Mathematics & Statistics

Maths · Algebra · Topic 1 of 24

Basic Algebra: Expansion, Factorisation, Indices

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Think of it like…

Making a square farm plot bigger

Your father owns a square plot of land, a metres on each side. He buys an extra strip b metres wide along the east side and another along the north side, so the plot is still a square, now of side a + b. Walk the new plot and count the land: the old square (a²), two long strips of a × b each, and one small square piece in the corner where the strips meet (b × b). So (a + b)² = a² + 2ab + b². If you write only a² + b², you have forgotten the two strips you paid for.

Factorisation is the reverse job. The 7/12 land record shows only the total area, and you must find the length and the width that multiply to give it. Expansion breaks the plot into pieces; factorisation puts the pieces back into one rectangle.

In everyday life	In the exam topic
Old square plot of side a	a²
Two new strips along the east and north sides	2ab
Small corner piece where the two strips meet	b²
Total area of the enlarged square plot	(a + b)² = a² + 2ab + b²
Finding length × width from the total area on the land record	Factorisation

The concept

Expansion means multiplying out brackets: every term of the first bracket multiplies every term of the second. The standard identities below are just expansions done once and remembered, so you do not redo them in the exam.

Factorisation is the reverse: writing an expression as a product. Always try in this order: (1) take out the common factor, (2) look for an identity such as a² − b², (3) for a quadratic, split the middle term.

Indices (powers) follow a few laws that all come from “aⁿ means a multiplied n times”. In CET, most index questions become easy once every number is written as a power of the same base (for example 4 = 2², 8 = 2³, 27 = 3³).

Key points & formulas

(a ± b)² = a² ± 2ab + b²; a² − b² = (a − b)(a + b)
(a ± b)³ = a³ ± 3a²b + 3ab² ± b³ = a³ ± b³ ± 3ab(a ± b)
a³ + b³ = (a + b)(a² − ab + b²); a³ − b³ = (a − b)(a² + ab + b²)
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
If a + b + c = 0, then a³ + b³ + c³ = 3abc
aᵐ × aⁿ = aᵐ⁺ⁿ; aᵐ ÷ aⁿ = aᵐ⁻ⁿ; (aᵐ)ⁿ = aᵐⁿ; (ab)ⁿ = aⁿbⁿ
a⁰ = 1 (a ≠ 0); a⁻ⁿ = 1/aⁿ; a^p/q = (q-th root of a)^p
x + 1/x = k ⇒ x² + 1/x² = k² − 2 and x³ + 1/x³ = k³ − 3k

Worked example

If x + 1/x = 5, find x³ + 1/x³.

Show solution

Use (a + b)³ = a³ + b³ + 3ab(a + b) with a = x, b = 1/x, so ab = 1.
(x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x), so 125 = x³ + 1/x³ + 15.
x³ + 1/x³ = 125 − 15.

Answer: 110

Exam shortcut

When the options contain algebraic expressions, put a small number such as x = 1 or x = 2 into the question and into each option. The option that gives the same number is the answer. This checks a factorisation in five seconds.

Common traps

(a + b)² is not a² + b². The middle term 2ab is the most commonly forgotten piece.
aᵐ + aⁿ cannot be combined into one power. Only multiplication and division of the same base add or subtract the powers.
a⁻ⁿ is not negative: 2⁻³ = 1/8, a positive number.

Practice · tap an option to check

Q1. If a + b = 7 and ab = 10, then a² + b² equals:

Q2. If 2^{x + 3} = 32^{x − 1}, then x equals:

Q3. One factor of x³ − 8 is:

Q4. The value of 27^2/3 + 16^3/4 − 32^1/5 is:

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Maths · Algebra · Topic 2 of 24

Quadratic Equations

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Think of it like…

A cricket ball thrown upward

Throw a cricket ball straight up. Its height rises, stops for an instant at the top, then falls back. Plot height against time and you get a curve shaped like an arch: a parabola. The equation ax² + bx + c = 0 asks one question: at what times is the height zero? Usually two moments (the throw and the catch), sometimes one (the ball just touches the ground), and sometimes none (you threw it from a rooftop terrace and it never comes down to road level in your graph). The discriminant is how you tell which case you are in without solving.

In everyday life	In the exam topic
Height of the ball at time t	f(x) = ax² + bx + c
Moments the ball is at ground level	Roots of ax² + bx + c = 0
Ball touches ground exactly once	D = 0, equal roots
Ball never reaches ground level	D < 0, no real roots
Highest point of the throw	Vertex at x = −b/2a

The concept

A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0. It has exactly two roots (real or complex), found by factorising or by the quadratic formula.

The discriminant D = b² − 4ac decides the nature of the roots before you solve anything. Many CET questions ask only for the nature of the roots, or for the sum and product, so you can skip solving entirely.

Key points & formulas

Roots: x = (−b ± √D) / 2a, where D = b² − 4ac
D > 0: two distinct real roots; D = 0: two equal real roots; D < 0: complex conjugate roots
D a perfect square (with a, b, c rational): roots are rational
Sum of roots α + β = −b/a; product αβ = c/a
Equation from roots: x² − (sum)x + (product) = 0
α² + β² = (α + β)² − 2αβ

Worked example

If α and β are roots of x² − 5x + 6 = 0, find α² + β².

Show solution

Sum α + β = −(−5)/1 = 5. Product αβ = 6/1 = 6.
α² + β² = (α + β)² − 2αβ = 25 − 12.

Answer: 13

Exam shortcut

For “find α² + β²” or “1/α + 1/β” questions, never solve for the roots. Write the expression in terms of sum and product first: 1/α + 1/β = (α + β)/αβ.

Common traps

Sign of the sum: it is −b/a, not b/a. For x² − 5x + 6 the sum is +5.
If the equation is not in standard form (e.g. 2x² = 3x − 1), move everything to one side before reading a, b, c.

Practice · tap an option to check

Q1. The roots of 2x² − 4x + 5 = 0 are:

Q2. The quadratic equation whose roots are 3 and −2 is:

Q3. For what value of k does x² + kx + 9 = 0 have equal roots (k > 0)?

Q4. If one root of x² − 7x + k = 0 is 2, then the other root and k are:

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Maths · Algebra · Topic 3 of 24

Logarithms

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Think of it like…

A savings scheme that doubles your money

Aai puts ₹1,000 in a post office scheme that doubles the money every period. After one period ₹2,000, after two ₹4,000, after three ₹8,000. Now ask the question backwards: the money became 8 times bigger; how many periods did that take? The answer, 3, is exactly log₂ 8. A logarithm does not ask “what is the answer of the multiplication”; it asks “how many times did we multiply by the base?”

This explains the laws. Wait 3 periods (money × 8) and then 2 more periods (money × 4): the money grows 8 × 4 = 32 times, but the periods simply add, 3 + 2 = 5. Multiplying growths means adding counts, which is log(xy) = log x + log y. And growth of 8 × 8 × 8 = 8³ needs 3 × 3 = 9 periods, which is log(8³) = 3 log 8.

In everyday life	In the exam topic
Money doubles every period	Base 2 (the number we keep multiplying by)
Money became 8 times bigger	The number, 8
Number of periods it took (3)	log₂ 8 = 3, because 2³ = 8
3 periods then 2 more: growth 8 × 4, periods 3 + 2	log(xy) = log x + log y
Growth of 8³ takes 3 × (periods for 8)	log(xⁿ) = n log x

The concept

log_a x = y means exactly the same as a^y = x. The logarithm is the power. It is defined only when x > 0, a > 0 and a ≠ 1. Base 10 logs are called common logs (written log x); base e logs are natural logs (ln x).

Every log law is an index law in disguise: multiplying numbers adds their powers, so the log of a product is the sum of the logs. Most CET questions are solved by (1) writing every number as a power of a small base, (2) bringing all logs to one base using change of base, then (3) solving a simple linear or quadratic equation.

After solving any log equation, put the answer back and check that every log has a positive argument. Values that make an argument zero or negative must be rejected.

Key points & formulas

log_a x = y ⟺ aʸ = x (x > 0, a > 0, a ≠ 1)
log_a 1 = 0; log_a a = 1; a^{log_a x} = x
log(xy) = log x + log y; log(x/y) = log x − log y; log(xⁿ) = n log x
Change of base: log_a x = log_b x / log_b a; so log_a b × log_b a = 1
log_aᵐ xⁿ = (n/m) log_a x
log₁₀ 2 ≈ 0.3010, log₁₀ 3 ≈ 0.4771, log₁₀ 5 = 1 − log₁₀ 2 ≈ 0.6990
Number of digits in N (N ≥ 1) = (integer part of log₁₀ N) + 1

Worked example

Given log₁₀ 2 = 0.3010, how many digits does 2⁵⁰ have?

Show solution

log₁₀(2⁵⁰) = 50 × log₁₀ 2 = 50 × 0.3010 = 15.05.
The integer part (characteristic) is 15, so 2⁵⁰ lies between 10¹⁵ and 10¹⁶.
Number of digits = 15 + 1.

Answer: 16 digits

Exam shortcut

Mixed bases? Convert all to the smallest base. For log₂ x + log₄ x + log₁₆ x = 7: log₄ x = ½ log₂ x and log₁₆ x = ¼ log₂ x, so (1 + ½ + ¼) log₂ x = 7, giving log₂ x = 4 and x = 16 in two lines.

Common traps

log(x + y) is NOT log x + log y. The sum law is for a product inside the log.
log x / log y is NOT log(x/y). log x / log y = log_y x (change of base).
Forgetting to reject roots that make an argument ≤ 0, e.g. x = −5 in log₂(x + 3).

Practice · tap an option to check

Q1. The value of log₂ 64 − log₃ 81 + log₅ 1 is:

Q2. If log₁₀ 2 = 0.3010 and log₁₀ 3 = 0.4771, then log₁₀ 72 is:

Q3. The solution of log₂(x + 3) + log₂(x − 3) = 4 is:

Q4. The value of log₃ 5 × log₂₅ 27 is:

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Maths · Algebra · Topic 4 of 24

Progressions: AP, GP, HP

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Think of it like…

Gokulashtami: the Dahi Handi pyramid and the WhatsApp invite

On Gokulashtami your chawl’s govinda pathak builds a human pyramid. The bottom tier has 20 govindas and each tier above has 4 fewer: 20, 16, 12, 8, 4. The change is always the same difference, so the tiers form an AP. To count everyone quickly, pair the bottom tier with the top tier (20 + 4 = 24), the second with the fourth (16 + 8 = 24): every pair gives the same total, so the sum is (number of tiers ÷ 2) × (first + last).

Meanwhile the mandal’s invite goes out on WhatsApp. One person sends it to 3 people, each of them to 3 more: 1, 3, 9, 27… Here each step multiplies by the same ratio, so it is a GP, and it grows much faster than the pyramid. Finally, you ride your bike to the handi venue at 30 km/h and back at 20 km/h. Your average speed for the trip is the harmonic mean 24 km/h, not 25, because you spend more time on the slow ride.

