Friday, July 17, 2026

Guide to Percentages for Digital SAT Math, GCSE, IGCSE, ACT, and High School Algebra

Master Percentages with Step by Step Explanations, Worked Examples

Introduction

Percentages are one of the most useful mathematical concepts you will ever learn. Whether you are calculating a discount while shopping, comparing examination scores, interpreting statistical reports, reading graphs, analysing scientific data, or solving algebra problems, percentages appear almost everywhere.

In mathematics examinations, percentage questions often look simple, but they frequently test several concepts at the same time. A single problem may combine percentages with fractions, decimals, ratios, equations, graphs, probability, data analysis, or financial mathematics. Learning to recognise these connections is an important step towards becoming a confident problem solver.

A solid understanding of percentages is valuable for students preparing for the Digital SAT Math, PSAT, ACT Math, GCSE Mathematics, IGCSE Mathematics, Cambridge IGCSE Mathematics, Edexcel GCSE Mathematics, AQA GCSE Mathematics, OCR GCSE Mathematics, Scottish National 5 Mathematics, and many other secondary school mathematics courses around the world. Although examination styles may differ, the mathematical principles remain exactly the same.

This guide has been written from first principles. Every method is explained carefully, every algebraic step is shown, and every worked example follows a logical sequence so that you understand why each step works instead of simply memorising a formula.

After studying this chapter, you will be able to

• Understand the meaning of a percentage.

• Convert between percentages, fractions and decimals.

• Find the percentage of any quantity.

• Determine what percentage one number is of another.

• Solve percentage increase and percentage decrease problems.

• Apply percentage concepts to algebra and word problems.

• Develop the mathematical reasoning required for college entrance examinations and secondary school mathematics.

What Does the Word Percentage Mean?

The word percentage comes from the Latin phrase meaning per hundred.

Therefore,

1% means 1 out of every 100 equal parts.

Similarly,

10% means 10 parts out of 100.

25% means 25 parts out of 100.

75% means 75 parts out of 100.

100% means the entire quantity.

Understanding this simple idea makes every percentage calculation much easier.

Writing Percentages as Fractions

Every percentage can be written as a fraction whose denominator is 100.

Examples

25%

= 25/100

= 1/4

50%

= 50/100

= 1/2

75%

= 75/100

= 3/4

80%

= 80/100

= 4/5

125%

= 125/100

= 5/4

Notice that percentages greater than 100% are perfectly possible. They simply represent quantities larger than the original amount.

Writing Percentages as Decimals

Many Digital SAT, ACT, GCSE and IGCSE questions require changing percentages into decimals.

The rule is simple.

Divide the percentage by 100.

Examples

45%

= 45 ÷ 100

= 0.45

= 8 ÷ 100

= 0.08

150%

= 150 ÷ 100

= 1.5

0.5%

= 0.5 ÷ 100

= 0.005

Moving the decimal point two places to the left produces exactly the same result.

Converting Decimals into Percentages

To change a decimal into a percentage,

multiply by 100.

Examples

0.6

= 0.6 × 100

= 60%

0.08

= 0.08 × 100

= 8%

1.25

= 1.25 × 100

= 125%

Always remember to write the percentage symbol after multiplying by 100.

Converting Fractions into Percentages

There are two common methods.

Method 1

Convert the fraction into a decimal first.

Example

3/5

Divide.

3 ÷ 5

= 0.6

Multiply by 100.

0.6 × 100

= 60%

Method 2

Multiply the fraction directly by 100.

Example

3/5 × 100

= 300/5

= 60%

Both methods produce the same answer.

Choose whichever method you find easier.

Finding the Percentage of a Number

One of the most common examination questions asks you to calculate a certain percentage of a quantity.

The general rule is

Percentage of a number = Percentage × Number ÷ 100

Example 1

Find 25% of 80.

Step 1

Write the formula.

Percentage of a number

= Percentage × Number ÷ 100

Step 2

Substitute the values.

25 × 80 ÷ 100

Step 3

Multiply.

25 × 80

= 2000

Step 4

Divide by 100.

2000 ÷ 100

= 20

Therefore,

25% of 80 is 20.

Example 2

Find 18% of 250.

Step 1

Write the formula.

Percentage × Number ÷ 100

Step 2

Substitute.

18 × 250 ÷ 100

Step 3

Multiply.

18 × 250

= 4500

Step 4

Divide.

4500 ÷ 100

= 45

Therefore,

18% of 250 equals 45.

Example 3

Find 12.5% of 96.

Step 1

Write the formula.

Percentage × Number ÷ 100

Step 2

Substitute.

12.5 × 96 ÷ 100

Step 3

Multiply.

12.5 × 96

= 1200

Step 4

Divide.

1200 ÷ 100

= 12

Therefore,

12.5% of 96 is 12.

Using Fractions Instead of Percentages

Sometimes converting the percentage into a fraction makes the calculation much faster.

Example

Find 50% of 240.

50%

= 1/2

Half of 240

= 120

No multiplication is necessary.

Find 25% of 64.

25%

= 1/4

One quarter of 64

= 16

Find 75% of 80.

75%

= 3/4

First find one quarter.

80 ÷ 4

= 20

Now multiply by 3.

20 × 3

= 60

This approach is often quicker during timed examinations.

What Percentage Is One Number of Another?

Another common examination question asks

"What percentage is one quantity of another?"

The formula is

Percentage

= (Part ÷ Whole) × 100

Example 4

A class contains 40 students.

Twenty-eight students passed an examination.

What percentage passed?

Step 1

Identify the part.

Step 2

Identify the whole.

Step 3

Use the formula.

(28 ÷ 40) × 100

Step 4

Divide.

28 ÷ 40

= 0.7

Step 5

Multiply.

0.7 × 100

= 70%

Therefore,

70% of the students passed the examination.

Example 5

A football team won 18 matches out of 24.

What percentage of matches did they win?

Step 1

Write the formula.

(Part ÷ Whole) × 100

Step 2

Substitute.

