MHT CET LAW Portal

Module 05 — Basic Mathematics · 5-Year LLB

Module 05 · ⭐ 5-Year LLB Exclusive

Basic Mathematics

Arithmetic, Algebra, Data Interpretation — 10th Standard Level. Score all 8 marks with targeted formula practice and smart shortcuts.

8 Questions 10 Topics ~10 min ⭐ 5-Yr LLB Only
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Basic Mathematics — Module Overview

Mathematics appears only in the 5-Year LLB paper with 8 questions at Class 10 level (Maharashtra Board). This is the most scoring section per time invested — 8 marks in just ~10 minutes if you practice the key formulas!

The topics are straightforward and predictable. With focused preparation using this module, you can reliably score 6–8 out of 8 marks.

Topic-wise Breakdown

📊
Profit & Loss
CP, SP, Profit%, Loss%, Discount, Marked Price
★ 2–3 Questions
💰
SI & CI
Simple Interest, Compound Interest, Rate & Time
★ 1–2 Questions
🚗
Speed, Distance & Time
Relative speed, average speed, train problems
★ 1–2 Questions
Percentage & Average
% change, weighted average, Ratio & Proportion
★ 1 Question
🔣
Algebra
Linear equations, basic identities, word problems
★ 1 Question
Venn Diagrams
Set Theory, Union, Intersection, n(A∪B) formula
★ 1 Question

Exam Strategy for Mathematics (5-Year LLB)

1
Attempt Maths first or last — it is calculation-based; solving it with a clear head avoids careless errors. Most toppers recommend solving GK → Maths → Legal in that order.
2
Memorize 10 core formulas — Profit/Loss, SI/CI, Speed-Distance-Time, and Percentage cover 80% of marks. Practice 5 problems each per formula.
3
Use approximation — For MCQs, options are spread enough that rounding off saves time. If options differ by more than 5%, approximate boldly.
4
No negative marking — Never leave a Mathematics question blank. Eliminate 2 options and make a calculated guess if needed.
5
Target: 10 minutes for all 8 Qs — That's 75 seconds per question. If stuck, skip & return. Never spend more than 2 minutes on a single Maths question.
All formulas are from Class 9–10 Maharashtra Board syllabus. No higher mathematics is tested. Trigonometry, Calculus, and Coordinate Geometry are NOT in scope.

Profit & Loss · Percentage

This is the highest-weightage topic — expect 2–3 questions. Master these formulas and you've guaranteed yourself 2 easy marks every paper.

Core Formulas — Profit & Loss

Profit
SP − CP
when SP > CP
Loss
CP − SP
when CP > SP
Profit %
(Profit / CP) × 100
always on Cost Price
Loss %
(Loss / CP) × 100
always on Cost Price
SP from CP
CP × (100 ± P%) / 100
+ for profit, − for loss
CP from SP
SP × 100 / (100 ± P%)
+ if profit given, − if loss
Discount
MP − SP
Marked Price − Selling Price
Discount %
(Discount / MP) × 100
always on Marked Price
Most Common Mistake: Profit/Loss % is ALWAYS on CP (Cost Price), NOT on Selling Price. Discount % is on Marked Price. Remember this — it's tested every year!

Core Formulas — Percentage & Average

x% of y
(x × y) / 100
% Increase
(Increase / Original) × 100
% Decrease
(Decrease / Original) × 100
Average
Sum / Count
Ratio a:b
a / b
simplify to lowest terms
Proportion
a/b = c/d → ad = bc
cross-multiplication
Solved Example 1 — Profit & Loss

Q. A shopkeeper bought a shirt for ₹500 and sold it for ₹650. Find the profit percentage.

Step 1: Profit = SP − CP = 650 − 500 = ₹150

Step 2: Profit % = (150 / 500) × 100 = 30%

Answer: 30% — This is a classic Type-1 question. Direct formula application.

Solved Example 2 — Finding CP

Q. A book is sold at ₹240, gaining 20% profit. What was the Cost Price?

Step: CP = SP × 100 / (100 + Profit%) = 240 × 100 / 120 = ₹200

Answer: ₹200 — Remember: when profit% is given and you need CP, divide by (100 + P%).

