Mathematics Part 1 (Algebra) गणित भाग 1 (बीजगणित)

Quick summary

A set is a well-defined collection of objects. This chapter teaches how to write sets, the types of sets, subsets, the universal set, Venn diagrams, union and intersection of sets, and how to count the number of elements in a set.

Key points

• A collection is a set only if we can clearly decide whether an object belongs to it. "Odd numbers less than 10" is a set; "good students" is not.
• Sets are named with capital letters (A, B, C) and elements with small letters. a ∈ A means a belongs to A; a ∉ A means it does not.
• Two ways to write a set: listing (roster) method – A = {2, 3, 5, 7}; rule (set-builder) method – A = {x | x is a prime number less than 10}.
• Standard sets: N = natural numbers, W = whole numbers, I = integers, Q = rational numbers, R = real numbers.
• Types: singleton set (one element), empty set ∅ or { } (no element), finite set, infinite set.
• Equal sets: A = B if both have exactly the same elements (order and repetition do not matter).
• Subset: A ⊆ B if every element of A is in B. Every set is a subset of itself; ∅ is a subset of every set.
• Universal set (U): a bigger set that contains all the sets under discussion. Complement A′ = elements of U not in A.
• Intersection A ∩ B = elements common to both. If A ∩ B = ∅, the sets are disjoint.
• Union A ∪ B = all elements in A or B or both.
• Properties: A ∩ B = B ∩ A; A ∪ B = B ∪ A; if A ⊆ B then A ∩ B = A and A ∪ B = B; A ∩ A′ = ∅; A ∪ A′ = U.
• Venn diagrams show sets as circles inside a rectangle (U); they make union and intersection easy to see.

Important terms

• Set (संच) – a well-defined collection of objects.
• Element – a member of a set.
• Subset (उपसंच) – a set whose every element is in another set.
• n(A) – the number of elements in set A.

Formulas

• n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
• If A and B are disjoint: n(A ∪ B) = n(A) + n(B)
• n(A′) = n(U) − n(A)

5 important questions

Q1. Write in listing method: B = {x | x is a prime number between 1 and 20}.

B = {2, 3, 5, 7, 11, 13, 17, 19}.

Q2. If U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 3, 5, 7} and B = {2, 3, 5, 8}, find A ∩ B, A ∪ B and A′.

A ∩ B = {3, 5}; A ∪ B = {1, 2, 3, 5, 7, 8}; A′ = {2, 4, 6, 8}.

Q3. In a class of 40 students, 25 like cricket, 20 like football and 10 like both. How many like at least one game? How many like neither?

n(C ∪ F) = n(C) + n(F) − n(C ∩ F) = 25 + 20 − 10 = 35. Students who like neither = 40 − 35 = 5.

Q4. In a survey, 70 people like tea, 50 like coffee and 15 like both. Using a Venn diagram, how many like only tea and only coffee?

Only tea = 70 − 15 = 55. Only coffee = 50 − 15 = 35. Total who like at least one = 55 + 15 + 35 = 105.

Q5. Write all subsets of P = {1, 2}.

∅, {1}, {2}, {1, 2}. (A set with n elements has 2ⁿ subsets; here 2² = 4.)

Exam tip

In word problems, first fill the intersection in the Venn diagram, then the "only" parts – this avoids double counting.

Quick summary

This chapter revises rational numbers, introduces irrational numbers, and shows that rational and irrational numbers together form the real numbers. It also covers surds, rationalising the denominator and the absolute value of a number.