In everyday life	In the exam topic
Each tier has 4 fewer govindas than the tier below	AP with common difference d = −4
Pairing bottom tier with top tier (20 + 4 = 16 + 8)	Sₙ = (n/2)(first term + last term)
Each person forwards the invite to 3 people	GP with common ratio r = 3
Total phones reached after n rounds	Sₙ = a(rⁿ − 1)/(r − 1)
Same distance there at 30 km/h and back at 20 km/h	Average speed = HM = 2ab/(a + b) = 24

The concept

An AP (arithmetic progression) adds a fixed number d each time: a, a + d, a + 2d, … A GP (geometric progression) multiplies by a fixed ratio r each time: a, ar, ar², … An HP (harmonic progression) is a sequence whose reciprocals form an AP, such as 1, 1/2, 1/3, 1/4.

There is no direct sum formula for an HP: always convert it to the AP of reciprocals, work there, and flip back at the end.

For two positive numbers a and b, the three means satisfy AM ≥ GM ≥ HM (equal only when a = b), and GM² = AM × HM. CET loves questions that give two of these means and ask for the third.

Key points & formulas

AP: aₙ = a + (n − 1)d; Sₙ = (n/2)[2a + (n − 1)d] = (n/2)(a + l)
GP: aₙ = arⁿ⁻¹; Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1; infinite sum = a/(1 − r) when |r| < 1
HP: reciprocals are in AP; nth term of HP = 1/(nth term of that AP)
AM = (a + b)/2; GM = √(ab); HM = 2ab/(a + b)
AM ≥ GM ≥ HM for positive numbers, and GM² = AM × HM
Three terms in AP: a − d, a, a + d; three terms in GP: a/r, a, ar
aₙ = Sₙ − Sₙ₋₁ (find any term from the sum formula)

Worked example

Find the sum of all two-digit numbers divisible by 7.

Show solution

The numbers are 14, 21, …, 98: an AP with a = 14, d = 7, l = 98.
Number of terms: 98 = 14 + (n − 1)7, so n − 1 = 12 and n = 13.
Sum = (13/2)(14 + 98) = (13/2)(112) = 13 × 56.

Answer: 728

Exam shortcut

For an AP with an odd number of terms, sum = (number of terms) × (middle term). If the 4th term of a 7-term AP is 10, the sum is 7 × 10 = 70 without finding a or d.

Common traps

The infinite GP sum a/(1 − r) works only when |r| < 1. For r = 2 the sum does not exist.
There is no “sum of HP” formula. Convert to the AP of reciprocals first.
Counting terms: from 14 to 98 in steps of 7 is (98 − 14)/7 + 1 = 13 terms, not 12.

Practice · tap an option to check

Q1. The 10th term of the AP 3, 7, 11, … is:

Q2. The sum of the infinite GP 6 + 2 + 2/3 + … is:

Q3. The AM of two positive numbers is 10 and their GM is 8. Their HM is:

Q4. The sum of the first n terms of an AP is 3n² + 2n. Its 5th term is:

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Maths · Algebra · Topic 5 of 24

Binomial Theorem

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Think of it like…

Five friends ordering in the college canteen

Five friends walk into the college canteen. Each one orders either poha (call it a) or upma (call it b). The waiter writes only the totals on the slip, for example “3 poha, 2 upma”. How many different ways can that slip happen? You just choose which 2 of the 5 friends took upma: ⁵C₂ = 10 ways.

That is exactly what happens when you expand (a + b)⁵ = (a + b)(a + b)(a + b)(a + b)(a + b). Each bracket is one friend who must pick a or b. The term a³b² is the slip “3 poha, 2 upma”, and its coefficient 10 is the number of ways to choose which brackets gave b. The possible slips go from 0 upma to 5 upma, so there are 6 terms, and the total number of ways the five friends can order is 2⁵ = 32, which is the sum of all the coefficients.

In everyday life	In the exam topic
Each friend chooses poha or upma	Each bracket (a + b) gives either a or b
Order slip “3 poha, 2 upma”	The term a³b²
Ways to choose which 2 friends took upma	Coefficient ⁵C₂ = 10
Possible slips: 0, 1, 2, 3, 4 or 5 upma	n + 1 = 6 terms
All possible ways the 5 friends can order	Sum of coefficients = 2⁵ = 32

The concept

The binomial theorem gives the expansion of (a + b)ⁿ for a positive integer n without multiplying the bracket n times. The coefficients ⁿC₀, ⁿC₁, …, ⁿCₙ are the binomial coefficients (the rows of Pascal’s triangle).

Almost every CET question uses the general term T_r+1 = ⁿCᵣ aⁿ⁻ʳ bʳ. For “coefficient of xᵏ” or “term independent of x”, write the general term, collect the power of x, set it equal to what you want and solve for r.

To get the sum of coefficients of any expansion in x, simply put x = 1.

Key points & formulas

(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ (r = 0 to n); it has n + 1 terms
General term: T_r+1 = ⁿCᵣ aⁿ⁻ʳ bʳ
n even: one middle term T_{n/2 + 1}; n odd: two middle terms T_(n+1)/2 and T_(n+3)/2
ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ; sum of odd-position coefficients = sum of even-position = 2ⁿ⁻¹
ⁿCᵣ = ⁿCₙ₋ᵣ; ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ
In (a − b)ⁿ the general term has (−1)ʳ: signs alternate
Term independent of x: put the power of x in T_r+1 equal to 0

Worked example

Find the term independent of x in (x² + 1/x)⁹.

Show solution

T_r+1 = ⁹Cᵣ (x²)⁹⁻ʳ (1/x)ʳ = ⁹Cᵣ x^{18 − 2r − r} = ⁹Cᵣ x^{18 − 3r}.
Independent of x: 18 − 3r = 0, so r = 6. It is the 7th term.
T₇ = ⁹C₆ = ⁹C₃ = (9 × 8 × 7)/(3 × 2 × 1).

Answer: 84 (the 7th term)

Exam shortcut

In (xᵖ + 1/x^q)ⁿ the power of x in T_r+1 is p(n − r) − qr. Set it to the power you need and solve; no need to write the whole term first.

Common traps

The general term is T_r+1, not Tᵣ. r = 6 means the 7th term.
“Coefficient” includes the constants: in (1 + 3x)⁵ the coefficient of x² is ⁵C₂ × 3² = 90, not just ⁵C₂ = 10.
Forgetting the (−1)ʳ sign in expansions like (2x − 1/x)ⁿ.

Practice · tap an option to check

Q1. The coefficient of x² in the expansion of (1 + 3x)⁵ is:

Q2. The middle term in the expansion of (x − 2/x)⁸ is:

Q3. The sum of the coefficients in the expansion of (3x − 2)⁵ is:

Q4. The term independent of x in (2x − 1/x)⁶ is:

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Maths · Algebra · Topic 6 of 24

Permutations and Combinations

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Think of it like…

Planning the sweets at a wedding

Your cousin’s wedding caterer can make 8 sweets. The family must choose 3 for the menu card. Jalebi, gulab jamun and basundi is the same menu whichever way you list it, so order does not matter: that is a combination, ⁸C₃ = 56 menus.

Now the head cook must decide the order in which the servers bring those sweets to the pangat: first, second, third. Jalebi-then-basundi is different from basundi-then-jalebi, so order matters: that is a permutation, ⁸P₃ = 336. Notice 336 = 56 × 6, because each chosen menu can be served in 3! = 6 orders. Then the caterer also offers 4 rice dishes: picking one sweet and one rice multiplies the choices, while picking a sweet or a farsan as the single extra item adds them.

In everyday life	In the exam topic
Choosing 3 sweets out of 8 for the menu card	Combination ⁸C₃ = 56 (order does not matter)
Deciding the serving order of the 3 sweets	Permutation ⁸P₃ = 336 (order matters)
Each chosen menu can be served in 3! = 6 orders	ⁿPᵣ = ⁿCᵣ × r!
One sweet AND one rice dish	Multiply the counts (fundamental principle)
One sweet OR one farsan as the single extra item	Add the counts

The concept

Ask one question first: does order matter? If yes (ranks, seating, digits of a number, passwords, arrangements of letters), it is a permutation. If no (committees, teams, selections, handshakes), it is a combination.

The fundamental principle of counting: if one job can be done in m ways and, after that, another in n ways, both together can be done in m × n ways (AND = multiply). If you do either one job or the other, there are m + n ways (OR = add).

When some objects are identical (like the four A’s in MAHARASHTRA), divide by the factorial of each repeat count, because swapping identical letters does not create a new word.

Key points & formulas

ⁿPᵣ = n!/(n − r)!; ⁿCᵣ = n!/[r!(n − r)!]; ⁿPᵣ = r! × ⁿCᵣ
Arrangements of n objects with p alike, q alike, …: n!/(p! q! …)
Circular arrangements of n different objects: (n − 1)!; necklace or garland (can be flipped): (n − 1)!/2
ⁿCᵣ = ⁿCₙ₋ᵣ; if ⁿCₓ = ⁿCᵧ then x = y or x + y = n
Selections of at least one from n different things: 2ⁿ − 1
Diagonals of an n-sided polygon: ⁿC₂ − n; triangles from n points (no three collinear): ⁿC₃
“Together”: tie the group into one block, arrange, then multiply by the arrangements inside the block

Worked example

How many different arrangements can be made of the letters of the word MAHARASHTRA?

Show solution

There are 11 letters: A appears 4 times, H 2 times, R 2 times, and M, S, T once each.
Arrangements = 11!/(4! × 2! × 2!) = 39916800/(24 × 2 × 2) = 39916800/96.

Answer: 415800

Exam shortcut

“Never together” = (total arrangements) − (arrangements where they are together). Counting the together case is usually much easier.

Common traps

Using ⁿPᵣ for a committee or team. A committee of A, B, C is the same as C, B, A, so use ⁿCᵣ.
Forgetting that a number cannot start with 0 when forming numbers from digits that include 0.
Forgetting to divide by p! for repeated letters.

Practice · tap an option to check

Q1. A committee of 3 men and 2 women is to be formed from 6 men and 4 women. The number of ways is:

Q2. In how many ways can 6 people sit around a round table?

Q3. How many 3-digit numbers with all digits different can be formed using 0, 1, 2, 3, 4, 5?

Q4. If ⁿC₈ = ⁿC₆, then ⁿC₂ equals:

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Maths · Co-ordinate Geometry · Topic 7 of 24

Straight Lines

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Think of it like…

An ST bus climbing a ghat road

The ST bus from Pune to Mahabaleshwar climbs the ghat. A board says “Gradient 1 in 10”: for every 10 m the bus moves forward, the road rises 1 m. That ratio, rise ÷ forward distance, is exactly the slope of a line. A flat stretch on the plateau has slope 0; a steeper ghat has a bigger slope.

Put the district map on graph paper and each village becomes a point (x, y). The straight-line distance between two villages comes from Pythagoras. The tea stop exactly halfway is the midpoint; a stop two-thirds of the way along divides the route in the ratio 2 : 1, which is the section formula. Two roads with the same gradient run side by side and never meet (parallel), a road that crosses at a right angle has m₁m₂ = −1, and the junction where two roads cross is found by solving their two equations together.

In everyday life	In the exam topic
Road rises 1 m for every 10 m forward	Slope m = (y₂ − y₁)/(x₂ − x₁) = 1/10
Tea stop exactly halfway between two villages	Midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2)
Bus stop two-thirds of the way along the route	Section formula with ratio 2 : 1
Two roads with the same gradient that never meet	Parallel lines: m₁ = m₂
Junction where two roads cross	Intersection point: solve both equations together

The concept

Every point in the plane is a pair (x, y). From two points you can get the distance between them, the midpoint, any point dividing the join in a given ratio (section formula), and the slope of the line through them.

