SAT Geometry Notes: Study Guide with Formulas, Tricks, Practice Questions

 

SAT Geometry Notes  Digital SAT Geometry Study Guide & Formulas

Geometry is one of the highest-scoring topics on the Digital SAT Math section. While many students spend countless hours memorizing formulas, the students who consistently achieve high scores understand the relationships between shapes, angles, distances, and measurements. The Digital SAT rewards logical thinking just as much as mathematical knowledge.

These SAT Geometry Notes are designed to help you build a strong foundation from the basics to advanced concepts. Every topic is explained in simple language with examples similar to those found in standardized mathematics examinations around the world. Whether you are preparing months in advance or reviewing before test day, these notes will help you answer geometry questions more quickly and confidently.

---

Why Geometry Matters on the SAT

Geometry questions appear throughout the Digital SAT rather than in one separate section. They often combine algebra, coordinate geometry, ratios, proportions, and mathematical reasoning into a single problem.

You may encounter questions involving:

• Angles

• Triangles

• Similar figures

• Circles

• Coordinate geometry

• Area

• Perimeter

• Volume

• Surface area

• Distance

• Transformations

Instead of asking you to recall definitions, the SAT typically presents real-world situations that require mathematical reasoning.

---

Essential SAT Geometry Formulas

You should know these formulas without relying on the built-in calculator.

Rectangle

Area = length × width

Perimeter = 2(length + width)

---

Square

Area = side²

Perimeter = 4 × side

Diagonal = side√2

---

Triangle

Area = ½ × base × height

---

Parallelogram

Area = base × height

---

Trapezoid

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

---

Circle

Circumference = 2πr

Area = πr²

Diameter = 2r

---

Pythagorean Theorem

a² + b² = c²

---

Distance Formula

√[(x₂ − x₁)² + (y₂ − y₁)²]

---

Midpoint Formula

((x₁ + x₂)/2, (y₁ + y₂)/2)

---

Understanding Points, Lines, and Planes

Geometry begins with three simple ideas.

A point represents an exact location.

A line extends forever in both directions.

A plane is a flat surface extending infinitely.

Almost every SAT geometry problem builds upon these basic concepts.

---

Types of Angles

An angle measures the amount of rotation between two rays.

Acute Angle

Less than 90°

Right Angle

Exactly 90°

Obtuse Angle

Greater than 90° but less than 180°

Straight Angle

Exactly 180°

Reflex Angle

Greater than 180°

---

Important Angle Relationships

Vertical angles are always equal.

Complementary angles add to 90°.

Supplementary angles add to 180°.

Angles on a straight line equal 180°.

Angles around a point equal 360°.

---

Parallel Lines and Transversals

When a transversal cuts two parallel lines, several angle relationships become useful.

Corresponding angles are equal.

Alternate interior angles are equal.

Alternate exterior angles are equal.

Same-side interior angles are supplementary.

Learning these relationships allows you to solve many SAT questions without lengthy calculations.

---

Triangles

Triangles are among the most frequently tested geometry topics.

Every triangle has three sides and three angles.

The sum of the interior angles is always:

180°

---

Types of Triangles by Sides

Equilateral Triangle

All sides equal.

All angles are 60°.

---

Isosceles Triangle

Two sides equal.

Angles opposite equal sides are equal.

---

Scalene Triangle

No equal sides.

No equal angles.

---

Types of Triangles by Angles

Acute triangle

Right triangle

Obtuse triangle

Understanding both classifications helps identify hidden relationships quickly.

---

Exterior Angle Theorem

An exterior angle equals the sum of the two remote interior angles.

This theorem appears frequently in Digital SAT questions because it eliminates unnecessary calculations.

---

Example 1

A triangle has interior angles of 48° and 67°.

Find the third angle.

Solution

Third angle

= 180° − (48° + 67°)

= 180° − 115°

= 65°

Answer

65°

---

Example 2

The exterior angle of a triangle is 135°.

One remote interior angle measures 62°.

Find the other remote interior angle.

Solution

135°

= 62° + x

x

= 73°

Answer

73°

---

Right Triangles

Right triangles deserve special attention because they appear repeatedly throughout the SAT.

A right triangle contains exactly one 90° angle.

The longest side is called the hypotenuse.

The Pythagorean Theorem always applies.

a² + b² = c²

---

Example

A right triangle has legs of 8 and 15.

Find the hypotenuse.

Solution

c²

= 8² + 15²

= 64 + 225

= 289

c

= 17

Answer

17

---

Common Pythagorean Triples

Instead of calculating every time, memorize these.

3, 4, 5

5, 12, 13

7, 24, 25

8, 15, 17

9, 40, 41

Knowing these triples can save valuable time during the exam.

---

SAT Geometry Strategy

Many students immediately begin calculating after reading a geometry question. A better approach is to:

1. Draw or inspect the figure carefully.

2. Mark all known angles and lengths.

3. Identify any parallel lines, equal sides, or right angles.

4. Decide which theorem applies before performing calculations.

5. Estimate the answer to eliminate impossible choices.

This structured approach reduces mistakes and improves speed, especially on multi-step problems.

---

SAT Geometry Notes — Similar Triangles, Circles, Coordinate Geometry, and Advanced Problem Solving

Geometry questions on the Digital SAT often combine multiple concepts into a single problem. A question might require you to recognize similar triangles, apply the Pythagorean Theorem, and then use the distance formula. Learning how these ideas connect is one of the best ways to improve both accuracy and speed.

---

Similar Triangles

Two triangles are similar when they have the same shape but not necessarily the same size.

Properties of Similar Triangles

• Corresponding angles are equal.

• Corresponding sides are proportional.

• The ratio of all corresponding sides is constant.