Key points

• A rational number can be written as p/q where p, q are integers and q ≠ 0.
• The decimal form of a rational number is either terminating (3/8 = 0.375) or non-terminating recurring (1/3 = 0.333…).
• A fraction p/q in lowest form has a terminating decimal if the prime factors of q are only 2 and/or 5.
• Recurring decimals can be changed to p/q form (see Q1).
• An irrational number cannot be written as p/q; its decimal form is non-terminating and non-recurring. Examples: √2 = 1.41421…, √3 = 1.732…, π.
• Rational and irrational numbers together make the set of real numbers (R). Every real number has a unique point on the number line, and every point on the number line is a real number.
• For any two real numbers a and b, exactly one of a = b, a < b or a > b is true.
• The square of any real number is never negative, so √(−5) is not a real number.
• Surds: if a is a positive rational number and ⁿ√a is irrational, it is a surd. √a is a surd of order 2 (quadratic surd). Like surds have the same number under the root (2√3 and 5√3) and can be added or subtracted.
• Rationalising factor: multiplying a surd by another to get a rational number. For 1/(√a + √b), multiply by (√a − √b) – this is the conjugate.
• Absolute value: |x| = x if x ≥ 0, and |x| = −x if x < 0. It is the distance of x from zero, so it is never negative. |−7| = 7.

Important terms

• Real numbers (वास्तव संख्या) – all rational and irrational numbers.
• Irrational number (अपरिमेय संख्या) – number that cannot be written as p/q.
• Surd (करणी) – an irrational root of a positive rational number.
• Conjugate pair – (√a + √b) and (√a − √b); their product is a − b.

Formulas

• √a × √b = √(ab); √a ÷ √b = √(a/b)
• (√a + √b)(√a − √b) = a − b
• |x| = x when x ≥ 0; |x| = −x when x < 0

5 important questions

Q1. Write 0.777… in p/q form.

Let x = 0.777… Then 10x = 7.777… Subtract: 10x − x = 7, so 9x = 7 and x = 7/9.

Q2. Write 7.529529… in p/q form.

Let x = 7.529529… Three digits recur, so 1000x = 7529.529529… Subtract: 999x = 7522, so x = 7522/999.

Q3. Rationalise the denominator of 1/(√5 + √3).

Multiply by (√5 − √3)/(√5 − √3): (√5 − √3)/(5 − 3) = (√5 − √3)/2.

Q4. Simplify: 3√8 + 5√2 − √50.

3√8 = 3 × 2√2 = 6√2; √50 = 5√2. So 6√2 + 5√2 − 5√2 = 6√2.

Q5. Prove that √2 is irrational.

Suppose √2 = p/q in lowest form. Then 2 = p²/q², so p² = 2q² and p is even; let p = 2t. Then 4t² = 2q², so q² = 2t² and q is also even. Then p and q have common factor 2, which contradicts lowest form. So √2 is irrational.

Exam tip

To compare surds, bring them to the same order or square both sides – for example 2√3 = √12 and 3√2 = √18, so 3√2 is bigger.

Quick summary

A polynomial is an algebraic expression in which the powers of the variable are whole numbers. This chapter covers the degree and types of polynomials, their standard and coefficient forms, operations on polynomials, synthetic division, value of a polynomial and the remainder theorem.

Key points

• In a polynomial, the index (power) of each variable is a whole number. So √x or 1/x terms are not allowed. y² + 5 is a polynomial; y + 1/y is not.
• Degree of a polynomial in one variable = highest power of the variable. For more than one variable, the degree is the highest sum of powers in any term. The degree of a non-zero constant (like 10) is 0. The zero polynomial has no degree.
• Types by number of terms: monomial (1), binomial (2), trinomial (3). By degree: linear (1), quadratic (2), cubic (3).
• Standard form: terms in descending powers, e.g. x⁴ − 3x² + x + 5.
• Coefficient form: write only the coefficients, putting 0 for missing powers. x³ − 3x² − 8 → (1, −3, 0, −8).
• Index form: write all powers, with zero coefficients for missing terms: 4y⁴ + 0y³ − 5y² + 0y + 1.
• We can add, subtract, multiply and divide polynomials by working with like terms.
• Division rule: Dividend = Divisor × Quotient + Remainder; degree of the remainder is less than that of the divisor.
• Synthetic division is a short method to divide a polynomial by (x − a), using the coefficient form.
• Value of a polynomial: p(a) is found by putting x = a.
• Remainder theorem: if p(x) is divided by (x − a), the remainder is p(a).
• If p(a) = 0, then (x − a) is a factor of p(x).

