Comprehensive notes covering Arithmetic, Algebra, Geometry, Mensuration and Data Interpretation for the Bihar PSC Combined Competitive Examination. All formulas are exam-verified; worked examples follow Bihar PSC previous-year question patterns.
1. Number System
Classification of Numbers
A thorough understanding of the number system is the foundation for all quantitative aptitude topics. The BPSC CCE paper regularly tests divisibility rules, LCM/HCF problems, and prime factorisation. Questions typically carry 2–3 marks and are solvable in under a minute if formulas are memorised.
Types of Numbers
Natural Numbers (N): 1, 2, 3, 4, … — counting numbers; do not include 0.
Integers (Z): …–3, –2, –1, 0, 1, 2, 3… — includes positives, negatives and zero.
Rational Numbers (Q): Numbers expressible as p/q where p and q are integers and q ≠ 0. Examples: 3/4, –5/2, 0.75, 2.333…
Irrational Numbers: Cannot be expressed as p/q. Their decimal expansion is non-terminating and non-repeating. Examples: √2 = 1.41421…, √3, π = 3.14159…, e = 2.71828…
Real Numbers (R): Union of rational and irrational numbers — every point on the number line is a real number.
Divisibility Rules
Divisibility rules allow you to test whether a number is divisible by a given divisor without performing full division. These are extremely common in BPSC paper I.
Divisor
Rule
Quick Example
2
Last digit is 0, 2, 4, 6 or 8 (even)
348 → last digit 8 → YES
3
Sum of all digits divisible by 3
123 → 1+2+3=6 → YES
4
Last two digits form a number divisible by 4
1312 → 12÷4=3 → YES
5
Last digit is 0 or 5
225 → YES
6
Divisible by both 2 AND 3
132: even + digit sum 6 → YES
8
Last three digits form a number divisible by 8
3816 → 816÷8=102 → YES
9
Sum of all digits divisible by 9
729 → 7+2+9=18 → YES
10
Last digit is 0
340 → YES
11
(Sum of digits at odd positions) − (Sum at even positions) = 0 or multiple of 11
2728: (2+2)−(7+8) = −11 → YES
HCF and LCM
HCF (Highest Common Factor) is the largest number that divides all given numbers exactly. LCM (Least Common Multiple) is the smallest number that is divisible by all given numbers.
─── HCF AND LCM CORE FORMULAS ───
HCF(a, b) x LCM(a, b) = a x b (for any two numbers)
LCM = (a x b) / HCF(a, b)
HCF = (a x b) / LCM(a, b)
Prime Factorisation Method:
HCF = product of COMMON prime factors with LOWEST powers
LCM = product of ALL prime factors with HIGHEST powers
Example: HCF and LCM of 12 and 18
12 = 2^2 x 3^1
18 = 2^1 x 3^2
HCF = 2^1 x 3^1 = 6
LCM = 2^2 x 3^2 = 36
Verify: HCF x LCM = 6 x 36 = 216 = 12 x 18 = 216 ✓
Key Points for BPSC
HCF of fractions = HCF of numerators / LCM of denominators
LCM of fractions = LCM of numerators / HCF of denominators
If HCF of two numbers is h, both numbers are multiples of h.
The largest number that divides a, b, c leaving remainders r1, r2, r3 is HCF(a–r1, b–r2, c–r3).
The smallest number divisible by a, b, c and leaving remainder r is LCM(a,b,c) + r.
2. Percentage, Profit & Loss
Percentage Fundamentals
Percentage means "per hundred." It is one of the most heavily tested arithmetic topics in BPSC CCE. Mastery of percentage change and its chain application directly helps in profit/loss, discount, and data interpretation questions.
─── PERCENTAGE FORMULAS ───
Percentage of a number:
x% of A = (x / 100) x A
Percentage change (increase or decrease):
% Change = [(New Value - Original Value) / Original Value] x 100
If a value increases by x% then decreases by y%:
Net % change = x - y - (xy/100)
(positive = net increase; negative = net decrease)
If price rises by x%, % reduction in consumption to keep expenditure same:
Reduction = [x / (100 + x)] x 100
If price falls by x%, % increase in consumption to keep expenditure same:
Increase = [x / (100 - x)] x 100
Converting fractions to common percentages:
1/2 = 50% 1/3 = 33.33% 1/4 = 25%
1/5 = 20% 1/6 = 16.67% 1/8 = 12.5%
1/10 = 10% 1/20 = 5% 3/4 = 75%
Profit and Loss
In BPSC CCE, profit and loss questions often combine with percentage change. Always identify Cost Price (CP) and Selling Price (SP) first, then apply the relevant formula.