(18 ÷ 24) × 100

Step 3

Simplify.

18 ÷ 24

= 0.75

Step 4

Multiply.

0.75 × 100

= 75%

Therefore,

The team won 75% of its matches.

SAT Strategy

Many Digital SAT, ACT, GCSE and IGCSE questions disguise percentage problems inside word problems, graphs, tables or algebraic expressions. Before beginning any calculation, identify whether the question is asking you to find a percentage of a quantity, what percentage one quantity is of another, or how much a quantity changes by a given percentage. Recognising the type of problem before performing any arithmetic often saves valuable time during an examination.

Practice Questions

Find 35% of 240.
Find 12% of 350.
Find 62.5% of 160.
Express 7/20 as a percentage.
Express 0.84 as a percentage.
What percentage is 45 out of 60?
What percentage is 18 out of 48?
Find 5% of 640.
Find 125% of 48.
A school has 600 students. If 456 students attend on a particular day, what percentage attended?

Answers

84
42
100
35%
84%
75%
37.5%
32
60
76%

Solve the following initial value differential equation (x − 1) dy/dx = 2xy, when y(2) = 1

Solve the following initial value differential equation

(x − 1) dy/dx = 2xy, when y(2) = 1.

This is a variable separable differential equation.

Separate variables

(x − 1) dy/dx = 2xy

dy/y = [2x / (x − 1)] dx

Integrate both sides

∫ dy/y = ∫ [2x / (x − 1)] dx

2x/(x−1) = 2 + 2/(x−1) using long division or manipulation of the numerator

∫ dy/y =∫ [2 + 2/(x − 1)] dx

ln|y| = 2x + 2ln|x − 1| + C

Apply initial condition y(2) = 1

When x = 2, y = 1

ln|1| = 2(2) + 2ln|2 − 1| + C

0 = 4 + 2ln(1) + C

0 = 4 + 0 + C

⇒ C = −4

ln|y| = 2x + 2ln|x − 1| − 4

ln|y| - 2ln|x − 1| = 2x − 4

using property of loagarithms

ln|y| - ln|x − 1|² =2x − 4

ln [|y| / |x − 1|² ] =2x − 4

y = (x − 1)² e^(2x − 4)

see this video for more explanation

cbse 12th applied mathematics variable separable differential equation previous year question papers 2025 2026

Thursday, July 16, 2026

The Coordinates of the Centre of a Circle Are (x − 7, 2x): Find the Value of x if the Circle Passes Through (−9, 11) and Has Radius 5√2 | Step-by-Step Solution

The coordinates of the centre of a circle are (x − 7, 2x). Find the value(s) of ‘x’, if the circle passes through the point (−9, 11) and has radius 5√2 units.

For a circle,

distance between centre and any point on circle = radius

using square of distance formula:

(x₂ − x₁)² + (y₂ − y₁)² = r²

Given

Centre = (x − 7, 2x)

Point on circle = (−9, 11)

Radius r = 5√2

r² = (5√2)² = 25 × 2 = 50

(-9 - (x - 7))² + (11 - 2x)² = 50

(-9 - x + 7)² + (11 - 2x)² = 50

(-x - 2)² + (11 - 2x)² = 50

(x + 2)² + (11 - 2x)² = 50

Expand using identities

(x² + 4x + 4) + (121 - 44x + 4x²) = 50

5x² - 40x + 125 = 50

5x² - 40x + 125 - 50 =0

5x² - 40x + 75 = 0

Divide by 5

x² - 8x + 15 = 0

Factorise:

x² - 5x - 3x + 15 = 0

x(x - 5) - 3(x - 5) = 0

(x - 5)(x - 3) = 0

x = 5 or x = 3

for more explanation watch the video

cbse 10th maths coordinate geometry distance formula previous year question paper 2025 2026

Friday, July 10, 2026

Three pipes A, B and C can together fill a tank in 8 hours. After working at it together for 2 hours, B is closed and A and C fill the remaining part in 9 hours. Determine the time in which pipe B alone can fill the tank.

Three pipes A, B and C can together fill a tank in 8 hours. After working at it together for 2 hours, B is closed and A and C fill the remaining part in 9 hours. Determine the time in which pipe B alone can fill the tank.

A + B + C together fill the tank in 8 hours

So, rate of (A + B + C) = [1/8] tank per hour

Work done by A + B + C in 2 hours = 2 × 1/8 =[ 1/4 ]tank

Remaining work = 1 − 1/4 = [3/4 ]tank

Remaining [3/4] tank is filled by A + C in 9 hours

So, rate of (A + C) = 3/4 × 1/9 = 1/12 tank per hour

Rate of B = Rate of (A + B + C) − Rate of (A + C)

= 1/8 − 1/12

= 3/24 − 2/24 =[ 1/24 [tank per hour

Time taken by B alone = 24 hours

see this video for more explanation

pipes problem, cbse 12th applied maths old board exam question paper 2025 2026

Solving Quadratic Equations by Completing the Square The Digital SAT Math Guide to Quadratic Equations (Part 2)

The Digital SAT Math Guide to Quadratic Equations (Part 2)

Solving Quadratic Equations by Completing the Square

In the previous chapter, you learned how to solve quadratic equations by factoring. Factoring is often the quickest method, but many quadratic equations on the Digital SAT cannot be factored easily. Some have large coefficients, some produce fractional values, and others have no integer factors at all.

For these equations, completing the square provides a systematic method that always works. Unlike factoring, you do not have to guess factor pairs or recognize patterns. Instead, you follow the same sequence of algebraic steps every time.

Before learning the procedure, remember one important rule.

The coefficient of x² should be 1 before you begin completing the square.

A quadratic equation whose coefficient of x² is 1 is called a monic quadratic equation.

If the equation is not monic, divide every term on both sides of the equation by the coefficient of x². This makes the remaining steps much easier and reduces mistakes.

Universal Method for Completing the Square

Always follow these steps.

