Mathematics — Complete Formula Sheet

$x,y$: Variables (unknowns) · $a,b$: Coefficients of x and y (not both zero) · $c$: Constant term

a, b, c are real numbers and a, b are not both zero. Its graph is a straight line; every point on the line is a solution of the equation.

Two equations in the same two variables. Bring both to this form (constants on the right) before using elimination or Cramer's rule.

For ax + by = c (a, b ≠ 0), putting x = 0 gives the point on the Y-axis and putting y = 0 gives the point on the X-axis. Plot at least four points in all, as the textbook advises.

Draw both lines on the same axes. The coordinates of the point where they meet satisfy both equations.

Express one variable in terms of the other from one equation and substitute it in the second equation. You get an equation in one variable.

In practice, multiply the equations so that the coefficients of one variable become equal, then add or subtract to remove it.

Adding the equations gives (a + b)(x + y) = c + d; subtracting gives (a − b)(x − y) = c − d. Valid when a ≠ b and a + b ≠ 0.

Product along the diagonal from top-left to bottom-right minus product along the other diagonal. Put negative entries in brackets.

First column = coefficients of x, second column = coefficients of y.

Replace the x-column (first column) of D by the constants c₁, c₂.

Replace the y-column (second column) of D by the constants c₁, c₂.

$D$: Determinant of the coefficients of x and y · $D_x$: D with the x-column replaced by the constants · $D_y$: D with the y-column replaced by the constants

Works only when the equations are in the form a₁x + b₁y = c₁, a₂x + b₂y = c₂ and D ≠ 0.

If D = 0, the two lines are either parallel (no solution) or the same line (infinitely many solutions), so x = Dx/D cannot be found.

Solve the linear equations in m and n, then x = 1/m and y = 1/n. Here x ≠ 0 and y ≠ 0.

After finding p and q, solve the two new linear equations x + y = 1/p and x − y = 1/q for x and y.

$x$: Digit in tens place · $y$: Digit in units place

x = tens digit, y = units digit. Number + reversed number = 11(x + y); number − reversed number = 9(x − y).

$a$: Coefficient of x² (must be non-zero) · $b$: Coefficient of x · $c$: Constant term

Degree-2 polynomial equation. Real roots exist when D ≥ 0.

Works for every quadratic. Gives both roots at once.

D > 0 two distinct real roots; D = 0 two equal real roots; D < 0 no real roots.

Vieta's relation. Useful for verifying answers and forming new quadratics.

Together with the sum, uniquely determines both roots.

Take k = 1 for the simplest form. Multiply by any k ≠ 0 for a general quadratic.

Find two numbers that multiply to a·c and add to b — the splitting-the-middle-term trick.

Manual route to the quadratic formula. Works when factorisation fails.

Applies when coefficients are rational and D is not a perfect square.

Handy identity to compute α² + β² without solving the equation.

$a_n$: nth term of the AP · $a$: First term · $d$: Common difference · $n$: Position of the term

a = first term, d = common difference, n = position.

Also written as S_n = (n/2)(a + l), where l is the last term.

Use when last term l is known — the quickest route.

d is constant for a valid AP. Check at least three consecutive pairs before concluding it is an AP.

Useful when specific term positions and values are given.

If a, b, c are in AP, then b is the arithmetic mean of a and c.

Special AP with a = 1, d = 1.

Special AP with a = 1, d = 2. Neat perfect-square pattern.

Twice the sum of first n naturals.

If S_n formula is given, this recovers the nth term.

Taxable value is the price of the goods or service before tax (after any discount).

Within one state, half the GST goes to the Centre (CGST) and half to the State (SGST). Example: 18% GST = 9% CGST + 9% SGST. In a Union Territory without its own legislature, UTGST takes the place of SGST.

For a sale from one state to another, the whole GST is charged as IGST at the full rate. It is not split.

For an inter-state invoice, Total = Taxable value + IGST.

GST is charged on the discounted price, not on the marked price.

Use when the price including GST is given. Example: Rs 5900 including 18% GST means taxable value 5900 × 100/118 = Rs 5000.