─── PROFIT AND LOSS FORMULAS ───
Profit = SP - CP (when SP > CP)
Loss = CP - SP (when CP > SP)
Profit% = (Profit / CP) x 100
Loss% = (Loss / CP) x 100
Selling Price:
SP = CP x (1 + Profit%/100) [when profit]
SP = CP x (1 - Loss%/100) [when loss]
Cost Price:
CP = SP / (1 + Profit%/100) [when profit known]
CP = SP / (1 - Loss%/100) [when loss known]
Marked Price (MP) and Discount:
Discount = MP - SP
Discount% = (Discount / MP) x 100
SP = MP x (1 - Discount%/100)
Successive Discounts of d1% and d2%:
Equivalent single discount = d1 + d2 - (d1 x d2)/100
If CP of n articles = SP of m articles:
Profit% = [(n - m) / m] x 100
Profit% when false weight is used (claiming W but giving W'):
Profit% = [(W - W') / W'] x 100
BPSC Exam Tip
Memorise the "multiplying factor" approach: a 20% profit means SP = 1.2 × CP. A 15% discount means SP = 0.85 × MP. Chain these for consecutive operations — e.g., 20% profit after 10% discount: SP = MP × 0.9; profit on CP means CP × 1.2 = MP × 0.9, so MP = 1.2CP/0.9 = (4/3)CP.
3. Simple Interest & Compound Interest
Simple Interest (SI)
Simple interest is calculated only on the original principal throughout the period. BPSC questions on SI usually ask for missing variables — P, R, T, or SI itself — given the other three.
─── SIMPLE INTEREST FORMULAS ───
SI = (P x R x T) / 100
Where:
P = Principal (initial amount)
R = Rate of interest per annum (%)
T = Time in years
Amount A = P + SI = P (1 + RT/100)
Derived formulas:
P = (SI x 100) / (R x T)
R = (SI x 100) / (P x T)
T = (SI x 100) / (P x R)
If amount becomes n times at rate R% SI:
T = [(n - 1) x 100] / R years
Compound Interest (CI)
In compound interest, interest is added to the principal at the end of each period and then itself earns interest. BPSC CCE commonly tests CI for 2 or 3 years, population growth, and depreciation.
─── COMPOUND INTEREST FORMULAS ───
Amount after n years (compounded annually):
A = P(1 + R/100)^n
CI = A - P = P[(1 + R/100)^n - 1]
Compounding frequency:
Half-yearly: A = P(1 + R/200)^(2n)
Quarterly: A = P(1 + R/400)^(4n)
Monthly: A = P(1 + R/1200)^(12n)
Difference between CI and SI for 2 years:
CI - SI = P(R/100)^2
Difference between CI and SI for 3 years:
CI - SI = P(R/100)^2 x (3 + R/100)
Population Growth (same formula as CI):
P_n = P_0 x (1 + r/100)^n
Depreciation (value decreases each year):
V_n = V_0 x (1 - r/100)^n
Effective annual rate when compounded half-yearly at R%:
Effective rate = (1 + R/200)^2 - 1 (expressed as %)
CI Quick Values to Memorise
CI on Rs 1000 at 10% for 2 years = Rs 210 (vs SI = Rs 200; difference = Rs 10)
CI on Rs 1000 at 10% for 3 years = Rs 331 (vs SI = Rs 300; difference = Rs 31)
At 20% for 2 years: A = P × 1.44; CI = 0.44P; SI = 0.40P
If CI for 2nd year − CI for 1st year = x, then x = (first year CI × R)/100
4. Ratio, Proportion & Mixture
Ratio and Proportion
A ratio a:b compares two quantities of the same kind. A proportion states that two ratios are equal. BPSC tests partnership problems, dividing amounts in given ratios, and mixture problems.