Write the equation in standard form.
If the coefficient of x² is not 1, divide every term on both sides by that coefficient.
Move the constant term to the opposite side.
Take half of the coefficient of x.
Square that number.
Add the squared value to both sides.
Rewrite the left side as the square of a binomial.
Take the square root of both sides.
Remember both the positive and negative square roots.
Solve the resulting linear equations.
Check every solution in the original equation.

Example 1

Solve

2x² + 12x + 4 = 0

Step 1

The coefficient of x² is 2.

Divide every term on both sides by 2.

2x² ÷ 2 + 12x ÷ 2 + 4 ÷ 2 = 0 ÷ 2

Simplify.

x² + 6x + 2 = 0

The equation is now monic.

Step 2

Subtract 2 from both sides.

x² + 6x + 2 − 2 = 0 − 2

Simplify.

x² + 6x = −2

Step 3

Take half of 6.

6 ÷ 2 = 3

Square it.

3² = 9

Step 4

Add 9 to both sides.

x² + 6x + 9 = −2 + 9

Simplify.

x² + 6x + 9 = 7

Step 5

Rewrite the left side.

(x + 3)² = 7

Step 6

Take square roots.

√((x + 3)²) = ±√7

Simplify.

x + 3 = ±√7

Step 7

Subtract 3 from both sides.

Positive solution:

x = −3 + √7

Negative solution:

x = −3 − √7

Example 2

Now solve a question that produces fractions immediately after making the quadratic monic.

4x² + 10x − 3 = 0

Step 1

The coefficient of x² is 4.

Divide every term by 4.

4x² ÷ 4 + 10x ÷ 4 − 3 ÷ 4 = 0 ÷ 4

Simplify.

x² + ⁵⁄₂x − ³⁄₄ = 0

Notice that fractions are perfectly acceptable. Do not convert them to decimals because exact fractions make later calculations more accurate.

Step 2

Move the constant term.

Add ³⁄₄ to both sides.

x² + ⁵⁄₂x = ³⁄₄

Step 3

Take half of the coefficient of x.

The coefficient is ⁵⁄₂.

Half of ⁵⁄₂ is

⁵⁄₂ ÷ 2 = ⁵⁄₄

Now square the result.

(⁵⁄₄)² = ²⁵⁄₁₆

Step 4

Add ²⁵⁄₁₆ to both sides.

x² + ⁵⁄₂x + ²⁵⁄₁₆ = ³⁄₄ + ²⁵⁄₁₆

Convert ³⁄₄ to sixteenths.

³⁄₄ = ¹²⁄₁₆

Now add.

¹²⁄₁₆ + ²⁵⁄₁₆ = ³⁷⁄₁₆

The equation becomes

x² + ⁵⁄₂x + ²⁵⁄₁₆ = ³⁷⁄₁₆

Step 5

Rewrite the left side.

(x + ⁵⁄₄)² = ³⁷⁄₁₆

Step 6

Take square roots.

√((x + ⁵⁄₄)²) = ±√(³⁷⁄₁₆)

Simplify.

x + ⁵⁄₄ = ±√37⁄4

Step 7

Subtract ⁵⁄₄ from both sides.

x = −⁵⁄₄ ± √37⁄4

These are the exact solutions.

Notice that completing the square works just as well with fractions as it does with whole numbers. On the Digital SAT, leaving answers in exact fractional or radical form is often the correct approach unless the question specifically asks for a decimal approximation.

Thursday, July 9, 2026

The Digital SAT Math Guide to Quadratic Equations (Part 1)

Master Quadratic Equations for the Digital SAT with Step-by-Step Explanations

Quadratic equations are one of the most important algebra topics on the Digital SAT. They appear in many forms, from straightforward equation-solving questions to graph interpretation, mathematical modeling, and real-world word problems. A strong understanding of quadratics also makes it much easier to learn functions, parabolas, coordinate geometry, and polynomial expressions.

Unlike linear equations, which produce straight lines when graphed, quadratic equations create curved graphs called parabolas. Learning how these equations behave will help you answer a wide variety of SAT Math questions quickly and accurately.

This guide is written for students who want to build a solid understanding of quadratics from the ground up. Every solution is explained one step at a time, with no skipped steps or unexplained shortcuts. By the time you finish this chapter, you'll understand what quadratic equations are, how to recognize them, and how to solve many of them by factoring.

Learning Goals

In this chapter, you will learn how to:

Recognize a quadratic equation.
Understand why quadratic equations are different from linear equations.
Identify the standard form of a quadratic equation.
Understand quadratic expressions and quadratic functions.
Solve simple quadratic equations by factoring.
Apply the Zero Product Property.
Check your answers correctly.
Avoid common mistakes made on the Digital SAT.

These concepts form the foundation for more advanced methods such as completing the square and using the quadratic formula, which will be covered in later chapters.

What Is a Quadratic Equation?

A quadratic equation is an equation in which the highest exponent of the variable is 2.

Examples include:

x² = 49

x² + 5x + 6 = 0

2x² − 7x + 3 = 0

4x² = 100

Notice that each equation contains x².

That squared variable is what makes the equation quadratic.

Compare these two equations.

Linear equation:

2x + 7 = 13

Highest exponent = 1

Quadratic equation:

x² + 2x − 15 = 0

Highest exponent = 2

The difference may seem small, but it changes how the equation behaves. A linear equation usually has one solution, while a quadratic equation can have two solutions, one solution, or no real solutions.

The Standard Form of a Quadratic Equation

Most quadratic equations on the SAT are written in standard form:

ax² + bx + c = 0

Each letter has a meaning.

a is the coefficient of x².

b is the coefficient of x.

c is the constant term.

For example,

3x² + 8x − 11 = 0

Here,

a = 3

b = 8

c = −11

Learning to identify these three values is important because later methods, especially the quadratic formula, use them directly.

Understanding the Parts of a Quadratic Equation

Consider

2x² + 9x − 18 = 0

This equation has three terms.

First term:

2x²

This is called the quadratic term because it contains x².

