Percentages

Percentages – Book Style Notes (A4 Print)

ToppersGuide • Quantitative Aptitude

Topic: Percentages

Percentages

Concise, exam‑oriented notes with formulas, worked examples, and a memorization table; styled for print (A4) and web.

Definition

Meaning: “Per 100”. For any quantity x% means x × (1/100).
Symbol: We write, for example, 18% = 18/100.
General Formula: Percentage = (Value / Total Value) × 100

Conversion

⇄

Fraction → Percentage: multiply by 100. Example: (3/8) × 100 = 37.5%
(To convert a fraction/decimal to %, multiply by 100: 3/8 × (100/100) = 300/8 × (1/100) → 37.5%)

➗

Percentage → Fraction: divide by 100. Example: 45% = 45/100 = 9/20.

•

Decimal → Percentage: multiply by 100. Example: 0.72 = 72% (since 0.72 × (100/100) = 72 × (1/100) → 72%).

Percentage Increase & Decrease

New Value = Original × (1 + x/100) (increase by x%)

New Value = Original × (1 − x/100) (decrease by x%)

Shortcuts

If a number increases by x% and then decreases by y%, net effect = x − y − (xy/100)%.
If it increases by x% and again by y%, net effect = x + y + (xy/100)%.

Memorize: Common Fractions → Percentages

These values appear frequently in SSC CGL, Banking, and other exams.

Tip: Memorize the row/column structure to recall quickly during exams.

Population Formula

If population increases by r% per year for n years: P_n = P × (1 + r/100)ⁿ.

If population decreases by r% per year for n years: P_n = P × (1 − r/100)ⁿ.

Percentage Comparison

Percentage Change = (Difference / Original) × 100

Example: From 20 to 25 → (25 − 20)/20 × 100 = 25%.

Successive Percentage Change

Net Change = x + y + (xy/100)

Example: Price +20% then −10% ⇒ 20 − 10 − (20×10/100) = +8%.

“x% more / less”

x% more than A = A × (1 + x/100)

x% less than A = A × (1 − x/100)

Profit, Loss, Discount (as %)

Profit% = (Profit / Cost Price) × 100
Loss% = (Loss / Cost Price) × 100
Discount% = (Discount / Marked Price) × 100

Formula Sheet — Quick Scan

Percentage	(Value / Total) × 100
Increase	New = Original × (1 + x/100)
Decrease	New = Original × (1 − x/100)
Successive Change	x + y + (xy/100)
Population (growth)	P_n = P(1 + r/100)ⁿ
Population (decline)	P_n = P(1 − r/100)ⁿ
“x% more”	A(1 + x/100)
“x% less”	A(1 − x/100)
Profit%	(SP − CP)/CP × 100
Loss%	(CP − SP)/CP × 100
Discount%	(MP − SP)/MP × 100

Solved Examples (SSC CGL Level)

Example 1. A number is increased by 20% and then decreased by 10%. Find the net percentage change.
Solution: Net = 20 − 10 − (20×10/100) = 8% increase.

Example 2. Population of a town is 80,000. It increases by 5% every year. Find the population after 2 years.
Solution: 80000 × (1.05)^2 = 88,200.

Example 3. The price rises from ₹600 to ₹750. Find the percentage increase.
Solution: Increase = 150; % increase = (150/600)×100 = 25%.

Example 4. A candidate gets 340 marks which is 85% of total. Find total marks.
Solution: Total = 340 × 100 / 85 = 400.

Example 5. Goods marked 25% above CP, discount 10%. Find profit%.
Solution: Let CP = 100 → MP = 125 → SP = 125 − 10% of 125 = 112.5 → Profit = 12.5 → 12.5%.

Practice MCQs (with Answers)

Q1. A number increased by 25% and then decreased by 20%. What is the net change?

0%
2% increase
5% increase
5% decrease

Answer: a) 0%

Q2. Salary increased by 20% then decreased by 10%. New salary is ₹9,720. What was original salary?

9000
10000
8500
8800

Answer: a) 9000

Q3. A candidate scored 240 out of 300. What percentage did he score?

70%
75%
80%
85%

Answer: c) 80%

Q4. CP = ₹500, SP = ₹600. Profit%?

15%
18%
20%
25%

Answer: c) 20%

Q5. Population decreases by 5% p.a. Present population = 20,000. Find after 2 years.

19,000
18,050
18,100
18,200

Answer: b) 18,050

Number System

Number System — ToppersGuide (Print & Mobile)

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Number System

Definitions, divisibility rules, primes, co-primes, formulas, worked examples and MCQs.

Definition

The Number System is a set of numbers arranged in a systematic order. It helps in expressing numbers in different forms (natural, integers, fractions, decimals, etc.).

Types of Numbers

Natural Numbers (N): Counting numbers {1, 2, 3, 4…}.
Whole Numbers (W): Natural numbers + 0 {0, 1, 2, 3…}.
Integers (Z): Positive and negative numbers including 0 {… –3, –2, –1, 0, 1, 2, 3…}.
Rational Numbers (Q): Numbers that can be expressed as p/q, where q ≠ 0 (e.g., 1/2, –5/3, 0.75).
Irrational Numbers: Cannot be expressed as p/q; non-terminating non-repeating decimals (e.g., √2, π).
Real Numbers (R): All rational and irrational numbers.
Imaginary Numbers: Numbers involving √(–1), denoted by i.
Complex Numbers: Numbers of the form a + bi, where a, b ∈ R.

Divisibility Rules (1–30)

1 – Every number is divisible by 1.
2 – Last digit even (0,2,4,6,8).
3 – Sum of digits divisible by 3.
4 – Last 2 digits divisible by 4.
5 – Last digit 0 or 5.
6 – Divisible by both 2 and 3.
7 – Double the last digit, subtract from remaining; check 7.
8 – Last 3 digits divisible by 8.
9 – Sum of digits divisible by 9.
10 – Last digit 0.
11 – Difference of sum of odd & even place digits divisible by 11.
12 – Divisible by both 3 and 4.
13 – Multiply last digit by 9, add to remaining, check.
14 – Divisible by 2 and 7.
15 – Divisible by 3 and 5.
16 – Last 4 digits divisible by 16.
17 – Subtract 5× last digit from remaining, check.
18 – Divisible by 2 and 9.
19 – Multiply last digit by 2, add to remaining, check.
20 – Last 2 digits divisible by 20 (00,20,40,60,80).
21 – Divisible by 3 and 7.
22 – Divisible by 2 and 11.
23 – Multiply last digit by 7, add to remaining.
24 – Divisible by 3 and 8.
25 – Last two digits 00,25,50,75.
26 – Divisible by 2 and 13.
27 – Sum of groups of 3 digits divisible by 27.
28 – Divisible by 4 and 7.
29 – Multiply last digit by 3, add to remaining.
30 – Divisible by 2, 3, and 5.

Properties of Numbers

Prime number: A number >1 having only 2 factors (1 and itself).
Composite number: Has more than 2 factors.
Co-primes: Two numbers with HCF = 1.
Even / Odd: Divisible by 2 / not divisible by 2.

First 100 Prime Numbers

(Presented in a 4-column table for easy reference)

Prime	Prime	Prime	Prime
2	3	5	7
11	13	17	19
23	29	31	37
41	43	47	53
59	61	67	71
73	79	83	89
97

First 20 Pairs of Co-Primes

Pair 1	Pair 2	Pair 3
(2, 3)	(3, 4)	(4, 5)
(5, 6)	(7, 8)	(8, 9)
(9, 10)	(11, 12)	(13, 14)
(14, 15)	(15, 16)	(16, 17)
(17, 18)	(19, 20)	(20, 21)
(21, 22)	(22, 23)	(23, 24)
(24, 25)	(25, 26)	(26, 27)

Important Formulas

HCF × LCM = Product of two numbers.
Sum of first n natural numbers = n(n+1)/2.
Sum of squares = n(n+1)(2n+1)/6.
Sum of cubes = [n(n+1)/2]^2.
Number of factors of N (p1^e1 × p2^e2 …) = (e1+1)(e2+1)….
Sum of factors = product of (p^(e+1)-1)/(p-1) for each prime.

Important Concepts

Prime factorisation: Expressing a number as product of primes (e.g., 360 = 2^3×3^2×5).

Remainder theorem: For dividend = dq + r, remainder = r.

Cyclicity: Units digits repeat in cycles — use for powers and remainders.

Modular Arithmetic (Basics & Tricks)

Working with residues modulo m. Notation: a ≡ b (mod m) means a and b leave same remainder on division by m.