─── RATIO AND PROPORTION FORMULAS ───
Ratio: a : b = a/b
Equivalent ratio: a:b = ka:kb (multiply/divide both by same non-zero k)
Proportion: a:b = c:d means a/b = c/d
Product of extremes = Product of means
a x d = b x c
Fourth proportional: if a:b = c:x, then x = bc/a
Third proportional: if a:b = b:x, then x = b^2/a
Mean proportional between a and b: x = sqrt(a x b)
Partnership:
Profit shared proportional to (Capital x Time)
Partner A's profit / Partner B's profit = (C_A x T_A) / (C_B x T_B)
Componendo and Dividendo:
If a/b = c/d, then (a+b)/(a-b) = (c+d)/(c-d)
Alligation / Mixture Rule
Alligation is a shortcut method to find the ratio in which two ingredients must be mixed to achieve a desired mean value (price, concentration, etc.).
Alligation (Cross Method) Diagram
Ratio of cheaper to dearer = (D − M) : (M − C)
─── ALLIGATION FORMULA ───
Cheaper quantity : Dearer quantity = (D - M) : (M - C)
Where:
C = price/strength of cheaper ingredient
D = price/strength of dearer ingredient
M = desired mean price/strength of mixture
Mixture removal and replacement (n operations):
Final quantity of original ingredient / Initial quantity
= [(V - v) / V]^n
Where V = total volume, v = volume removed each time, n = number of operations
5. Time, Work & Speed
Time and Work
BPSC regularly includes problems on combined work, pipes & cisterns, and efficiency. The golden rule: work = rate × time. Express rates as fractions of the total work done per unit time.
─── TIME AND WORK FORMULAS ───
If A completes work in 'a' days, A's 1-day work = 1/a
A and B working together:
Combined 1-day work = 1/a + 1/b
Time to finish together = ab / (a + b)
A, B and C together:
Time = abc / (ab + bc + ca)
If A is twice as efficient as B:
A takes half the time B takes.
Ratio of work rates A:B = 2:1 (A does 2 units while B does 1)
Pipes and Cisterns:
Inlet pipe fills tank in 'a' hours => rate = +1/a
Outlet pipe empties in 'b' hours => rate = -1/b
Net rate when both open = 1/a - 1/b
Time to fill = ab / (b - a) [provided b > a]
Work and wages:
Wages are distributed in ratio of work done
i.e., in ratio of (daily rate x days worked)
Speed, Distance and Time
Distance-speed-time is foundational for trains, boats & streams, and circular track problems. Keep units consistent (km/h or m/s).
─── SPEED, DISTANCE, TIME FORMULAS ───
D = S x T
S = D / T
T = D / S
Unit conversion:
x km/h = x x (5/18) m/s
x m/s = x x (18/5) km/h
Average speed for equal DISTANCES at speeds u and v:
Avg speed = 2uv / (u + v) [Harmonic mean — NOT arithmetic mean!]
Average speed for equal TIMES at speeds u and v:
Avg speed = (u + v) / 2
Trains:
Time to cross stationary pole = Length of train / Speed of train
Time to cross platform = (Length of train + Length of platform) / Speed
Two trains SAME direction : Relative speed = |S1 - S2|
Two trains OPPOSITE direction: Relative speed = S1 + S2
Boats and Streams:
Speed of boat in still water = u
Speed of stream = v
Downstream speed = u + v
Upstream speed = u - v
u = (Downstream + Upstream) / 2
v = (Downstream - Upstream) / 2
Time to travel distance D:
Downstream: T_d = D / (u + v)
Upstream: T_u = D / (u - v)
6. Algebra
Linear Equations
A linear equation in one variable has the form ax + b = 0. In two variables: ax + by = c. BPSC typically gives two simultaneous equations to solve for two unknowns using substitution or elimination.
Quadratic Equations
Standard form: ax² + bx + c = 0, where a ≠ 0. Roots can be found by factoring, completing the square, or the quadratic formula. The discriminant (b² − 4ac) determines the nature of the roots.
─── QUADRATIC EQUATION ───
Standard form: ax^2 + bx + c = 0 (a ≠ 0)
Quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / (2a)
Discriminant D = b^2 - 4ac
D > 0 => two distinct real roots
D = 0 => two equal (repeated) real roots
D < 0 => no real roots (complex roots)
Sum of roots (alpha + beta) = -b/a
Product of roots (alpha x beta) = c/a
Quadratic from roots:
x^2 - (sum of roots)x + (product of roots) = 0
Key Algebraic Identities
These identities appear repeatedly in simplification questions. Memorise all of them.