Second term:

This is called the linear term because it contains x.

Third term:

−18

This is the constant term because it contains no variable.

Recognizing these parts helps you identify which solving method to use.

What Does It Mean to Solve a Quadratic Equation?

Solving a quadratic equation means finding every value of the variable that makes the equation true.

For example,

x² = 25

Which numbers produce 25 when squared?

5² = 25

(−5)² = 25

Therefore,

x = 5

and

x = −5

Unlike linear equations, quadratic equations often have more than one correct answer.

Why Are There Two Answers?

Many students are surprised to discover that one equation can have two solutions.

The reason is simple.

Squaring removes the negative sign.

Positive example:

5 × 5 = 25

Negative example:

−5 × −5 = 25

Both calculations produce the same answer.

Whenever you solve an equation involving x², always ask yourself whether both a positive and a negative solution are possible.

Introduction to Factoring

Factoring x² + bx + c type when a = 1

Factoring is one of the fastest methods for solving many quadratic equations on the Digital SAT.

Factoring means rewriting an expression as the product of two smaller expressions.

Example:

x² + 5x + 6

can be written as

(x + 2)(x + 3)

When multiplied together,

(x + 2)(x + 3)

= x² + 3x + 2x + 6

= x² + 5x + 6

The original expression and its factored form are mathematically identical.

The Zero Product Property

Factoring works because of an important algebra rule.

A × B = 0

then

A = 0

B = 0

or both.

This rule is called the Zero Product Property.

Example:

(x + 4)(x − 7) = 0

Either

x + 4 = 0

x − 7 = 0

Solve each equation separately.

First equation:

x + 4 = 0

Subtract 4 from both sides.

x + 4 − 4 = 0 − 4

Simplify.

x = −4

Second equation:

x − 7 = 0

Add 7 to both sides.

x − 7 + 7 = 0 + 7

Simplify.

x = 7

Therefore,

the two solutions are

x = −4

and

x = 7

Example 1

Solve

x² + 7x + 12 = 0

Step 1

Write the equation.

x² + 7x + 12 = 0

Step 2

Find two numbers whose product is 12 and whose sum is 7.

Possible factor pairs of 12 are

1 and 12

2 and 6

3 and 4

Only

3 and 4

add to 7.

Step 3

Write the factors.

(x + 3)(x + 4) = 0

Step 4

Apply the Zero Product Property.

Either

x + 3 = 0

x + 4 = 0

Step 5

Solve the first equation.

Subtract 3 from both sides.

x + 3 − 3 = 0 − 3

Simplify.

x = −3

Step 6

Solve the second equation.

Subtract 4 from both sides.

x + 4 − 4 = 0 − 4

Simplify.

x = −4

Final Answer

x = −3

x = −4

Example 2

Solve

x² − 9x + 20 = 0

Step 1

Find two numbers whose product is 20.

1 and 20

2 and 10

4 and 5

Step 2

Which pair adds to −9?

Since the product is positive and the sum is negative,

both numbers must be negative.

−4 and −5

Step 3

Write the factors.

(x − 4)(x − 5) = 0

Step 4

Set each factor equal to zero.

x − 4 = 0

x − 5 = 0

Step 5

Solve.

Add 4 to both sides.

x = 4

Add 5 to both sides.

x = 5

Check

4² − 9(4) + 20

16 − 36 + 20

Correct.

Now check 5.

25 − 45 + 20

Correct.

Both answers satisfy the equation.

Example 3

Solve

x² + x − 12 = 0

Step 1

Find two numbers whose product is −12.

Possible pairs include

1 and −12

2 and −6

3 and −4

Step 2

Find the pair whose sum equals 1.

4 and −3

Step 3

Write the factors.

(x + 4)(x − 3) = 0

Step 4

Set each factor equal to zero.

x + 4 = 0

x − 3 = 0

Step 5

Solve.

Subtract 4 from both sides.

x = −4

Add 3 to both sides.

x = 3

Final Answer

x = −4

x = 3

A Quick Factoring Strategy

Whenever you see

x² + bx + c

ask yourself two questions.

Question 1

Which two numbers multiply to give c?

Question 2

Do those same numbers add to give b?

If the answer is yes,

you have found the correct factors.

With practice, this process becomes much faster.

Forgetting to Check

Substitute every solution back into the original equation.

If the equation balances,

your solution is correct.

Practice Questions

Solve by factoring.

x² + 5x + 6 = 0
x² − 8x + 15 = 0
x² + 9x + 20 = 0
x² − 7x + 10 = 0
x² + 2x − 15 = 0
x² − x − 12 = 0
x² + 11x + 24 = 0
x² − 10x + 24 = 0

Answers

x = −2, −3
x = 3, 5
x = −4, −5
x = 2, 5
x = 3, −5
x = 4, −3
x = −3, −8
x = 4, 6

A quadratic equation is an equation whose highest exponent is two. Before attempting to solve it, identify whether it is already in standard form and determine the values of a, b, and c. When the equation can be factored, rewriting it as the product of two binomials often provides the quickest solution. The Zero Product Property then allows each factor to be solved separately, producing all possible solutions. As you continue practicing, you'll begin to recognize common factor patterns quickly, an essential skill for success on the Digital SAT Math section.

Wednesday, July 8, 2026

SAT Algebra Linear Equations: Step-by-Step Guide to Solving Linear Equations on the Digital SAT (Part 2)

Solving Equations with Variables on Both Sides, Fractions, Decimals, Ratios, and Proportions

In Part 1, you learned how to solve one-step and two-step linear equations by isolating the variable one operation at a time. Those skills form the foundation of almost every algebra problem on the Digital SAT.

In this chapter, you'll solve equations that look more complicated because variables appear on both sides of the equation. You'll also learn how to work confidently with fractions, decimals, ratios, and proportions—topics that frequently appear in SAT Math questions. Although these problems may seem challenging at first, they all follow the same golden rule of algebra:

Perform the same operation on both sides of the equation while keeping the equation balanced.