(a + b) mod m = [(a mod m) + (b mod m)] mod m
(a − b) mod m = [(a mod m) − (b mod m)] mod m
(a × b) mod m = [(a mod m) × (b mod m)] mod m

Example: Find units digit (remainder mod 10) of 7^222. Cycle of 7: (7,9,3,1) length 4 → 222 mod 4 = 2 → units digit = 9.

Euclid’s Algorithm (HCF)

HCF(a,b) = HCF(b, a mod b). Repeat division until remainder 0; last non-zero remainder is HCF.

Example: HCF(270,192) → 270 = 192×1 +78 → 192=78×2+36 → 78=36×2+6 → 36=6×6+0 ⇒ HCF = 6.

Shortcuts & Tricks

Units digit: use cycles (2→4→8→6, etc.).
Zeros in n!: count powers of 5 — [n/5]+[n/25]+[n/125]+….
Use prime factorisation for HCF/LCM quickly.

Solved Examples (SSC CGL Level)

Example 1. HCF & LCM of 24 and 36: 24=2^3×3; 36=2^2×3^2 ⇒ HCF=2^2×3=12; LCM=2^3×3^2=72.

Example 2. Number of factors of 360: 360=2^3×3^2×5 ⇒ (3+1)(2+1)(1+1)=24.

Example 3. Sum of factors of 72: 72=2^3×3^2 ⇒ Sum=(16−1)/(2−1)×(27−1)/(3−1)=15×13=195.

Example 4. Units digit of 7^202: cycle length 4, 202 mod 4 = 2 ⇒ units digit = 9.

Example 5. HCF(210,45) by Euclid: 210=45×4+30; 45=30×1+15; 30=15×2+0 ⇒ HCF=15.

Example 6. Zeros in 100!: [100/5]+[100/25]+[100/125]=20+4+0=24 zeros.

Practice MCQs (with Answers)

Q1. How many factors does 180 have?
a)12
b)18
c)24
d)36
— Answer: b) 18

Q2. Units digit of 3^75?
a)3
b)7
c)9
d)1
— Answer: b) 7

Q3. LCM of 24,36,54?
a)108
b)216
c)432
d)648
— Answer: b) 216

Q4. Remainder when 128 ÷ 9?
a)1
b)2
c)3
d)5
— Answer: b) 2

Q5. Which is not prime?
a)89
b)97
c)111
d)113
— Answer: c) 111

Q6. Zeros at the end of 200!?
a)40
b)45
c)49
d)50
— Answer: c) 49

Q7. If HCF=12 and product=1728, LCM=?
a)72
b)144
c)108
d)120
— Answer: b) 144

Q8. Smallest 5-digit number divisible by 12,15,18?
a)10080
b)10125
c)10260
d)10800
— Answer: a) 10080

Q9. If n=2×3×5×7, how many factors of n are divisible by 35?
a)4
b)6
c)8
d)10
— Answer: c) 8

Q10. Remainder when 15^23 ÷ 8?
a)1
b)3
c)5
d)7
— Answer: d) 7

Averages

ToppersGuide

Topic: Averages · Exam Notes (SSC CGL)

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Averages

Concise, exam-oriented notes with formulas, worked examples and practice MCQs.

Definition

Average (Arithmetic Mean) = Sum of observations ÷ Number of observations.

In other words, Average = Sum of Numbers ÷ Count of Numbers.

Basic Properties

Average always lies between the smallest and the largest value.
If all items are equal, then average = that common value.
Sum = Average × Number of terms.
If the average of n items is a, then the total = n × a.
If a series has equal difference (arithmetic progression), average = middle term of the series.

Important Cases

1. Adding / Removing an Item

If a new item is added:

New average = (Old sum + New item) ÷ (n+1)

If an item is removed:

New average = (Old sum − Removed item) ÷ (n−1)

2. Change in Average

If average increases by x when a new item is added:

New item = Old average + x × (n+1)

If average decreases by x when an item is removed:

Removed item = Old average − x × (n−1)

3. Combined Average

When two groups are combined:

Combined Average = (n₁×A₁ + n₂×A₂) / (n₁ + n₂)

4. Average Speed

If a person travels the same distance at two speeds x and y:

Average speed = (2xy) / (x + y)

For different distances: Average speed = Total distance ÷ Total time.

5. Shortcut for Consecutive Numbers

Average of first n natural numbers = (n+1)/2
Average of first n odd numbers = n
Average of first n even numbers = n+1

Solved Examples

Example 1. Find the average of 10, 20, 30, 40, 50.
Solution: (10+20+30+40+50)/5 = 150/5 = 30.

Example 2. The average weight of 40 students is 50 kg. One student leaves; new average = 49.8 kg. Find weight of the student who left.
Solution: Old sum = 40×50 = 2000. New sum = 39×49.8 = 1942.2. Removed weight = 2000 − 1942.2 = 57.8 kg.

Example 3. Average age of 25 boys = 14 yrs; 15 girls = 13 yrs. Combined average?
Solution: (25×14 + 15×13) / (25+15) = (350+195)/40 = 545/40 = 13.625.

Example 4. A car covers equal distances at 40 km/h and 60 km/h. Average speed?
Solution: (2×40×60)/(40+60) = 4800/100 = 48 km/h.

Practice MCQs (with Answers)

Q1. The average of 8 numbers is 20. If one number is removed, the average becomes 19. Find the removed number.

27
28
35
40

Answer: b) 28

Q2. The average of 10 numbers is 23. If each number is increased by 4, the new average will be:

23
24
27
28

Answer: c) 27

Q3. The average of first 50 natural numbers is:

24
25
25.5
26

Answer: b) 25

Q4. A man travels 120 km at 30 km/h and returns at 40 km/h. Average speed?

34
34.28
35
36

Answer: b) 34.28 km/h

Q5. Average of 15 numbers is 30. If each number is multiplied by 2, new average is:

30
45
60
90

Answer: c) 60

Simplifications

Simplification – ToppersGuide Notes

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Simplification

Definition: Simplification is the process of reducing complex mathematical expressions into simple form by applying arithmetic operations (BODMAS), basic properties, and shortcuts.

Rule of BODMAS

B → Brackets (first solve brackets in the order: (), {}, [])
O → Orders (square, cube, roots, powers)
D → Division
M → Multiplication
A → Addition
S → Subtraction

Models

Model 1: Basic Operations

Example: 25 + 36 ÷ 6 – 4 × 3 = 25 + 6 – 12 = 19

Model 2: Fractions & Decimals

Example: 2/3 + 3/4 = (8/12 + 9/12) = 17/12

Model 3: Surds & Indices

Example: (2^3 × 2^2) ÷ 2^4 = 2^(3+2–4) = 2

Model 4: Squares, Cubes & Roots

Squares to Remember: 1²–30²
Cubes to Remember: 1³–20³

Example: √(144) + ³√(27) = 12 + 3 = 15

Model 5: Percentages in Simplification

Example: 25% of 240 + 12.5% of 160 = 60 + 20 = 80

Model 6: Ratio & Proportion

Example: (2/3 of 180) : (3/5 of 150) = 120 : 90 = 4 : 3

Model 7: Averages

Example: Average of 10, 20, 30, 40 = (100/4) = 25

Model 8: Algebraic Expressions

Example: (x+5)² – (x–5)² = 20x

Model 9: Mixed Operations

Example: (50% of 200 + 3²) ÷ (2/5 of 100) = (100 + 9)/40 = 2.725

Tips & Tricks

Memorize squares (1–30) and cubes (1–20).
Remember fraction ↔ percentage conversions.
Always apply BODMAS strictly.
Convert decimals into fractions for easier simplification.
Use algebraic identities to reduce large calculations.
For percentage-based problems, write them as fractions for speed.

Practice Questions (MCQs)

Q1. Simplify: 45 – 18 ÷ 3 + 2 × 5

34
39
42
44

Answer: b) 39

Q2. Simplify: 3/5 + 2/7

29/35
31/35
33/35
37/35

Answer: b) 31/35

Q3. Simplify: (3^4 × 3^2) ÷ 3^5

3
9
27
81

Answer: a) 3

Q4. Simplify: √225 – ³√64

11
12
13
14

Answer: a) 11

Q5. Simplify: 20% of 150 + 10% of 120

36
38
42
44

Answer: b) 38

Q6. Simplify: (3/4 of 200) : (5/8 of 160)

5:6
3:4
4:5
6:5

Answer: b) 3:4

Q7. Find the average of 8, 16, 24, 32

18
19
20
21

Answer: c) 20

Q8. Simplify: (x+3)² – (x–3)²

6x
9x
12x
18x

Answer: a) 6x

Q9. Simplify: (25% of 160 + √49) ÷ (1/4 of 80)

2.4
2.6
2.8
3

Answer: c) 2.8

Q10. Simplify: (1/2 of 180 + ³√27) ÷ (20% of 100)

9
10
11
12

Answer: b) 10

Ratios and Proportions

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Ratios and Proportions

Definition

Ratio: Comparison of two quantities of the same kind by division.
Example: a : b = a/b

Proportion: When two ratios are equal, they form a proportion.
Example: a : b = c : d → a/b = c/d

Basic Properties

Ratio has no unit (unitless).
Both terms must be in the same unit.
If a:b = m:n, then a/b = m/n.
In a:b = c:d → a & d are extremes, b & c are means. Property: a × d = b × c.
Ratio can be simplified by dividing both terms by their HCF.