A straight line is any equation of the form ax + by + c = 0. Choose the form that matches the data given: slope and intercept → y = mx + c; a point and slope → y − y₁ = m(x − x₁); two intercepts → x/a + y/b = 1.

Two lines are parallel when their slopes are equal and perpendicular when the product of slopes is −1. Their point of intersection is found by solving the two equations simultaneously.

Key points & formulas

Distance: √[(x₂ − x₁)² + (y₂ − y₁)²]
Section formula (internal, ratio m : n): ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)); external: replace + by − in the ratio part
Centroid of a triangle: ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)
Slope m = tan θ = (y₂ − y₁)/(x₂ − x₁); for ax + by + c = 0, m = −a/b
Forms: y = mx + c; y − y₁ = m(x − x₁); x/a + y/b = 1
Parallel: m₁ = m₂; perpendicular: m₁m₂ = −1; angle: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
Distance of (x₁, y₁) from ax + by + c = 0: |ax₁ + by₁ + c| / √(a² + b²); between parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0: |c₁ − c₂| / √(a² + b²)
Area of triangle: ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|; area 0 means the points are collinear

Worked example

Find the equation of the line through (2, 3) that is perpendicular to 3x − 4y + 5 = 0.

Show solution

Slope of the given line = −a/b = −3/(−4) = 3/4.
Perpendicular slope = −1 ÷ (3/4) = −4/3.
Point-slope form: y − 3 = (−4/3)(x − 2), so 3y − 9 = −4x + 8.
Rearranging: 4x + 3y − 17 = 0. Check: 4(2) + 3(3) − 17 = 0. ✓

Answer: 4x + 3y − 17 = 0

Exam shortcut

Line perpendicular to ax + by + c = 0: write bx − ay + k = 0. Line parallel to it: ax + by + k = 0. Then find k by putting in the given point. Above: perpendicular to 3x − 4y + 5 = 0 is 4x + 3y + k = 0; (2, 3) gives k = −17.

Common traps

A vertical line (x = constant) has an undefined slope, not slope 0. A horizontal line has slope 0.
Section formula order: for ratio m : n from A towards B, m multiplies B’s coordinates. Swapping gives the wrong point.
For the distance between parallel lines, first make the x and y coefficients identical (e.g. 6x + 8y = … must be halved to match 3x + 4y).

Practice · tap an option to check

Q1. The distance between the points (−1, 2) and (5, −6) is:

Q2. The point dividing the join of A(1, −2) and B(7, 4) internally in the ratio 2 : 1 is:

Q3. The perpendicular distance of the point (3, −2) from the line 3x + 4y − 6 = 0 is:

Q4. The lines 2x + ky − 3 = 0 and 3x − 6y + 1 = 0 are perpendicular when k equals:

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Maths · Co-ordinate Geometry · Topic 8 of 24

Circles

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Think of it like…

A goat tied to a peg

Ajoba ties the goat to a peg in the angan with a 3 m rope. Pulling the rope tight, the goat can just reach every point exactly 3 m from the peg, so the edge of the grass it eats is a circle. If the peg is at the point (h, k) on our graph, any point (x, y) on that edge satisfies (x − h)² + (y − k)² = 3², which is just the distance formula squared.

Is a particular tuft of grass reachable? Put its position into the left side minus 9. If the result is negative the tuft is inside (the goat eats it), zero means it is exactly on the edge, positive means out of reach. A straight compound wall that the stretched rope just brushes is a tangent: the distance from the peg to that wall is exactly the rope length.

In everyday life	In the exam topic
Position of the peg	Centre (h, k), or (−g, −f) in the general form
Length of the rope	Radius r = √(g² + f² − c)
Edge of the grazed patch	(x − h)² + (y − k)² = r²
Is a tuft of grass reachable? Put its position in	S₁ < 0 inside, S₁ = 0 on, S₁ > 0 outside
Straight wall that the tight rope just brushes	Tangent: distance from centre to line = r

The concept

A circle is the set of points at a fixed distance r (radius) from a fixed point (centre). Writing the distance formula for this gives the centre-radius form (x − h)² + (y − k)² = r².

Expanding gives the general form x² + y² + 2gx + 2fy + c = 0, with centre (−g, −f) and radius √(g² + f² − c). An equation is a circle only when the x² and y² coefficients are equal and there is no xy term.

For a line and a circle, compare the perpendicular distance d from the centre to the line with r: d < r cuts in two points, d = r touches (tangent), d > r misses.

Key points & formulas

Centre-radius form: (x − h)² + (y − k)² = r²
General form x² + y² + 2gx + 2fy + c = 0: centre (−g, −f), radius √(g² + f² − c)
Diameter form with ends (x₁, y₁), (x₂, y₂): (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0
Tangent at (x₁, y₁) on x² + y² = a²: xx₁ + yy₁ = a²
y = mx + c touches x² + y² = a² if and only if c² = a²(1 + m²)
Length of tangent from (x₁, y₁): √S₁, where S₁ = x₁² + y₁² + 2gx₁ + 2fy₁ + c
Two circles touch externally if C₁C₂ = r₁ + r₂, internally if C₁C₂ = |r₁ − r₂|

Worked example

Find the centre and radius of x² + y² − 6x + 4y − 12 = 0.

Show solution

Compare with x² + y² + 2gx + 2fy + c = 0: 2g = −6, 2f = 4, c = −12.
So g = −3, f = 2 and the centre is (−g, −f) = (3, −2).
Radius = √(g² + f² − c) = √(9 + 4 + 12) = √25.

Answer: Centre (3, −2), radius 5

Exam shortcut

Centre = (−½ × coefficient of x, −½ × coefficient of y), but only after the x² and y² coefficients are made 1. For 2x² + 2y² − 8x + … divide by 2 first.

Common traps

Not dividing by the x² coefficient first. For 2x² + 2y² − 8x + 12y − 6 = 0 the centre is (2, −3), not (4, −6).
Sign of the centre: + 2gx means the centre’s x-coordinate is −g.
If g² + f² − c is negative, there is no real circle at all.

Practice · tap an option to check

Q1. The radius of the circle 2x² + 2y² − 8x + 12y − 6 = 0 is:

Q2. The equation of the circle with (1, 2) and (5, 6) as ends of a diameter is:

Q3. The length of the tangent from (5, 3) to the circle x² + y² − 2x + 4y − 4 = 0 is:

Q4. The line y = x + c touches the circle x² + y² = 8 when c equals:

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Maths · Co-ordinate Geometry · Topic 9 of 24

Conics: Parabola, Ellipse, Hyperbola

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Think of it like…

A torch beam on the wall during a power cut

Evening load-shedding: you switch on the battery torch and point it at the wall. The beam is a cone of light and the wall slices through it. Point the torch straight at the wall and the patch is a circle. Tilt it a little and the patch stretches into a closed oval, an ellipse. Keep tilting until the top edge of the beam runs exactly parallel to the wall: now the patch never closes at the far end. That open U-shape is a parabola. Tilt even more and the patch opens out even wider: a hyperbola.

That is literally why they are called conic sections: each is a slice of a cone. The single number that tells you how far you tilted is the eccentricity e. Every conic is the path of a point whose distance from a fixed point (the focus) is e times its distance from a fixed line (the directrix).

In everyday life	In the exam topic
Torch beam sliced by the wall	A cone cut by a plane; the light patch is the conic
Torch pointing straight at the wall	Circle, e = 0
Tilted a little: closed oval patch	Ellipse, 0 < e < 1
Beam edge parallel to the wall: patch never closes	Parabola, e = 1
Tilted even more: patch opens out wider	Hyperbola, e > 1

The concept

A conic is the path of a point P that moves so that PF = e × PM, where F is a fixed point (focus), PM is the distance to a fixed line (directrix) and e is the eccentricity. e = 1 gives a parabola, e < 1 an ellipse, e > 1 a hyperbola.

In CET, almost every question gives a standard equation and asks for the focus, directrix, eccentricity, latus rectum or axes. So first bring the equation to standard form (divide to get 1 on the right side), then read off a and b.

Two facts worth remembering: for an ellipse, the sum of distances of any point from the two foci is constant (= 2a); for a hyperbola, the difference is constant (= 2a).

Key points & formulas

Parabola y² = 4ax: vertex (0, 0), focus (a, 0), directrix x = −a, latus rectum 4a
Parabola x² = 4ay: focus (0, a), directrix y = −a
Ellipse x²/a² + y²/b² = 1 (a > b): b² = a²(1 − e²), foci (±ae, 0), directrices x = ±a/e
Ellipse: latus rectum 2b²/a; PF₁ + PF₂ = 2a (major axis)
Hyperbola x²/a² − y²/b² = 1: b² = a²(e² − 1), foci (±ae, 0), latus rectum 2b²/a, |PF₁ − PF₂| = 2a
Rectangular hyperbola (a = b, e.g. x² − y² = a² or xy = c²): e = √2
Focus distance from centre: ellipse ae = √(a² − b²); hyperbola ae = √(a² + b²)

Worked example

Find the eccentricity, foci and latus rectum of the ellipse 9x² + 25y² = 225.

Show solution

Divide by 225: x²/25 + y²/9 = 1, so a = 5, b = 3.
e² = 1 − b²/a² = 1 − 9/25 = 16/25, so e = 4/5.
Foci (±ae, 0) = (±4, 0). Latus rectum = 2b²/a = 18/5.

Answer: e = 4/5, foci (±4, 0), latus rectum 18/5

Exam shortcut

Ellipse: minus (ae = √(a² − b²)). Hyperbola: plus (ae = √(a² + b²)). For x²/25 + y²/9 = 1, ae = √16 = 4; for x²/16 − y²/9 = 1, ae = √25 = 5. Then e = (ae)/a.

Common traps

In y² = 12x, 4a = 12 so a = 3: the focus is (3, 0), not (12, 0).
If the larger denominator is under y² (b > a), the major axis is vertical and the foci are on the y-axis.
Mixing up the ellipse and hyperbola relations: ellipse uses 1 − e², hyperbola uses e² − 1.

Practice · tap an option to check

Q1. The focus of the parabola y² = 12x is:

Q2. The eccentricity of the hyperbola x²/16 − y²/9 = 1 is:

Q3. The length of the latus rectum of the ellipse x²/36 + y²/16 = 1 is:

Q4. For any point on the ellipse 4x² + 9y² = 36, the sum of its distances from the two foci is:

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Maths · Co-ordinate Geometry · Topic 10 of 24

Pair of Straight Lines and Transformations

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Think of it like…

Drawing a Diwali rangoli with a stencil

Before Diwali you draw rangoli on the floor. First you draw two chalk guide-lines crossing at the centre. They are two separate lines, but one single equation ax² + 2hxy + by² = 0 can hold both of them, because it is just the two line equations multiplied together. If the two guide-lines cross at a right angle, the coefficients satisfy a + b = 0.

Then you use a plastic stencil. Sliding it to the next tile without turning it is a translation: every point moves by the same amount. Turning it about its centre is a rotation: shapes and sizes stay the same, only the direction changes. Picking the bigger stencil from the same packet is scaling: the shape stays the same but every length is multiplied by the same factor, so the area is multiplied by the square of that factor.