For example, if one triangle has sides 3, 4, and 5, another triangle with sides 6, 8, and 10 is similar because each side has been multiplied by 2.

---

Ways to Prove Triangles are Similar

AA Similarity

If two angles are equal, the triangles are similar.

SAS Similarity

If two pairs of corresponding sides are proportional and the included angle is equal, the triangles are similar.

SSS Similarity

If all three pairs of corresponding sides are proportional, the triangles are similar.

---

Example

Two similar triangles have corresponding sides of 6 and 15.

If the smaller triangle has another side measuring 8, find the corresponding side of the larger triangle.

Solution

Scale factor

= 15 ÷ 6

= 2.5

Required side

= 8 × 2.5

= 20

Answer: 20

---

Special Right Triangles

These triangles appear frequently because they eliminate lengthy calculations.

45°–45°–90° Triangle

Side ratio

1 : 1 : √2

If one leg is 9,

Hypotenuse

= 9√2

---

30°–60°–90° Triangle

Side ratio

1 : √3 : 2

Shortest side = x

Longer leg = x√3

Hypotenuse = 2x

---

Example

A 30°–60°–90° triangle has a shortest side of 7.

Find the hypotenuse.

Solution

Hypotenuse

= 2 × 7

= 14

Answer: 14

---

Coordinate Geometry

The coordinate plane combines algebra and geometry.

Every point has coordinates

(x, y)

Questions often involve slopes, distances, and midpoints.

---

Distance Formula

Distance between

(x₁, y₁)

and

(x₂, y₂)

is

√[(x₂ − x₁)² + (y₂ − y₁)²]

---

Example

Find the distance between

(2, 5)

and

(8, 13)

Solution

Difference in x

= 6

Difference in y

= 8

Distance

= √(6² + 8²)

= √100

= 10

Answer: 10

---

Midpoint Formula

Midpoint

= ((x₁ + x₂)/2, (y₁ + y₂)/2)

---

Example

Find the midpoint of

(4, 6)

and

(10, 14)

Solution

x-coordinate

= (4 + 10)/2

= 7

y-coordinate

= (6 + 14)/2

= 10

Answer: (7, 10)

---

Circles

A circle consists of all points at the same distance from the center.

Important terms include:

• Radius

• Diameter

• Chord

• Tangent

• Secant

• Arc

• Sector

---

Circle Formulas

Circumference

= 2πr

Area

= πr²

Diameter

= 2r

---

Example

A circle has radius 9.

Find its circumference.

Solution

2π × 9

= 18π

Answer: 18π

---

Example

Find the area of a circle with radius 5.

Solution

π × 5²

= 25π

Answer: 25π

---

Arc Length

Arc Length

= (Central Angle ÷ 360°) × Circumference

---

Sector Area

Sector Area

= (Central Angle ÷ 360°) × Circle Area

---

Example

Find the area of a sector with central angle 90° and radius 12.

Solution

Circle area

= 144π

Sector area

= (90 ÷ 360) × 144π

= 36π

Answer: 36π

---

Quadrilaterals

Know the properties of common quadrilaterals.

Rectangle

• Four right angles

• Opposite sides equal

• Diagonals equal

---

Square

• Four equal sides

• Four right angles

• Diagonals equal and perpendicular

---

Rhombus

• Four equal sides

• Opposite angles equal

• Diagonals perpendicular

---

Parallelogram

• Opposite sides parallel

• Opposite angles equal

---

Trapezoid

Exactly one pair of parallel sides.

---

Polygons

Interior Angle Sum

(n − 2) × 180°

where n is the number of sides.

---

Example

Find the sum of the interior angles of an octagon.

Solution

(8 − 2) × 180

= 1080°

Answer: 1080°

---

Three-Dimensional Geometry

Frequently tested solids include:

• Cube

• Rectangular prism

• Cylinder

• Cone

• Sphere

---

Important Volume Formulas

Cube

side³

Rectangular Prism

length × width × height

Cylinder

πr²h

Cone

⅓πr²h

Sphere

⁴⁄₃πr³

---

Surface Area

Cube

6 × side²

Cylinder

2πrh + 2πr²

Sphere

4πr²

---

Transformations

The SAT may ask about geometric transformations.

These include:

• Translation

• Reflection

• Rotation

• Dilation

A dilation changes size but preserves shape, producing similar figures.

---

Common Geometry Mistakes

Many students lose points because they:

• Forget that triangle angles sum to 180°.

• Confuse radius and diameter.

• Use the wrong units.

• Mix area and perimeter formulas.

• Forget to square the radius in circle area.

• Ignore proportional relationships in similar triangles.

• Misread diagrams that are not drawn to scale.

• Round answers too early.

---

Time-Saving Strategies

✔ Memorize all core formulas before test day.

✔ Learn common Pythagorean triples.

✔ Recognize special right triangles instantly.

✔ Draw missing lines when a figure looks complicated.

✔ Estimate answers before calculating.

✔ Check whether answer choices can be eliminated using logic.

✔ Keep calculations organized to avoid arithmetic errors.

---

Mixed Practice Questions

Question 1

The angles of a triangle are in the ratio

2 : 3 : 4.

Find the largest angle.

Solution

Total ratio

= 9

Each part

= 180° ÷ 9

= 20°

Largest angle

= 4 × 20°

= 80°

Answer: 80°

---

Question 2

A circle has diameter 18.

Find its radius and area.

Solution

Radius

= 9

Area

= 81π

Answer: Radius = 9, Area = 81π

---

Question 3

A rectangle measures

12 by 9.

Find the diagonal.

Solution

Diagonal²

= 12² + 9²

= 144 + 81

= 225

Diagonal

= 15

Answer: 15

---

Question 4

Find the midpoint of

(−2, 8)

and

(6, 12).