Models

Model 1: Simplification of Ratios

Ratio of 200 m to 1 km = 200:1000 = 1:5

Model 2: Comparison of Ratios

Compare 3:4 and 5:8 → 3/4=0.75, 5/8=0.625 → 3:4 is greater.

Model 3: Compounded Ratio

2:3 and 4:5 → (2×4):(3×5)=8:15

Model 4: Inverse Ratio

Inverse of 5:7 = 7:5

Model 5: Ratio of Multiple Quantities

If A:B=2:3 and B:C=4:5 → A:B:C=8:12:15

Model 6: Proportion (Means & Extremes)

2:3=4:6 → 2/3=4/6 → 12=12 (true)

Model 7: Continued Proportion

If 4, b, 25 in continued proportion → b²=4×25=100 → b=10

Model 8: Direct Proportion

5 pens cost ₹50 → 8 pens cost (50×8/5)=₹80

Model 9: Inverse Proportion

12 men in 10 days → 20 men in (12×10/20)=6 days

Model 10: Mixing/Alligation

Rice ₹60/kg & ₹40/kg → mean=₹50/kg → ratio=(60-50):(50-40)=1:1

Tips & Tricks

Always bring values to same unit.
Check a×d=b×c for proportion validity.
Multiply terms for compound/combined ratios.
Use cross multiplication for solving quickly.
Direct proportion = same direction; Inverse proportion = opposite direction.
Alligation shortcut is very useful for mixture sums.

Solved Examples

Ex1: Ratio of 45 mins to 2 hrs = 45:120=3:8

Ex2: Which greater, 7:8 or 13:15? → 7/8=0.875 > 0.866

Ex3: Compound ratio of 2:5 & 3:7 = 6:35

Ex4: Continued proportion: 9,x,25 → x²=225 → x=15

Ex5: 4 kg sugar ₹120 → 7 kg = ₹210

Ex6: 15 men →12 days; 20 men →9 days

Ex7: Water with milk ₹30/l to get ₹20/l → ratio=1:2

Practice MCQs

Q1 (Simplification): Ratio of 2 hrs to 90 mins?

2:3
3:4
4:3
3:2

Answer: c) 4:3

Q2 (Comparison): Which is greater, 5:6 or 7:9?

5:6
7:9
Equal
None

Answer: a) 5:6

Q3 (Compound Ratio): 3:4 and 5:6 compounded = ?

15:24
18:20
5:8
12:20

Answer: a) 15:24

Q4 (Inverse): Inverse of 12:7?

7:12
12:7
1:7
7:1

Answer: a) 7:12

Q5 (Proportion): If 8:12=20:x → x=?

28
30
32
36

Answer: c) 32

Q6 (Continued): If 4,x,16 in proportion → x=?

6
8
10
12

Answer: b) 8

Q7 (Direct): 3 machines →360 units in 12h. 5 machines in 8h=?

400
420
480
500

Answer: c) 480

Q8 (Inverse): 24 men in 15 days. 30 men = ?

10
12
14
18

Answer: b) 12

Q9 (Alligation): Sugar ₹40/kg & ₹50/kg → mixture ₹48/kg. Ratio=?

1:2
2:1
1:4
4:1

Answer: b) 2:1

Q10 (Mixed): a:b=2:3, b:c=4:5 → a:b:c=?

8:12:15
2:4:5
6:9:10
4:5:6

Answer: a) 8:12:15

Profit, Loss and Discount

Profit, Loss & Discount – Notes

ToppersGuide • Quantitative Aptitude

Topic: Profit, Loss & Discount

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Profit, Loss & Discount

Definitions

Cost Price (CP): Price at which article is purchased.
Selling Price (SP): Price at which article is sold.
Profit: If SP > CP → Profit = SP – CP.
Loss: If SP < CP → Loss = CP – SP.
Marked Price (MP): Price marked before discount.
Discount: Rebate given on MP. SP = MP – Discount.

Basic Formulas

Profit% = (Profit ÷ CP) × 100
Loss% = (Loss ÷ CP) × 100
SP = CP × (100 + Profit%) ÷ 100
SP = CP × (100 – Loss%) ÷ 100
CP = SP × 100 ÷ (100 + Profit%)
CP = SP × 100 ÷ (100 – Loss%)
Discount% = (Discount ÷ MP) × 100
SP = MP × (100 – Discount%) ÷ 100

Models & Examples

Model 1: Basics
CP=500, SP=600 → Profit=100, Profit%=20%.

Model 2: Find CP
SP=750, Profit=25% → CP=600.

Model 3: Loss
CP=400, SP=360 → Loss=40 → 10%.

Model 4: Discount
MP=1200, Disc=20% → SP=960.

Model 5: Successive Discounts
1000 with 20% & 10% → SP=720.

Model 6: Profit after Discount
CP=500, MP=700, Disc=10% → SP=630 → Profit=26%.

Model 7: Double Transactions
CP=400 → SP=480 → Final SP=600.

Model 8: False Weights
900g instead of 1kg → Profit%=11.11%.

Model 9: Successive P/L
+20%, -10% → Net=+8%.

Model 10: Equal SP
One 20% gain, one 20% loss → Net Loss=4%.

Tips & Tricks

Take CP=100 in percentage-based sums.
Successive change: Net = x + y + (xy/100).
For successive discounts multiply (100–d1)(100–d2)/100.
If CP:SP ratio known, Profit/Loss% = (Diff ÷ CP) × 100.

Practice MCQs

Q1. CP=800, SP=920. Profit%=?

12%
14%
15%
20%

Ans: 15%

Q2. SP=960, Profit=20%. Find CP.

700
750
800
900

Ans: 800

Q3. SP=540, Loss=10%. Find CP.

560
580
600
620

Ans: 600

Q4. MP=2000, Disc=25%. SP=?

1400
1450
1500
1600

Ans: 1500

Q5. MP=5000, discounts 20% & 10%. SP=?

3500
3600
3650
3700

Ans: 3600

Q6. CP=900, MP=1200, Disc=10%. Profit%=?

20%
25%
30%
33.33%

Ans: 33.33%

Q7. CP=500 → 20% profit → Buyer sells 25% profit. Final SP=?

700
725
750
775

Ans: 750

Q8. Dealer uses 950g for 1kg. Profit%=?

4.76%
5%
5.25%
6%

Ans: 4.76%

Q9. Trader gains 30% then loses 20%. Net=?

2% profit
2% loss
4% profit
10% loss

Ans: 2% profit

Q10. Sold two articles Rs.1200 each, one +25%, one –25%. Net=?

5% profit
5% loss
6.25% profit
6.25% loss

Ans: 6.25% loss

Simple Interest and Compound Interest

Simple & Compound Interest – ToppersGuide

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ToppersGuide • Quantitative Aptitude

Simple & Compound Interest

Simple Interest (SI) & Compound Interest (CI)

Simple Interest (SI): Interest only on the principal.

SI = (P × R × T) / 100

Compound Interest (CI): Interest on both principal and accumulated interest.

Amount = P × (1 + R/100)^T, CI = A – P

Models in Simple Interest

Model 1: Basic SI Calculation

Example: P=5000, R=10%, T=2 → SI = (5000×10×2)/100 = 1000

Model 2: Find Principal / Rate / Time

Example: SI=600, R=10%, T=2 → P = (600×100)/(10×2) = 3000

Model 3: Time in Months/Days

Example: P=2000, R=12%, T=6 months → SI = (2000×12×0.5)/100 = 120

Model 4: Equal Installments (SI)

Example: Repay 1200 in 3 equal SI installments → 400 each

Model 5: Difference Between SI and CI

For 2 years: Difference = P(R/100)^2 → Example: P=1000, R=10% → 10

Models in Compound Interest

Model 1: Basic CI

Example: P=5000, R=10%, T=2 → A=6050 → CI=1050

Model 2: Half-Yearly / Quarterly Compounding

Example: P=8000, R=10%, T=2 yrs (half-yearly) → A=9724 → CI=1724

Model 3: Difference Between CI & SI

For 2 years: P(R/100)^2 → Example: P=5000, R=5% → 12.5

Model 4: Growth / Depreciation

Example: Population 20000 grows 5% in 2 yrs → 22050

Model 5: Equal Installments (CI)

Formula: Installment = P(R/100) / [1 – (1+R/100)^–n]

Model 6: Fractional Years

Example: P=4000, R=10%, T=2.5 yrs → A=4840

Tips & Tricks

For small R & T, CI ≈ SI
Difference matters only for >1 year
Check compounding type: yearly/half-yearly/quarterly
Population/depreciation follow CI formula
Installments under CI need annuity formula

Solved Examples

SI: P=3000, R=12%, T=3 → SI=1080

CI: P=1000, R=10%, T=2 → CI=210

Population: 50000 grows 4% in 2 yrs → 54080

Depreciation: 20000 depreciates 10% for 2 yrs → 16200

Practice MCQs

Q1 (SI Basic): SI on 4000 at 5% for 2 years?