In everyday life	In the exam topic
Two chalk lines crossing at the centre	ax² + 2hxy + by² = 0 (two lines through the origin)
Chalk lines crossing at a right angle	a + b = 0 (coefficient of x² + coefficient of y²)
Sliding the stencil to the next tile without turning	Translation: (x, y) → (x + p, y + q)
Turning the stencil about its centre by θ	Rotation: (x, y) → (x cos θ − y sin θ, x sin θ + y cos θ)
Using the bigger stencil from the same packet	Scaling: (x, y) → (kx, ky), area × k²

The concept

The homogeneous equation ax² + 2hxy + by² = 0 represents two straight lines through the origin. If they are y = m₁x and y = m₂x, dividing by x² gives a quadratic in y/x whose roots are m₁ and m₂; so sum and product of slopes come straight from the coefficients.

The general second-degree equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines only when the determinant condition abc + 2fgh − af² − bg² − ch² = 0 holds.

Transformations move points. Translation shifts every point by the same amount; rotation turns points about the origin by an angle θ; scaling stretches by factors along the axes. Be careful to note whether a question moves the point or moves the axes: moving the axes one way is the same as moving the point the opposite way.

Key points & formulas

ax² + 2hxy + by² = 0: two lines through the origin; real and distinct if h² > ab, coincident if h² = ab
Slopes m₁, m₂: m₁ + m₂ = −2h/b, m₁m₂ = a/b
Angle between them: tan θ = 2√(h² − ab) / |a + b|; perpendicular if a + b = 0
ax² + 2hxy + by² + 2gx + 2fy + c = 0 is a pair of lines if abc + 2fgh − af² − bg² − ch² = 0
Shift origin to (h, k) (axes moved, not turned): new coordinates X = x − h, Y = y − k
Rotate a point anticlockwise by θ about the origin: (x cos θ − y sin θ, x sin θ + y cos θ); by 90°: (x, y) → (−y, x)
Rotate the axes by θ: X = x cos θ + y sin θ, Y = −x sin θ + y cos θ
Scaling by sₓ and sᵧ: (x, y) → (sₓx, sᵧy); areas are multiplied by sₓsᵧ

Worked example

Find the angle between the lines represented by x² − 4xy + y² = 0.

Show solution

Compare with ax² + 2hxy + by² = 0: a = 1, 2h = −4 so h = −2, b = 1.
tan θ = 2√(h² − ab) / |a + b| = 2√(4 − 1) / 2 = √3.
tan θ = √3 gives θ = 60°.

Answer: 60°

Exam shortcut

To get the two separate lines, factorise like a quadratic: x² − 5xy + 6y² = (x − 2y)(x − 3y), so the lines are x = 2y and x = 3y. If it factorises easily, slopes and angle follow at once.

Common traps

h is HALF the coefficient of xy. In 3x² + 2xy − y² = 0, h = 1, not 2.
Shifting the origin to (2, −1) makes the new coordinates (x − 2, y + 1), not (x + 2, y − 1).
Rotating the axes and rotating the point by the same θ give opposite results. Read which one is asked.

Practice · tap an option to check

Q1. The equation kx² + 5xy − 3y² = 0 represents a pair of perpendicular lines if k equals:

Q2. The point (2, 3) is rotated through 90° anticlockwise about the origin. Its new position is:

Q3. The origin is shifted to (2, −1) without rotating the axes. The new coordinates of the point (5, 3) are:

Q4. The sum of the slopes of the lines given by 3x² + 2xy − y² = 0 is:

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Maths · Differential Equations · Topic 11 of 24

Differential Equations

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Think of it like…

A glass of cutting chai cooling at the tapri

The tapri-wala pours cutting chai at 80°C on a 30°C afternoon. At first it cools fast; as it gets close to room temperature the cooling slows down. Say it in one sentence: the rate at which the chai cools is proportional to how much hotter it is than the room. Written in maths, that sentence is dT/dt = −k(T − 30), a differential equation. It does not tell you the temperature directly; it tells you how the temperature is changing.

Solving it gives T = 30 + Ce^−kt. This is a whole family of curves, one for every possible starting temperature, because the constant C is not yet fixed: that is the general solution. The moment you know the chai was poured at 80°C (T = 80 at t = 0), C = 50 is fixed and you get one particular curve: the particular solution.

In everyday life	In the exam topic
Temperature of the chai at time t	Dependent variable T (or y)
How fast the chai is cooling right now	Derivative dT/dt
Hotter chai cools faster	The equation dT/dt = −k(T − 30)
All cooling curves for chai poured at any temperature	General solution T = 30 + Ce^−kt (arbitrary constant C)
Knowing it was poured at 80°C	Initial condition fixes C = 50: particular solution

The concept

A differential equation (DE) connects a function with its derivatives. Its order is the order of the highest derivative present; its degree is the power of that highest derivative once the equation is free of radicals and fractions in the derivatives. The general solution of an order-n equation contains exactly n arbitrary constants.

First-order methods you must know: variables separable (bring all y to one side, all x to the other, integrate); homogeneous (dy/dx is a function of y/x, so put y = vx); and linear dy/dx + Py = Q, solved with the integrating factor IF = e^{∫P dx}.

For linear equations with constant coefficients such as y″ + ay′ + by = 0, try y = e^mx. This turns the DE into the auxiliary equation m² + am + b = 0, and the type of roots decides the form of the solution.

Key points & formulas

Order = highest derivative; degree = its power (after clearing radicals/fractions); number of arbitrary constants = order
Variables separable: f(y) dy = g(x) dx, integrate both sides
Homogeneous dy/dx = F(y/x): put y = vx, dy/dx = v + x dv/dx
Linear dy/dx + Py = Q: IF = e^{∫P dx}; solution y × IF = ∫(Q × IF) dx + C
y″ + ay′ + by = 0: solve m² + am + b = 0
Distinct real roots m₁, m₂: y = c₁e^m₁x + c₂e^m₂x; equal roots m: y = (c₁ + c₂x)e^mx; roots α ± iβ: y = e^αx(c₁ cos βx + c₂ sin βx)
Growth or decay dy/dt = ky gives y = y₀e^kt

Worked example

Solve dy/dx + y/x = x² (x > 0).

Show solution

This is linear with P = 1/x and Q = x².
IF = e^{∫(1/x) dx} = e^{ln x} = x.
y × x = ∫x² × x dx = ∫x³ dx = x⁴/4 + C.
Divide by x: y = x³/4 + C/x.

Answer: y = x³/4 + C/x

Exam shortcut

To form the DE of a family of curves with n arbitrary constants, differentiate n times and eliminate the constants. The order equals the number of constants: y = A cos x + B sin x has 2 constants, so its DE is second order: y″ + y = 0.

Common traps

Degree is not defined if a derivative sits inside sin, e^… or log, e.g. sin(dy/dx) + y = 0.
Read P only after making the coefficient of dy/dx equal to 1. For x dy/dx + 2y = x³, divide by x first: P = 2/x.
Equal roots of the auxiliary equation need (c₁ + c₂x)e^mx, not c₁e^mx + c₂e^mx.

Practice · tap an option to check

Q1. The order and degree of (d²y/dx²)³ + (dy/dx)⁴ + y = 0 are:

Q2. The integrating factor of dy/dx + (2/x)y = x is:

Q3. The general solution of y″ − 5y′ + 6y = 0 is:

Q4. The general solution of the homogeneous equation dy/dx = (x + y)/x is:

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Maths · Trigonometry · Topic 12 of 24

Trigonometric Identities and Equations

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Think of it like…

Riding the giant wheel at the village jatra

At the village jatra you sit in the giant wheel. Take its radius as 1 unit. When the wheel has turned through an angle θ from the horizontal, your seat’s height above the axle is sin θ and its sideways distance from the axle is cos θ. However the wheel turns, your seat is always exactly 1 unit from the axle, so by Pythagoras (cos θ)² + (sin θ)² = 1. That is the most important identity, and most other identities are built from it.

The wheel also explains trigonometric equations. Ask “when is my seat at height ½?” You pass that height twice in each turn: once going up (θ = 30°) and once coming down (θ = 180° − 30° = 150°). And after a full turn of 360° everything repeats. So a trig equation does not have one answer; it has a family of answers, written with an integer n. That is the general solution.

In everyday life	In the exam topic
Angle the wheel has turned	θ
Seat’s height above the axle (radius 1)	sin θ
Seat’s sideways distance from the axle	cos θ
Seat always stays 1 unit from the axle	sin²θ + cos²θ = 1
Same height twice per turn, and again every turn	sin θ = sin α ⇒ θ = nπ + (−1)ⁿα

The concept

A trigonometric identity is true for every angle; an equation is true only for some angles. Identities are tools to simplify or to prove; equations are to be solved.

Most simplification uses three families: the Pythagorean identities (sin²θ + cos²θ = 1 and its cousins), compound-angle formulas (sin(A ± B), cos(A ± B)), and multiple-angle formulas (sin 2A, cos 2A, sin 3A …). The cos 2A formula has three forms; choose the one that removes the term you do not want.

Because sin, cos and tan repeat (period 2π for sin and cos, π for tan), every trig equation has infinitely many solutions. Reduce the equation to one of sin θ = sin α, cos θ = cos α or tan θ = tan α, then write the general solution.

Key points & formulas

sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ
sin(A ± B) = sin A cos B ± cos A sin B; cos(A ± B) = cos A cos B ∓ sin A sin B; tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
sin 2A = 2 sin A cos A; cos 2A = cos²A − sin²A = 1 − 2 sin²A = 2 cos²A − 1; tan 2A = 2 tan A/(1 − tan²A)
sin 3A = 3 sin A − 4 sin³A; cos 3A = 4 cos³A − 3 cos A
sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2); cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2)
General solutions: sin θ = sin α ⇒ θ = nπ + (−1)ⁿα; cos θ = cos α ⇒ θ = 2nπ ± α; tan θ = tan α ⇒ θ = nπ + α
sin θ = 0 ⇒ θ = nπ; cos θ = 0 ⇒ θ = (2n + 1)π/2; n is any integer

Worked example

Find the general solution of 2 cos²θ + 3 sin θ = 0.

Show solution

Replace cos²θ by 1 − sin²θ: 2 − 2 sin²θ + 3 sin θ = 0, i.e. 2 sin²θ − 3 sin θ − 2 = 0.
Factorise: (2 sin θ + 1)(sin θ − 2) = 0. sin θ = 2 is impossible, so sin θ = −1/2.
sin θ = −1/2 = sin(−π/6), so θ = nπ + (−1)ⁿ(−π/6).

Answer: θ = nπ + (−1)ⁿ(−π/6), n ∈ ℤ

Exam shortcut

For “simplify” questions with options, put a convenient angle such as θ = 30° or 45° into the question and into each option. For (1 − cos 2θ)/sin 2θ at θ = 30°: (1 − ½)/(√3/2) = 1/√3, and only tan θ gives 1/√3 among the options. (Avoid 45° here: tan and cot are both 1 there, so it cannot separate them.)

Common traps

Never cancel sin θ (or cos θ) from both sides of an equation; you lose the solutions where it is 0. Factor it out instead.
Reject values like sin θ = 2: sin and cos always lie between −1 and 1.
The general solution of cos θ = cos α is 2nπ ± α, not nπ + (−1)ⁿα. Do not mix the sin and cos forms.