Once you understand that principle, every new type of equation becomes much easier to solve.

Solving Equations with Variables on Both Sides

Many SAT questions contain variables on both sides of the equation.

For example,

5x + 8 = 2x + 20

At first glance, students often wonder which variable to solve first. The answer is simple:

Move all the variables to one side and all the numbers to the other side.

Example 1

Solve

5x + 8 = 2x + 20

Step 1: Look at both sides.

The left side contains:

5x + 8

The right side contains:

2x + 20

Both sides contain a variable.

Our first goal is to move all the x terms to the same side.

Step 2: Remove the smaller variable.

Subtract 2x from both sides.

5x + 8 − 2x = 2x + 20 − 2x

Step 3: Simplify.

On the right side,

+2x and −2x cancel.

3x + 8 = 20

Now every variable is on the left side.

Step 4: Remove the constant.

Subtract 8 from both sides.

3x + 8 − 8 = 20 − 8

Step 5: Simplify.

The +8 and −8 cancel.

3x = 12

Step 6: Remove the multiplication.

Divide both sides by 3.

3x ÷ 3 = 12 ÷ 3

Step 7: Simplify.

x = 4

Step 8: Check.

Original equation:

5(4) + 8 = 2(4) + 20

20 + 8 = 8 + 20

28 = 28

The answer is correct.

Example 2

Solve

7x − 5 = 4x + 16

Step 1

Variables appear on both sides.

Subtract 4x from both sides.

7x − 5 − 4x = 4x + 16 − 4x

Step 2

Simplify.

3x − 5 = 16

Step 3

Add 5 to both sides.

3x − 5 + 5 = 16 + 5

Step 4

Simplify.

3x = 21

Step 5

Divide both sides by 3.

3x ÷ 3 = 21 ÷ 3

Step 6

Simplify.

x = 7

Step 7

Check.

7(7) − 5 = 4(7) + 16

49 − 5 = 28 + 16

44 = 44

Correct.

Example 3

Solve

9x + 12 = 6x + 30

Step 1

Subtract 6x from both sides.

9x + 12 − 6x = 6x + 30 − 6x

Step 2

Simplify.

3x + 12 = 30

Step 3

Subtract 12 from both sides.

3x + 12 − 12 = 30 − 12

Step 4

Simplify.

3x = 18

Step 5

Divide both sides by 3.

3x ÷ 3 = 18 ÷ 3

Step 6

Simplify.

x = 6

Step 7

Check.

9(6)+12=6(6)+30

54+12=36+30

66=66

Correct.

Which Variable Should You Move?

Students often ask:

"Should I move the variable on the left or the one on the right?"

Either method works.

However, it is usually easier to move the smaller coefficient.

Example:

9x = 4x + 20

Subtracting 4x produces

5x = 20

which is simpler than subtracting 9x.

Working with Fractions

Fractions appear frequently on the SAT.

Fortunately, the solving process is exactly the same.

Example 4

Solve

x/4 = 9

Step 1

The variable has been divided by 4.

We must undo the division.

Step 2

Multiply both sides by 4.

(x/4) × 4 = 9 × 4

Step 3

The 4 cancels.

x = 36

Step 4

Check.

36 ÷ 4 = 9

Correct.

Example 5

Solve

x/5 + 6 = 14

Step 1

The variable has been divided by 5.

Before removing the division, remove the addition.

Subtract 6 from both sides.

x/5 + 6 − 6 = 14 − 6

Step 2

Simplify.

x/5 = 8

Step 3

Undo the division.

Multiply both sides by 5.

(x/5) × 5 = 8 × 5

Step 4

Simplify.

x = 40

Step 5

Check.

40/5 + 6

8 + 6

Correct.

Example 6

Solve

2 + x/3 = 11

Step 1

Subtract 2 from both sides.

2 + x/3 − 2 = 11 − 2

Step 2

Simplify.

x/3 = 9

Step 3

Multiply both sides by 3.

(x/3) × 3 = 9 × 3

Step 4

Simplify.

x = 27

Step 5

Check.

2 + 27/3

2 + 9

Correct.

Clearing Fractions

Sometimes every term contains a fraction.

Instead of solving with fractions, remove them first.

Example 7

Solve

x/2 + x/3 = 10

Step 1

Find the Least Common Denominator (LCD).

The denominators are:

2 and 3

The LCD is:

Step 2

Multiply every term by 6.

6(x/2)+6(x/3)=6(10)

Step 3

Simplify.

3x+2x=60

Step 4

Combine like terms.

5x=60

Step 5

Divide both sides by 5.

5x÷5=60÷5

Step 6

Simplify.

x=12

Step 7

Check.

12/2 +12/3

6+4

Correct.

Working with Decimals

SAT questions sometimes contain decimals instead of fractions.

Many students become nervous when they see decimals, but decimals follow exactly the same algebra rules.

Example 8

Solve

0.5x = 9

Step 1

The variable has been multiplied by 0.5.

Undo the multiplication.

Divide both sides by 0.5.

0.5x ÷0.5 =9÷0.5

Step 2

Simplify.

x=18

Step 3

Check.

0.5 ×18

Correct.

Example 9

Solve

2.4x +1.2 =13.2

Step 1

Subtract 1.2 from both sides.

2.4x +1.2−1.2 =13.2−1.2

Step 2

Simplify.

2.4x=12

Step 3

Divide both sides by 2.4.

2.4x÷2.4 =12÷2.4

Step 4

Simplify.

x=5

Step 5

Check.

2.4(5)+1.2

12+1.2

13.2

Correct.

Ratios and Proportions

Many SAT word problems involve ratios.

A proportion is simply two equal fractions.

Example:

x/8 = 6/12

Example 10

Solve

x/8 =6/12

Step 1

Notice that

6/12 simplifies to

1/2

However, we can solve directly.

Step 2

Cross multiply.

12 × x =8 ×6

12x=48

Step 3

Divide both sides by 12.

12x÷12=48÷12

Step 4

Simplify.

x=4

Step 5

Check.