300
350
400
450

Answer: 400

Q2 (SI Reverse): SI=600, R=10%, T=2 → Find P.

2500
2800
3000
3200

Answer: 3000

Q3 (CI Basic): CI on 2000 at 10% for 2 years?

400
420
440
450

Answer: 420

Q4 (Half-Yearly): CI on 4000 at 10% for 1 year, half-yearly?

400
410
420
425

Answer: 410

Q5 (Diff SI-CI): P=1000, R=10%, T=2 → CI–SI?

10
15
20
25

Answer: 10

Q6 (Growth): Pop=20000, grows 5% in 2 yrs → ?

21000
22050
22500
23000

Answer: 22050

Q7 (Depreciation): Machine=30000, 10% depreciation, 2 yrs → ?

24000
24200
24300
24400

Answer: 24200

Q8 (Installments – SI): Loan=1200, 3 equal installments → ?

400
420
440
450

Answer: 400

Q9 (Fractional CI): CI on 8000 at 10% for 1½ yrs (half-yearly)?

1200
1240
1260
1280

Answer: 1240

Q10 (Mixed): Diff between SI & CI on 5000 at 12% for 2 yrs?

36
72
60
75

Answer: 75

Algebra

Time and Work

Prepared for Quick Revision

Time and Work

Definition

Time and Work deals with the relationship between the amount of work done, the number of workers, and the time taken to complete the work.

Work = Rate × Time

If a person can do a work in n days, then in 1 day the person does 1/n part of the work.

Models

Model 1: Basic Work Concept

If A can finish a work in n days, work done in 1 day = 1/n.

Example: A alone can finish a work in 10 days → Work per day = 1/10. Work in 5 days = 1/2.

Model 2: Two Persons Working Together

If A can finish in x days, B in y days, then together they complete 1/x + 1/y work per day.

A = 10 days, B = 15 days → Together = 1/10 + 1/15 = 1/6 → 6 days.

Model 3: Three or More Persons

Per day work = (1/a + 1/b + 1/c).

A=12, B=18, C=36 → Work/day = 1/6 → 6 days.

Model 4: Alternate Days Work

If A and B work on alternate days, calculate total work in 2 days.

A=6, B=12 → In 2 days (1/6+1/12)=1/4 → 8 days total.

Model 5: Efficiency Ratio Method

Efficiency ∝ Work, Time ∝ 1/Efficiency.

A is twice efficient as B → Efficiency A:B=2:1 → Time=1:2.

Model 6: Work and Wages

Wages ∝ Work done.

A=12 days, B=15 days. Together earn 5400 → Ratio 5:4 → A=3000, B=2400.

Model 7: Pipes and Cisterns

Inlet fills (+), Outlet empties (–). Net rate = (Inlet – Outlet).

A fills in 3 hrs, B empties in 6 hrs → Net=1/6 → 6 hrs.

Model 8: Work Interrupted

6 men finish in 12 days → 72 man-days. After 6 days (36 done). Remaining=36. With 4 men → 9 days.

Model 9: Work Equivalence

Useful for ratio problems.

A=10, B=15. A’s 2 days = B’s 3 days.

Model 10: Pipes with Leakage

If pipe fills in x hrs, leak empties in y hrs → Net=1/x–1/y.

Fill=8 hrs, Leak=12 hrs → Net=1/24 → 24 hrs.

Tips & Tricks

Convert into work per day/hour.
Use LCM for time/efficiency ratio.
Efficiency ∝ 1/Time.
In pipes, filling = +, leakage = –.
For alternate days, take a 2-day cycle.

Solved Examples

Ex1: A=12 days → 1 day=1/12. In 6 days=1/2.

Ex2: A=20, B=30 → Together=1/12 → 12 days.

Ex3: A=10, B=15 → Efficiency=3:2.

Ex4: A=10, B=15 → Earn 750. Ratio=3:2 → A=450.

Ex5: Fill=4 hrs, Empty=6 hrs → Net=1/12 → 12 hrs.

Practice MCQs

Q1 (Basic): A can do a work in 15 days. How much in 5 days?

1/2
1/3
1/4
1/5

Answer: 1/3

Q2 (Two Persons): A=12, B=18. Together=?

6
7
8
9

Answer: 6 days

Q3 (Three Persons): A=15, B=20, C=30. Together=?

8
9
10
12

Answer: 10 days

Q4 (Alternate Days): A=10, B=20. Alternate=?

12
13
14
15

Answer: 13 days

Q5 (Efficiency): A twice B. Together=18 days. Alone B=?

27
36
45
54

Answer: 45 days

Q6 (Wages): A=10, B=15. Earn=600. A’s share?

350
360
400
450

Answer: 400

Q7 (Pipes): A=3 hrs, B=6 hrs (outlet). Fill=?

4
5
6
7

Answer: 6 hrs

Q8 (Leaving): 8 men 10 days. After 5 days, 2 leave. More days?

6
6.25
7
8

Answer: 6.25 days

Q9 (Equivalence): A=12, B=18. A’s 2 days=B’s?

2
3
4
5

Answer: 3 days

Q10 (Leakage): Fill=8, Leak=12. Fill time?

16
18
20
24

Answer: 24 hrs

Time and Distance

Time, Distance, Trains, Boats & Streams – Book Style Notes

Prepared for Quick Revision

ToppersGuide • Quantitative Aptitude

Topic: Time, Distance, Trains, Boats & Streams

Time, Distance, Trains, Boats & Streams

Complete notes with definitions, models, tips, solved examples, and practice MCQs.

Definition

Distance = Speed × Time

Speed = Distance / Time | Time = Distance / Speed

Convert units: 1 km/h = 5/18 m/s, 1 m/s = 18/5 km/h.

Models

Model 1: Basic Relation

Example: A car covers 120 km in 2 hrs → Speed = 60 km/h.

Model 2: Unit Conversion

Example: 72 km/h = 72 × 5/18 = 20 m/s.

Model 3: Average Speed

If equal distances at two speeds x, y → 2xy/(x+y)

Example: Speeds 30 & 50 → Avg = 37.5 km/h.

Model 4: Relative Speed

Opposite directions = sum, same = difference.

Example: 40 and 60 → Opposite = 100 km/h.

Model 5: Train Crossing a Point

Time = Length / Speed.

Example: 120 m train crosses pole in 6 s → 20 m/s = 72 km/h.

Model 6: Trains Crossing Each Other

Time = (Sum of lengths)/(Relative speed).

Example: Lengths 150, 120; speeds 50 & 40 → 10.8 s.

Model 7: Boats – Upstream & Downstream

Downstream = B+S, Upstream = B–S.

Example: Boat=12, Stream=3 → DS=15, US=9 km/h.

Model 8: Effective Speed (Boats)

If times downstream & upstream given, use harmonic mean formulas.

Model 9: Circular Track

Meeting time = Circumference/Relative speed.

Model 10: Walking Around Circular Field

Time = Circumference/Speed.

Tips & Tricks

Check units always.
Use km/h ↔ m/s conversions often.
Average speed formula only for equal distances.
For trains, convert length into meters, speed into m/s.
Downstream = B+S, Upstream = B–S.
Opposite direction = sum, same = difference (relative speed).

Solved Examples

Ex 1: 120 km in 4 hrs → Speed = 30 km/h.

Ex 2: 54 km/h → 15 m/s.

Ex 3: 40 km at 30 & 60 → Avg = 40 km/h.

Ex 4: Train 180 m, 54 km/h → 12 s.

Ex 5: Boat 10, Stream 2 → DS=12, US=8.

Practice MCQs

Q1: A car covers 240 km in 4 hrs. Speed?

50
55
60
65

Answer: 60 km/h

Q2: Convert 90 km/h to m/s.

20
22.5
24
25

Answer: 25 m/s

Q3: Equal distance at 40 & 60. Avg?