Practice · tap an option to check

Q1. The value of sin²30° + cos²60° + tan²45° is:

Q2. (1 − cos 2θ)/sin 2θ is equal to:

Q3. The value of cos 15° − sin 15° is:

Q4. The general solution of tan 2θ = √3 is:

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Maths · Trigonometry · Topic 13 of 24

Solution of Triangles, Heights and Distances

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Think of it like…

Finding the height of the school flagpole on Republic Day

On 26 January the PT sir asks: how tall is our school flagpole? Nobody is allowed to climb it. So you stand 20 m from its base and look at the flag on top. The angle your line of sight makes with the ground, the angle of elevation, is 30°. Your eye, the base of the pole and the top of the pole form a right-angled triangle: the 20 m is the side next to the angle, and the height is the side opposite it. So height = 20 × tan 30°.

If you walk closer, the angle grows; using two such readings you can find the height even without knowing how far the pole is. And when the three points are not at a right angle, like three neem trees in the school ground, you use the sine rule and cosine rule, which work in any triangle.

In everyday life	In the exam topic
Your eye, the base and the top of the pole	The three vertices of a right-angled triangle
Distance you stand from the pole	Adjacent side (horizontal distance d)
How far you tilt your head to see the top	Angle of elevation θ
Height of the flagpole	Opposite side h = d tan θ
Three neem trees in the ground, not at right angles	Sine rule and cosine rule for any triangle

The concept

In any triangle ABC, side a is opposite angle A, b opposite B and c opposite C. The sine rule connects each side with the angle opposite it; the cosine rule connects three sides with one angle. Use the cosine rule when you know all three sides (or two sides and the angle between them); use the sine rule when you know an angle and the side opposite it.

For heights and distances, always draw the figure first. The angle of elevation is measured upwards from the horizontal; the angle of depression is measured downwards from the horizontal. Because the two horizontals are parallel, the angle of depression from A to B equals the angle of elevation from B to A.

Learn tan 30° = 1/√3, tan 45° = 1, tan 60° = √3 by heart; almost every height question uses only these.

Key points & formulas

Sine rule: a/sin A = b/sin B = c/sin C = 2R (R = circumradius)
Cosine rule: cos A = (b² + c² − a²)/(2bc), i.e. a² = b² + c² − 2bc cos A
Projection rule: a = b cos C + c cos B
Area Δ = ½ bc sin A = √[s(s − a)(s − b)(s − c)], s = (a + b + c)/2
R = abc/(4Δ); inradius r = Δ/s
Height h = d tan θ; angle of depression from the top = angle of elevation from the ground point
Elevations α and β (α > β) from two points d apart in line with the base: h = d tan α tan β/(tan α − tan β)

Worked example

The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 20 m towards the tower, it becomes 60°. Find the height of the tower.

Show solution

Let the height be h and the nearer distance x. Then h = x tan 60° = x√3, and h = (x + 20) tan 30° = (x + 20)/√3.
Equate: x√3 = (x + 20)/√3, so 3x = x + 20 and x = 10 m.
h = 10√3 ≈ 17.32 m.

Answer: 10√3 m (about 17.32 m)

Exam shortcut

For the 30° → 60° walk-towards case, the height is always (√3/2) × distance walked. Here (√3/2) × 20 = 10√3. Also, in that case the distance walked equals the distance from the second point to the top of the tower.

Common traps

The d in h = d tan θ is the horizontal distance along the ground, not the slant line of sight.
The angle of depression is from the horizontal, not from the vertical wall of the building.
In the sine rule, pair each side with the angle opposite it, not the angle next to it.

Practice · tap an option to check

Q1. In ΔABC, a = 3, b = 5 and c = 7. Angle C is:

Q2. A 10 m ladder leans against a wall, making an angle of 60° with the ground. How high up the wall does it reach?

Q3. From the top of a 30 m high building, the angle of depression of a car on the road is 30°. How far is the car from the foot of the building?

Q4. The area of a triangle with sides 13 cm, 14 cm and 15 cm is:

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Maths · Trigonometry · Topic 14 of 24

Inverse Trigonometric Functions

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Think of it like…

Guessing the time from the temperature on a Nagpur summer day

On a May day in Nagpur the temperature climbs from the cool dawn to a burning afternoon peak and then falls again by night. If your friend says “it was 38°C”, you cannot tell the time: 38°C happened once in the morning while it was rising and once in the evening while it was falling, and again tomorrow, and the day after.

To make “time from temperature” a proper answer, you must agree on a rule: only look at the rising part of the day. Then every temperature matches exactly one time. That is exactly what an inverse trig function does. sin θ takes each value again and again, so sin⁻¹ is allowed to answer only from one stretch where sin keeps rising, from −π/2 to π/2. That chosen answer is called the principal value.

In everyday life	In the exam topic
Time of the day	Angle θ
Temperature at that time	sin θ
Same temperature in morning and evening, every day	sin is not one-one: many θ give the same value
Agreeing to read only the rising morning hours	Principal range of sin⁻¹: [−π/2, π/2]
“At what morning hour was it 38°C?”	sin⁻¹x: the one angle in that range with sine x

The concept

sin, cos and tan are not one-one, so they have no inverse over all angles. We restrict each to a principal range where it is one-one, and define the inverse only there. sin⁻¹x means “the angle in [−π/2, π/2] whose sine is x”; it is NOT 1/sin x.

Because of these ranges, sin⁻¹(sin θ) equals θ only when θ is already in [−π/2, π/2]. Otherwise first rewrite sin θ as the sine of an angle inside the range (using sin(π − θ) = sin θ, etc.) and then remove the sin⁻¹.

Most CET questions use the complementary pairs (sin⁻¹x + cos⁻¹x = π/2), the negative-argument rules, and the tan⁻¹ addition formula.

Key points & formulas

sin⁻¹x: domain [−1, 1], range [−π/2, π/2]
cos⁻¹x: domain [−1, 1], range [0, π]; tan⁻¹x: domain all real x, range (−π/2, π/2)
sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2; sec⁻¹x + cosec⁻¹x = π/2
sin⁻¹(−x) = −sin⁻¹x; tan⁻¹(−x) = −tan⁻¹x; but cos⁻¹(−x) = π − cos⁻¹x
tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] when xy < 1
2 tan⁻¹x = tan⁻¹[2x/(1 − x²)] for |x| < 1
sin⁻¹(sin θ) = θ only for θ in [−π/2, π/2]; cos⁻¹(cos θ) = θ only for θ in [0, π]

Worked example

Find the value of sin⁻¹(sin 5π/6).

Show solution

5π/6 is not in [−π/2, π/2], so the answer is not 5π/6.
sin(5π/6) = sin(π − 5π/6) = sin(π/6).
π/6 is in the principal range, so sin⁻¹(sin π/6) = π/6.

Answer: π/6

Exam shortcut

Remember three famous results: tan⁻¹(1/2) + tan⁻¹(1/3) = π/4; tan⁻¹1 + tan⁻¹2 + tan⁻¹3 = π; and if sin⁻¹x + sin⁻¹y = θ then cos⁻¹x + cos⁻¹y = π − θ (because each pair adds to π/2).

Common traps

sin⁻¹x is not (sin x)⁻¹ = cosec x.
Writing sin⁻¹(sin 2π/3) = 2π/3. It is π/3, because 2π/3 is outside [−π/2, π/2].
cos⁻¹(−1/2) is 2π/3, not −π/3: the range of cos⁻¹ is [0, π], which contains no negative angles.

Practice · tap an option to check

Q1. The principal value of cos⁻¹(−1/2) is:

Q2. tan⁻¹(1/2) + tan⁻¹(1/3) equals:

Q3. The value of sin⁻¹(sin 2π/3) is:

Q4. If sin⁻¹x + sin⁻¹y = 2π/3, then cos⁻¹x + cos⁻¹y equals:

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Maths · Probability and Statistics · Topic 15 of 24

Probability: Basics, Dependent and Independent Events

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Think of it like…

The lucky draw at the Ganpati mandal

On the last day of Ganeshotsav, the mandal holds a lucky draw. The box has 100 chits and 5 of them say “prize”. Your chance on one draw is 5 out of 100. That is all probability is: favourable outcomes ÷ total outcomes, provided every chit is equally likely.

Now two prizes are drawn. If the volunteer keeps the first chit aside, the box changes: after a prize chit is out, only 4 prize chits remain in 99. The second draw depends on the first, so you multiply by the changed chance: P(both) = 5/100 × 4/99. But if the chit is put back and the box shaken (or the second draw is from a different mandal’s box), the first draw tells you nothing about the second: the events are independent, and you simply multiply 5/100 × 5/100.

In everyday life	In the exam topic
All the chits in the box	Sample space S
Prize chits	Event A; P(A) = n(A)/n(S)
Chit kept aside before the next draw	Dependent events: P(A ∩ B) = P(A) × P(B \| A)
Chit put back and box shaken	Independent events: P(A ∩ B) = P(A) × P(B)
Chance of winning at least one of the two prizes	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

The concept

For equally likely outcomes, P(A) = (number of outcomes favourable to A) ÷ (total number of outcomes). Probabilities lie between 0 and 1, and P(not A) = 1 − P(A).

Conditional probability P(B | A) is the probability of B once you know A has happened: P(B | A) = P(A ∩ B)/P(A). Events A and B are independent when knowing A does not change the chance of B, i.e. P(A ∩ B) = P(A) P(B). Without replacement, draws are dependent; with replacement, they are independent.

Bayes’ theorem turns the question around: given that a result happened, which cause was it likely to come from? Find each cause’s share of the result, then divide by the total.

Key points & formulas

P(A) = n(A)/n(S); 0 ≤ P(A) ≤ 1; P(A′) = 1 − P(A)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B); mutually exclusive means P(A ∩ B) = 0
P(B | A) = P(A ∩ B)/P(A)
Independent: P(A ∩ B) = P(A) P(B); then A′ and B′ are independent too
P(at least one) = 1 − P(none)
Bayes: P(Eᵢ | A) = P(Eᵢ) P(A | Eᵢ) / Σ P(Eⱼ) P(A | Eⱼ)
Two dice: 36 outcomes; a sum of 7 has 6 ways (the most likely sum)
Odds in favour a : b means P = a/(a + b)

Worked example

A bag has 5 red and 3 black balls. Two balls are drawn one after the other without replacement. Find the probability that both are red.

Show solution

First ball red: 5/8.
Given the first was red, 4 red remain in 7 balls: P(second red | first red) = 4/7.
P(both red) = 5/8 × 4/7 = 20/56.

Answer: 5/14

Exam shortcut

Whenever you see “at least one”, use 1 − P(none). For two shooters who hit with probabilities 1/3 and 1/4: P(target hit) = 1 − (2/3)(3/4) = 1/2, no case-by-case adding.

Common traps

Mutually exclusive is not the same as independent. If two events with non-zero probabilities cannot happen together, they are dependent.
Without replacement, both the favourable count and the total drop by one for the second draw.
For independent events, P(A ∪ B) is still P(A) + P(B) − P(A)P(B), not P(A) + P(B).