Important terms

• Polynomial (बहुपदी) – expression with whole-number powers of variables.
• Degree (कोटी) – highest power in the polynomial.
• Coefficient (सहगुणक) – the number multiplying the variable.
• Synthetic division – shortcut division by (x − a).

Formulas

• Dividend = Divisor × Quotient + Remainder
• Remainder when p(x) ÷ (x − a) = p(a)
• (a + b)³ = a³ + 3a²b + 3ab² + b³; (a − b)³ = a³ − 3a²b + 3ab² − b³
• a³ + b³ = (a + b)(a² − ab + b²); a³ − b³ = (a − b)(a² + ab + b²)

5 important questions

Q1. Find the degree of 3m⁵ − 7m + 5m³ + 2 and write it in standard form.

Standard form: 3m⁵ + 5m³ − 7m + 2. Highest power is 5, so the degree is 5.

Q2. Divide 2x² + 2 by (x + 2) and verify.

2x² + 0x + 2 ÷ (x + 2): quotient 2x − 4, remainder 10. Check: (x + 2)(2x − 4) + 10 = 2x² − 8 + 10 = 2x² + 2. ✓

Q3. Using synthetic division, divide x³ − 2x² + 3x + 4 by (x − 1).

Coefficients 1, −2, 3, 4; a = 1. Bring down 1 → 1×1 = 1, −2 + 1 = −1 → −1×1 = −1, 3 − 1 = 2 → 2×1 = 2, 4 + 2 = 6. Quotient = x² − x + 2, remainder = 6.

Q4. Find the remainder when p(x) = x³ − 3x² + 4 is divided by (x − 2).

By the remainder theorem, remainder = p(2) = 8 − 12 + 4 = 0. So (x − 2) is a factor of p(x).

Q5. If (x − 2) is a factor of x³ − mx² + 10x − 20, find m.

p(2) = 0: 8 − 4m + 20 − 20 = 0, so 8 − 4m = 0 and m = 2.

Exam tip

In synthetic division, always write 0 for a missing power, and for (x + a) use −a.

Quick summary

This chapter explains ratio, the properties of ratios and of equal ratios, the theorem on equal ratios, continued proportion and the "k method" for solving problems. It also links ratios with percentages.

Key points

• A ratio a : b (or a/b) compares two quantities of the same kind in the same units; b ≠ 0.
• A ratio does not change if both terms are multiplied or divided by the same non-zero number: a/b = ka/kb.
• Ratio can be written as a percentage: 3/4 = 75%.
• To compare ratios a/b and c/d: if ad > bc then a/b > c/d.
• Proportion: if a/b = c/d, then a, b, c, d are in proportion; ad = bc (product of extremes = product of means).
• Properties of equal ratios (if a/b = c/d):
• Invertendo: b/a = d/c
• Alternando: a/c = b/d
• Componendo: (a + b)/b = (c + d)/d
• Dividendo: (a − b)/b = (c − d)/d
• Componendo-dividendo: (a + b)/(a − b) = (c + d)/(c − d)
• Theorem of equal ratios: if a/b = c/d = e/f, then each ratio = (a + c + e)/(b + d + f) = (la + mc + ne)/(lb + md + nf) (if the denominator ≠ 0).
• Continued proportion: a, b, c are in continued proportion if a/b = b/c, i.e. b² = ac. Then b is the geometric mean (mean proportional) of a and c.
• k method: if a/b = c/d = k, then a = bk and c = dk. Substitute to prove results.

Important terms

• Ratio (गुणोत्तर) – comparison of two quantities by division.
• Proportion (प्रमाण) – equality of two ratios.
• Mean proportional – b in a : b = b : c.

