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

Class 11 Commerce · 9 chapters · 9 with notes

📘 Official textbook PDF

Chapters

1Partition ValuesNotes
2Measures of DispersionNotes
3SkewnessNotes
4Bivariate Frequency Distribution and Chi Square StatisticNotes
5CorrelationNotes
6Permutations and CombinationsNotes
7ProbabilityNotes
8Linear InequationsNotes
9Commercial MathematicsNotes

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 1

Partition Values

Quick summary

Partition values divide an arranged set of data into equal parts. The median divides it into 2 parts, quartiles into 4, deciles into 10 and percentiles into 100. This chapter shows how to calculate quartiles, deciles and percentiles for raw data, discrete frequency distributions and grouped (continuous) data, and how to locate quartiles on a graph using ogives.

Key points

First arrange the data in ascending order.
There are 3 quartiles (Q₁, Q₂, Q₃), 9 deciles (D₁ to D₉) and 99 percentiles (P₁ to P₉₉).
Q₂ = D₅ = P₅₀ = the median.
Raw (ungrouped) data, with n observations:
Qᵢ = the value of the i(n + 1)/4th observation.
Dᵢ = the value of the i(n + 1)/10th observation.
Pᵢ = the value of the i(n + 1)/100th observation.
If the position is a fraction, interpolate. For example, for position 6.25: the 6th value + 0.25 × (7th value − 6th value).
Discrete frequency distribution: find the cumulative frequencies, and Qᵢ is the value whose c.f. is just greater than or equal to i(N + 1)/4.
Continuous (grouped) data: first find the class containing the iN/4th observation, then use
Qᵢ = l + (h/f)(iN/4 − c.f.)
Dᵢ = l + (h/f)(iN/10 − c.f.) and Pᵢ = l + (h/f)(iN/100 − c.f.).
Here l is the lower limit of the class, h the class width, f the frequency of the class, c.f. the cumulative frequency of the class before it, and N the total frequency.
Graphical location: draw a 'less than' ogive. Mark iN/4 on the Y-axis, draw a horizontal line to the curve, then drop a perpendicular to the X-axis to read Qᵢ. The 'less than' and 'more than' ogives intersect at the median.

Important terms

Quartiles – values that divide the data into 4 equal parts.
Deciles – values that divide the data into 10 equal parts.
Percentiles – values that divide the data into 100 equal parts.
Ogive – a cumulative frequency curve.

Formats and formulas

Data type	Formula for Qᵢ
Raw data	Value of the i(n + 1)/4 th observation
Discrete series	Value with c.f. ≥ i(N + 1)/4
Continuous series	l + (h/f)(iN/4 − c.f.)

5 important questions

Q1. Find Q₁ for 3, 5, 7, 9, 11, 13, 15.

n = 7, so Q₁ is the 1 × (7 + 1)/4 = 2nd observation, which is 5.

Q2. Find D₆ for 10, 12, 15, 18, 20, 25, 30, 35, 40.

n = 9, so D₆ is the 6 × 10/10 = 6th observation, which is 25.

Q3. Write the formula for Q₃ in a continuous series.

Q₃ = l + (h/f)(3N/4 − c.f.).

Q4. Which partition value equals the median?

Q₂ = D₅ = P₅₀.

Q5. How do you locate quartiles on a graph?

Draw a 'less than' ogive, mark N/4, N/2 and 3N/4 on the Y-axis, and read the corresponding x-values on the X-axis.

Exam tip

Always show the cumulative frequency column and the position iN/4 clearly. Most marks are given for the steps.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

Measures of Dispersion →

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 2

Measures of Dispersion

Quick summary

Two sets of data can have the same average but different spread. Dispersion measures how widely the values are scattered around the average. This chapter covers the meaning of dispersion, absolute and relative measures, the range, the quartile deviation (semi-interquartile range), the variance and standard deviation, the combined standard deviation, and the coefficient of variation (C.V.) used to compare consistency.