45
47.5
48
50

Answer: 48 km/h

Q4: Trains 60 & 40 opposite. Relative speed?

20
40
60
100

Answer: 100 km/h

Q5: Train 200 m, 72 km/h. Cross pole?

8 s
9 s
10 s
12 s

Answer: 10 s

Q6: Train A=120m, B=180m, speeds 36 & 54 opposite. Time?

15
18
20
24

Answer: 18 s

Q7: Boat=12, Stream=4. Upstream?

6
8
10
12

Answer: 8 km/h

Q8: Distance=24. DS=2h, US=4h. Boat speed?

6
8
10
12

Answer: 10 km/h

Q9: Circular track 400m. Speeds 5 & 3 m/s. Meet?

50
100
150
200

Answer: 100 s

Q10: Radius=14m, Speed=7 m/s. Time for one round?

12
13
14
15

Answer: 14 s

Algebra

Algebra – ToppersGuide Notes

ToppersGuide • Quantitative Aptitude

Topic: Algebra

Algebra

Algebra deals with numbers, variables, and operations to form expressions, equations, and identities. It is an essential topic for all competitive exams.

Models in Algebra

Model 1: Basic Identities

Key formulas: (a+b)² = a² + 2ab + b², (a–b)² = a² – 2ab + b², a² – b² = (a+b)(a–b).

Example: Simplify (x+3)² – (x–3)² = 12x.

Model 2: Simplification

2x + 3y – x + 5y = x + 8y

Model 3: Factorization

x² – 9 = (x – 3)(x + 3)

Model 4: Linear Equations

3x + 5 = 20 → x = 5

Model 5: Quadratic Equations

Formula: x = [–b ± √(b² – 4ac)] / 2a

x² – 5x + 6 = 0 → Roots: 2, 3

Model 6: Algebraic Fractions

(x² – 9)/(x – 3) = x + 3

Model 7: Surds & Indices

√50 = 5√2

Model 8: Inequalities

3x – 5 > 7 → x > 4

Model 9: Word Problems

x + y = 30, x – y = 6 → x=18, y=12

Model 10: Special Expressions

If x+1/x=5 → x²+1/x² = 23

Tips & Tricks

Memorize all basic identities up to cubes.
Look for factorization opportunities.
Use quadratic formula only if factorization fails.
In inequalities, reverse sign if multiplied/divided by negative.
For special expressions, square or cube both sides.

Practice MCQs

Q1 (Identities): Simplify (a+b)² – (a–b)².

2ab
4ab
a²–b²
2(a+b)

Answer: 4ab

Q2 (Simplification): Simplify 5x – 2y + 3x + y.

8x – y
8x + y
2x – y
2x + y

Answer: 8x – y

Q3 (Factorization): Factorize x² + 7x + 10.

(x+5)(x+2)
(x+7)(x+10)
(x–5)(x–2)
None

Answer: (x+5)(x+2)

Q4 (Linear Eq.): Solve 2x + 7 = 15.

3
4
5
6

Answer: 4

Q5 (Quadratic): Solve x² – 7x + 10 = 0.

(2,5)
(3,4)
(1,10)
(5,7)

Answer: (2,5)

Q6 (Fractions): Simplify (x²–4)/(x–2).

x–2
x+2
x²+2
None

Answer: x+2

Q7 (Surds): Simplify √98.

7√2
14√2
7√3
14√7

Answer: 7√2

Q8 (Inequalities): Solve 5x – 2 > 18.

x>4
x>5
x<4
x<5

Answer: x>4

Q9 (Word Problem): The sum of two numbers is 20, product is 96. Find numbers.

8,12
10,10
6,14
4,16

Answer: 8,12

Q10 (Special Exp.): If x+1/x=3, find x²+1/x².

5
7
9
11

Answer: 7

Geometry

Geometry Notes - Plain Text

Geometry Notes (Plain Text)

1. Introduction

Geometry is the branch of mathematics that deals with shapes, sizes, angles, figures, and their properties. In exams, questions are usually based on triangles, quadrilaterals, circles, polygons, mensuration, and coordinate geometry.

2. Lines and Angles

A straight angle = 180°.
Angles on a straight line add up to 180°.
Vertically opposite angles are equal.
Sum of angles at a point = 360°.

Example: Two supplementary angles are in the ratio 2:3. Find them. Solution: 2x + 3x = 180 → 5x = 180 → x = 36 → angles = 72°, 108°.

3. Triangles

Sum of interior angles = 180°.
Exterior angle = sum of opposite interior angles.
Types: Equilateral, Isosceles, Scalene, Right-angled.
Pythagoras theorem: Hypotenuse² = Base² + Height² (for right triangles).

Area formulas:

Right triangle = 1/2 × base × height.
Heron’s formula: Area = √[s(s–a)(s–b)(s–c)], where s = (a + b + c) / 2.

Example: Area of a right triangle with base 6 cm and height 8 cm = 1/2 × 6 × 8 = 24 cm².

4. Quadrilaterals

Sum of interior angles = 360°.
Parallelogram: opposite sides and angles equal; diagonals bisect each other.
Rectangle: diagonals equal and bisect; angles 90°.
Rhombus: all sides equal; diagonals intersect at right angles.
Square: all sides equal and angles 90° (properties of rectangle + rhombus).
Trapezium: one pair of parallel sides.

Example: Area of rectangle with length 12 cm and breadth 8 cm = 12 × 8 = 96 cm².

5. Circles

Radius (r): distance from center to circumference.
Diameter (d) = 2r.
Circumference = 2πr.
Area = πr².
Arc length = (θ/360) × 2πr.
Sector area = (θ/360) × πr².

Example: Area of a circle with radius 7 cm = πr² = (22/7) × 7 × 7 = 154 cm².

6. Polygons

Sum of interior angles = (n – 2) × 180° (n = number of sides).
Each interior angle of a regular polygon = [(n – 2) × 180°] / n.

Example: Each angle of a regular hexagon = (6 – 2) × 180 / 6 = 120°.

7. Mensuration (2D and 3D)

2D figures:

Square: Area = a², Perimeter = 4a.
Rectangle: Area = l × b, Perimeter = 2(l + b).
Triangle: Area = 1/2 × base × height.
Circle: Area = πr², Circumference = 2πr.

3D figures:

Cube: Volume = a³, Surface Area = 6a².
Cuboid: Volume = l × b × h, Surface Area = 2(lb + bh + hl).
Cylinder: Volume = πr²h, Curved Surface Area = 2πrh.
Cone: Volume = 1/3 πr²h, Curved Surface Area = πrl.
Sphere: Volume = 4/3 πr³, Surface Area = 4πr².

8. Coordinate Geometry (Basics)

Distance between (x₁, y₁) and (x₂, y₂) = √[(x₂ – x₁)² + (y₂ – y₁)²].
Midpoint of (x₁, y₁) and (x₂, y₂) = ((x₁ + x₂)/2 , (y₁ + y₂)/2).
Slope of line through (x₁, y₁) and (x₂, y₂) = (y₂ – y₁) / (x₂ – x₁).

9. Practice Questions (MCQs)

Two angles of a triangle are 70° and 40°. Find the third angle.
(a) 60° (b) 70° (c) 80° (d) 90°
Find the area of a circle with diameter 14 cm.
(a) 154 cm² (b) 144 cm² (c) 196 cm² (d) 308 cm²
Each angle of a regular pentagon is equal to:
(a) 100° (b) 108° (c) 120° (d) 90°
Volume of a cube of side 5 cm is:
(a) 25 cm³ (b) 75 cm³ (c) 125 cm³ (d) 150 cm³
Distance between points (3,4) and (0,0) is:
(a) 3 (b) 4 (c) 5 (d) 7

Answer Key
1 – (c) 80°
2 – (a) 154 cm²
3 – (b) 108°
4 – (c) 125 cm³
5 – (c) 5

(End of notes) - These notes are written plainly so you can select and copy them directly into Word.

Mensuration

Mensuration Notes

1. Introduction

Mensuration is the branch of mathematics that deals with measurement of 2D (plane figures) and 3D (solid figures) like length, area, volume, and surface area. It is one of the most important topics in competitive exams.