Formulas

• a/b = c/d ⇔ ad = bc
• b² = ac (continued proportion); b = √(ac)
• Componendo-dividendo: (a + b)/(a − b) = (c + d)/(c − d)

5 important questions

Q1. The ratio of length and breadth of a rectangle is 3 : 1 and its perimeter is 36 cm. Find the length and breadth.

Let length = 3x, breadth = x. 2(3x + x) = 36, so 8x = 36 and x = 4.5. Length = 13.5 cm, breadth = 4.5 cm.

Q2. Find the mean proportional of 4 and 25.

b² = 4 × 25 = 100, so b = 10.

Q3. If a/b = 7/3, find (5a + 3b)/(5a − 3b).

Let a = 7k, b = 3k. (35k + 9k)/(35k − 9k) = 44k/26k = 22/13.

Q4. Solve using componendo-dividendo: (x + 3)/(x − 3) = 7/5.

Apply componendo-dividendo: [(x + 3) + (x − 3)] / [(x + 3) − (x − 3)] = (7 + 5)/(7 − 5). So 2x/6 = 12/2 = 6, giving x = 18. Check: 21/15 = 7/5. ✓

Q5. If a/b = c/d, prove that (a + b)/(c + d) = b/d.

Let a/b = c/d = k, so a = bk, c = dk. LHS = (bk + b)/(dk + d) = b(k + 1)/d(k + 1) = b/d = RHS.

Exam tip

For proofs, the k method almost always works: put a = bk, c = dk and simplify both sides.

Quick summary

A linear equation in two variables has the form ax + by + c = 0. This chapter shows how to solve a pair of such equations (simultaneous equations) by the elimination and substitution methods, and how to use them to solve word problems.

Key points

• Linear equation in two variables: ax + by + c = 0, where a, b, c are real numbers and a and b are not both zero. The highest power of each variable is 1.
• One such equation has infinitely many solutions. Two equations together (simultaneous equations) usually have one common solution (x, y).
• Elimination method: make the coefficients of one variable equal (multiply the equations by suitable numbers), then add or subtract to remove that variable. Solve for the other and substitute back.
• Substitution method: from one equation express one variable in terms of the other (e.g. y = 4 − x), and put it in the second equation.
• Always check the answer by putting the values in both original equations.
• When the coefficients of x and y are interchanged in two equations (e.g. 49x + 51y = 499 and 51x + 49y = 501), add and subtract the equations to get x + y and x − y quickly.
• Steps for word problems: (1) choose variables for the unknowns, (2) form two equations from the given conditions, (3) solve, (4) answer in words with units.
• Typical word problems: ages, digits of a two-digit number (number = 10x + y), money (notes/coins), fractions, perimeter, speed and time.

Important terms

• Simultaneous equations (एकसामयिक समीकरणे) – two equations solved together.
• Solution – the pair (x, y) that satisfies both equations.
• Elimination (विलोपन) – removing one variable by adding or subtracting.
• Substitution (आदेश) – replacing one variable by its value in terms of the other.

Formulas

• General form: ax + by + c = 0
• Two-digit number with tens digit x and units digit y = 10x + y; reversed number = 10y + x

5 important questions

Q1. Solve by elimination: x + y = 4 and 2x + 3y = 3.

Multiply the first by 2: 2x + 2y = 8. Subtract from the second: (2x + 3y) − (2x + 2y) = 3 − 8, so y = −5. Then x = 4 − (−5) = 9. Solution (9, −5). Check: 18 − 15 = 3 ✓.

Q2. Solve by substitution: 3x − y = 7 and x + 4y = 11.

From the first, y = 3x − 7. Put in the second: x + 4(3x − 7) = 11, so 13x − 28 = 11, 13x = 39, x = 3. Then y = 9 − 7 = 2. Solution (3, 2).

Q3. Solve: 49x + 51y = 499 and 51x + 49y = 501.

Adding: 100x + 100y = 1000, so x + y = 10. Subtracting (second − first): 2x − 2y = 2, so x − y = 1. Adding these: 2x = 11, x = 5.5; y = 4.5.