Key points

Example from the textbook: three batsmen each average 50 runs, but the spread of their scores is very different.
Absolute measures are in the same units as the data (range, Q.D., S.D.). Relative measures have no units and are used to compare distributions (such as the C.V.).
Range = L − S (largest minus smallest). It is affected by extreme values (outliers).
Quartile deviation (Q.D.) = (Q₃ − Q₁)/2. It is also called the semi-interquartile range and is not affected by extreme values.
Variance σ² = Σ(x − x̄)²/n = Σx²/n − x̄². For a frequency distribution: σ² = Σf(x − x̄)²/N.
Standard deviation σ = √variance. It is the best measure of dispersion.
Combined S.D. of two groups: σc² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/(n₁ + n₂), where d₁ = x̄₁ − x̄c and d₂ = x̄₂ − x̄c.
Combined mean x̄c = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂).
Coefficient of variation: C.V. = (σ/x̄) × 100.
The distribution with the smaller C.V. is more consistent, steady and homogeneous. The larger C.V. means more variability.
The S.D. is not affected by a change of origin (adding or subtracting a constant) but is affected by a change of scale.

Important terms

Dispersion – the scatter of values around an average.
Standard deviation – the square root of the mean of squared deviations from the mean.
Variance – the square of the S.D.
Coefficient of variation – the S.D. as a percentage of the mean.

Formats and formulas

Measure	Formula
Range	L − S
Q.D.	(Q₃ − Q₁)/2
Variance	Σx²/n − x̄²
C.V.	(σ/x̄) × 100
Combined mean	(n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)

5 important questions

Q1. Find the range of 12, 18, 25, 30, 40. Why is the range not always a good measure?

Range = 40 − 12 = 28. It depends only on the two extreme values, so outliers distort it.

Q2. If Q₁ = 20 and Q₃ = 40, find the quartile deviation.

Q.D. = (40 − 20)/2 = 10.

Q3. Find the S.D. of 2, 4, 6, 8, 10.

x̄ = 6, and Σ(x − x̄)² = 16 + 4 + 0 + 4 + 16 = 40. σ² = 8, so σ = 2.83.

Q4. Player A has mean 50 and S.D. 10. Player B has mean 40 and S.D. 6. Who is more consistent?

C.V. of A = 20% and C.V. of B = 15%. B is more consistent.

Q5. Distinguish between absolute and relative measures of dispersion.

Absolute measures have units (S.D., range). Relative measures have no units and are used for comparison (C.V.).

Exam tip

For 'who is more consistent' questions, compare the C.V., not the S.D.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

← Partition Values Skewness →

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 3

Skewness

Quick summary

Skewness means a lack of symmetry in a distribution. Two distributions can have the same mean and the same spread but different shapes. This chapter covers symmetric, positively skewed and negatively skewed distributions, the positions of the mean, median and mode in each, and two measures of skewness: Karl Pearson's coefficient and Bowley's coefficient.

Key points

Symmetric distribution: values at equal distances from the mean have equal frequencies. Mean = median = mode, and the curve is bell-shaped.
Positively skewed: the right tail is longer, and mean > median > mode.
Negatively skewed: the left tail is longer, and mode > median > mean. The textbook's example is lifespans, which are often negatively skewed.
Karl Pearson's coefficient of skewness: Skp = (Mean − Mode)/S.D.
When the mode is ill-defined, use Mean − Mode = 3(Mean − Median), which gives Skp = 3(Mean − Median)/S.D.
Skp = 0 for a symmetric distribution, Skp > 0 for a positively skewed one and Skp < 0 for a negatively skewed one.
Most Skp values lie between −1 and 1, and all lie between −3 and 3.
Skp has no units, so different distributions can be compared.
Mode for grouped data: Mode = L + [(fm − f1)/(2fm − f1 − f2)] × h, where fm is the frequency of the modal class, f1 of the class before it and f2 of the class after it.
Bowley's coefficient of skewness (based on quartiles): Skb = (Q₃ + Q₁ − 2Q₂)/(Q₃ − Q₁).
Skb = 0 when Q₁, Q₂ and Q₃ are equally spaced (symmetric).
Skb > 0 when Q₃ − Q₂ > Q₂ − Q₁ (positive skew), and Skb < 0 when Q₃ − Q₂ < Q₂ − Q₁ (negative skew).
Skb lies between −1 and 1.
Skb can be used for distributions with open-ended classes, because open ends do not affect the quartiles.