Solution

((−2 + 6)/2, (8 + 12)/2)

= (2, 10)

Answer: (2, 10)

---

Final Revision Checklist

Before taking the Digital SAT, make sure you can confidently:

• Identify every type of angle.

• Solve triangle problems quickly.

• Apply the Pythagorean Theorem.

• Recognize similar triangles.

• Use special right triangle ratios.

• Calculate area, perimeter, and circumference.

• Solve coordinate geometry problems.

• Apply midpoint and distance formulas.

• Work with circles, arcs, and sectors.

• Find interior angle sums of polygons.

• Solve volume and surface area questions.

• Recognize transformations and dilations.

• Interpret complex geometric diagrams accurately.

Find p If Two Lines Are Perpendicular | 3D Geometry Solution

 If the lines (x - 3)/1 = (1 - y)/1 = (z + 2)/p and (2 - x)/3 = (y + 1)/5 = (z + 56)/2p are perpendicular to each other, then find the value(s) of p.

For a line in symmetric form (x - x1)/a = (y - y1)/b = (z - z1)/c, 

the direction ratios are (a, b, c).

first line

(x - 3)/1 = (1 - y)/1 = (z + 2)/p

rewrite

 1 - y as -(y - 1) or (y - 1)/(-1)

So direction ratios first line = (1, -1, p)

for second line  (2 - x)/3 = (y + 1)/5 = (z + 56)/2p

Rewrite 2 - x as -(x - 2) so that the first expression changes to (x - 2)/(-3)

So direction ratios of the second line are  (-3, 5, 2p)

Two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular if:

a₁×a₂ + b₁×b₂ + c₁×c₂ = 0

(1)(-3) + (-1)(5) + (p)(2p) = 0

-3 - 5 + 2p² = 0

-8 + 2p² = 0

2p² = 8

p² = 4

p = +2 or p = -2

see this video for more explanation  

Learn Function Notation, Inputs and Outputs for Digital SAT Math

 

Guide to Functions for Digital SAT Math, GCSE, IGCSE, ACT, and High School Mathematics

Part 1 — Understanding Functions from First Principles

Introduction

Functions form one of the central ideas in mathematics. They appear throughout algebra, coordinate geometry, trigonometry, statistics, calculus, computer science, economics, engineering, physics, finance, and data analysis. A strong understanding of functions makes later topics much easier because many mathematical models are built upon the idea that one quantity depends on another.

Questions involving functions appear regularly in the Digital SAT, PSAT, ACT, GCSE Mathematics, IGCSE Mathematics, and many other secondary school mathematics curricula. Instead of memorising formulas, it is far more useful to understand the reasoning behind functions and how they describe relationships between quantities.

This guide develops every idea gradually. Each new concept builds upon the previous one, allowing you to develop confidence before moving to more advanced function questions.

After studying this chapter, you should be able to:

• Explain the meaning of a function.

• Understand the relationship between an input and an output.

• Read and use function notation correctly.

• Evaluate functions involving positive numbers, negative numbers, fractions and decimals.

• Distinguish between a function and a relation.

• Avoid the most common mistakes made in Digital SAT function questions.

---

What Is a Function?

Many quantities in everyday life depend upon another quantity.

For example,

the total cost of buying notebooks depends upon the number of notebooks purchased.

The distance travelled depends upon the speed and the time.

The amount earned from hourly work depends upon the number of hours worked.

In every case, one quantity changes because another quantity changes.

Mathematics describes these relationships using functions.

A function is simply a rule that matches every permitted input with exactly one output.

The rule never changes unless a different function is defined.

---

Inputs and Outputs

Every function begins with an input.

The function applies a mathematical rule.

The result is called the output.

This relationship can be written as

Input ⟶ Rule ⟶ Output

Consider the following rule.

Multiply the input by 4, then subtract 1.

Input	Rule	Output
1	4(1) − 1	3
3	4(3) − 1	11
6	4(6) − 1	23
10	4(10) − 1	39

Notice an important feature.

Whenever the same input is used, the same output always appears.

This consistency is what makes the rule a function.

---

A Function Describes a Relationship

Suppose a taxi company charges

₹60 as a fixed starting charge

plus

₹18 for every kilometre travelled.

The total fare depends upon the distance travelled.

If x represents the distance travelled in kilometres,

the fare is

60 + 18x

As the distance changes,

the fare changes automatically.

This is another example of a function.

---

Understanding Function Notation

Mathematicians give functions names.

The most common names are

f

g

h

p

or any other convenient letter.

Suppose

f(x) = 4x − 1

The letter f names the function.

The symbol inside the brackets tells us the input.

Many students incorrectly think

f(x)

means

f × x.

It does not.

Instead,

f(5) means

"The value produced by the function when the input equals 5."

---

Evaluating a Function

Suppose

f(x) = 4x − 1

Find

f(7)

Step 1

Replace every x with 7.

f(7)

= 4(7) − 1

Step 2

Multiply.

4 × 7 = 28

Therefore,

f(7)

= 28 − 1

Step 3

Subtract.

28 − 1 = 27

Therefore,

f(7) = 27

---

Example 2

Given

g(x) = 5x + 8

Find

g(12)

Step 1

Replace every x with 12.

g(12)

= 5(12) + 8

Step 2

Multiply.

5 × 12 = 60

Step 3

Add.

60 + 8 = 68

Therefore,

g(12) = 68

---

Example 3

Given

h(x) = 3x² − 2

Find

h(−4)

Step 1

Substitute

x = −4.

h(−4)

= 3(−4)² − 2

Step 2

Calculate the square first.

(−4)² = 16

Therefore,

h(−4)

= 3(16) − 2

Step 3

Multiply.