Solved Example 3 — Discount

Q. A TV has a Marked Price of ₹15,000 and is sold at 15% discount. Find the Selling Price.

Step: Discount = 15% of 15,000 = ₹2,250 → SP = 15,000 − 2,250 = ₹12,750

Answer: ₹12,750 — Shortcut: SP = MP × (100 − D%) / 100 = 15000 × 85/100 = ₹12,750

Quick Shortcuts to Remember
  • If an item is sold at x% profit and y% loss on two identical items → Net effect = (x − y − xy/100)%
  • If two items sold at same SP with x% profit and x% loss → Always a net loss of (x²/100)%
  • Successive Discount of a% & b% = Single discount of (a + b − ab/100)%
  • Dishonest shopkeeper using false weight → Gain% = [(True Weight − False Weight) / False Weight] × 100

Simple Interest & Compound Interest

SI & CI contribute 1–2 questions per paper. The formulas are fixed — memorize them once and you'll always get these marks.

Simple Interest (SI)

Simple Interest
SI = (P × R × T) / 100
P = Principal, R = Rate%, T = Time (years)
Total Amount
A = P + SI
Amount = Principal + Interest
Principal (P)
P = (SI × 100) / (R × T)
Rate (R)
R = (SI × 100) / (P × T)
Time (T)
T = (SI × 100) / (P × R)

Compound Interest (CI)

CI Amount
A = P × (1 + R/100)ⁿ
n = number of years
CI
CI = A − P
CI Half-Yearly
A = P × (1 + R/200)²ⁿ
R halved, n doubled
CI for 2 years
CI = P × [2R/100 + (R/100)²]
Quick formula for 2 years
SI vs CI — Key Differences
SIInterest computed on original principal for every period. Linear growth.
CIInterest computed on principal + accumulated interest. Exponential growth.
DiffFor same P, R, T → CI > SI. Difference = P × (R/100)² for 2 years.
Rule of 72Money doubles in ≈ 72/R years at compound interest. Quick mental estimate.
Solved Example — SI

Q. ₹5,000 is invested at 8% per annum for 3 years. Find SI and Total Amount.

SI = (5000 × 8 × 3) / 100 = ₹1,200

Amount = 5000 + 1200 = ₹6,200

Solved Example — CI for 2 Years

Q. ₹10,000 invested at 10% p.a. compound interest for 2 years. Find CI.

A = 10,000 × (1.1)² = 10,000 × 1.21 = ₹12,100

CI = 12,100 − 10,000 = ₹2,100

Quick check: SI for 2 years = ₹2,000. CI > SI ✅. Difference = ₹100 = P × (R/100)² = 10000 × 0.01 ✅

Speed, Distance & Time

A consistent source of 1–2 questions. The golden triangle formula applies to almost every variant of this question type.

The Golden Triangle Formula

Distance
D = S × T
Speed × Time
Speed
S = D / T
Distance ÷ Time
Time
T = D / S
Distance ÷ Speed
km/h to m/s
× 5/18
60 km/h = 60 × 5/18 = 16.67 m/s
m/s to km/h
× 18/5
10 m/s = 10 × 18/5 = 36 km/h
Average Speed
2ab / (a + b)
for same distance at a and b speed

Relative Speed — Key Rules

  • Same direction: Relative Speed = S₁ − S₂ (subtract speeds)
  • Opposite direction: Relative Speed = S₁ + S₂ (add speeds)
  • Train crossing a pole/person: Distance = Length of Train
  • Train crossing a platform: Distance = Length of Train + Length of Platform
  • Two trains crossing each other: Distance = Sum of their lengths; Relative Speed = Sum (opposite) or Difference (same direction)
Solved Example — Train Problem

Q. A 200 m long train runs at 72 km/h. In how many seconds will it cross a 100 m platform?

Step 1: Convert speed → 72 km/h = 72 × 5/18 = 20 m/s

Step 2: Total distance = 200 + 100 = 300 m

Step 3: Time = 300 / 20 = 15 seconds

Solved Example — Average Speed

Q. A car travels from A to B at 60 km/h and returns at 40 km/h. Find average speed.

Average Speed = 2 × 60 × 40 / (60 + 40) = 4800 / 100 = 48 km/h

Note: Average speed ≠ arithmetic mean (60+40)/2 = 50. The harmonic mean formula must be used when same distance at two speeds.