4/8=6/12

1/2=1/2

Correct.

SAT Strategy

Whenever fractions appear,

ask yourself,

"Can I remove the fractions first?"

Whenever decimals appear,

ask yourself,

"Would converting them to fractions make this easier?"

Many difficult SAT algebra questions become much simpler after removing fractions or decimals.

Practice Questions

Solve each equation.

6x + 9 = 3x + 24
8x − 7 = 5x + 20
x/6 = 8
x/4 + 7 = 15
2 + x/5 = 10
0.25x = 12
3.5x + 7 = 28
x/3 + x/6 = 15
x/10 = 7/14
4x + 18 = 2x + 34

Answers

x = 5
x = 9
x = 48
x = 32
x = 40
x = 48
x = 6
x = 30
x = 5
x = 8

As equations become more complex, the underlying algebra never changes. Whether variables appear on both sides, fractions need to be cleared, decimals are involved, or ratios must be solved using proportions, the objective is always to isolate the variable while keeping the equation balanced. By practicing these methods carefully and checking each solution, you'll develop the accuracy and confidence needed for more challenging SAT algebra, Digital SAT math, linear equation solving, and SAT word problem questions. In Part 3, you'll apply these equation-solving skills to linear functions, slope, intercepts, graphs, and systems of linear equations—the concepts that connect algebra with coordinate geometry and mathematical modeling on the Digital SAT.

SAT Algebra Linear Equations Step-by-Step Guide to Solving Linear Equations on the Digital SAT (Part 1)

SAT Algebra Linear Equations: Step-by-Step Guide to Solving Linear Equations on the Digital SAT (Part 1)

Build a Strong Algebra Foundation for a High SAT Math Score

Success on the Digital SAT Math section begins with mastering algebra. Among all the algebra topics tested, linear equations are the most fundamental because they appear directly in equation-solving questions and indirectly in linear functions, graph interpretation, systems of equations, coordinate geometry, mathematical modeling, and many real-world word problems.

Many students believe algebra is about memorizing formulas. In reality, algebra is about logical thinking. Every equation tells a mathematical story, and every solution follows a sequence of logical steps. Once you understand those steps, even difficult-looking SAT questions become manageable.

This guide explains every concept carefully, assuming no prior knowledge beyond basic arithmetic. Each example is solved one step at a time, with an explanation for every operation performed. Instead of simply showing the answer, you'll learn why each step works. By the end of this chapter, you'll be able to solve one-step and two-step linear equations confidently, avoid common mistakes, and build the foundation needed for more advanced SAT algebra topics.

What Is a Linear Equation?

A linear equation is an equation in which every variable has an exponent of one. When the equation is represented on a graph, it forms a straight line rather than a curve.

Some examples of linear equations are:

x + 5 = 12
3x − 7 = 20
4y = 36
2a + 9 = 19

Although these equations look different, they all follow exactly the same mathematical principles.

Every linear equation contains an unknown value called a variable. Your goal is to determine the value of the variable that makes the equation true.

Understanding Variables

A variable is simply a symbol that represents an unknown number.

Instead of writing

□ + 8 = 15

mathematicians write

x + 8 = 15

The letter x can represent any number.

If x equals 7, then

7 + 8 = 15

Both sides are equal, so the equation is true.

The variable does not always have to be x.

It may also be

The letter changes, but the method of solving the equation never changes.

What Does It Mean to Solve an Equation?

To solve an equation means to find the value of the variable that makes both sides exactly equal.

Imagine an old-fashioned balance scale.

If both sides contain the same weight, the scale remains perfectly balanced.

If you remove weight from one side only, the balance tips.

To keep the balance level, whatever you do to one side must also be done to the other side.

This simple idea is the foundation of all algebra.

The Golden Rule of Algebra

Whatever operation you perform on one side of an equation must also be performed on the other side.

This rule never changes.

If you add 6 to one side, add 6 to the other side.

If you subtract 10 from one side, subtract 10 from the other side.

If you multiply one side by 4, multiply the other side by 4.

If you divide one side by 7, divide the other side by 7.

Following this rule ensures that both sides remain equal throughout the solution.

Understanding Inverse Operations

An inverse operation is an operation that reverses another operation.

Operation	Inverse Operation
Addition	Subtraction
Subtraction	Addition
Multiplication	Division
Division	Multiplication

For example,

if 8 has been added,

subtract 8.

If a number has been multiplied by 5,

divide by 5.

Inverse operations allow us to remove numbers one by one until only the variable remains.

Solving One-Step Linear Equations

One-step equations require only one operation to isolate the variable.

Although these questions are among the easiest on the SAT, learning them thoroughly makes later topics much easier.

Example 1

Solve

x + 9 = 18

Step 1: Identify the operation attached to the variable.

The variable x has 9 added to it.

Since our goal is to leave x by itself, we must remove the +9.

Step 2: Choose the inverse operation.

The opposite of adding 9 is subtracting 9.

Therefore, subtract 9 from both sides of the equation.

x + 9 − 9 = 18 − 9

Step 3: Simplify both sides.

On the left side,

+9 and −9 cancel each other.

x = 9

On the right side,

18 − 9 = 9

Therefore,

x = 9

Step 4: Check the answer.

Substitute 9 into the original equation.

9 + 9 = 18

18 = 18

Both sides are equal.

The solution is correct.

Example 2

Solve

x − 14 = 23

Step 1

The variable has 14 subtracted from it.

We must remove the −14.

Step 2

The opposite of subtracting 14 is adding 14.

Add 14 to both sides.

x − 14 + 14 = 23 + 14

Step 3

The −14 and +14 cancel.

x = 37

Step 4

Check.

37 − 14 = 23

23 = 23

The answer is correct.

Example 3

Solve

6x = 48

Step 1

The variable has been multiplied by 6.

Our goal is to remove the multiplication.

Step 2

The opposite of multiplying by 6 is dividing by 6.

Divide both sides by 6.