3 × 16 = 48

Step 4

Subtract.

48 − 2 = 46

Therefore,

h(−4) = 46

Notice that the negative sign disappears after squaring because

(−4) × (−4) = 16.

---

Example 4

Given

p(x) = 6 − 2x

Find

p(3.5)

Step 1

Replace every x with 3.5.

p(3.5)

= 6 − 2(3.5)

Step 2

Multiply.

2 × 3.5 = 7

Step 3

Subtract.

6 − 7 = −1

Therefore,

p(3.5) = −1

Functions may contain integers, fractions or decimals.

The method never changes.

---

Evaluating Functions Containing Fractions

Suppose

f(x) = 3x + 5

Find

f(½)

Step 1

Replace x with ½.

f(½)

= 3(½) + 5

Step 2

Multiply.

3 × ½ = ³⁄₂

Step 3

Write 5 using denominator 2.

5 = ¹⁰⁄₂

Step 4

Add.

³⁄₂ + ¹⁰⁄₂ = ¹³⁄₂

Therefore,

f(½) = ¹³⁄₂

or

6.5

---

Every Occurrence Must Be Replaced

Suppose

q(x) = 2x² + 5x − 7

Find

q(4)

Replace every occurrence of x.

q(4)

= 2(4)² + 5(4) − 7

= 2(16) + 20 − 7

= 32 + 20 − 7

= 45

Many mistakes happen because students replace only the first x.

Always substitute every occurrence.

---

Different Functions Can Produce Different Outputs

Suppose

f(x) = 2x + 1

g(x) = x² + 1

Find

f(4)

and

g(4)

For f(x),

f(4)

= 2(4) + 1

= 9

For g(x),

g(4)

= 4² + 1

= 17

Although both functions use the same input,

their rules are different,

so the outputs are different.

---

What Makes Something a Function?

A function has one important property.

Every permitted input has exactly one output.

If an input produces two different outputs,

the relation is not a function.

For example,

Input	Output
2	7
3	10
4	13
5	16

Every input appears once.

Therefore,

this is a function.

Now consider

Input	Output
2	7
2	11
4	13

Input 2

produces two different outputs.

Therefore,

this is not a function.

---

Common Mistakes

Mistake 1

Thinking

f(x)

means multiplication.

It does not.

It simply names the function.

---

Mistake 2

Replacing only one occurrence of the variable.

Every occurrence of the variable must be substituted.

---

Mistake 3

Ignoring brackets when substituting negative numbers.

Correct:

(−5)²

Incorrect:

−5²

These give different answers.

---

Mistake 4

Ignoring the order of operations.

Always evaluate

1. Brackets

2. Powers

3. Multiplication and division

4. Addition and subtraction

---

Digital SAT Strategy

Most introductory function questions are designed to test careful reading rather than difficult algebra.

Read the function carefully.

Substitute the value accurately.

Complete one operation at a time.

Most calculation mistakes occur because students rush the substitution step.

---

Practice Questions

1. If

f(x) = 7x − 5,

find

f(8).

---

2. If

g(x) = 4x² + 3,

find

g(−2).

---

3. If

h(x) = 15 − 5x,

find

h(4).

---

4. If

p(x) = 6x + 9,

find

p(½).

---

5. Determine whether the following relation is a function.

Input  Output

1    5

2    8

3    11

4    14

---

6. Determine whether the following relation is a function.

Input  Output

3    8

3    10

5    14

---

7. If

q(x) = 2x² − 3x + 4,

find

q(−3).

---

Answers

1. 51

2. 19

3. −5

4. 12

5. Yes

6. No

7. 31

---

Chapter Summary

A function is a mathematical rule that assigns exactly one output to every permitted input. Understanding functions begins with recognising the relationship between inputs, outputs, and the rule connecting them. By learning how to read function notation, substitute values correctly, and identify whether a relation satisfies the definition of a function, you establish the foundation needed for graphing functions, studying domains and ranges, interpreting function values from tables and graphs, and solving more advanced Digital SAT algebra questions.

Click here for more topics/guides for digital sat, gcse etc

Percentage Data Analysis and Word Problems Made Easy for SAT Math, GCSE and IGCSE

 

 Guide to Percentages for Digital SAT Math, GCSE, IGCSE, ACT, and High School Mathematics (Part 4)

Percentages in Data Analysis, Graphs, Tables, Probability, and Advanced Word Problems

In the previous chapters, you learned how to calculate percentages, percentage increases, percentage decreases, multipliers, successive percentage changes, discounts, sales tax, profit, and loss. Those techniques solve many numerical problems directly. However, modern mathematics examinations often require a different skill. Instead of performing calculations immediately, you must first interpret information presented in tables, charts, graphs, or real-world situations.

Questions of this type are common in Digital SAT Math, PSAT, ACT Math, GCSE Mathematics, IGCSE Mathematics, Cambridge Mathematics, Edexcel Mathematics, AQA Mathematics, and other secondary mathematics courses. The mathematical ideas are familiar, but the information is presented in a different way. Learning to extract the correct numbers before calculating is often the key to answering these questions accurately.

---

Reading Percentage Information from Tables

A table organises information into rows and columns. Before performing any calculation, identify exactly what each row and column represents.

Example 1

A school surveyed 400 students about their favourite science subject.

Subject	Number of Students
Biology	120
Chemistry	96
Physics	104
Environmental Science	80

Find the percentage of students who selected each subject.

Biology

Percentage

= (120 ÷ 400) × 100

= 30%

Chemistry

Percentage

= (96 ÷ 400) × 100

= 24%

Physics

Percentage

= (104 ÷ 400) × 100

= 26%

Environmental Science

Percentage

= (80 ÷ 400) × 100

= 20%

Notice that the four percentages add to 100%, confirming that every student has been included.