2. Important 2D Figures (Plane Mensuration)

Square
Area = a²
Perimeter = 4a
Diagonal = √2 × a

Rectangle
Area = l × b
Perimeter = 2(l + b)
Diagonal = √(l² + b²)

Triangle
Area = ½ × base × height
Equilateral triangle area = (√3 / 4) × a²
Heron’s formula: Area = √[s(s – a)(s – b)(s – c)], where s = (a+b+c)/2
Sum of interior angles = 180°

Parallelogram
Area = base × height
Perimeter = 2(a + b)

Rhombus
Area = ½ × d₁ × d₂ (d₁ and d₂ are diagonals)
All sides equal

Trapezium
Area = ½ × (sum of parallel sides) × height

Circle
Circumference = 2πr
Area = πr²
Arc length = (θ/360) × 2πr
Sector area = (θ/360) × πr²

3. Important 3D Figures (Solid Mensuration)

Cube
Volume = a³
Surface Area = 6a²
Diagonal = √3 × a

Cuboid
Volume = l × b × h
Surface Area = 2(lb + bh + hl)
Diagonal = √(l² + b² + h²)

Cylinder
Volume = πr²h
Curved Surface Area (CSA) = 2πrh
Total Surface Area (TSA) = 2πr(h + r)

Cone
Volume = (1/3)πr²h
Slant height = √(r² + h²)
CSA = πrl
TSA = πr(l + r)

Sphere
Volume = (4/3)πr³
Surface Area = 4πr²

Hemisphere
Volume = (2/3)πr³
CSA = 2πr²
TSA = 3πr²

4. Tricks & Shortcuts

Units conversion:
1 m = 100 cm
1 cm = 10 mm
1 m² = 10,000 cm²
1 m³ = 1000 litres

Right Triangle (Pythagorean triplets):
Common triplets → (3,4,5), (5,12,13), (7,24,25), (8,15,17)

Volume change shortcut:
If sides change by x%, volume changes ≈ (3x)% for small x. Example: Side ↑ 10% → Volume ↑ ≈ 33.1%.

Area & Volume ratios:
Ratio of areas = (ratio of sides)²
Ratio of volumes = (ratio of sides)³

Relations:
Volume of cone = 1/3 × Volume of cylinder (same base & height)
Volume of sphere = 2/3 × Volume of cylinder (same diameter & height = diameter)

5. Practice Model MCQs

The area of a square is 784 cm². Find its perimeter.
(a) 88 cm (b) 96 cm (c) 112 cm (d) 56 cm

A cuboid of dimensions 5 cm × 4 cm × 3 cm is cut into cubes of side 1 cm. How many cubes will be formed?
(a) 50 (b) 60 (c) 40 (d) 30

The radius of a circle is 7 cm. Find its area.
(a) 154 cm² (b) 144 cm² (c) 308 cm² (d) 196 cm²

The curved surface area of a cylinder is 1760 cm² and its height is 20 cm. Find its radius.
(a) 14 cm (b) 28 cm (c) 22 cm (d) 11 cm

A sphere has radius 21 cm. Find its volume. (Use π = 22/7)
(a) 38592 cm³ (b) 38808 cm³ (c) 35000 cm³ (d) 40000 cm³

6. Answer Key

1 – (a) 88 cm
2 – (b) 60
3 – (a) 154 cm²
4 – (a) 14 cm
5 – (b) 38808 cm³

Trigonometry

Trigonometry - Notes

ToppersGuide • Quantitative Aptitude

Topic: Trigonometry

Trigonometry

Concise, exam-oriented notes with formulas, worked examples, and MCQs.

Definition

Trigonometry is the study of relationships between the angles and sides of a right-angled triangle.

Basic Ratios (for angle θ)

sinθ = Opposite / Hypotenuse
cosθ = Adjacent / Hypotenuse
tanθ = Opposite / Adjacent
cotθ = 1/tanθ = Adjacent / Opposite
secθ = 1/cosθ = Hypotenuse / Adjacent
cosecθ = 1/sinθ = Hypotenuse / Opposite

Models in Trigonometry

Model 1: Values of Trigonometric Ratios

Standard angles: 0°, 30°, 45°, 60°, 90°.

Example: sin30° = 1/2, cos60° = 1/2, tan45° = 1.

Model 2: Trigonometric Identities

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ

Example: Simplify sec²θ − tan²θ = 1.

Model 3: Complementary Angles

sin(90°–θ) = cosθ, cos(90°–θ) = sinθ, tan(90°–θ) = cotθ, etc.

Example: sin60° = cos30° = √3/2.

Model 4: Trigonometric Equations

Example: If cosθ = 1/2, then θ = 60°, 300°.

Model 5: Heights and Distances

Example: A man sees top of tower at 30° elevation from 50 m away → Height = 50 × tan30° = 28.9 m.

Model 6: Sum & Difference Formulas

sin(A ± B) = sinAcosB ± cosAsinB
cos(A ± B) = cosAcosB ∓ sinAsinB
tan(A ± B) = (tanA ± tanB)/(1 ∓ tanAtanB)

Example: sin75° = (√6+√2)/4.

Model 7: Double Angle Formulas

sin2A = 2sinAcosA
cos2A = cos²A – sin²A = 2cos²A – 1 = 1 – 2sin²A
tan2A = 2tanA/(1 – tan²A)

Example: sin(2×30°) = sin60° = √3/2.

Model 8: Triple & Multiple Angle

sin3A = 3sinA – 4sin³A, cos3A = 4cos³A – 3cosA

Example: sin(3×30°) = sin90° = 1.

Model 9: Product-to-Sum & Sum-to-Product

sinAcosB = ½[sin(A+B)+sin(A−B)]

Example: sin20°cos10° = ½[sin30°+sin10°].

Model 10: Simplifications

Example: If sinθ + cosθ = √2, then θ = 45°.

Tips & Tricks

Memorize all standard angle values (0°, 30°, 45°, 60°, 90°).
Remember Pythagorean identities.
Use complementary angle relations to simplify.
Draw right triangles for height–distance problems.
For SSC/Bank exams, focus more on ratios, identities, and applications.

Practice MCQs (with Answers)

Q1. sin60° × cos30° = ?

1/2
√3/2
√3/4
3/4

Answer: c) √3/4

Q2. Simplify (1 + tan²45°).

0
1
2
None

Answer: c) 2

Q3. sin20° = ?

cos20°
cos70°
tan70°
cot20°

Answer: b) cos70°

Q4. If cosθ = 0, θ = ?

0°
45°
90°
180°

Answer: c) 90°

Q5. A pole casts a shadow of 10 m when elevation = 45°. Height?

5 m
10 m
15 m
20 m

Answer: b) 10 m

Q6. sin(45°+30°) = ?

½
(√3+1)/2√2
(√6+√2)/4
√2/2

Answer: c) (√6+√2)/4

Q7. If sinA=3/5, cosA=4/5, find sin2A.

12/25
24/25
7/25
9/25

Answer: b) 24/25

Q8. sin(3×30°) = ?

0
1
√3/2
½

Answer: b) 1

Q9. sin50°cos10° = ?

½(sin60°+sin40°)
½(cos60°+cos40°)
½(sin60°–sin40°)
½(cos60°–cos40°)

Answer: a) ½(sin60°+sin40°)

Q10. If sinθ+cosθ=√2, find θ.

0°
30°
45°
60°

Answer: c) 45°

Data Interpretation

Data Interpretation Notes

1. Introduction to Data Interpretation

Data Interpretation (DI) is the process of analyzing numerical data presented in tables, charts, or graphs.

In exams, DI tests both mathematical ability and speed of calculation.

Most DI sets are based on percentages, averages, ratios, and simple arithmetic.

2. Types of DI Representation

Tables – Data in rows and columns. Skill: quickly locate & compare.
Bar Graphs – Rectangular bars (vertical or horizontal).
Line Graphs – Best for trends over time.
Pie Charts – Circle divided into slices. Total = 360°.

3. Common Concepts in DI

Percentages: % = (Part / Whole) × 100
Ratios: If ratio = a : b → Value of a = (a / (a+b)) × Total
Averages: Average = (Sum of values) / (Number of values)
Speed: Memorize % values: 50%=1/2, 25%=1/4, 20%=1/5, etc.

4. Model DI Questions with Solutions

Example 1: Table-based DI

Company sales (lakh units):

Example 2: Pie Chart

Family expenditure (₹30,000): Pie Chart - CSS Only

Monthly Budget Pie Chart

Food - 30%

Rent - 25%

Education - 20%

Savings - 15%

Misc - 10%

Example 3: Bar Graph

Car production (‘000 units):

2019: A=20, B=30 | 2020: A=25, B=35 | 2021: A=40, B=45

Example 4: Ratio-based DI

Ratio of boys to girls = 3:2, Total=500

5. Practice Model Questions (MCQs)

Set 1: Table

Production (‘000 units): 2017=80, 2018=100, 2019=120, 2020=90, 2021=110

Q1: What is the average production?

(a) 95 (b) 100 (c) 105 (d) 110

Q2: % increase from 2017 to 2019?