Important terms

Skewness – a lack of symmetry in a distribution.
Positive skew – a longer tail towards higher values.
Negative skew – a longer tail towards lower values.
Open-ended class – a class with no lower or no upper limit.

Formats and formulas

Measure	Formula	Range
Karl Pearson	(Mean − Mode)/S.D.	−3 to 3
Karl Pearson (alternative)	3(Mean − Median)/S.D.	−3 to 3
Bowley	(Q₃ + Q₁ − 2Q₂)/(Q₃ − Q₁)	−1 to 1

5 important questions

Q1. Distinguish between positive and negative skewness.

Positive skew has a longer right tail and mean > median > mode. Negative skew has a longer left tail and mode > median > mean.

Q2. Mean = 45, mode = 40 and S.D. = 10. Find Skp.

(45 − 40)/10 = 0.5, which is a positive skew.

Q3. Mean = 32.5, median = 34 and S.D. = 6. Find Skp.

3(32.5 − 34)/6 = −0.75, which is a negative skew.

Q4. Q₁ = 20, Q₂ = 30 and Q₃ = 45. Find Skb.

(45 + 20 − 60)/(45 − 20) = 5/25 = 0.2, which is a positive skew.

Q5. When is Bowley's coefficient preferred?

When the distribution has open-ended classes, or when the mean and S.D. cannot be found reliably, because Bowley's coefficient uses only the quartiles.

Exam tip

Always write the conclusion after the number, for example 'Skp < 0, so the distribution is negatively skewed'.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

← Measures of Dispersion Bivariate Frequency Distribution and Chi Square Statistic →

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 4

Bivariate Frequency Distribution and Chi Square Statistic

Quick summary

Bivariate data has two variables measured on each item, such as the age of a husband and the age of his wife, or marks in two subjects. This chapter covers how to classify and tabulate bivariate data in a two-way frequency table, marginal and conditional frequency distributions, contingency tables for categorical data, and the chi-square (χ²) statistic for measuring the relationship between two categorical variables.

Key points

Univariate data: one variable. Bivariate data: pairs of values (x, y).
Bivariate frequency table (two-way table): the values or classes of X go along one side and those of Y along the other. Tally marks are entered in the cells, and each cell shows the frequency of that (x, y) pair. If X has m values and Y has n values, the table has m × n cells.
Marginal frequency distribution of X: the values of X with their totals (fx), read from the row or column totals. Similarly for Y (fy).
Conditional frequency distribution: the distribution of one variable when the other is fixed at a given value or class, for example the marks in Statistics of students who scored 20–30 in English.
Contingency table: a two-way table for categorical variables, for example gender × food preference. A 2 × 2 table has cells a, b, c and d with row and column totals.
Chi-square statistic χ² (read 'kai'): measures the relationship (association) between two categorical variables.
Expected frequency of a cell: Eᵢⱼ = (Rᵢ × Cⱼ)/N, where Rᵢ is the row total, Cⱼ the column total and N the grand total.
χ² = Σ (Oᵢⱼ − Eᵢⱼ)²/Eᵢⱼ, where O is the observed frequency and E the expected frequency.
χ² is always non-negative. χ² = 0 when the observed and expected frequencies are equal, which shows no association. A larger χ² shows a stronger association.
A statistic is any quantity calculated from data, for example the mean.

Important terms

Bivariate data – data on two variables for the same items.
Marginal frequency – the total frequency of one variable's value.
Contingency table – a two-way table of categorical data.
Chi-square (χ²) – a statistic that measures association between categorical variables.

Formats and formulas

Item	Formula
Expected frequency	Eᵢⱼ = (Rᵢ × Cⱼ)/N
Chi-square	χ² = Σ (O − E)²/E
Number of cells	m × n

5 important questions

Q1. What is a bivariate frequency distribution?

A two-way table that shows the frequencies of pairs of values (or classes) of two variables.