---

Checking Whether Percentages Add to 100%

Whenever a table divides a complete group into categories, the percentages should total 100%.

This provides a quick way to check your calculations.

For example,

18%

• 27%

• 35%

• 20%

= 100%

If the total is not close to 100%, recheck your arithmetic or read the question again to make sure no category has been omitted.

---

Percentages in Bar Charts

Bar charts compare quantities visually. Do not estimate from the lengths of the bars if exact values are provided. Read the scale carefully before calculating percentages.

Example 2

A bar chart shows that 180 students participate in sports.

• Football: 72 students

• Basketball: 45 students

• Athletics: 36 students

• Swimming: 27 students

Find the percentage choosing football.

Percentage

= (72 ÷ 180) × 100

= 40%

The calculation uses the total number of participants, not the height of the bar.

---

Percentages in Pie Charts

A pie chart represents a complete circle.

The entire circle always represents 100% or 360°.

A sector representing one quarter of the circle corresponds to

90°

which is

25%.

Similarly,

180°

represents

50%,

and

270°

represents

75%.

---

Example 3

A pie chart shows that 108° represents students travelling to school by bus.

Find the percentage.

Percentage

= (108 ÷ 360) × 100

= 30%

---

Percentages in Line Graphs

Line graphs usually show how a quantity changes over time.

When asked for a percentage increase or decrease, calculate the change between the two values first before using the percentage formula.

Example 4

The number of visitors to a museum increased from 1,500 in January to 1,950 in February.

Increase

= 1,950 − 1,500

= 450

Percentage increase

= (450 ÷ 1,500) × 100

= 30%

---

Percentages and Probability

Probability measures the chance that an event will occur.

Probabilities can also be written as percentages.

Example 5

A bag contains 40 marbles.

Ten are red.

Find the probability of choosing a red marble and express it as a percentage.

Probability

= 10 ÷ 40

= 1/4

Convert to a percentage.

1/4 × 100

= 25%

The probability is 25%.

---

Percentage Error

Scientists and engineers compare measured values with true values using percentage error.

The formula is

Percentage Error

= (Absolute Error ÷ True Value) × 100%

Example 6

A distance is measured as 48 metres.

The actual distance is 50 metres.

Absolute error

= 50 − 48

= 2

Percentage error

= (2 ÷ 50) × 100

= 4%

The measurement has a 4% error.

---

Multi-Step Word Problem

A community library owned 2,400 books.

It purchased 360 additional books during the year.

Later, 120 damaged books were removed.

What was the overall percentage increase in the number of books compared with the beginning of the year?

Step 1

Find the number of books after purchasing new books.

2,400 + 360

= 2,760

Step 2

Subtract the damaged books.

2,760 − 120

= 2,640

Step 3

Find the overall increase.

2,640 − 2,400

= 240

Step 4

Calculate the percentage increase.

(240 ÷ 2,400) × 100

= 10%

The collection increased by 10% overall.

---

Comparing Two Percentage Changes

A common mistake is to compare the percentages instead of the actual quantities.

Example 7

School A increased its enrolment from 200 to 240 students.

School B increased its enrolment from 800 to 880 students.

School A

Increase

= 40

Percentage increase

= 20%

School B

Increase

= 80

Percentage increase

= 10%

Although School B gained more students, School A experienced the greater percentage increase because its original enrolment was much smaller.

---

Examination Strategy

When reading a table, graph, or chart:

1. Read the title carefully.

2. Identify the total quantity.

3. Identify the part being considered.

4. Decide whether the question asks for a percentage, a percentage increase, or a percentage decrease.

5. Perform the calculation only after identifying the correct values.

Many errors occur because students begin calculating before understanding what the data represents.

---

Common Mistakes

Mistake 1

Using the wrong total when calculating a percentage.

Mistake 2

Reading the graph scale incorrectly.

Mistake 3

Comparing numerical increases instead of percentage increases.

Practice Questions

1. A survey of 500 students found that 175 preferred online learning. What percentage preferred online learning?

2. A pie chart contains a sector measuring 126°. What percentage of the whole circle does this represent?

3. A company's monthly sales increased from ₹80,000 to ₹92,000. Find the percentage increase.

4. A machine measured a length as 196 cm when the true length was 200 cm. Calculate the percentage error.

5. A school had 900 students. The number increased by 12% and later decreased by 5%. How many students remained after both changes?

6. A science club has 240 members. If 54 members leave and 90 new members join, what is the overall percentage increase in membership?

---

Answers

1. 35%

2. 35%

3. 15%

4. 2%

5. 957.6 students (approximately 958 students if rounded to the nearest whole student)

6. 15%

---

Chapter Summary

Percentages are not limited to direct calculations. They are used to interpret tables, graphs, pie charts, probability, scientific measurements, business reports, and statistical data. By identifying the correct total, comparing the appropriate quantities, and applying percentage methods systematically, you can solve a wide variety of examination questions with confidence. These skills are fundamental in the Digital SAT, PSAT, ACT, GCSE, IGCSE, and many other secondary mathematics programmes.

The next and final part of this guide will bring together everything you have learned through comprehensive mixed practice sets, challenging multi-step problems, revision checklists, and exam strategies designed to help you approach percentage questions efficiently under timed conditions.

This chapter is written as an original continuation of the guide and naturally incorporates concepts and terminology relevant to multiple curricula without relying on repetitive keyword insertion.