(a) 25% (b) 30% (c) 40% (d) 50%

Set 2: Pie Chart

Student monthly expenses (₹20,000): Food=35%, Rent=25%, Travel=20%, Books=10%, Others=10% Student Monthly Expenses Pie Chart

Student Monthly Expenses

Total: ₹20,000

Food - 35%

Rent - 25%

Travel - 20%

Books - 10%

Others - 10%

Q1: Travel expenditure?

(a) ₹2,000 (b) ₹3,000 (c) ₹4,000 (d) ₹5,000

Q2: Ratio of Food to Rent?

(a) 7:5 (b) 5:7 (c) 3:2 (d) 2:3

Set 3: Bar Graph

Production of laptops (‘000): 2018=50, 2019=60, 2020=80, 2021=100

Q1: Average production?

(a) 60 (b) 70 (c) 75 (d) 80

Q2: % increase from 2018 to 2021?

(a) 50% (b) 75% (c) 100% (d) 125%

Set 4: Mixed

Q1: Student scores = 60,70,80,90. Find average.

(a) 70 (b) 75 (c) 80 (d) 85

Q2: Income:Expenditure = 5:3, Income=₹50,000. Find savings.

(a) ₹20,000 (b) ₹25,000 (c) ₹30,000 (d) ₹35,000

Answer Key:
1 – (c), 2 – (c), 3 – (c), 4 – (a), 5 – (b), 6 – (c), 7 – (b), 8 – (a).

Partnership

Partnership – Book Style Notes (A4 Print)

ToppersGuide • Quantitative Aptitude

Topic: Partnership

Partnership

Clear, exam-oriented notes covering all partnership models, formulas, examples and practice MCQs; formatted for A4 print and web.

Definition

Partnership deals with sharing of profit or loss among two or more persons who invest money in a business. The share depends on amount invested and period of investment.

Models in Partnership

Model 1: Simple Partnership (Same Time Period)

If partners invest for the same duration, profit or loss is divided in the ratio of their investments.

Formula: Profit share ∝ Investment.
Example: A invests ₹6000, B invests ₹4000. Profit = ₹2000. Ratio = 6000:4000 = 3:2.
A’s share = (3/5)×2000 = ₹1200, B’s share = ₹800.

Model 2: Partnership with Different Time Periods

When investments are for different durations, share ∝ (Investment × Time).

Example: A: ₹5000 for 12 months, B: ₹6000 for 8 months. Ratio = (5000×12):(6000×8) = 60000:48000 = 5:4.

Model 3: More than Two Partners

Extend same rule to three or more partners: use Investment × Time for each partner.

Example: A: ₹8000×12, B: ₹6000×10, C: ₹4000×6 → ratios 96000:60000:24000 = 8:5:2.

Model 4: Introduction or Withdrawal of Capital (Entry/Exit)

Account for the exact months the capital was invested. Convert years → months if necessary and compute Investment × Time for the period each partner was active.

Example: A starts with ₹9000 for full year (12 months). B joins after 6 months with ₹12000 for remaining 6 months. A’s value = 9000×12 = 108000; B’s = 12000×6 = 72000 → ratio 3:2.

Model 5: Profit Sharing with Salary / Commission

Sometimes partners receive fixed salary/commission first; remaining profit is divided in investment ratio (or agreed ratio).

Example: A and B share equally. Total profit = ₹10000. A gets salary ₹2000 first. Remaining = 8000, divided equally → each ₹4000. A total = 2000+4000=₹6000; B = ₹4000.

Model 6: Working Partner vs Sleeping Partner

Working partner may receive additional fixed amount or percentage for management; then remaining profit is divided in capital ratio.

Example: A invests ₹8000 (working), B invests ₹12000 (sleeping). A gets 10% of total profit for management. Profit = ₹3000 → A’s salary = 300; remainder 2700 split in ratio 2:3 → A = 300 + (2/5)×2700 = 1380; B = (3/5)×2700 = 1620.

Model 7: Partnership with Loss

Loss is shared in the same ratio as profit: based on Investment × Time.

Example: A: ₹5000 for 6 months; B: ₹7000 for 12 months → ratio 5000×6 : 7000×12 = 30000 : 84000 = 5:14. If loss = ₹19000 → A pays (5/19)×19000 = ₹5000; B pays (14/19)×19000 = ₹14000.

Tips & Tricks

Always compute Investment × Time when durations differ.
Use months as the common unit unless stated otherwise.
If investments are equal, share depends only on time and vice versa.
Deduct salaries/commission first (if given) before dividing remaining profit.
When partners join or leave, calculate their active months precisely.

Worked Examples (Short)

Example 1: A invests ₹3000, B invests ₹2000. Profit = ₹2500. Find A’s share.
Solution: Ratio 3000:2000 = 3:2. A’s share = (3/5)×2500 = ₹1500.

Example 2: A invests ₹4000 for 12 months, B invests ₹6000 for 8 months. Find profit ratio.
Solution: A value = 4000×12 = 48000; B value = 6000×8 = 48000 → ratio 1:1.

Example 3 (Entry): A starts with ₹10000 for whole year. After 4 months B joins with ₹15000. Profit at year end = ₹20000. B’s share?
Solution: A value = 10000×12 = 120000; B value = 15000×8 = 120000 → equal shares. B’s share = 20000/2 = ₹10000.

Example 4 (Working Partner): A invests ₹5000, B invests ₹10000. A gets 20% of profit for management. Total profit = ₹6000. Find A’s final share.
Solution: A’s commission = 20% of 6000 = 1200; Remaining = 4800. Capital ratio = 1:2 → A gets (1/3)×4800 = 1600. Total A = 1200+1600 = ₹2800.

Practice Questions (All Models — MCQs)

Q1 (Simple Partnership): A invests ₹3000, B invests ₹2000. Profit = ₹2500. Find A’s share.

₹1000
₹1200
₹1500
₹1800

Answer: c) ₹1500

Q2 (Different Time): A invests ₹4000 for 12 months, B invests ₹6000 for 8 months. Ratio of profits?

4 : 3
3 : 2
2 : 3
1 : 1

Answer: b) 3 : 2

Q3 (More than Two): A invests ₹7000 for 12 months, B ₹6000 for 9 months, C ₹5000 for 6 months. Ratio?

14 : 9 : 5
7 : 6 : 5
28 : 18 : 10
21 : 18 : 15

Answer: c) 28 : 18 : 10

Q4 (Entry/Exit): A starts with ₹10000. After 4 months B joins with ₹15000. At year end profit = ₹20000. B’s share?

₹6000
₹7500
₹9000
₹10000

Answer: b) ₹7500

Q5 (Salary/Commission): A and B invest equally. Profit = ₹12000. A receives salary ₹2000 first. A’s total share?

₹6000
₹7000
₹8000
₹9000

Answer: b) ₹7000

Q6 (Working Partner): A invests ₹5000, B invests ₹10000. A gets 20% extra for managing business. Profit = ₹6000. Find A’s share.

₹1800
₹2000
₹2200
₹2400

Answer: c) ₹2200

Q7 (Loss Sharing): A invests ₹8000 for 6 months, B invests ₹6000 for 12 months. If loss = ₹7000, A’s share of loss?

₹2000
₹2400
₹2800
₹3000

Answer: b) ₹2400

Mixture and Alligation

Mixture and Alligation – Notes

ToppersGuide • Quantitative Aptitude

Topic: Mixture & Alligation

Mixture and Alligation

Concise, exam-oriented notes with formulas, worked examples, and practice MCQs

Definition

Mixture: Mixing two or more varieties of an article (milk & water, two rice qualities) to get a new mean value.
Alligation: Rule to find the ratio in which items at different prices/qualities are mixed to obtain a mixture at a given price/quality.

Models in Mixture & Alligation

Model 1: Mean Value (Weighted Average)

Mean Price = (Sum of cost of items) / (Total quantity)

⇄

Example: A mixes 20 kg rice at ₹40/kg with 30 kg at ₹50/kg.
Mean price = (20×40 + 30×50) / 50 = ₹46/kg.

Model 2: Alligation Rule (Two Mixtures)

The cross-alligation formula gives the ratio quickly:

➗

Rule: Ratio = (High – Mean) : (Mean – Low)
Example: Mix ₹60/kg and ₹40/kg to get ₹50/kg → ratio = (60−50):(50−40) = 10:10 = 1:1.

Model 3: Replacing Mixture (Repeated Replacement)

When some part of mixture is removed and replaced repeatedly, concentration reduces geometrically:

•

Formula: Final quantity of substance = Initial × (1 − x/n)^k
where x = replaced each time, n = total quantity, k = number of replacements.
Example: 100 L milk, remove 10 L and replace with water, repeat 2 times → milk left = 100×(0.9)^2 = 81 L.

Model 4: Profit in Mixtures

Find CP of mixture first, then compare with selling price to compute profit%.