Q4. The sum of two numbers is 50 and their difference is 14. Find the numbers.

x + y = 50 and x − y = 14. Adding: 2x = 64, x = 32; y = 18. The numbers are 32 and 18.

Q5. A two-digit number is 4 times the sum of its digits. If 18 is added, the digits are reversed. Find the number.

Let the number be 10x + y. 10x + y = 4(x + y) gives 6x = 3y, so y = 2x. Also 10x + y + 18 = 10y + x gives 9x − 9y = −18, so y − x = 2. Then 2x − x = 2, x = 2, y = 4. The number is 24.

Exam tip

Write the equations clearly as (I) and (II) and show which one you multiply – step marks are given for each step.

Quick summary

This chapter teaches how to plan income and expenses, the importance of saving, different ways of investing money (bank deposits, post office schemes, mutual funds, shares), and how income tax is calculated in India.

Key points

• Savings = Income − Expenditure. A family should plan its budget so that some money is saved every month.
• Planned expenditure includes insurance premium, provident fund contribution and money kept aside for emergencies.
• Investment options: savings account and fixed deposits in banks, post office schemes (NSC, recurring deposit, Sukanya Samruddhi for girls), Public Provident Fund (PPF), life insurance, gold, property, mutual funds and shares. Bank and post office deposits are safer; shares and mutual funds can give more gain but carry risk.
• Interest earned on deposits can be simple or compound. Compound interest: A = P(1 + R/100)ⁿ.
• Taxes: taxes paid directly by the taxpayer to the government (income tax) are direct taxes; taxes collected through goods and services (GST) are indirect taxes.
• The financial year in India is 1 April to 31 March. The year after it, in which tax is assessed, is the assessment year.
• PAN (Permanent Account Number) is a 10-character number given by the Income Tax Department and is needed for filing returns and big transactions.
• Under section 80C, savings like PPF, GPF, LIC premium, NSC, tuition fees of 2 children, housing loan principal and Sukanya Samruddhi can be deducted from income, generally up to ₹1,50,000. Donations under 80G (PM/CM Relief Fund) and health insurance premium (80D) are also deductible. Agricultural income is tax-free.
• Taxable income = total income − permitted deductions. No tax is charged up to the basic exemption limit.
• Textbook sample slabs (individuals below 60 years): up to ₹2,50,000 – nil; ₹2,50,001–5,00,000 – 5% of income above ₹2.5 lakh; ₹5,00,001–10,00,000 – ₹12,500 + 20% of income above ₹5 lakh; above ₹10,00,000 – ₹1,12,500 + 30% of income above ₹10 lakh. For senior citizens (60–80) the exemption limit is ₹3,00,000. Plus 2% education cess and 1% secondary and higher education cess on the tax.
• Tax rules change every year in the budget; the textbook slabs are for practice only.

Important terms

• Budget (अंदाजपत्रक) – plan of income and expenditure.
• Direct tax / indirect tax – paid directly by the person / collected through goods and services.
• Taxable income (करपात्र उत्पन्न) – income on which tax is charged after deductions.
• Mutual fund – a fund that pools money from many people and invests it in shares and bonds.

Formats and formulas

• Savings = Income − Expenditure
• Percentage gain = (gain ÷ amount invested) × 100
• Compound interest = P(1 + R/100)ⁿ − P
• Total income tax = income tax + 2% education cess + 1% higher education cess

5 important questions

Q1. Mr Mhatre's yearly income is ₹12,00,000 and his 80C savings are ₹1,50,000. Find his income tax (age below 60).

Taxable income = 12,00,000 − 1,50,000 = ₹10,50,000. Tax = 1,12,500 + 30% of 50,000 = 1,12,500 + 15,000 = ₹1,27,500. Education cess 2% = ₹2,550; higher education cess 1% = ₹1,275. Total tax = ₹1,31,325.

Q2. ₹3,20,000 is invested at 10% compound interest for 2 years. Find the compound interest.