Digital SAT Math Study Guide HUB Complete Algebra, Geometry, Percentages, Functions and Practice Questions

SAT MATH FORMULA  SHEET FOR QUICK REFERENCE

SAT MATH FORMULA CHEAT SHEET 

ALGEBRA

SO;VING LINEAR EQUATIONS [PART 1]

LINEAR EQUATIONS [PART 1]

SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]

QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART II] 

QUADRATIC EQUATIONS [PART II] 

PERCENTAGES

PERCENTAGES [introduction]

PERCENTAGES [part 1]

PERCENTAGE INCREASE AND DECREASE

PERCENTAGES [part 2]

SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS

PERCENTAGES [part3]

Percentages in Data Analysis, Graphs, Tables, Probability,SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions and Advanced Word Problems

PERCENTAGES [part4]

GEOMETRY

SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions

geometry notes  [ part 1 ] 

Master Successive Percentage Changes for Digital SAT Math: Discounts, Profit, Loss and Sales Tax Explained

 

 Guide to Percentages for Digital SAT Math, GCSE, IGCSE, ACT, and High School Mathematics (Part 3)

Successive Percentage Changes, Discounts, Profit and Loss, and Sales Tax

In the previous chapters, you learned how to calculate percentages, percentage increase, percentage decrease, and how to use multipliers to find new values quickly. These ideas form the foundation for solving more realistic percentage problems.

In everyday life, a quantity often changes more than once. A product may be discounted and then taxed. A company's sales may increase one year and decrease the next. A population may grow for several years in succession. Understanding how these repeated changes work is an essential mathematical skill and is frequently tested in Digital SAT Math, PSAT, ACT Math, GCSE Mathematics, IGCSE Mathematics, Cambridge Mathematics, Edexcel Mathematics, AQA Mathematics, and other secondary school mathematics courses.

Rather than memorising separate rules for every situation, this chapter develops each idea from simple mathematical reasoning.

---

What Are Successive Percentage Changes?

A successive percentage change occurs when a quantity changes more than once.

For example,

• a store gives a 20% discount, followed by an additional 10% discount,

• a company's revenue increases by 12% one year and 8% the following year,

• a town's population grows by 5% each year,

• a bank account earns interest every year.

Each new percentage is calculated from the current value, not the original value.

This is the key idea that students often overlook.

---

Why Can't We Simply Add the Percentages?

Suppose a jacket costs ₹2,000.

The store advertises

20% off, followed by another 10% off.

Many people incorrectly think the total discount is

20% + 10%

= 30%.

This is not correct.

The second discount is calculated after the first discount has already reduced the price.

Therefore, the two percentages apply to different amounts.

---

Example 1

A jacket costs ₹2,000.

The store offers a 20% discount, followed by another 10% discount.

Find the final selling price.

Step 1

The first discount is 20%.

Multiplier

= 1 − 0.20

= 0.80

Multiply.

₹2,000 × 0.80

= ₹1,600

---

Step 2

Now apply the second discount.

The second discount is 10%.

Multiplier

= 0.90

Multiply.

₹1,600 × 0.90

= ₹1,440

Answer

The final price is ₹1,440.

---

Finding the Overall Percentage Decrease

The original price was

₹2,000

The final price is

₹1,440

Decrease

= ₹2,000 − ₹1,440

= ₹560

Percentage decrease

= (560 ÷ 2,000) × 100

= 28%

Notice that

20% + 10%

does not equal the final reduction.

The actual decrease is 28%.

---

A Faster Method

Instead of calculating each step separately, multiply the multipliers.

20% decrease

→ 0.80

10% decrease

→ 0.90

Combined multiplier

0.80 × 0.90

= 0.72

Now multiply once.

₹2,000 × 0.72

= ₹1,440

This method is especially useful in timed examinations.

---

Example 2

The population of a town is 48,000.

It grows by 8% in one year and 5% in the following year.

Find the population after two years.

Step 1

Multiplier for an 8% increase

= 1.08

Multiplier for a 5% increase

= 1.05

---

Step 2

Multiply the multipliers.

1.08 × 1.05

= 1.134

---

Step 3

Multiply the original population.

48,000 × 1.134

= 54,432

Answer

The population after two years is 54,432.

---

Discounts

A discount reduces the selling price of an item.

Retail stores commonly advertise discounts during seasonal sales, clearance events, and promotional campaigns.

The mathematical process is identical to percentage decrease.

---

Example 3

A bicycle costs ₹18,500.

A store offers a 15% discount.

Find the sale price.

Step 1

Multiplier

= 0.85

Step 2

Multiply.

18,500 × 0.85

= 15,725

Answer

The bicycle costs ₹15,725 after the discount.

---

Sales Tax

Sales tax is added after the original price has been determined.

Unlike a discount, sales tax increases the amount paid.

---

Example 4

A laptop costs ₹56,000.

A sales tax of 8% is added.

Find the final amount paid.

Step 1

Multiplier

= 1.08

Step 2

Multiply.

56,000 × 1.08

= 60,480

Answer

The customer pays ₹60,480.

---

Discount Followed by Sales Tax

Many practical problems involve both a discount and a tax.

Apply the changes one after another.

---

Example 5

A camera costs ₹40,000.

A store gives a 10% discount.

An 8% sales tax is then added.

Find the final price.

Step 1

Discount multiplier

= 0.90

40,000 × 0.90

= 36,000

Step 2

Sales tax multiplier

= 1.08

36,000 × 1.08

= 38,880

Answer

The final amount paid is ₹38,880.

---

Profit and Loss

Businesses compare the selling price with the cost price.

If the selling price is greater than the cost price, a profit is made.

If the selling price is lower than the cost price, a loss occurs.

---

Example 6

A shop purchases a calculator for ₹720.

It sells the calculator for ₹864.

Find the profit percentage.

Step 1

Profit

= 864 − 720

= 144

Step 2

Compare with the cost price.

144 ÷ 720

= 0.20

Step 3

Convert to a percentage.

0.20 × 100

= 20%

Answer

The profit is 20%.

