrix.
• A matrix is singular if |A| = 0 (no inverse) and non-singular if |A| ≠ 0.
• Cramer's rule solves n equations in n unknowns using determinants, provided D ≠ 0.

महत्त्वाच्या संज्ञा

• Element – each number in a matrix, written aᵢⱼ (row i, column j).
• Minor – determinant left after deleting the row and column of an element.
• Cofactor – minor with sign (−1)^(i+j).
• Adjoint (adj A) – transpose of the cofactor matrix.
• Inverse A⁻¹ – matrix such that A × A⁻¹ = I.

नमुने व सूत्रे

• For A = [a b; c d] (rows separated by ;): |A| = ad − bc
• Inverse of 2 × 2: A⁻¹ = (1/|A|) × [d −b; −c a]
• General inverse: A⁻¹ = adj A ÷ |A|
• 3 × 3 determinant (expansion along row 1): |A| = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁)
• Cramer's rule (2 unknowns): x = Dx ÷ D, y = Dy ÷ D

Worked examples with A = [2 3; 1 4] and B = [1 0; 2 5]:

Check: A × A⁻¹ row 1 = (2×0.8 + 3×(−0.2), 2×(−0.6) + 3×0.4) = (1, 0). Correct.

५ महत्त्वाचे प्रश्न

Q1. Find the determinant of [1 2 3; 0 1 4; 5 6 0].

|A| = 1(1×0 − 4×6) − 2(0×0 − 4×5) + 3(0×6 − 1×5) = 1(−24) − 2(−20) + 3(−5) = −24 + 40 − 15 = 1.

Q2. When is a matrix singular? Can it have an inverse?

A square matrix is singular when its determinant is zero. A singular matrix has no inverse, because A⁻¹ = adj A ÷ |A| needs division by |A|.

Q3. Solve by Cramer's rule: 2x + 3y = 13 and x + 4y = 14.

D = |2 3; 1 4| = 8 − 3 = 5. Dx = |13 3; 14 4| = 52 − 42 = 10. Dy = |2 13; 1 14| = 28 − 13 = 15. x = 10 ÷ 5 = 2, y = 15 ÷ 5 = 3. Check: 2(2) + 3(3) = 13 and 2 + 4(3) = 14.

Q4. Two shops sell pens and notebooks. Shop A sells 10 pens and 5 notebooks; Shop B sells 8 pens and 12 notebooks. A pen costs ₹10 and a notebook ₹40. Find each shop's revenue using matrices.

Quantity matrix Q = [10 5; 8 12], price matrix P = [10; 40] (2 × 1). Revenue = QP = [10×10 + 5×40; 8×10 + 12×40] = [300; 560]. Shop A earns ₹300 and Shop B earns ₹560.

Q5. Find the inverse of A = [3 2; 4 3] and verify.

|A| = 9 − 8 = 1. A⁻¹ = (1/1) × [3 −2; −4 3] = [3 −2; −4 3]. Verify: row 1 of A × A⁻¹ = (3×3 + 2×(−4), 3×(−2) + 2×3) = (1, 0); row 2 = (4×3 + 3×(−4), 4×(−2) + 3×3) = (0, 1). So A × A⁻¹ = I.

परीक्षेसाठी टीप

Before multiplying, write the orders (2 × 2 · 2 × 1 = 2 × 1). It tells you instantly whether the product exists and what shape the answer has.