Output tax = GST collected on sale; input tax = GST paid on purchase. A trader pays the government only the difference. For an intra-state trader, CGST payable = SGST payable = half of this.

$\text{FV}$: Face value of a share (printed value, Rs) · $\text{MV}$: Market value of a share (price on the stock market, Rs)

Premium = MV − FV; discount = FV − MV. The face value is fixed by the company and printed on the share certificate; the market value keeps changing with demand.

Dividend is declared as a percentage of the FACE value, whatever the market price.

This is the investor's income from the shares.

If brokerage and GST are included, divide by the purchase value of one share instead of the MV.

Charged by the broker on the market value, both when buying and when selling.

Brokerage is a service, so 18% GST is charged on the brokerage amount only. If a question does not mention GST on brokerage, the textbook leaves it out.

When buying, brokerage and GST are ADDED to the market value.

When selling, brokerage and GST are SUBTRACTED from the market value — this is what the seller actually receives.

Answer in percent. Use the amount actually invested (based on MV), not the face value. Ignoring brokerage, it equals dividend per share ÷ MV × 100.

A negative result is a loss. Use the purchase value and sale value including brokerage and GST. If a dividend was received before selling, the textbook adds it to the money received: Profit = (sale value + dividend) − sum invested.

NAV = net asset value of one unit. In a SIP (Systematic Investment Plan) a fixed amount is invested at regular intervals, so more units are bought when the NAV is low.

Example: one die, S = {1, 2, 3, 4, 5, 6}, n(S) = 6. List outcomes in a fixed order so none is missed.

An event is a subset of the sample space. n(A) is the number of outcomes favourable to A.

$P(A)$: Probability of event A · $n(A)$: Number of outcomes favourable to A · $n(S)$: Total number of outcomes in the sample space

Valid when all outcomes are equally likely. Count carefully — most errors are in n(A).

Since 0 ≤ n(A) ≤ n(S). A probability can be written as a fraction, decimal or percentage, but never negative and never more than 1.

An event that always happens, e.g. getting a number less than 7 on a die.

An event with no favourable outcome (empty set, written { } or φ in the textbook), e.g. getting 7 on an ordinary die.

For a fair coin each face has probability 1/2; for a fair die each face has probability 1/6.

Useful as a check when you write a table of probabilities.

A' (also written Ā) means 'A does not happen'. P(A) + P(A') = 1. The Std X textbook only names the complement of an event under 'For more information', so counting the outcomes of 'not A' directly is equally acceptable.

For n coins tossed together (or one coin tossed n times). Two coins: S = {HH, HT, TH, TT}; three coins: 8 outcomes.

One die: 6 outcomes; two dice: 36 ordered pairs (1, 1) to (6, 6).

Not a textbook formula — only a quick check against your list of pairs; in the exam, write the pairs. Sum 7 has the most ways (6); sums 2 and 12 have one way each.

4 suits of 13 cards: spades and clubs (black), hearts and diamonds (red). Face cards are J, Q, K: 3 per suit, 12 in all. Aces are not face cards.

Tens place has k choices; without repetition, the units place has k − 1. If 0 is one of the digits, it cannot be in the tens place — list the numbers instead.

The class mark represents the whole class in mean calculations. For 20–30 the class mark is 25.

$\bar{X}$: Mean of the data · $x_i$: Class mark of the i-th class · $f_i$: Frequency of the i-th class

Multiply each class mark by its frequency, add, and divide by the total frequency. Fine for small numbers; heavy arithmetic for large ones.

A is the assumed mean, usually a class mark near the middle. dᵢ is negative for classes below A.

$A$: Assumed mean · $d_i$: Deviation of class mark from A · $\bar{d}$: Mean of the deviations

Smaller numbers than the direct method, and the answer is the same whatever A you choose.

The textbook takes g as the G.C.D. of all the dᵢ; for equal classes with A at a class mark, g is the class width. Then the uᵢ are small integers like −2, −1, 0, 1, 2.