Algebra — Basic Equations

Algebra questions test linear equations, basic identities, and word-problem translation. Class 9–10 level; no quadratic formula required in the exam.

Key Algebraic Identities

(a + b)²
a² + 2ab + b²
(a − b)²
a² − 2ab + b²
(a + b)(a − b)
a² − b²
(a + b)³
a³ + 3a²b + 3ab² + b³
a² + b²
(a+b)² − 2ab
useful when a+b and ab given
a³ + b³
(a + b)(a² − ab + b²)

Solving Linear Equations — Steps

  • Step 1: Translate the word problem into an equation. Assign a variable (x) to the unknown.
  • Step 2: Collect like terms on both sides of the equation.
  • Step 3: Isolate the variable by performing inverse operations.
  • Step 4: Verify by substituting the value back into the original equation.
Solved Example — Age Problem (Word Equation)

Q. The sum of ages of A and B is 45 years. If A is 5 years older than B, find their ages.

Let B's age = x, then A's age = x + 5

Equation: x + (x + 5) = 45 → 2x = 40 → x = 20

✅ B = 20 years, A = 25 years. Verify: 20 + 25 = 45 ✓

Solved Example — Simultaneous Equations

Q. If 2x + y = 7 and x + 2y = 8, find x and y.

Multiply eq. 1 by 2: 4x + 2y = 14

Subtract eq. 2: 3x = 6 → x = 2

Substitute: 2(2) + y = 7 → y = 3

Common Word Problem Types in MHT CET
  • Age Problems: "A is twice as old as B…", "ratio of ages 3:5…"
  • Number Problems: "Sum of two numbers is 50, difference is 10…"
  • Mixture Problems: "Mix 20L of 30% solution with 30L of 20% solution…"
  • Work Problems: "A can do a work in 6 days, B in 9 days. Together…" → Combined rate = 1/6 + 1/9

Venn Diagrams & Set Theory

1 question reliably from Set Theory. The n(A∪B) formula is all you need for MHT CET. Practice reading Venn diagram diagrams and translating the word problem into set notation.

Core Set Theory Formulas

Union
n(A∪B) = n(A) + n(B) − n(A∩B)
Only A (not B)
n(A) − n(A∩B)
Only B (not A)
n(B) − n(A∩B)
Neither A nor B
n(U) − n(A∪B)
U = Universal Set
Three Sets
n(A∪B∪C) = n(A)+n(B)+n(C) −n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)
U Only A A∩B (Both) Only B Set A Set B

Set Notation Glossary

n(A)Number of elements in Set A
A∩BIntersection — elements common to both A and B
A∪BUnion — all elements in A or B (or both)
A'Complement of A — elements NOT in A (but in Universal Set)
Empty set / Null set — no elements
UUniversal Set — contains all elements under consideration
Solved Example — Classic Survey Problem

Q. In a group of 100 students, 60 like cricket, 45 like football, and 25 like both. How many like at least one sport?

n(A∪B) = n(A) + n(B) − n(A∩B) = 60 + 45 − 25 = 80 students

How many like neither? = 100 − 80 = 20 students

How many like only cricket? = 60 − 25 = 35 students

How many like only football? = 45 − 25 = 20 students

Exam Tip: Draw the Venn diagram on your rough sheet before solving! Fill in values starting from the innermost region (intersection) outward. This visual approach eliminates calculation errors in 30 seconds.

Formula Flashcards

Click each card to reveal the formula or answer. Review all 12 cards to mark your progress. ✅ Cards turn green when reviewed.