6x ÷ 6 = 48 ÷ 6

Step 3

On the left side,

the 6 in the numerator and denominator cancel.

x = 8

On the right side,

48 ÷ 6 = 8

Therefore,

x = 8

Step 4

Check.

6 × 8 = 48

48 = 48

The answer is correct.

Example 4

Solve

x ÷ 5 = 12

Step 1

The variable has been divided by 5.

We need to undo the division.

Step 2

The opposite of division is multiplication.

Multiply both sides by 5.

(x ÷ 5) × 5 = 12 × 5

Step 3

The 5 in the numerator and denominator cancel.

x = 60

Step 4

Check.

60 ÷ 5 = 12

12 = 12

The solution is correct.

Solving Two-Step Linear Equations

Two-step equations require removing one operation before removing another.

Always work from the outside toward the variable.

Never try to remove the multiplication before removing the addition or subtraction.

Example 5

Solve

3x + 7 = 25

Step 1

The variable x has first been multiplied by 3.

After that,

7 has been added.

Since addition happened last,

we remove the addition first.

Step 2

Subtract 7 from both sides.

3x + 7 − 7 = 25 − 7

Step 3

Simplify.

+7 and −7 cancel.

3x = 18

Step 4

The variable is still multiplied by 3.

Undo the multiplication by dividing both sides by 3.

3x ÷ 3 = 18 ÷ 3

Step 5

Simplify.

The 3 cancels.

x = 6

Step 6

Check.

3(6) + 7 = 25

18 + 7 = 25

25 = 25

The solution is correct.

Example 6

Solve

5x − 20 = 35

Step 1

The variable has 20 subtracted.

Remove the subtraction first.

Add 20 to both sides.

5x − 20 + 20 = 35 + 20

Step 2

Simplify.

−20 and +20 cancel.

5x = 55

Step 3

The variable is multiplied by 5.

Divide both sides by 5.

5x ÷ 5 = 55 ÷ 5

Step 4

Simplify.

x = 11

Step 5

Check.

5 × 11 − 20 = 35

55 − 20 = 35

35 = 35

The answer is correct.

Example 7

Solve

8x + 16 = 64

Step 1

Subtract 16 from both sides.

8x +16 −16 =64 −16

Step 2

Simplify.

8x =48

Step 3

Divide both sides by 8.

8x ÷8 =48 ÷8

Step 4

Simplify.

x =6

Step 5

Check.

8(6)+16=64

48+16=64

64=64

Correct.

If an equation says

4x + 12

the multiplication happened first,

then the addition.

When solving,

remove the addition first,

then the multiplication.

Thinking this way makes multi-step equations much easier.

Working with Negative Numbers

Negative numbers scare many students, but the solving process never changes.

Treat them exactly like positive numbers while paying close attention to the signs.

Example

Solve

−4x = 28

Step 1

The variable is multiplied by −4.

Undo the multiplication by dividing both sides by −4.

−4x ÷ −4 = 28 ÷ −4

Step 2

Simplify.

The −4 cancels.

x = −7

Step 3

Check.

−4(−7)=28

28=28

Correct.

Tuesday, July 7, 2026

Show that f: R → R defined as f(x) = x / sqrt(1 + x^2) is one-one but not onto.

Show that f: R → R defined as f(x) = x / √(1 + x²) is one-one but not onto.

f: R → R, f(x) = x / √(1 + x²)

To check One-One / Injective

Let x₁, x₂ ∈ R

such that f(x₁) = f(x₂)

x₁ / √(1 + x₁²) = x₂ / √(1 + x₂²) --------[1]

Squaring both sides:

x₁² / (1 + x₁²) = x₂² / (1 + x₂²)

x₁²(1 + x₂²) = x₂²(1 + x₁²)

x₁² + x₁²x₂² = x₂² + x₁²x₂²

x₁² = x₂²

x₁ = ± x₂

[1] is possible only if x₁, x₂ have the same sign

x₁ = -x₂, is rejected

we have to conclude that x₁ =x₂,

Therefore f is one-one.

To check Onto / Surjective

Let y = x / √(1 + x²)

squaring

y² = x² / (1 + x²)

y²(1 + x²) = x²

y² + y²x² = x²

y² = x² - y²x² = x²(1 - y²)

x² = y² / (1 - y²)

For x to be real, RHS ≥ 0

Since y² ≥ 0, we need 1 - y² > 0

⇒ y² < 1

⇒ -1 < y < 1

Range of f = (-1, 1) ≠ R

Therefore f is not onto.

see this video for more explanation

cbse 12th maths old board exam question paper 2025 2026 one to one injective function onto function

SAT Math Formula Cheat Sheet for Quick Reference

SAT Math Formulas

The Digital SAT includes a reference sheet with some geometry formulas, but it does not include everything you'll need. Knowing the most common formulas before test day helps you solve problems more quickly and reduces the chance of making simple mistakes.

Exponent Rules

For any nonzero number a:

a⁰ = 1

a¹ = a

aᵐ × aⁿ = aᵐ⁺ⁿ

aᵐ ÷ aⁿ = aᵐ⁻ⁿ

(aᵐ)ⁿ = aᵐⁿ

(ab)ⁿ = aⁿbⁿ

(a/b)ⁿ = aⁿ/bⁿ

a⁻ⁿ = 1/aⁿ

Remember

Multiply → add exponents.
Divide → subtract exponents.
A negative exponent means take the reciprocal.

Radicals

√a × √b = √(ab)

√a ÷ √b = √(a/b)

Examples

√49 = 7

√81 = 9

∛125 = 5

Perfect squares worth memorizing:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

121, 144, 169, 196, 225, 256, 289, 324, 361, 400

Linear Equations

Slope

m = (y₂ − y₁)/(x₂ − x₁)

Slope-intercept form

y = mx + b

Point-slope form

y − y₁ = m(x − x₁)

Standard form

Ax + By = C

Parallel lines have the same slope.

Perpendicular lines have negative reciprocal slopes.