Q2. Distinguish between marginal and conditional frequency distributions.

A marginal distribution gives one variable's totals regardless of the other. A conditional distribution gives one variable's frequencies for a fixed value of the other.

Q3. A 2 × 2 table has R₁ = 80, R₂ = 120, C₁ = 75, C₂ = 125 and N = 200. Find the expected frequencies.

E₁₁ = 80 × 75/200 = 30, E₁₂ = 50, E₂₁ = 45 and E₂₂ = 75.

Q4. Write the formula for χ² and explain its terms.

χ² = Σ(O − E)²/E, where O is the observed frequency and E is the expected frequency (R × C/N).

Q5. Can χ² be negative? Why?

No. It is a sum of squared terms divided by positive expected frequencies, so it is always zero or positive.

Exam tip

Check that the expected frequencies add up to the same row and column totals as the observed ones. This catches arithmetic errors.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

← Skewness Correlation →

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 5

Correlation

Quick summary

Correlation is the relationship between two variables: when one changes, the other tends to change too. This chapter covers the concept and types of correlation, the scatter diagram, covariance and its properties, Karl Pearson's coefficient of correlation (r), its properties and how to interpret its value.

Key points

Positive correlation: both variables move in the same direction (for example, height and weight). Negative correlation: they move in opposite directions (for example, price and demand). No correlation: no relationship.
Scatter diagram: plot the pairs (x, y) as points.
Points rising from left to right show positive correlation. Points falling from left to right show negative correlation.
Points spread with no pattern show no correlation. All points on a straight line show perfect correlation.
Covariance: Cov(X, Y) = Σ(x − x̄)(y − ȳ)/n = Σxy/n − x̄ȳ.
Properties of covariance:
Cov(X, Y) = Cov(Y, X).
Cov(X, C) = 0 for a constant C.
Cov(X, X) = Var(X).
It can be positive, negative or zero.
It does not change with a change of origin but is affected by a change of scale.
Karl Pearson's coefficient of correlation: r = Cov(X, Y)/(σx·σy).
Alternative formula: r = [nΣxy − ΣxΣy] / √{[nΣx² − (Σx)²][nΣy² − (Σy)²]}.
Properties of r:
−1 ≤ r ≤ 1.
r has no units.
Corr(X, X) = 1.
r does not change with a change of origin and scale. If U = (x − a)/h and V = (y − b)/k, then r(uv) = r(xy) when h and k have the same sign, and r(uv) = −r(xy) when they have opposite signs.
Interpretation (as given in the textbook):
r = 1 is perfect positive and r = −1 is perfect negative.
r > 0.8 is high positive.
0.3 < r < 0.8 is moderate positive.
r between −0.3 and 0.3 is insignificant (poor).
r = 0 means no correlation.

Important terms

Correlation – an association between two variables.
Covariance – the average product of the deviations from the means.
Scatter diagram – a graph of the paired values.
Correlation coefficient (r) – a number between −1 and 1 that measures linear association.

Formats and formulas

Item	Formula
Covariance	Σxy/n − x̄ȳ
σx	√(Σx²/n − x̄²)
r	Cov(X, Y)/(σx·σy)
Range of r	−1 ≤ r ≤ 1

5 important questions

Q1. What is a scatter diagram? What does it show?

A graph of the paired values. The pattern of the points shows whether the correlation is positive, negative or zero, and roughly how strong it is.

Q2. State the properties of covariance.

Cov(X, Y) = Cov(Y, X), Cov(X, C) = 0, Cov(X, X) = Var(X), and it is unaffected by a change of origin but affected by a change of scale.

Q3. If Cov(X, Y) = 12, σx = 4 and σy = 5, find r.

r = 12/(4 × 5) = 0.6, which is a moderate positive correlation.

Q4. Given n = 5, Σx = 15, Σy = 25 and Σxy = 85, find Cov(X, Y).

85/5 − 3 × 5 = 17 − 15 = 2.

Q5. Interpret r = −0.9.

There is a high negative correlation. As one variable increases, the other decreases strongly.