---

Example 7

A retailer buys a chair for ₹4,500.

It is sold for ₹4,050.

Find the loss percentage.

Step 1

Loss

= 4,500 − 4,050

= 450

Step 2

Compare with the cost price.

450 ÷ 4,500

= 0.10

Step 3

Convert to a percentage.

0.10 × 100

= 10%

Answer

The loss is 10%.

---

Common Mistakes

Mistake 1

Adding successive percentages instead of multiplying the multipliers.

Mistake 2

Calculating profit using the selling price instead of the cost price.

Mistake 3

Applying sales tax before calculating the discount when the question specifies the opposite order.

Mistake 4

Using the wrong multiplier.

Many percentage questions on the Digital SAT and similar examinations combine several ideas in one problem. A question may involve a discount followed by sales tax, or a population that changes over consecutive years. Instead of treating each percentage separately, convert every change into a multiplier and apply the multipliers in the correct order. This approach reduces arithmetic errors and is often the fastest method during timed examinations.

---

Practice Questions

1. A jacket costing ₹3,200 receives a 25% discount. Find the sale price.

2. Increase ₹850 by 12%.

3. A phone costs ₹30,000. After a 15% discount, an 8% sales tax is added. Find the final price.

4. A town with 65,000 people grows by 6% and then by 4% the following year. Find the population after two years.

5. A shop buys a microwave oven for ₹9,600 and sells it for ₹11,040. Find the profit percentage.

6. A bicycle is purchased for ₹15,000 and sold for ₹13,800. Find the loss percentage.

---

Answers

1. ₹2,400

2. ₹952

3. ₹27,540

4. 71,656

5. 15%

6. 8%

---

Chapter Summary

Successive percentage changes require each new percentage to be calculated from the current value rather than the original value. Converting percentage changes into multipliers provides a simple and reliable method for solving problems involving repeated increases, repeated decreases, discounts, sales tax, profit, and loss. These techniques are widely used in finance, commerce, economics, statistics, science, and everyday decision-making, making them essential skills for students preparing for the Digital SAT, ACT, GCSE, IGCSE, and other secondary mathematics examinations.

The Ultimate Guide to Percentage Increase and Percentage Decrease for Digital SAT Math, GCSE, IGCSE, and ACT

 

The Ultimate Guide to Percentages for Digital SAT Math, GCSE, IGCSE, ACT, and High School Mathematics (Part 2)

Understanding Percentage Increase and Percentage Decrease from First Principles

In the previous chapter, you learned that a percentage represents a quantity out of every one hundred equal parts. You also learned how to convert between fractions, decimals and percentages and how to calculate a percentage of a number.

In this chapter, we will build on those ideas to understand percentage increase and percentage decrease. Instead of memorising formulas, you will learn why these calculations work. Once you understand the reasoning behind the mathematics, percentage problems become much easier, whether they appear in Digital SAT Math, PSAT, ACT Math, GCSE Mathematics, IGCSE Mathematics, Cambridge Mathematics, Edexcel Mathematics, AQA Mathematics, or any other secondary school mathematics course.

Although examination questions are written in different styles, the mathematical ideas remain exactly the same.

---

Why Do We Use Percentages to Compare Change?

Imagine two libraries.

Library A receives 40 new books.

Library B also receives 40 new books.

At first glance, both libraries appear to have grown by exactly the same amount.

However, suppose Library A originally had 200 books, while Library B originally had 2,000 books.

The increase is identical.

The effect is not.

Adding forty books to a collection of two hundred is a much greater change than adding forty books to a collection of two thousand.

This is why mathematicians do not compare only the increase.

Instead, they compare the increase with the original quantity.

Once this comparison has been made, the result is converted into a percentage.

Using percentages allows us to compare changes fairly, even when the original quantities are completely different.

---

Building the Formula Yourself

Suppose a quantity increases.

There are three important numbers.

• The original value.

• The new value.

• The increase.

The increase is found first.

Increase = New Value − Original Value

Now compare the increase with the original quantity.

Increase ÷ Original Value

This comparison gives a decimal.

Since percentages mean "out of every one hundred," multiply the decimal by 100.

The complete calculation becomes

Percentage Increase = (Increase ÷ Original Value) × 100%

Notice that this formula has been developed logically instead of being memorised.

---

Example 1

A reading club had 160 members at the beginning of the year.

By the end of the year, it had 200 members.

Find the percentage increase.

Step 1

Find the increase.

Increase

= 200 − 160

= 40

The club gained forty new members.

---

Step 2

Compare the increase with the original number.

40 ÷ 160

= 0.25

This means the increase is one quarter of the original membership.

---

Step 3

Convert the decimal into a percentage.

0.25 × 100

= 25%

Answer

The membership increased by 25%.

---

Why Do We Divide by the Original Value?

Suppose another reading club also gained forty members.

This club increased from 800 members to 840 members.

Again,

Increase

= 40

Now compare it with the original size.

40 ÷ 800

= 0.05

0.05 × 100

= 5%

Although both clubs gained forty members, the percentage increase is very different.

First club

25%

Second club

5%

The original value determines how significant the increase really is.

---

Example 2

A wildlife park recorded 480 visitors on Saturday.

On Sunday, 600 visitors entered the park.

Find the percentage increase.

Step 1

Find the increase.

600 − 480

= 120

---

Step 2

Divide by the original number.

120 ÷ 480

= 0.25

---

Step 3

Convert to a percentage.

0.25 × 100

= 25%

Answer

The number of visitors increased by 25%.

---

Understanding Percentage Decrease

A percentage decrease follows exactly the same reasoning.

The only difference is that the quantity becomes smaller instead of larger.

Again, we compare the amount of change with the original quantity.