─── ALGEBRAIC IDENTITIES ───
Square identities:
(a + b)^2 = a^2 + 2ab + b^2
(a - b)^2 = a^2 - 2ab + b^2
(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)
(a + b)^2 - (a - b)^2 = 4ab
Difference of squares:
(a + b)(a - b) = a^2 - b^2
Cube identities:
(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3
(a - b)^3 = a^3 - 3a^2 b + 3ab^2 - b^3
Sum and difference of cubes:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Three-variable identities:
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
If a + b + c = 0, then a^3 + b^3 + c^3 = 3abc
Useful substitutions:
If a + b + c = 0 => a^2 + b^2 + c^2 = -2(ab + bc + ca)
If x + 1/x = k => x^2 + 1/x^2 = k^2 - 2
x^3 + 1/x^3 = k^3 - 3k
7. Geometry
Lines, Angles and Triangles
Geometry questions in BPSC test angle relationships (especially with parallel lines), triangle properties, and circle theorems. Pythagoras theorem and its triples are essential.
Triangle angle sum: sum of interior angles = 180°; exterior angle = sum of two non-adjacent interior angles
─── TRIANGLE FORMULAS ───
Pythagoras Theorem (right triangle):
a^2 + b^2 = c^2 (c = hypotenuse)
Common Pythagorean Triples:
3-4-5, 5-12-13, 8-15-17, 7-24-25, 9-40-41
Area of triangle:
= (1/2) x base x height
= sqrt[s(s-a)(s-b)(s-c)] (Heron's formula, s = perimeter/2)
= (1/2) x a x b x sin(C) (using two sides and included angle)
Centroid: divides median in 2:1 ratio from vertex
Circumradius R = (a x b x c) / (4 x Area)
Inradius r = Area / s
For equilateral triangle (side a):
Height = (sqrt(3)/2) x a
Area = (sqrt(3)/4) x a^2
Perimeter = 3a
Circles
Circle Theorems
Angle at centre = 2 × angle at circumference (same arc)
Angles in the same segment are equal
Angle in a semicircle = 90°
Tangent to a circle is perpendicular to the radius at the point of contact
Two tangents from an external point are equal in length
Alternate segment theorem: angle between tangent and chord = inscribed angle in alternate segment
Area and Perimeter of 2D Shapes
Common 2D Shapes with Formulas
All major 2D shapes: Area and Perimeter at a glance
─── 2D AREA AND PERIMETER FORMULAS ───
Shape | Area | Perimeter
----------- | ----------------------- | -----------------
Square (a) | a^2 | 4a
Rectangle | l x b | 2(l + b)
Triangle | (1/2) x b x h | a + b + c
Circle (r) | pi x r^2 | 2 x pi x r
Trapezium | (1/2)(a + b) x h | sum of all sides
Rhombus | (1/2) x d1 x d2 | 4a
Parallelogram | base x height | 2(a + b)
Sector (r,th) | (1/2) x r^2 x theta | r*theta + 2r (theta in radians)
pi ≈ 3.14159 ≈ 22/7 (use 22/7 unless told otherwise)
8. Mensuration (3D Solids)
Three-dimensional mensuration questions frequently appear in BPSC CCE. You must memorise volume and surface area formulas for all standard solids. Watch for combined solids (cone on cylinder, hemisphere on cylinder).
3D Solids — Labelled Diagram
Key 3D solids used in BPSC mensuration problems
─── 3D MENSURATION FORMULAS ───
CUBE (side = a)
Volume = a^3
Total Surface Area (TSA) = 6a^2
Face diagonal = a x sqrt(2)
Space diagonal = a x sqrt(3)
CUBOID (length l, breadth b, height h)
Volume = l x b x h
TSA = 2(lb + bh + lh)
LSA = 2(l + b) x h
Space diagonal = sqrt(l^2 + b^2 + h^2)
CYLINDER (radius r, height h)
Volume = pi x r^2 x h
CSA = 2 x pi x r x h
TSA = 2 x pi x r x (r + h)
CONE (radius r, height h, slant height l)
l = sqrt(r^2 + h^2)
Volume = (1/3) x pi x r^2 x h
CSA = pi x r x l
TSA = pi x r x (r + l)
SPHERE (radius r)
Volume = (4/3) x pi x r^3
SA = 4 x pi x r^2
HEMISPHERE (radius r)
Volume = (2/3) x pi x r^3
CSA = 2 x pi x r^2
TSA = 3 x pi x r^2
FRUSTUM of Cone (radii R and r, height h, slant l)
l = sqrt[h^2 + (R-r)^2]
Volume = (pi x h / 3)(R^2 + r^2 + Rr)
CSA = pi x l x (R + r)
TSA = pi[l(R + r) + R^2 + r^2]
BPSC Common Trap
When water from a cone/cylinder is poured into another shape, volume is conserved — set volumes equal and solve for the unknown dimension. Always check whether the question asks for CSA or TSA; confusing them is a common error.