Practice · tap an option to check

Q1. Box I has 3 red and 2 white balls; Box II has 2 red and 3 white balls. A box is chosen at random and a ball drawn from it is red. The probability that it came from Box I is:

Q2. A and B are independent events with P(A) = 0.4 and P(B) = 0.5. Then P(A ∪ B) is:

Q3. The probability that A hits a target is 1/3 and that B hits it is 1/4. If both fire once independently, the probability that the target is hit is:

Q4. Two cards are drawn without replacement from a pack of 52. The probability that both are kings is:

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Maths · Probability and Statistics · Topic 16 of 24

Averages, Dispersion, Skewness and Kurtosis

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Think of it like…

Evening milk collection at the village dairy

Every evening farmers bring milk to the village co-operative dairy. One day five farmers bring 4, 5, 5, 6 and 30 litres. The secretary adds it up, 50 litres, and divides by 5: the mean is 10 L. But nobody brought 10 L! Line the farmers up by litres and the one in the middle brought 5 L: that is the median. The most common amount, also 5 L, is the mode.

The one big farmer with 30 L pulls the mean far above the typical farmer. When a long tail stretches to the right like this, the mean is pulled above the median and the mode, and the data is positively skewed. Another dairy may also average 10 L with every farmer bringing 9 to 11 L. Same average, very different spread: the standard deviation measures that spread. And kurtosis tells whether most farmers bunch tightly around the middle (a sharp peak) or are spread flat.

In everyday life	In the exam topic
Litres brought by each farmer	Observations x₁, x₂, …, xₙ
Total milk ÷ number of farmers	Mean x̄ = Σx/n
The farmer in the middle when lined up by litres	Median
How much farmers differ from the average	Standard deviation σ (dispersion)
One big 30 L farmer pulls the average above the typical farmer	Positive skew: mean pulled above median and mode

The concept

Averages (measures of central tendency) give one typical value: mean, median, mode. Dispersion (range, quartile deviation, mean deviation, standard deviation) tells how spread out the data is. Two data sets can have the same mean but very different spreads.

Skewness describes lack of symmetry. In a symmetric distribution mean = median = mode. A long right tail gives positive skew (mean is pulled up); a long left tail gives negative skew. Kurtosis describes how peaked the distribution is compared with the normal curve.

Two rules save time: adding a constant to every value changes the mean but not the SD; multiplying every value by k multiplies both mean and SD by k (and the variance by k²).

Key points & formulas

Mean x̄ = Σfx/Σf; grouped median = l + [(N/2 − cf)/f] × h; grouped mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
Empirical relation (moderately skewed data): Mode = 3 Median − 2 Mean
Variance σ² = Σ(x − x̄)²/n = Σx²/n − x̄²; SD σ = √variance
Coefficient of variation CV = (σ/x̄) × 100; lower CV means more consistent
Range = max − min; quartile deviation = (Q₃ − Q₁)/2
y = ax + b ⇒ ȳ = ax̄ + b and σᵧ = |a|σₓ (adding b does not change SD)
Karl Pearson’s skewness Sk = (Mean − Mode)/σ = 3(Mean − Median)/σ
Kurtosis β₂ = μ₄/μ₂²: β₂ = 3 mesokurtic (normal), β₂ > 3 leptokurtic (peaked), β₂ < 3 platykurtic (flat)

Worked example

Find the mean and standard deviation of 2, 4, 4, 4, 5, 5, 7, 9.

Show solution

Mean = (2 + 4 + 4 + 4 + 5 + 5 + 7 + 9)/8 = 40/8 = 5.
Σx² = 4 + 16 + 16 + 16 + 25 + 25 + 49 + 81 = 232.
Variance = Σx²/n − x̄² = 232/8 − 25 = 29 − 25 = 4, so σ = √4.

Answer: Mean = 5, SD = 2

Exam shortcut

For the first n natural numbers: mean = (n + 1)/2 and variance = (n² − 1)/12. So the variance of 1, 2, 3, 4, 5 is (25 − 1)/12 = 2 straight away.

Common traps

Adding a constant to every value does not change the SD or variance.
Multiplying by k multiplies the variance by k², not k.
CET uses the population formula (divide by n) unless the question says otherwise. Dividing by n − 1 gives a different answer.

Practice · tap an option to check

Q1. A distribution has β₂ = 2.4. It is:

Q2. For a moderately skewed distribution, mean = 40 and median = 42. The mode is approximately:

Q3. The variance of the observations 1, 2, 3, 4, 5 (population variance, dividing by n) is:

Q4. The SD of a set of observations is 4. Each observation is multiplied by 3 and then 5 is added. The new SD is:

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Maths · Probability and Statistics · Topic 17 of 24

Random Variables and Probability Distributions

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Think of it like…

Packing Devgad hapus mangoes

A Devgad trader packs Alphonso mangoes in boxes of 12. Each mango, independently, has a 10% chance of being spoilt. Open a box and count the spoilt ones: you might get 0, 1, 2 or more. That count is a random variable X, and because each of the 12 mangoes is a separate yes/no trial with the same chance, X follows the Binomial distribution with n = 12 and p = 0.1. Over a whole season the average spoilt per box settles at np = 1.2. That long-run average is the expectation E(X).

Now think of a very rare pest that hits only 1 mango in 2,500. In a truck of 5,000 mangoes, n is huge and p is tiny, but np = 2 is a sensible number. Counts of such rare events follow the Poisson distribution with λ = np. Finally, weigh the mangoes: most are near 250 g, fewer are much lighter or heavier, and the graph is a symmetric bell. That is the Normal distribution.

In everyday life	In the exam topic
Number of spoilt mangoes in a box	Random variable X
12 mangoes, each spoilt or not with the same chance 0.1	Binomial: n = 12, p = 0.1
Average spoilt mangoes per box over the season	Expectation E(X) = np = 1.2
Rare pest in a huge truckload	Poisson with λ = np; mean = variance = λ
Mango weights bunching around 250 g	Normal distribution: bell curve centred at μ

The concept

A random variable assigns a number to each outcome of an experiment. A discrete random variable has a probability for each value; the probabilities must add to 1. Its expectation E(X) = Σx p(x) is the long-run average, and its variance measures the spread around that average.

The Binomial distribution counts successes in n independent trials with the same success probability p. The Poisson distribution counts rare events in a fixed interval; it is the limit of the binomial when n is large and p is small, with λ = np. For Poisson, mean and variance are both λ.

The Normal distribution is a continuous, symmetric bell curve. Mean = median = mode, the total area under the curve is 1, and probabilities are areas. Any normal X is converted to the standard normal Z = (X − μ)/σ to use tables.

Key points & formulas

Discrete RV: Σp(x) = 1; E(X) = Σx p(x); Var(X) = E(X²) − [E(X)]²
E(aX + b) = aE(X) + b; Var(aX + b) = a² Var(X)
Binomial B(n, p): P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, q = 1 − p; mean np, variance npq (variance < mean)
Poisson: P(X = r) = e^−λ λʳ / r!; mean = variance = λ
Normal N(μ, σ²): symmetric; mean = median = mode; total area 1; P(X = a) = 0 for a single point
Standard normal Z = (X − μ)/σ has mean 0 and SD 1
About 68%, 95% and 99.7% of values lie within 1, 2 and 3 SDs of the mean
Continuous RV: f(x) ≥ 0 and the total area ∫f(x) dx = 1

Worked example

A binomial distribution has mean 4 and variance 3. Find n and p.

Show solution

np = 4 and npq = 3, so q = 3/4.
p = 1 − q = 1/4.
n = 4 ÷ p = 4 ÷ (1/4) = 16.

Answer: n = 16, p = 1/4

Exam shortcut

For a binomial, variance ÷ mean = q. So if a question says “binomial with mean 3 and variance 4”, it is impossible (q would be more than 1). Use this to reject options quickly.

Common traps

Poisson: mean = variance (not mean = standard deviation).
Binomial variance is npq, not np.
For a continuous distribution the probability of one exact value is 0; only intervals have probability.

Practice · tap an option to check

Q1. A random variable X takes values 0, 1, 2 with probabilities 1/4, 1/2, 1/4. Its variance is:

Q2. For a binomial distribution with n = 10 and p = 0.4, the variance is:

Q3. For a Poisson variable X, P(X = 1) = P(X = 2). The variance of X is:

Q4. Marks are normally distributed with mean 50 and SD 10. About what percentage of students score between 40 and 60?

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Maths · Probability and Statistics · Topic 18 of 24

Correlation, Regression and Curve Fitting

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Think of it like…

Study hours and marks in a Pune coaching class

Sir at a coaching class in Pune notes, for each of 30 students, the hours studied per day (x) and the mock test marks (y). Plotted as dots, they lean upward: more hours, more marks. How tightly the dots hug a straight line is the correlation coefficient r: close to +1 means tight and rising, close to 0 means a shapeless cloud, close to −1 tight and falling.

Now Sir places a ruler through the cloud. The “best” position is the one where the vertical misses from the dots, squared and added, are as small as possible: that is the least squares line. He uses it to predict marks from hours: the regression line of y on x. But a parent who knows only the marks and wants to guess the hours needs a different line, x on y. The two lines always cross at the point (x̄, ȳ), the “average student”.

In everyday life	In the exam topic
Each student’s (hours, marks)	A point (x, y) on the scatter diagram
How tightly the dots follow a line	Correlation coefficient r, −1 ≤ r ≤ 1
Ruler placed to make the total squared misses smallest	Least squares line of best fit
Predicting marks from hours vs hours from marks	Two lines: y on x (slope b_yx) and x on y (b_xy)
The “average student” where both lines meet	The point (x̄, ȳ)

The concept

Correlation measures the strength and direction of a linear relationship between two variables. Karl Pearson’s r = cov(x, y)/(σₓσᵧ) lies between −1 and +1, and it does not change if you shift the origin or change the scale of either variable (except for a sign flip with a negative scale).

Regression gives the line used for prediction. There are two lines: y on x (to predict y) and x on y (to predict x). Their slopes are the regression coefficients b_yx and b_xy; their product is r², and r takes the common sign of the two coefficients.

Curve fitting by least squares chooses the constants so that the sum of squared errors is least. For a straight line y = a + bx this gives two normal equations, which you solve for a and b.

Key points & formulas

r = cov(x, y)/(σₓσᵧ), cov(x, y) = Σ(x − x̄)(y − ȳ)/n; −1 ≤ r ≤ 1
y on x: y − ȳ = b_yx(x − x̄), b_yx = r σᵧ/σₓ
x on y: x − x̄ = b_xy(y − ȳ), b_xy = r σₓ/σᵧ
b_yx × b_xy = r²; r = ±√(b_yx b_xy) with the same sign as the coefficients
Both regression lines pass through (x̄, ȳ); they coincide if r = ±1
Least squares line y = a + bx: Σy = na + bΣx and Σxy = aΣx + bΣx²
Spearman’s rank correlation: R = 1 − 6Σd²/[n(n² − 1)]

Worked example

The two regression lines are 3x + 2y = 26 and 6x + y = 31. Find x̄, ȳ and r.

Show solution

Means: solve together. From 6x + y = 31, y = 31 − 6x; then 3x + 62 − 12x = 26, so x = 4 and y = 7.
Try 3x + 2y = 26 as y on x: y = 13 − 1.5x, so b_yx = −1.5. Then 6x + y = 31 as x on y: x = 31/6 − y/6, so b_xy = −1/6.
r² = (−1.5)(−1/6) = 0.25 ≤ 1, so this choice is valid. Both coefficients are negative, so r = −0.5.

Answer: x̄ = 4, ȳ = 7, r = −0.5

Exam shortcut

If you are not told which line is which, guess. If b_yx × b_xy comes out greater than 1, your guess was wrong; swap the lines.