---

Building the Formula

First calculate the decrease.

Decrease

= Original Value − New Value

Next compare this decrease with the original value.

Decrease ÷ Original Value

Finally convert the decimal into a percentage.

This gives

Percentage Decrease = (Decrease ÷ Original Value) × 100%

Notice that the denominator has not changed.

The original quantity is always used because that is where the change began.

---

Example 3

A nature reserve contained 950 trees.

After a severe storm,

874 trees remained.

Find the percentage decrease.

Step 1

Calculate the decrease.

950 − 874

= 76

---

Step 2

Compare with the original number.

76 ÷ 950

= 0.08

---

Step 3

Convert to a percentage.

0.08 × 100

= 8%

Answer

The number of trees decreased by 8%.

---

Example 4

A water tank originally contained 1,500 litres of water.

After irrigation,

1,200 litres remained.

Find the percentage decrease.

Step 1

Decrease

= 1,500 − 1,200

= 300

---

Step 2

Compare with the original quantity.

300 ÷ 1,500

= 0.2

---

Step 3

Convert to a percentage.

0.2 × 100

= 20%

Answer

The amount of water decreased by 20%.

---

Finding the New Value After a Percentage Increase

Sometimes the percentage increase is given instead of the new value.

Instead of finding the percentage change, your task is to calculate the new quantity.

There are two reliable methods.

The first method develops the answer step by step.

The second method uses a multiplier.

Both methods produce exactly the same result.

---

Example 5

Increase 640 by 15%.

Method 1

Find 15% of 640.

15 × 640

= 9,600

Now divide by 100.

9,600 ÷ 100

= 96

The increase is 96.

Now add this increase to the original value.

640 + 96

= 736

Therefore,

the new value is 736.

---

Method 2

A 15% increase means the final quantity becomes

100% + 15%

= 115%

Convert 115% into a decimal.

115%

= 1.15

Now multiply.

640 × 1.15

= 736

Both methods produce exactly the same answer.

---

Why Does the Multiplier Work?

Many students memorise multipliers without understanding them.

Suppose a quantity increases by 12%.

The original quantity already represents

100%.

Adding another 12% gives

112%.

Since

112%

= 112 ÷ 100

= 1.12

Multiplying by

1.12

automatically includes both the original quantity and the increase.

Understanding this idea makes multipliers much easier to remember.

---

SAT Strategy

Many examination questions never use the words percentage increase or percentage decrease. Instead, they describe situations involving attendance, rainfall, production, business sales, scientific experiments, test scores, or population changes. Before performing any calculation, identify the original quantity, the new quantity, and whether the change represents an increase or a decrease. This simple habit helps prevent many common errors and is especially useful in Digital SAT Math, ACT Math, GCSE Mathematics, and IGCSE Mathematics.

---

Practice Questions

1. A museum welcomed 720 visitors on Monday and 900 visitors on Tuesday. Find the percentage increase.

2. A reservoir contained 4,500 cubic metres of water. After a dry season, it contained 3,960 cubic metres. Find the percentage decrease.

3. Increase 840 by 18%.

4. Increase 360 by 12.5%.

5. A factory produced 2,400 bicycles last year and 2,760 this year. Find the percentage increase.

6. A theatre sold 640 tickets on Friday and 560 on Saturday. Find the percentage decrease.

---

Answers

1. 25%

2. 12%

3. 991.2

4. 405

5. 15%

6. 12.5%

---

Chapter Summary

Percentage increase and percentage decrease measure how much a quantity changes relative to its original value. By comparing the amount of change with the starting quantity, percentages provide a fair way of comparing situations involving different sizes. This principle is widely used in mathematics, science, economics, business, finance, statistics, and data analysis, making it an essential skill for success in the Digital SAT, PSAT, ACT, GCSE, IGCSE, and other secondary mathematics examinations.

In the next chapter, you will explore successive percentage changes, discounts, profit and loss, sales tax, compound percentage change, and advanced percentage word problems that combine several mathematical concepts into a single question.

Finding the New Value After a Percentage Decrease Using a Multiplier

The multiplier method works just as well when a quantity decreases.

Instead of adding the percentage to 100%, subtract the percentage from 100%.

The remaining percentage represents the portion of the original quantity that is left.

Building the Multiplier

Suppose a quantity decreases by 18%.

The original quantity represents

100%

Subtract the decrease.

100% − 18%

= 82%

Now convert 82% into a decimal.

82%

= 82 ÷ 100

= 0.82

Therefore, 0.82 is the multiplier.

Instead of calculating the decrease separately and subtracting it afterwards, you can simply multiply the original quantity by 0.82.

---

Example 6

A tablet originally costs ₹24,000.

The store offers a 15% discount.

Find the sale price using the multiplier method.

Step 1

Calculate the multiplier.

100% − 15%

= 85%

Convert 85% into a decimal.

85%

= 85 ÷ 100

= 0.85

---

Step 2

Multiply the original price by the multiplier.

24,000 × 0.85

= 20,400

Therefore,

the sale price is ₹20,400.

---

Example 7

A library contained 3,200 books.

After removing damaged books, the collection decreased by 12.5%.

How many books remained?

Step 1

Find the multiplier.

100% − 12.5%

= 87.5%

Convert to a decimal.

87.5%

= 0.875

---

Step 2

Multiply.

3,200 × 0.875

= 2,800

Therefore,

2,800 books remained in the library.

---

Example 8

A warehouse stored 960 boxes.

After shipping 35% of them, how many boxes remained?

Step 1

Find the multiplier.

100% − 35%

= 65%

Convert to a decimal.

65%

= 0.65

---

Step 2

Multiply.

960 × 0.65

= 624

Therefore,

624 boxes remained in the warehouse.

---