A = 3,20,000 × (1.1)² = 3,20,000 × 1.21 = ₹3,87,200. CI = 3,87,200 − 3,20,000 = ₹67,200.

Q3. Mr Shah invested ₹2,40,000 in a mutual fund and got ₹3,05,000 after 2 years. Find his percentage gain.

Gain = 3,05,000 − 2,40,000 = ₹65,000. Percentage gain = (65,000 ÷ 2,40,000) × 100 ≈ 27.08%.

Q4. Distinguish between direct and indirect taxes.

A direct tax (like income tax) is paid by the person directly to the government on his income. An indirect tax (like GST) is added to the price of goods and services and is collected by the seller from the buyer.

Q5. What is meant by financial year and assessment year?

The financial year is the period from 1 April to 31 March in which income is earned. The next year, in which the tax on that income is assessed and paid, is the assessment year.

Exam tip

In income tax sums, write the steps in order: total income → deductions → taxable income → tax by slab → cess → total.

Quick summary

Statistics is the study of collecting, organising, presenting and interpreting data. This chapter covers bar diagrams (sub-divided and percentage), collection and classification of data, frequency distribution tables, cumulative frequency and the three measures of central tendency – mean, median and mode.

Key points

• Sub-divided bar diagram: each bar shows a total, divided into parts for the items in it (e.g. wheat and jowar production in a year).
• Percentage bar diagram: every bar is of height 100 and divided into parts by the percentage of each item.
• Primary data is collected by the investigator directly. Secondary data is taken from records already collected by others (census, newspapers).
• Each number in the data is a score. Range = largest score − smallest score.
• Data is arranged in classes (groups). The end values are class limits (lower and upper). Class width = upper limit − lower limit. Class mark = (lower limit + upper limit)/2.
• Inclusive method: both class limits included (11–15, 16–20). Exclusive method: upper limit not included in that class (10–20, 20–30); a score of 20 goes into 20–30.
• Frequency = number of scores in a class; shown with tally marks.
• Less than type cumulative frequency: add frequencies from the top; it tells how many scores are less than the upper limit of a class.
• More than or equal to type: start from the total and subtract downwards; it tells how many scores are equal to or more than the lower limit.
• Mean = sum of all observations ÷ number of observations. For a frequency table, mean = Σfᵢxᵢ ÷ Σfᵢ.
• Median = the middle value when data is in ascending or descending order. If n is even, it is the average of the two middle values.
• Mode = the value that occurs most often. Data can have more than one mode.

Important terms

• Data (सामग्री) – information collected in numbers.
• Frequency (वारंवारता) – how many times a score or class occurs.
• Class mark – middle value of a class.
• Mean / median / mode (मध्य / मध्यक / बहुलक) – measures of central tendency.

Formulas

5 important questions

Q1. Find the mean of 25, 30, 27, 23 and 25.

Sum = 130, n = 5. Mean = 130/5 = 26.

Q2. Find the median of 30, 25, 32, 23, 42, 36, 40, 33, 21, 43.

Ascending order: 21, 23, 25, 30, 32, 33, 36, 40, 42, 43. n = 10 (even), so median = (5th + 6th)/2 = (32 + 33)/2 = 32.5.

Q3. Find the mode of 90, 55, 67, 55, 75, 75, 40, 35, 55, 95.

55 occurs 3 times, more than any other value. Mode = 55.

Q4. The mean of 100 observations is 40. One observation, 30, was replaced by 70. Find the new mean.

Old total = 100 × 40 = 4000. New total = 4000 − 30 + 70 = 4040. New mean = 4040/100 = 40.4.

Q5. For classes 0–10, 10–20, 20–30, 30–40 with frequencies 2, 12, 20, 16, write the less than type cumulative frequencies.

2, 2 + 12 = 14, 14 + 20 = 34, 34 + 16 = 50. So 34 students scored less than 30.

Exam tip

Always arrange data in order before finding the median – it is the most common mistake.