9. Data Interpretation
Data Interpretation (DI) questions require you to read and analyse data presented as tables, bar graphs, line graphs, or pie charts. BPSC CCE DI sets typically have 4–5 questions per set worth 2–3 marks each. Speed and accuracy in percentage calculations determine your score here.
Statistical Measures
─── MEAN, MEDIAN AND MODE ───
Arithmetic Mean (ungrouped data):
Mean = Sum of all observations / Number of observations
Mean (x-bar) = (x1 + x2 + ... + xn) / n
Arithmetic Mean (grouped data — frequency distribution):
Mean = (Sum of f_i x x_i) / (Sum of f_i)
where f_i = frequency, x_i = class midpoint
Weighted Mean:
= (w1*x1 + w2*x2 + ... + wn*xn) / (w1 + w2 + ... + wn)
Median (ungrouped, n observations sorted in order):
n odd: Median = value at position (n+1)/2
n even: Median = average of values at positions n/2 and (n/2)+1
Median (grouped data — continuous frequency distribution):
Median = L + [(n/2 - cf) / f] x h
where L = lower limit of median class
n = total frequency
cf= cumulative frequency before median class
f = frequency of median class
h = class width
Mode (ungrouped): most frequently occurring value
Mode (grouped):
Mode = L + [(f1 - f0) / (2f1 - f0 - f2)] x h
where f1 = highest frequency, f0 = freq before, f2 = freq after modal class
Empirical relation: Mode ≈ 3*Median - 2*Mean
Reading Data from Charts
Tips for Reading Charts Quickly
Bar Graphs: Each bar represents a category. Read height from the y-axis. For grouped bars, identify the legend colour before computing.
Pie Charts: Total = 360°. Sector angle = (value / total) × 360°. Each 1% corresponds to 3.6°. To compare two sectors, compare their angles or percentages directly.
Line Graphs: Read values at specific points and compute differences or percentage change between years: % change = (difference / base year value) × 100.
Tables: Be careful about units (thousands, lakhs, crores). Scan row and column totals before starting calculations. Use estimation to eliminate obviously wrong options.
Sample Pie Chart — Budget Allocation
Sample pie chart: total budget = Rs 500 crore. Education = 0.30 × 500 = Rs 150 cr
BPSC DI Strategy
Read the question before looking at the data — know what you need to extract.
Convert percentages to actual values (or vice versa) as the first step.
Use approximation: 33% of 480 ≈ 160, not 159.something — elimination works.
For "what fraction of total" questions, set up the ratio before computing.
Watch out for questions asking about a specific year or category — misreading the column is the biggest error source.
10. Solved Examples
The following 10 worked examples represent the type and difficulty level you will encounter in BPSC CCE. Study the method, not just the answer.
Example 1 — LCM/HCF (Number System)
Problem: The LCM and HCF of two numbers are 180 and 12 respectively. If one number is 36, find the other number.
Step 1: Use HCF × LCM = Product of two numbers ⇒ 12 × 180 = 2160
Step 2: Other number = 2160 / 36 = 60
Verify: HCF(36, 60) = 12 ✓; LCM(36, 60) = 180 ✓
Example 2 — Percentage (Profit & Loss)
Problem: A shopkeeper marks goods 40% above cost price and gives a 25% discount. Find his profit or loss percentage.
Step 2: Work done in 4 days = 4 × 3/20 = 12/20 = 3/5
Step 3: Remaining work = 1 − 3/5 = 2/5
Example 6 — Boats and Streams
Problem: A boat travels 24 km downstream in 2 hours and 20 km upstream in 4 hours. Find the speed of the boat in still water and the speed of the stream.