$u_i$: Step deviation of the i-th class · $g$: G.C.D. of all dᵢ (usually the class width) · $\bar{u}$: Mean of the step deviations

Do not forget to multiply ū by g at the end.

Subtract the correction from every lower limit and add it to every upper limit. For 10–19, 20–29, … the correction is 0.5, giving 9.5–19.5, 19.5–29.5, …

Running total of frequencies. The last cumulative frequency equals N.

Find N/2, then go down the cf column until it first reaches or crosses N/2.

$L$: Lower class limit of the median class · $N$: Total frequency (Σfᵢ) · $cf$: Cumulative frequency of the class preceding the median class · $f$: Frequency of the median class · $h$: Class width

cf is the cumulative frequency of the class BEFORE the median class, not of the median class itself.

Find it before applying the mode formula; f₀ and f₂ are the frequencies of the classes just before and just after it.

$L$: Lower class limit of the modal class · $f_1$: Frequency of the modal class · $f_0$: Frequency of the class preceding the modal class · $f_2$: Frequency of the class succeeding the modal class · $h$: Class width

If the modal class is the first class, take f₀ = 0; if it is the last class, take f₂ = 0.

Join the points (class mark, frequency) by straight lines. Close the polygon on the X-axis at the class marks of the classes just before the first class and just after the last class (frequency 0). It can also be drawn by joining the mid-points of the tops of the histogram rectangles.

Each component gets a sector whose angle is proportional to its share of the total.

Used when the pie diagram is given and you have to find the data.

If your angles do not add up to 360°, recheck the calculation before drawing.

$A(△ABC)$: Area of triangle ABC (cm²) · $BC,QR$: Bases of the two triangles (cm) · $AD,PS$: Heights drawn to those bases (cm)

Area of a triangle = ½ × base × height, so the ½ cancels and the ratio of areas is the ratio of the products of base and corresponding height. AD and PS are heights on bases BC and QR.

$A_1,A_2$: Areas of the two triangles · $b_1,b_2$: Their corresponding bases

Areas of triangles with equal heights are proportional to their bases. Typical case: two triangles with a common vertex and bases on the same line.

$A_1,A_2$: Areas of the two triangles · $h_1,h_2$: Their corresponding heights

Areas of triangles with equal (or common) bases are proportional to their heights.

$D,E$: Points on sides AB and AC where the parallel line cuts them

A line parallel to one side of a triangle, cutting the other two sides in distinct points D and E, divides those sides in the same ratio.

Follow from AD/DB = AE/EC by invertendo and componendo (e.g. DB/AD = EC/AE ⇒ AB/AD = AC/AE). Use them when the whole side is given instead of the second part.

If a line divides two sides of a triangle in the same ratio, it is parallel to the third side. Use it to PROVE that two segments are parallel.

$AD$: Bisector of ∠BAC meeting BC at D · $BD,DC$: Parts of side BC

The bisector of an angle of a triangle divides the opposite side in the ratio of the remaining two sides. The side next to BD is AB — keep them together.

If D on BC divides it in the ratio of the other two sides, then AD is the bisector of ∠A.

$AB,BC$: Intercepts on the first transversal · $PQ,QR$: Corresponding intercepts on the second transversal

Three parallel lines cut any two transversals in the same ratio. A, B, C lie on one transversal and P, Q, R on the other.

Corresponding angles are equal and corresponding sides are in proportion. The order of letters fixes which vertex matches which.

Two pairs of equal angles are enough, because the third pair is then equal automatically (angle sum 180°).

Two sides in proportion AND the angle INCLUDED between them equal. An angle that is not between the two sides does not work.

All three pairs of sides in the same ratio. Arrange both triangles' sides in increasing order before comparing.

$AB,PQ$: A pair of corresponding sides

Ratio of areas = square of the ratio of corresponding sides. If sides are in the ratio 2 : 3, areas are in the ratio 4 : 9 — not 2 : 3.

$a,b$: Natural numbers with a > b

Gives a Pythagorean triplet for any natural numbers a > b. Example: a = 2, b = 1 gives (5, 3, 4). Multiplying a triplet by any natural number gives another triplet.