Common traps

b_yx × b_xy = r², not r. Take the square root and attach the sign of the coefficients.
The two regression lines are different lines (unless r = ±1). Do not use y on x to predict x.
High correlation does not prove that one variable causes the other.

Practice · tap an option to check

Q1. If b_yx = 0.8 and b_xy = 0.45, the correlation coefficient r is:

Q2. The two regression lines are x + 2y = 5 and 2x + 3y = 8. The means (x̄, ȳ) are:

Q3. The least squares line y = a + bx for the points (1, 2), (2, 4), (3, 5) is:

Q4. For 5 pairs of ranks, Σd² = 4. Spearman’s rank correlation coefficient is:

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Maths · Arithmetic · Topic 19 of 24

Ratio, Proportion and Percentage

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Think of it like…

A day at Ramesh Kaka’s kirana shop

Ramesh Kaka sells two qualities of rice, ₹40/kg and ₹60/kg. A customer wants rice at ₹48/kg, so he mixes them. Each kg of cheap rice is ₹8 below the target and each kg of costly rice is ₹12 above it; to balance, he needs more of the cheap one: cheap : costly = 12 : 8 = 3 : 2. That balancing is the rule of alligation, and the answer is a ratio.

At Diwali his board says “10% off, plus extra 20% off”. A percentage just means “per hundred”. On a ₹100 item the first offer makes it ₹90, and the extra 20% is taken on ₹90, not on ₹100: ₹72 in the end, a 28% total cut, not 30%. And when sugar becomes 25% costlier, a family that wants the same monthly bill must buy 20% less sugar, not 25% less, because the 25% was calculated on the old price.

In everyday life	In the exam topic
Mixing ₹40 and ₹60 rice to sell at ₹48	Alligation: ratio = (60 − 48) : (48 − 40) = 3 : 2
“Per hundred rupees” on the discount board	Percentage = (part/whole) × 100
10% off and then an extra 20% off	Successive change: −10 − 20 + (10 × 20)/100 = −28%
Sugar 25% costlier, same monthly bill	Cut consumption by 25/(100 + 25) × 100 = 20%
Profit counted on cost price, discount on marked price	Always check the base of the percentage

The concept

A ratio a : b compares two quantities of the same kind; a proportion says two ratios are equal, a : b = c : d, which means ad = bc. To divide an amount in a given ratio, add the ratio parts and give each person their share of the total.

A percentage is a ratio with denominator 100. Every percentage is “of something”: profit and loss percentages are on cost price, discount percentage is on marked price, and a percentage change is on the original value.

When two percentage changes happen one after another, the second acts on the new value. So changes combine as a + b + ab/100 (use negative signs for decreases). Taking a starting value of 100 makes all such questions easy.

Key points & formulas

a : b = c : d ⇒ ad = bc; mean proportional of a and b = √(ab); third proportional = b²/a
Divide N in ratio a : b : c → shares aN/(a + b + c), bN/(a + b + c), cN/(a + b + c)
x% of y = y% of x
Successive changes a% and b%: net = a + b + ab/100
Price up r% ⇒ cut consumption by r/(100 + r) × 100% to keep spending the same
If A is r% more than B, then B is r/(100 + r) × 100% less than A
Profit% = (SP − CP)/CP × 100; Discount% = (MP − SP)/MP × 100
Alligation: cheaper : dearer = (dearer price − mean price) : (mean price − cheaper price)

Worked example

A shopkeeper marks his goods 40% above cost price and gives a 20% discount. Find his profit percentage.

Show solution

Let CP = ₹100. Marked price = ₹140.
20% discount on ₹140 = ₹28, so SP = ₹112.
Profit = 112 − 100 = ₹12 on CP ₹100. Check with the formula: 40 − 20 + (40 × −20)/100 = 12.

Answer: 12% profit

Exam shortcut

Start every percentage chain with 100. “Increase 20% then decrease 20%”: 100 → 120 → 96, so a 4% loss, read off in two seconds.

Common traps

Two successive discounts of 10% and 20% are 28%, not 30%.
If A is 25% more than B, B is 20% less than A, not 25% less (the base changes).
A 20% rise followed by a 20% fall is not “no change”; it is a 4% fall.

Practice · tap an option to check

Q1. A salary is first increased by 20% and then decreased by 20%. The net change is:

Q2. ₹1,800 is divided among A, B and C in the ratio 2 : 3 : 4. C’s share is:

Q3. In what ratio must rice at ₹30/kg be mixed with rice at ₹45/kg to get a mixture worth ₹36/kg?

Q4. A’s income is 25% more than B’s. By what percent is B’s income less than A’s?

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Maths · Arithmetic · Topic 20 of 24

Time and Work (with Pipes and Cisterns)

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Think of it like…

Filling the housing society’s overhead tank

Your housing society’s overhead tank is filled by two motors. Motor A alone fills it in 6 hours, motor B alone in 4 hours. How long with both? Do not add the hours; think per hour. In one hour A fills 1/6 of the tank and B fills 1/4. Together they fill 1/6 + 1/4 = 5/12 of the tank every hour, so the full tank takes 12/5 = 2.4 hours.

Then the secretary finds a leak at the bottom that would empty a full tank in 12 hours. The leak is simply a negative worker: −1/12 per hour. Now the net filling is 5/12 − 1/12 = 4/12 = 1/3 per hour, so the tank takes 3 hours. Every time-and-work question works the same way: men, women, taps or machines are all just rates that add or subtract.

In everyday life	In the exam topic
One full tank	Total work = 1 unit (or take the LCM of the times)
Motor A fills the tank in 6 hours	A’s rate = 1/6 of the work per hour
Both motors running together	Rates add: 1/6 + 1/4 = 5/12 per hour
A leak at the bottom	A negative rate: −1/12 per hour
Hours needed to fill	Time = work ÷ combined rate

The concept

If someone finishes a job in n days, their rate is 1/n of the job per day. Rates of people working together add; time is never added. Once you have the combined rate, time = 1 ÷ rate.

A convenient trick is to take the total work as the LCM of the given times, so every rate is a whole number of units per day. For 12 and 15 days, take 60 units: rates are 5 and 4 units per day.

Pipes and cisterns are the same idea: a filling pipe is a positive worker, an emptying pipe or leak is a negative worker. For man-days questions, the total work is proportional to men × days × hours.

Key points & formulas

A finishes in a days ⇒ A’s one-day work = 1/a
Two together: time T = ab/(a + b); three together: 1/T = 1/a + 1/b + 1/c
M₁D₁H₁/W₁ = M₂D₂H₂/W₂ (men × days × hours per unit of work)
Pipes: filling pipe +, emptying pipe or leak −; net rate × time = 1 tank
If A is k times as efficient as B, A takes 1/k of B’s time
Wages are shared in the ratio of work done
LCM method: total work = LCM of the given days; each rate = total ÷ own days

Worked example

A can do a piece of work in 12 days and B in 15 days. They work together for 4 days, then A leaves. In how many more days will B finish the work?

Show solution

Take total work = LCM(12, 15) = 60 units. A does 5 units/day, B does 4 units/day.
In 4 days together: 4 × (5 + 4) = 36 units. Remaining = 60 − 36 = 24 units.
B alone: 24 ÷ 4 = 6 days.

Answer: 6 more days (10 days in all)

Exam shortcut

For two workers, T = ab/(a + b), “product over sum”. 10 and 15 days together: 150/25 = 6 days. For a filling pipe a and an emptying pipe b: T = ab/(b − a).

Common traps

Adding times: 6 h + 4 h = 10 h is wrong. Add rates, never times.
Forgetting that a leak or outlet works against you (subtract its rate).
When someone leaves midway, first count the work already done before they left.

Practice · tap an option to check

Q1. A can do a work in 10 days and B in 15 days. Working together, they finish it in:

Q2. Pipe A fills a tank in 20 minutes and pipe B empties it in 30 minutes. With both open, an empty tank fills in:

Q3. 12 men can build a wall in 20 days. How many men are needed to build the same wall in 15 days?

Q4. A is twice as efficient as B. Together they finish a job in 14 days. A alone would take:

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Maths · Arithmetic · Topic 21 of 24

Time, Speed and Distance

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Think of it like…

Watching Mumbai locals at Dadar station

Stand on a platform at Dadar and watch. A local train passes the signal pole: the train is not “past” until its last coach crosses the pole, so it must travel its own full length. To clear the whole platform, its engine must go from one end of the platform to beyond the other end until the last coach is out: train length + platform length.

Now a fast local overtakes a slow local on the next track, in the same direction. From the slow train’s window, the fast one creeps past, because you see only the difference of their speeds. But when a train comes from the opposite direction it whooshes past in seconds, because the speeds add. And since lengths are in metres and speed boards are in km/h, you keep converting with 5/18.

In everyday life	In the exam topic
Train passing a signal pole	Distance = length of the train
Train clearing the whole platform	Distance = train length + platform length
Fast local overtaking a slow local (same direction)	Relative speed = difference of speeds
Two locals crossing in opposite directions	Relative speed = sum of speeds
Speed board in km/h, lengths in metres	Multiply km/h by 5/18 to get m/s

The concept

Everything rests on distance = speed × time. Keep units consistent: convert km/h to m/s by multiplying by 5/18, and m/s to km/h by multiplying by 18/5.

Average speed = total distance ÷ total time. It is not the average of the speeds. For two equal distances covered at speeds u and v, the average speed is 2uv/(u + v), the harmonic mean.

For two moving objects, use relative speed: subtract the speeds when they move in the same direction, add them when they move towards each other. Boats work the same way: the stream helps downstream and opposes upstream.

Key points & formulas

Distance = speed × time; 1 km/h = 5/18 m/s; 1 m/s = 18/5 km/h
Average speed = total distance ÷ total time; equal distances at u and v: 2uv/(u + v)
Train passing a pole or a person: time = train length ÷ speed
Train passing a platform, bridge or another train: distance = sum of the lengths
Relative speed: same direction u − v; opposite directions u + v
Boat speed b, stream s: downstream b + s, upstream b − s; b = (down + up)/2, s = (down − up)/2
For a fixed distance, time is inversely proportional to speed

Worked example

A 240 m long train running at 54 km/h crosses a 360 m long platform. How long does it take?

Show solution

Speed = 54 × 5/18 = 15 m/s.
Distance = train + platform = 240 + 360 = 600 m.
Time = 600 ÷ 15.

Answer: 40 seconds

Exam shortcut

Late/early problems: if at speed u you are late and at speed v you are early, distance = (u × v × total time difference)/(v − u). For 4 km/h (10 min late) and 5 km/h (5 min early): (4 × 5 × ¼)/1 = 5 km.

Common traps

Forgetting to convert km/h to m/s when lengths are in metres.
Adding only the platform length (or only the train length) when a train crosses a platform.
Taking (30 + 20)/2 = 25 as the average speed for equal distances; the correct answer is 2 × 30 × 20/50 = 24.

Practice · tap an option to check

Q1. Two trains 150 m and 100 m long run in opposite directions at 40 km/h and 50 km/h. How long do they take to cross each other completely?