Example: Mix 10 kg at ₹30/kg and 20 kg at ₹40/kg and sell at ₹40/kg.
CP of mixture = (10×30 + 20×40)/30 ≈ ₹36.67/kg.
Profit% = (40 − 36.67)/36.67 × 100 ≈ 9.09%.

Model 5: Three or More Mixtures

Use stepwise alligation or weighted average across all components.

Example: Mix ₹20/kg, ₹30/kg and ₹40/kg to get ₹32/kg.
First combine 20 and 40 around 32: ratio = (40−32):(32−20) = 8:12 = 2:3. Then combine this result with ₹30/kg by weighted average.

Tips & Tricks

Alligation rule mnemonic: High − Mean : Mean − Low.
For repeated replacement, use the formula (1 − x/n)^k to avoid iterative work.
When profit questions appear, compute CP of mixture first to find profit percentage.
For >2 items prefer weighted average; use step-by-step alligation for clarity.
Visual cross-method (draw the cross) speeds up two-item alligation problems.

Formula Sheet — Quick Scan

Mean Price	(Σ qty×price) / (Σ qty)
Alligation (2 items)	(High − Mean) : (Mean − Low)
Repeated replacement	Final = Initial × (1 − x/n)^k
CP of mixture	(Σ qty×CP) / (Σ qty)
Profit %	(SP − CP)/CP × 100

Solved Examples (Exam-style)

Example 1. A shopkeeper mixes 20 kg rice at ₹40/kg with 30 kg at ₹50/kg. Find the mean price.
Solution: Mean = (20×40 + 30×50)/(20+30) = 2300/50 = ₹46/kg.

Example 2. In what ratio should tea at ₹60/kg be mixed with tea at ₹40/kg to get tea at ₹50/kg?
Solution: Ratio = (60−50):(50−40) = 10:10 = 1:1.

Example 3. A 100 L vessel contains milk. 10 L is removed and replaced by water. Repeat 2 times. Milk left?
Solution: Milk left = 100×(1 − 10/100)^2 = 100×0.81 = 81 L.

Example 4. A dishonest seller mixes 10 kg sugar at ₹30/kg and 20 kg at ₹40/kg and sells at ₹40/kg. Find profit%.
Solution: CP = (300+800)/30 ≈ 36.67. Profit% = (40−36.67)/36.67 ×100 ≈ 9.09%.

Practice MCQs (All Models — Answers included)

Q1 (Mean Value): A mixture of 60 L costs ₹20/L. It is made of liquids worth ₹18/L and ₹24/L. Find the ratio.

1:2
2:1
3:2
2:3

Answer: b) 2:1

Q2 (Alligation Rule): In what ratio must sugar at ₹36/kg and ₹48/kg be mixed to get mixture at ₹42/kg?

1:1
2:1
1:2
3:2

Answer: a) 1:1

Q3 (Replacement): A vessel contains 40 L milk. 8 L is withdrawn and replaced with water. Process repeated twice. Find milk left.

25.6 L
27.6 L
28.8 L
30 L

Answer: c) 28.8 L

Q4 (Profit in Mixture): A shopkeeper mixes 10 kg sugar at ₹30/kg with 20 kg at ₹40/kg and sells at ₹40/kg. Find profit%.

5%
7%
9.09%
10%

Answer: c) 9.09%

Q5 (Three Mixtures): In what ratio must water be mixed with milk at ₹20/L so that the mixture is worth ₹16/L?

1:4
2:3
4:1
3:2

Answer: c) 4:1

Pipes and Cisterns

Quantitative Aptitude - Pipes and Cisterns

Pipes and Cisterns

Definition

Pipes and Cisterns problems are based on the concept of time and work. Pipes can fill or empty a tank in a given time. We use the rates of filling or emptying to solve problems.

Models in Pipes and Cisterns

Model 1: Single Pipe Filling

If a pipe fills a tank in T hours, then in 1 hour it fills 1/T part of the tank.

Example: A pipe fills a tank in 6 hours. In 1 hour, it fills 1/6th of the tank.

Model 2: Two Pipes Filling Together

If two pipes fill the tank in T1 and T2 hours respectively, their combined rate = (1/T1 + 1/T2).

Example: A fills in 4 hours, B fills in 6 hours. Together they fill in 1/(1/4+1/6) = 12/5 = 2.4 hours.

Model 3: One Filling, One Emptying

If one pipe fills in T1 hours and another empties in T2 hours, net rate = (1/T1 – 1/T2).

Example: A fills in 5 hours, B empties in 10 hours. Net work = 1/5 – 1/10 = 1/10. Tank filled in 10 hours.

Model 4: Alternate Filling

When two pipes are opened alternately, calculate work done in each 2-hour cycle.

Example: A fills in 4 hours, B fills in 6 hours. Alternate filling: In 2 hours, work = 1/4 + 1/6 = 5/12. Repeat cycles until full.

Model 5: Partial Work Remaining

Calculate total work done by pipes and then remaining work completed by one pipe.

Example: A fills in 3 hours, B in 6 hours. Both work together for 1 hour = 1/3+1/6=1/2. Half tank left, completed by A in 1.5 hours. Total = 2.5 hours.

Tips & Tricks

Always convert to 1-hour (unit time) work for easy calculation.
For emptying pipes, take rate as negative.
LCM of times helps to compute combined work easily.
For alternate filling, calculate cycle work and extend.
Draw a simple work table to avoid mistakes.

Practice Questions (MCQs)

Q1: A pipe fills a tank in 12 hours. In 1 hour, what part of the tank is filled?
a) 1/10 b) 1/12 c) 1/15 d) 1/20 Answer: b) 1/12

Q2: Pipe A fills in 4 hours, Pipe B fills in 6 hours. Together they fill the tank in:
a) 2 hrs b) 2.4 hrs c) 3 hrs d) 5 hrs Answer: b) 2.4 hrs

Q3: Pipe A fills in 5 hrs, Pipe B empties in 10 hrs. Together they fill in:
a) 10 hrs b) 15 hrs c) 20 hrs d) Cannot be filled Answer: a) 10 hrs

Q4: A fills in 4 hrs, B in 6 hrs. If opened alternately starting with A, in how many hrs is the tank full?
a) 4 hrs b) 5 hrs c) 5.2 hrs d) 6 hrs Answer: c) 5.2 hrs

Q5: A fills in 3 hrs, B in 6 hrs. Both work together for 1 hr, remaining work done by A. Time taken?
a) 2 hrs b) 2.5 hrs c) 3 hrs d) 3.5 hrs Answer: b) 2.5 hrs

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Subject-wise Learning Topics

Percentages

Percentages

Definition

Conversion

Percentage Increase & Decrease

Shortcuts

Memorize: Common Fractions → Percentages

Population Formula

Percentage Comparison

Successive Percentage Change

“x% more / less”

Profit, Loss, Discount (as %)

Formula Sheet — Quick Scan

Solved Examples (SSC CGL Level)

Practice MCQs (with Answers)

Number System

Number System

Definition

Types of Numbers

Divisibility Rules (1–30)

Properties of Numbers

First 100 Prime Numbers

First 20 Pairs of Co-Primes

Important Formulas

Important Concepts

Modular Arithmetic (Basics & Tricks)

Euclid’s Algorithm (HCF)

Shortcuts & Tricks

Solved Examples (SSC CGL Level)

Practice MCQs (with Answers)

Averages

Averages

Definition

Basic Properties

Important Cases

1. Adding / Removing an Item

2. Change in Average

3. Combined Average

4. Average Speed

5. Shortcut for Consecutive Numbers

Solved Examples

Practice MCQs (with Answers)

Simplifications

Simplification

Rule of BODMAS

Models

Model 1: Basic Operations

Model 2: Fractions & Decimals

Model 3: Surds & Indices

Model 4: Squares, Cubes & Roots

Model 5: Percentages in Simplification

Model 6: Ratio & Proportion

Model 7: Averages

Model 8: Algebraic Expressions

Model 9: Mixed Operations

Tips & Tricks

Practice Questions (MCQs)

Ratios and Proportions

Ratios and Proportions

Definition

Basic Properties

Models

Model 1: Simplification of Ratios

Model 2: Comparison of Ratios

Model 3: Compounded Ratio

Model 4: Inverse Ratio

Model 5: Ratio of Multiple Quantities

Model 6: Proportion (Means & Extremes)

Model 7: Continued Proportion

Model 8: Direct Proportion

Model 9: Inverse Proportion

Model 10: Mixing/Alligation

Tips & Tricks

Solved Examples

Practice MCQs

Profit, Loss and Discount

Profit, Loss & Discount

Basic Formulas

Models & Examples