$BD$: Perpendicular from the right-angle vertex B to the hypotenuse AC

The altitude to the hypotenuse splits a right triangle into two triangles, each similar to the whole triangle and to each other (AA test: a common angle plus a right angle).

$BD$: Altitude to the hypotenuse · $AD,DC$: Segments of the hypotenuse AC

In a right triangle, the perpendicular from the right-angle vertex to the hypotenuse is the geometric mean of the two segments of the hypotenuse.

$AC$: Hypotenuse (side opposite the right angle) · $AB,BC$: Perpendicular sides

In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

If the square of one side equals the sum of the squares of the other two, the angle opposite that side is a right angle.

Sides are in the ratio 1 : √3 : 2 (opposite 30° : opposite 60° : hypotenuse).

An isosceles right triangle. Sides are in the ratio 1 : 1 : √2.

$d$: Diagonal (cm) · $a$: Side of the square (cm)

The diagonal splits a square into two 45°-45°-90° triangles. Square of side 5 cm has diagonal 5√2 cm.

$h$: Height (altitude) (cm) · $a$: Side of the equilateral triangle (cm)

The altitude splits an equilateral triangle into two 30°-60°-90° triangles; the altitude is the side opposite 60°.

$D$: Foot of the perpendicular from A on BC

Opposite an acute angle, the square of the side is LESS than the sum of the squares of the other two.

Here D lies on BC produced beyond C. Opposite an obtuse angle, the square of the side is MORE than the sum of the squares of the other two.

$AM$: Median from A to side BC · $BM$: Half of BC (BM = MC)

Relates the two sides, the median AM and half the third side. Since BM = BC/2, it can also be written AB² + AC² = 2AM² + BC²/2.

Sum of squares of the sides of a parallelogram = sum of squares of its diagonals. Follows from Apollonius theorem because the diagonals bisect each other.

$O$: Centre of the circle · $P$: Point of contact of tangent l

A tangent at any point of a circle is perpendicular to the radius through the point of contact. Converse: a line through the end of a radius and perpendicular to it is a tangent.

$PA,PB$: Tangent segments from external point P touching the circle at A and B

Tangent segments drawn from an external point P to a circle are congruent. Also, OP bisects ∠APB.

$OP$: Distance of the external point from the centre (cm) · $r$: Radius (cm)

From the right angle at A in ΔOAP (tangent theorem) and Pythagoras. OP is the distance of P from the centre.

$d$: Distance between the centres (cm) · $r_1,r_2$: Radii of the two circles (cm)

If two circles touch each other, their point of contact lies on the line joining their centres. That is why the distances simply add or subtract.

$\angle AOB$: Central angle subtended by the arc

Arc measure equals the central angle. A semicircle measures 180°.

Arcs that share only an endpoint B add up. Useful for finding an unknown arc when the whole circle (360°) is split into pieces.

In the same circle (or congruent circles), chords of congruent arcs are congruent, and conversely.

The measure of an inscribed angle is half the measure of the arc it intercepts. B is on the circle; arc AXC is the arc inside the angle.

Angles inscribed in the same arc are congruent. An angle inscribed in a semicircle is a right angle, because its arc is 180°.

Opposite angles of a cyclic quadrilateral are supplementary. Converse: if a pair of opposite angles is supplementary, the quadrilateral is cyclic.

An exterior angle of a cyclic quadrilateral is congruent to the interior angle opposite to its adjacent interior angle. Both equal 180° − ∠ABC.

Vertex B on the circle, ray BC tangent at B and ray BA a secant (chord BA). The angle is half the arc it intercepts, arc AXB, which lies inside the angle.

Chords AB and CD intersect at E inside the circle. Arc AC is intercepted by ∠AEC and arc BD by its vertically opposite angle ∠BED. Proved by joining a pair of endpoints and using the exterior angle of a triangle.

Secants EBA and EDC meet at E outside the circle, with B and D the nearer points. Half the DIFFERENCE of the far arc and the near arc.