Q2. A student goes to college at 30 km/h and returns by the same road at 20 km/h. The average speed for the whole journey is:

Q3. A boat goes downstream at 16 km/h and upstream at 10 km/h. The speed of the stream is:

Q4. Walking at 4 km/h a student reaches the station 10 minutes late; walking at 5 km/h he is 5 minutes early. The distance to the station is:

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Maths · Basic Set Theory and Functions · Topic 22 of 24

Sets, Relations and Functions

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Think of it like…

The Aadhaar enrolment camp at the gram panchayat

An Aadhaar camp is held at the gram panchayat. On one side is the set of villagers; on the other, the pool of 12-digit numbers. The rule is that every villager gets exactly one number: nobody is left out and nobody gets two. A rule like that, which sends each element of the first set to exactly one element of the second, is a function.

Because no two villagers ever share a number, the function is one-one (injective). The pool has crores of unused numbers, so it is not onto; but if we look only at the numbers actually issued in the village, every one of them belongs to someone, and the function becomes onto too, a bijection, which is why it can be reversed: from the number you can find the person. Compare this with the relation “lives in the same village as”: it holds for yourself (reflexive), works both ways (symmetric) and passes along (transitive), so it is an equivalence relation.

In everyday life	In the exam topic
Villagers, and the pool of Aadhaar numbers	Domain A and codomain B
Every villager gets exactly one number	Function f : A → B
No two villagers share a number	One-one (injective)
Every issued number belongs to some villager	Onto (surjective); one-one + onto = bijection, so f⁻¹ exists
“Lives in the same village as”	Equivalence relation: reflexive, symmetric, transitive

The concept

A set is a well-defined collection. Union, intersection, difference and complement combine sets; Venn diagrams show them. The counting formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B) removes the double count of the common part.

A relation from A to B is any subset of A × B. A relation on A is an equivalence relation when it is reflexive (a R a for all a), symmetric (a R b ⇒ b R a) and transitive (a R b and b R c ⇒ a R c).

A function is a relation in which every element of the domain has exactly one image. It is one-one if different inputs give different outputs, onto if every element of the codomain is used, and invertible only if it is both. Composition f ∘ g means “apply g first, then f”.

Key points & formulas

n(A ∪ B) = n(A) + n(B) − n(A ∩ B); n(A ∪ B ∪ C) = Σn(A) − Σn(A ∩ B) + n(A ∩ B ∩ C)
De Morgan: (A ∪ B)′ = A′ ∩ B′; (A ∩ B)′ = A′ ∪ B′
A set with n elements has 2ⁿ subsets (2ⁿ − 1 proper subsets)
n(A × B) = mn; number of relations from A to B = 2^mn; relations on a set of n elements = 2^n²
Equivalence relation = reflexive + symmetric + transitive
Functions from A (m elements) to B (n elements): nᵐ in all; one-one: ⁿPₘ (m ≤ n); bijections when m = n: n!
(f ∘ g)(x) = f(g(x)); in general f ∘ g ≠ g ∘ f; f⁻¹ exists only if f is a bijection

Worked example

In a class of 60 students, 35 like cricket, 25 like kabaddi and 10 like both. How many like neither game?

Show solution

n(C ∪ K) = n(C) + n(K) − n(C ∩ K) = 35 + 25 − 10 = 50.
Neither = total − n(C ∪ K) = 60 − 50.

Answer: 10 students

Exam shortcut

For “how many functions or relations” questions, count choices element by element. Each of the m elements of A picks an image from n choices, so there are nᵐ functions. Each of the mn ordered pairs is either in or out of a relation, so there are 2^mn relations.

Common traps

n(A ∪ B) = n(A) + n(B) only when A and B are disjoint.
f ∘ g means g first, then f. (f ∘ g)(x) and (g ∘ f)(x) are usually different.
A relation can be symmetric and transitive without being reflexive. Check every element has (a, a).

Practice · tap an option to check

Q1. The number of relations on the set A = {1, 2, 3} is:

Q2. If f(x) = 2x + 3 and g(x) = x², then (f ∘ g)(2) is:

Q3. The number of one-one functions from {a, b, c} to {1, 2, 3, 4} is:

Q4. On A = {1, 2, 3}, the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} is:

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Maths · Mensuration · Topic 23 of 24

Mensuration 2D: Triangles, Quadrilaterals, Circles

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Think of it like…

Tiling the floor of a new flat

Your family has bought a new flat in Pune and the tile mistri comes to measure. The number of 1 ft × 1 ft tiles needed to cover the floor is the area. The skirting strip that runs along the bottom of every wall is the perimeter. They are measured in different units: area in square feet, perimeter in feet.

Now the mistri says: “The hall is double the length and double the width of the bedroom.” The skirting doubles, but the tiles become four times, because area grows with the square of the scale. For the round dining table, the decorative edge strip is the circumference 2πr, while the tablecloth must cover the area πr². And for the triangular corner piece he cuts, the area is half of the rectangle around it: ½ × base × height.

In everyday life	In the exam topic
Number of 1 ft × 1 ft tiles to cover the floor	Area (square units)
Skirting strip along the walls	Perimeter (units)
Hall with double the length and double the width	Perimeter × 2, but area × 4 (area scales as k²)
Edge strip vs tablecloth for a round table	Circumference 2πr vs area πr²
Triangular corner tile cut from a rectangle	Triangle area = ½ × base × height

The concept

Perimeter is the length of the boundary; area is the amount of surface enclosed. Always check units: 1 m = 100 cm, but 1 m² = 10,000 cm².

For triangles, use ½ × base × height when the height is known, Heron’s formula when all three sides are known, and (√3/4)a² for an equilateral triangle. Quadrilaterals reduce to rectangles and triangles; for a rhombus, the diagonals bisect each other at right angles.

For circles, the radius drives everything: area πr², circumference 2πr, and a sector of angle θ is the fraction θ/360 of the full circle. If every length is multiplied by k, the perimeter is multiplied by k and the area by k².

Key points & formulas

Triangle: ½ × base × height = ½ ab sin C; Heron: √[s(s − a)(s − b)(s − c)], s = (a + b + c)/2
Equilateral triangle of side a: area (√3/4)a², height (√3/2)a
Rectangle: area lb, perimeter 2(l + b), diagonal √(l² + b²); square: area a² = d²/2, diagonal a√2
Parallelogram: base × height; rhombus: ½ d₁d₂, side = ½√(d₁² + d₂²); trapezium: ½ (sum of parallel sides) × height
Circle: area πr², circumference 2πr; semicircle perimeter πr + 2r; ring: π(R² − r²)
Sector of angle θ°: area (θ/360)πr², arc length (θ/360) × 2πr
Wheel: distance in one revolution = circumference; revolutions = distance ÷ 2πr
Each length × k ⇒ perimeter × k, area × k² (r up 10% ⇒ area up 21%)

Worked example

The radius of a cycle wheel is 35 cm. How many revolutions does it make to cover 1.1 km? (π = 22/7)

Show solution

Circumference = 2πr = 2 × (22/7) × 35 = 220 cm.
1.1 km = 1,10,000 cm.
Revolutions = 1,10,000 ÷ 220.

Answer: 500 revolutions

Exam shortcut

Percentage change in area when every length changes by x%: use the successive-change formula x + x + x²/100. Radius +20% ⇒ area +(20 + 20 + 4) = 44%. Radius −10% ⇒ area −19%.

Common traps

Mixing units: convert everything to the same unit before multiplying, especially cm² and m².
Using diameter where the formula needs radius.
A rhombus’s area uses the diagonals (½ d₁d₂); its side comes from half-diagonals by Pythagoras, not from d₁ or d₂ directly.

Practice · tap an option to check

Q1. The diagonals of a rhombus are 16 cm and 12 cm. Its perimeter is:

Q2. If the radius of a circle is increased by 20%, its area increases by:

Q3. The area of an equilateral triangle of side 8 cm is:

Q4. The area of a sector of a circle of radius 14 cm with central angle 90° is (π = 22/7):

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Maths · Mensuration · Topic 24 of 24

Mensuration 3D: Cube, Cuboid, Sphere, Cylinder, Cone

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Think of it like…

An ice-cream parlour on a Sunday evening

At an ice-cream parlour you see three shapes: a cylindrical tub, cone cups and round scoops. How much ice-cream a container holds is its volume. The paper wrapper rolled around a cone is its curved surface area, which has nothing to do with how much it holds.

Try this with a cone and a tub that have the same base radius and the same height: it takes exactly three cone-fulls to fill the tub. That is why V_cone = ⅓ πr²h, one-third of the cylinder. The scoop on top of a cone is a hemisphere, ⅔ πr³. And when the shopkeeper melts a big block of ice-cream and refills it into small cones, no ice-cream is created or lost: the total volume before equals the total volume after. That single idea solves every “melted and recast” question.

In everyday life	In the exam topic
How much ice-cream a container holds	Volume (cubic units)
Paper wrapper rolled around a cone	Curved surface area πrl, with slant height l = √(r² + h²)
Three cone-fulls fill a tub of the same base and height	V_cone = ⅓ πr²h = ⅓ V_cylinder
Round scoop sitting on the cone	Hemisphere: volume ⅔ πr³
Melting a big block and refilling small cones	Recasting: total volume stays the same

The concept

Volume measures the space inside a solid (cubic units); surface area measures the material needed to cover it (square units). Curved surface area (CSA) covers only the curved part; total surface area (TSA) adds the flat ends too.

Cylinder, cone and sphere are linked: a cone is one-third of the cylinder with the same base and height, and a sphere of radius r has two-thirds the volume of the cylinder that just encloses it (radius r, height 2r). For a cone, keep the vertical height h and the slant height l separate: volume uses h, surface area uses l.

For melting and recasting, equate volumes. If every length is multiplied by k, surface areas are multiplied by k² and volumes by k³.

Key points & formulas

Cuboid: V = lbh; TSA = 2(lb + bh + hl); diagonal √(l² + b² + h²)
Cube: V = a³; TSA = 6a²; diagonal a√3
Cylinder: V = πr²h; CSA = 2πrh; TSA = 2πr(r + h)
Cone: V = ⅓πr²h; CSA = πrl; TSA = πr(l + r); l = √(r² + h²)
Sphere: V = (4/3)πr³; surface area 4πr²
Hemisphere: V = (2/3)πr³; CSA = 2πr²; TSA = 3πr²
Recasting: volume before = volume after; number of small pieces = big volume ÷ one small volume
Lengths × k ⇒ surface area × k², volume × k³

Worked example

A metal sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder.

Show solution

Volume of sphere = (4/3)π(6)³ = (4/3)π × 216 = 288π cm³.
Volume of cylinder = π(4)²h = 16πh.
Equate: 16πh = 288π, so h = 288/16.

Answer: 18 cm

Exam shortcut

In recasting questions π cancels from both sides, so never multiply out 22/7. Write both volumes with π, cancel it, and work with the plain numbers.

Common traps

Using the vertical height h in the cone’s CSA formula. CSA uses the slant height l.
Using TSA when the object is open (an open tank or a hemispherical bowl has no lid: bowl uses 2πr², not 3πr²).
Doubling the radius doubles the volume? No: volume grows 2³ = 8 times.

Practice · tap an option to check

Q1. How many cubes of edge 2 cm can be cut from a cuboid measuring 6 cm × 8 cm × 10 cm?

Q2. The curved surface area of a cone with radius 7 cm and height 24 cm is (π = 22/7):

Q3. If the radius of a sphere is doubled, its volume becomes:

Q4. The total surface area of a cube is 150 cm². Its volume is:

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