Exam tip

Always state your interpretation of r in words after calculating it (for example, 'high positive correlation'). It carries a mark.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

← Bivariate Frequency Distribution and Chi Square Statistic Permutations and Combinations →

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 6

Permutations and Combinations

Quick summary

Permutations and combinations let us count arrangements and selections without listing them all. This chapter covers the fundamental principles of counting (addition and multiplication), factorial notation, permutations (with and without repetition, with identical objects, and circular), combinations and their properties.

Key points

Addition principle: if a task can be done in m ways or in n ways (but not both together), it can be done in m + n ways.
Multiplication principle: if one task can be done in m ways and then another in n ways, both can be done in m × n ways.
Factorial: n! = n(n − 1)(n − 2)…3·2·1, with 0! = 1. Also n! = n·(n − 1)!.
Permutation (arrangement, order matters): ⁿPᵣ = n!/(n − r)!, and ⁿPₙ = n!.
With repetition allowed: the number of arrangements of r things from n is nʳ.
Identical objects: n objects of which p are alike, q are alike and r are alike can be arranged in n!/(p!·q!·r!) ways.
Circular permutations: n distinct objects can be arranged in a circle in (n − 1)! ways. When clockwise and anticlockwise arrangements are the same (for example, beads in a necklace), the number is (n − 1)!/2.
Combination (selection, order does not matter): ⁿCᵣ = n!/[r!(n − r)!].
Relation: ⁿPᵣ = r!·ⁿCᵣ.
Properties:
ⁿC₀ = ⁿCₙ = 1.
ⁿCᵣ = ⁿCₙ₋ᵣ.
If ⁿCₓ = ⁿCᵧ, then x = y or x + y = n.
ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.
Useful results: the number of diagonals of an n-sided polygon is ⁿC₂ − n, and the number of triangles from n points (no three collinear) is ⁿC₃.

Important terms

Permutation – an arrangement where order matters.
Combination – a selection where order does not matter.
Factorial – the product of all positive integers up to n.
Circular permutation – an arrangement around a circle.

Formats and formulas

Situation	Formula
Arrangements of r from n	n!/(n − r)!
Selections of r from n	n!/[r!(n − r)!]
With repetition	nʳ
Identical objects	n!/(p!q!r!)
Circular	(n − 1)!

5 important questions

Q1. Find ⁸P₃ and ⁸C₃.

⁸P₃ = 8 × 7 × 6 = 336. ⁸C₃ = 336/6 = 56.

Q2. How many arrangements of the letters of 'BANANA' are there?

6!/(3!·2!) = 720/12 = 60.

Q3. In how many ways can 6 people sit at a round table?

(6 − 1)! = 120.

Q4. How many 3-digit numbers can be formed from 1, 2, 3, 4, 5 without repetition?

⁵P₃ = 60.

Q5. If ⁿC₈ = ⁿC₆, find n.

8 + 6 = n, so n = 14.

Exam tip

First decide whether order matters (permutation) or not (combination). Most mistakes come from choosing the wrong one.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

← Correlation Probability →

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 7

Probability

Quick summary

Probability measures how likely an event is, as a number from 0 to 1. This chapter covers random experiments, the sample space, the types of events, the algebra of events, the classical definition of probability, the addition theorem, conditional probability, the multiplication theorem, independent events and odds.

Key points

Random experiment: its outcome cannot be predicted in advance. The sample space (S) is the set of all possible outcomes. An event is a subset of S.
Types of events: sure event (S), impossible event (∅), elementary (simple) event, compound event, equally likely events, mutually exclusive events (A ∩ B = ∅), exhaustive events (A ∪ B = S) and complementary events (A′).
Algebra of events: A ∪ B means A or B; A ∩ B means A and B; A′ means not A; A − B means A but not B.
Classical probability: P(A) = n(A)/n(S), and 0 ≤ P(A) ≤ 1.
P(S) = 1, P(∅) = 0 and P(A′) = 1 − P(A).
Addition theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For mutually exclusive events: P(A ∪ B) = P(A) + P(B).
For three events: P(A∪B∪C) = ΣP(A) − ΣP(A∩B) + P(A∩B∩C).
Conditional probability: P(A|B) = P(A ∩ B)/P(B), with P(B) ≠ 0.
Multiplication theorem: P(A ∩ B) = P(B)·P(A|B) = P(A)·P(B|A).
Independent events: P(A ∩ B) = P(A)·P(B), i.e. P(A|B) = P(A).
Odds: odds in favour of A = n(A) : n(A′). Odds against A = n(A′) : n(A). If the odds in favour are a : b, then P(A) = a/(a + b).