📊
What is the formula for Profit %?
Click to reveal
Profit % = (Profit / CP) × 100

Always on Cost Price (CP), never on SP!
🏷️
Discount % is calculated on which price?
Click to reveal
Marked Price (MP)

Discount % = (Discount / MP) × 100
SP = MP × (100 − D%) / 100
💰
Simple Interest formula?
Click to reveal
SI = (P × R × T) / 100

P = Principal, R = Rate%, T = Time in years
📈
Compound Interest Amount formula?
Click to reveal
A = P × (1 + R/100)ⁿ

CI = A − P
For 2 years: CI > SI; Diff = P(R/100)²
🚗
Average speed when same distance at speed a and b?
Click to reveal
Avg Speed = 2ab / (a + b)

NOT (a+b)/2 — common mistake!
This is the Harmonic Mean formula.
🚆
How to convert km/h to m/s?
Click to reveal
Multiply by 5/18

km/h → m/s: × 5/18
m/s → km/h: × 18/5
e.g., 72 km/h = 72 × 5/18 = 20 m/s
Venn Diagram: n(A∪B) = ?
Click to reveal
n(A∪B) = n(A) + n(B) − n(A∩B)

"At least one" = n(A∪B)
Neither = n(U) − n(A∪B)
🔣
Expand: (a + b)²
Click to reveal
(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²
(a + b)(a − b) = a² − b²
📉
Two items sold at x% profit and x% loss at same SP → Result?
Click to reveal
Always a Net Loss!

Loss % = x²/100
e.g., 10% profit + 10% loss → 1% net loss
📊
Formula for % increase?
Click to reveal
% Increase = (Increase / Original) × 100

% Decrease = (Decrease / Original) × 100
Original = New Value for reversal problems
🔀
Relative Speed — same vs opposite direction?
Click to reveal
Same direction → S₁ − S₂ (subtract)
Opposite direction → S₁ + S₂ (add)

Train crossing platform: D = L(train) + L(platform)
🧮
Work problem: A does work in 6 days, B in 9 days. Together in how many days?
Click to reveal
Days = 1 / (1/6 + 1/9) = 1 / (5/18) = 18/5 = 3.6 days

Combined rate = sum of individual rates

Practice Quiz — 8 Questions

MHT CET LAW style questions at exact difficulty. Select an answer to see instant explanation. Score 6/8 or above = Exam-Ready! 🏆

1
A shopkeeper buys goods for ₹800 and sells for ₹1,000. What is the profit percentage?
Answer: C — 25%
Profit = 1000 − 800 = ₹200. Profit % = (200/800) × 100 = 25%. Remember: always on CP!
2
₹6,000 is invested at 5% per annum Simple Interest for 4 years. What is the interest earned?
Answer: B — ₹1,200
SI = (P × R × T) / 100 = (6000 × 5 × 4) / 100 = 120000/100 = ₹1,200
3
A car travels 180 km in 3 hours. Find its speed in m/s.
Answer: C — 16.67 m/s
Speed = 180/3 = 60 km/h. Convert: 60 × 5/18 = 300/18 ≈ 16.67 m/s
4
In a survey of 200 people, 120 read newspaper A, 90 read newspaper B, and 40 read both. How many read neither?
Answer: B — 30
n(A∪B) = 120 + 90 − 40 = 170. Neither = 200 − 170 = 30.
5
If (a + b) = 10 and ab = 21, find (a − b).
Answer: C — 4
a² + b² = (a+b)² − 2ab = 100 − 42 = 58. (a−b)² = a²+b²−2ab = 58 − 42 = 16. a−b = 4.
6
₹5,000 at 10% CI for 2 years. Find the Compound Interest.
Answer: C — ₹1,050
A = 5000 × (1.1)² = 5000 × 1.21 = ₹6,050. CI = 6050 − 5000 = ₹1,050. (SI would be ₹1,000; CI > SI by ₹50 = 5000×(0.1)² = ₹50 ✓)
7
The sum of two numbers is 50 and their difference is 14. Find the larger number.
Answer: B — 32
Let x + y = 50, x − y = 14. Adding: 2x = 64, x = 32. Verify: 32 − y = 14, y = 18. Sum = 32+18 = 50 ✓
8
A merchant marks goods at 40% above cost price and allows a 20% discount. Find his profit %.
Answer: C — 12%
Let CP = 100. MP = 140 (40% markup). SP = 140 × 80/100 = 112 (20% discount). Profit = 112−100 = 12. Profit% = 12%.
🎯
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