$E$: Point of intersection of chords AB and CD (inside the circle)

Chords AB and CD intersect at E inside the circle. The product of the two parts of one chord equals that of the other.

$E$: Point of intersection of the secants (outside the circle)

The lines containing chords AB and CD intersect at E outside the circle. Each length is measured from E: AE and BE are both measured along the same secant.

$ET$: Tangent segment from E · $EA,EB$: Distances from E to the near and far points of the secant

From an external point E, a secant meets the circle at A and B, and a tangent touches it at T. The product of the secant segments equals the square of the tangent segment.

Length of segment joining (x₁, y₁) and (x₂, y₂). Always positive.

$m,n$: Ratio in which the point divides the segment · $(x_1,y_1)$: First endpoint · $(x_2,y_2)$: Second endpoint

Point dividing segment from (x₁,y₁) to (x₂,y₂) in ratio m:n internally.

Special case of section formula with m = n = 1.

Take absolute value. If area = 0, three points are collinear.

Three points collinear ⇔ area of triangle they form is zero.

Divides each median in the ratio 2:1 from vertex to midpoint.

Slope tells the inclination. Vertical line: slope undefined; horizontal: slope = 0.

Point lies on extended line outside the segment.

Special case of distance formula with (x₁, y₁) = (0, 0).

$\theta$: Angle of interest (0° < θ < 90°) · $Opp$: Side opposite to θ · $Adj$: Side adjacent to θ (not hypotenuse) · $Hyp$: Hypotenuse — side opposite the right angle

Only for a right-angled triangle. Ratios depend on the angle, not on triangle size.

Cosec, sec, cot are reciprocals of sin, cos, tan respectively.

Rewrite tan and cot in terms of sin and cos.

The fundamental identity. Rearranges to sin²θ = 1 − cos²θ or cos²θ = 1 − sin²θ.

Divide identity 1 by cos²θ.

Divide identity 1 by sin²θ.

Ratio of θ = co-ratio of (90° − θ). Also tan(90° − θ) = cot θ.

For cos, read the same values in reverse order. tan = sin/cos.

Measured upward from the horizontal to the line of sight to the object.

Measured downward from the horizontal. In a right triangle it equals the alternate angle of elevation from the object.

$r$: Radius of the circular base · $h$: Height of the cylinder

TSA includes the two circular bases; CSA is the curved lateral surface only.

$l$: Slant height · $h$: Perpendicular height

Always find slant height l before computing cone surface area.

Apply Pythagoras to the right triangle with legs r and h.

A sphere has no flat base — surface area formula covers the entire closed surface.

TSA of hemisphere = curved surface (2πr²) + flat circular base (πr²).

$R$: Radius of larger (bottom) base · $r$: Radius of smaller (top) base · $h$: Height between the two bases

A frustum is the piece of a cone cut off by a plane parallel to the base.

TSA = π(R + r)l + πR² + πr² (adds two circular bases).

a = edge length. Diagonal = a√3.

l, b, h are length, breadth, height. Diagonal = √(l² + b² + h²).

When a solid is melted and recast, total volume is preserved (no wastage stated).

Do NOT include internal surfaces where two solids meet.

$a,b,c$: Real coefficients with a ≠ 0

Degree = 2. If a > 0 parabola opens upward; if a < 0 it opens downward.

$\alpha ,\beta$: Zeros (roots) of the quadratic

Sum of the two roots equals -(coefficient of x)/(coefficient of x²).

Product of the two roots equals (constant term)/(leading coefficient).

k is any non-zero constant. Take k = 1 for the simplest form.

For p(x) = ax³ + bx² + cx + d. Same sign rule as quadratic.

Middle Vieta relation for a cubic polynomial.

Sign is negative for cubic — differs from the quadratic formula.

Dividend = Divisor × Quotient + Remainder, with deg r(x) < deg g(x).

Substitute x = a into p(x) to get the remainder directly. Bypasses long division.

Trial-check integer factors of the constant term for cubic factorisation.