Important terms

Sample space – the set of all outcomes.
Mutually exclusive – the events cannot happen together.
Exhaustive events – together they cover the whole sample space.
Conditional probability – the probability of A given that B has occurred.

Formats and formulas

Rule	Formula
Probability	n(A)/n(S)
Complement	P(A′) = 1 − P(A)
Addition	P(A∪B) = P(A) + P(B) − P(A∩B)
Conditional	P(A∩B)/P(B)
Independent events	P(A∩B) = P(A)·P(B)

5 important questions

Q1. Two coins are tossed. Find the probability of getting at least one head.

S = {HH, HT, TH, TT}, and 3 outcomes have at least one head, so P = 3/4.

Q2. A card is drawn from a pack of 52. Find P(king or heart).

4/52 + 13/52 − 1/52 = 16/52 = 4/13.

Q3. If P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2, find P(A|B) and say whether A and B are independent.

P(A|B) = 0.2/0.5 = 0.4 = P(A), so A and B are independent.

Q4. The odds in favour of an event are 3 : 5. Find its probability.

3/(3 + 5) = 3/8.

Q5. Two dice are thrown. Find the probability that the sum is 7.

The favourable outcomes are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), so P = 6/36 = 1/6.

Exam tip

Write the sample space and n(S) first. Most probability marks are lost through a wrong count of S.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

← Permutations and Combinations Linear Inequations →

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 8

Linear Inequations

Quick summary

A linear inequation (linear inequality) compares two linear expressions using <, >, ≤ or ≥. This chapter covers how to solve linear inequations in one variable, how to show their solutions on a number line, the graphical solution of a linear inequation in two variables, and the solution of a system of linear inequations in two variables.

Key points

Example from the textbook: Ramesh has ₹100 and each chocolate costs ₹15, so 15x ≤ 100 with x a whole number.
Rules for solving:
You can add or subtract the same number on both sides without changing the inequality sign.
Multiplying or dividing both sides by a positive number keeps the sign the same.
Multiplying or dividing by a negative number reverses the sign.
The solution set depends on the replacement set (N, W, Z or R).
One variable on a number line: use a filled (solid) dot for ≤ or ≥ (the point is included) and a hollow dot for < or > (the point is not included), then shade in the right direction.
Two variables (ax + by ≤ c, etc.):
Draw the line ax + by = c. Use a solid line for ≤ or ≥ and a dotted line for < or >.
Test a point not on the line, usually the origin (0, 0).
If the test point satisfies the inequation, shade the half-plane containing it. Otherwise shade the other side.
System of inequations: draw each one and shade it. The common shaded region is the solution set.
x ≥ 0 and y ≥ 0 restrict the solution to the first quadrant.
x ≥ a is the region to the right of the vertical line x = a. y ≤ b is the region below the horizontal line y = b.

Important terms

Inequation – a statement using <, >, ≤ or ≥.
Solution set – all values that satisfy the inequation.
Half-plane – the region on one side of a line.
Common region – the region that satisfies all the inequations of a system.

Formats and formulas

Sign	Line type	Point on line
< or >	Dotted	Not included
≤ or ≥	Solid	Included

5 important questions

Q1. Solve 3x − 5 < 10 for x ∈ R.

3x < 15, so x < 5. The solution set is (−∞, 5).

Q2. Solve −2x ≥ 8.

Divide by −2 and reverse the sign: x ≤ −4.

Q3. Solve 15x ≤ 100 for x ∈ W (the chocolate problem).

x ≤ 6.67, so x = 0, 1, 2, 3, 4, 5, 6.

Q4. Show 2x + 3y ≤ 6 graphically.

Draw the solid line through (3, 0) and (0, 2). The test point (0, 0) gives 0 ≤ 6, which is true, so shade the side containing the origin.

Q5. Solve graphically: x + y ≤ 4, x ≥ 0, y ≥ 0.

The common region is the triangle with vertices (0, 0), (4, 0) and (0, 4), including its boundary.

Exam tip

Remember to reverse the sign when you multiply or divide by a negative number. It is the most common error.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

← Probability Commercial Mathematics →

Class 11 Commerce · Mathematics and Statistics (Commerce) Part 2 · Chapter 9

Commercial Mathematics

Quick summary

Commercial mathematics applies arithmetic to everyday business. This chapter covers percentage, profit and loss, simple and compound interest, depreciation, partnership, GST, and shares and dividends.

Key points

Percentage: 'per cent' means 'for every hundred'. x% = x/100. Converting marks to percentages lets us compare students fairly.
Profit and loss: Profit = S.P. − C.P. and Loss = C.P. − S.P. Profit % = (Profit/C.P.) × 100. Discount is calculated on the marked price, and S.P. = M.P. − discount.
Simple interest: I = PNR/100. Amount A = P + I.
Compound interest: A = P(1 + R/100)ᴺ, and C.I. = A − P. For interest compounded half-yearly, use R/2 and 2N.
Depreciation: the fall in the value of an asset (machinery, vehicles, furniture) over time.
Present value after n years: V(1 − r/100)ⁿ, where V is the original value and r the rate of depreciation.
Partnership: two or more persons running a business jointly.
A working partner invests money and manages the business. A sleeping partner only invests money.
Profit is shared in the ratio of capital × time (for equal periods, in the ratio of capital).
GST: introduced through the GST Acts passed in 2017 (CGST, SGST/UTGST and IGST Acts).
For a sale within a state, GST = CGST + SGST, split equally. For a sale between states, IGST applies.
The textbook uses the GST slabs of 0%, 5%, 12%, 18% and 28%. (The slabs were restructured in September 2025, mainly to 5% and 18%, but use the textbook rates in exam sums.)
Input tax credit: the tax paid on purchases is set off against the tax on sales.
Shares and dividends:
Face value is the value printed on the share certificate. Market value is the price at which the share is bought or sold.
A share is at premium if M.V. > F.V., at par if they are equal, and at discount if M.V. < F.V.
Dividend is calculated on the face value.
Return % = (dividend/investment) × 100.
Brokerage is a percentage of the market value. It is added when buying and subtracted when selling.

Important terms

Principal – the sum lent or borrowed.
Depreciation – the decrease in the value of an asset.
Sleeping partner – a partner who invests but does not manage.
Face value – the nominal value of a share.

Formats and formulas

Topic	Formula
Simple interest	PNR/100
Compound amount	P(1 + R/100)ᴺ
Depreciated value	V(1 − r/100)ⁿ
Profit %	(Profit/C.P.) × 100
Return on investment	(Dividend/Investment) × 100

5 important questions

Q1. Find the simple interest on ₹8,000 at 6% p.a. for 3 years.

I = 8000 × 3 × 6/100 = ₹1,440.

Q2. Find the compound interest on ₹10,000 at 10% p.a. for 2 years.

A = 10000 × (1.1)² = ₹12,100, so C.I. = ₹2,100.

Q3. A machine worth ₹50,000 depreciates at 10% p.a. Find its value after 2 years.

50000 × (0.9)² = ₹40,500.

Q4. A and B invest ₹20,000 and ₹30,000 for a year and earn ₹15,000 profit. Find each partner's share.

The ratio is 2 : 3, so A gets ₹6,000 and B gets ₹9,000.

Q5. A ₹100 share is bought at ₹125 and pays a 10% dividend. Find the rate of return.

The dividend is ₹10 on an investment of ₹125, so the return is 8%.

Exam tip

In share problems, calculate the dividend on the face value but the return on the market value (investment). Mixing them up is the common error.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

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