Coordinate Geometry

Distance Formula

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Midpoint Formula

((x₁ + x₂)/2, (y₁ + y₂)/2)

Quadratic Equations

Standard form

ax² + bx + c = 0

Quadratic Formula

x = (−b ± √(b² − 4ac))/2a

Discriminant

b² − 4ac

Positive → two real solutions

Zero → one real solution

Negative → no real solutions

Vertex

x = −b/(2a)

Factoring Identities

Difference of Squares

a² − b² = (a + b)(a − b)

Perfect Square Trinomials

a² + 2ab + b² = (a + b)²

a² − 2ab + b² = (a − b)²

Circle Formulas

Circumference

C = 2πr

Area

A = πr²

Arc Length

(θ/360) × 2πr

Sector Area

(θ/360) × πr²

Diameter = 2r

Rectangles and Squares

Rectangle

Area = length × width

Perimeter = 2(length + width)

Square

Area = side²

Perimeter = 4 × side

Diagonal = side√2

Triangles

Area

½ × base × height

Pythagorean Theorem

a² + b² = c²

The angles inside every triangle add up to 180°.

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

Special Right Triangles

45°–45°–90°

1 : 1 : √2

30°–60°–90°

1 : √3 : 2

These ratios are tested regularly.

Trigonometry

SOH CAH TOA

sin θ = opposite/hypotenuse

cos θ = adjacent/hypotenuse

tan θ = opposite/adjacent

tan θ = sin θ/cos θ

Polygons

Interior Angle Sum

(n − 2) × 180°

Each Interior Angle of a Regular Polygon of n vertics

[(n − 2) × 180°]/n

Volume

Cube

s³

Rectangular Prism

lwh

Cylinder

πr²h

Cone

⅓πr²h

Sphere

⁴⁄₃πr³

Surface Area

Cube

6s²

Cylinder

2πr² + 2πrh

Sphere

4πr²

Mean

Sum of all values ÷ Number of values

Weighted Mean

Σ(value × weight) ÷ Σ(weights)

Probability

P(Event)

Number of Favorable Outcomes ÷ Total number of Outcomes

Complement Rule

P(complement event ) =1 − P(Event)

A probability is always between 0 and 1.

Percents

Increase

Original × (1 + rate)

Decrease

Original × (1 − rate)

Percent Change

(New − Original)/Original × 100%

Simple Interest

I = Prt

P = Principal

r = Interest Rate

t = Time

Exponential Growth and Decay

Growth

A = P(1 + r)ᵗ

Decay

A = P(1 − r)ᵗ

Functions

Example

f(x) = 2x + 3

f(5) = 13

Replace x with the given value.

Useful Constants

π ≈ 3.14

√2 ≈ 1.414

√3 ≈ 1.732

Monday, July 6, 2026

Find the mean of the following distribution :Class30 – 4040 – 5050 – 6060 – 7070 – 80Frequency61381211

Find the mean of the following distribution :

Class 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80

Frequency 6 13 8 12 11

calculate class mark, mid value or mid -x [x] then the product f*x

Class f x f*x

30-40 6 35 210

40-50 13 45 585

50-60 8 55 440

60-70 12 65 780

70-80 11 75 825

Total Σf =50, Σfx= 2840

Mean = Σfx / Σf = 2840 / 50 = 56.8

for more explanation watch this video

statistics arithmetic mean by direct method

cbse 10th maths previous years question papers 2025 2026 question in statistics

Thursday, July 2, 2026

A boy has a collection of balls of different colours. He has a total of 35 balls in his basket out of which seven are black in colour and eight are yellow in colour. Out of remaining balls, some are white and the rest are red. Based on the above, answer the following questions: (a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket. (b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball.

A boy has a collection of balls of different colours. He has a total of 35 balls in his basket out of which seven are black in colour and eight are yellow in colour. Out of remaining balls, some are white and the rest are red.

Based on the above, answer the following questions:

(a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket.

(b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball

Total balls = 35

Number of Black balls = 7

Number of Yellow balls = 8

Remaining balls = 35 − 7 − 8 = 20

Let number of white balls = w

Let number of red balls = r

Given that

Out of remaining balls, some are white and the rest are red.

Remaining balls = 20

w + r = 20 ---------[1]

Given

P(red) = 3 × P(white)

P(red) = r/35

P(white) = w/35

r/35 = 3 × w/35

r = 3w ---------(2)

Put (2) in (1)

w + r = 20

w + 3w = 20

4w = 20

w = 5

use r = 3w

r = 3 × 5 = 15

Number of red balls = 15

Number of Black balls = 7

Numbe r of White balls [w] = 5

Number of favourable outcomes = 7 + 5 = 12

P(black or white) = 12/35

foe more details use the video

probability, cbse 10th standard mathematics past years question papers 2025 2026

Wednesday, July 1, 2026

A survey was conducted on the patients who have undergone knee replacement surgeries. It was found that, Robotic Knee replacement surgeries have 90% success rate. On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that : (i) exactly one surgery is successful, (ii) at most two surgeries are successful. probability of independent events cbse 12th maths old board exam question paper 2025 2026 independent events success failure type

A survey was conducted on the patients who have undergone knee replacement surgeries. It was found that, Robotic Knee replacement surgeries have 90% success rate. On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that : (i) exactly one surgery is successful, (ii) at most two surgeries are successful. probability of independent events cbse 12th maths old board exam question paper 2025 2026 independent events success failure type

Given:

P(Success) = 90% = 0.9 = 9/10

P(Failure) = 1 − 0.9 = 0.1 = 1/10

let S debite Success,

F denote Failure

Probability that exactly one surgery is successful

possibilities SFF, FSF, FFS

P(SFF) = 0.9 × 0.1 × 0.1 = 0.009

P(FSF) = 0.1 × 0.9 × 0.1 = 0.009

P(FFS) = 0.1 × 0.1 × 0.9 = 0.009

P(exactly one auccess) = 0.009 + 0.009 + 0.009 = 0.027 = 27/1000

Probability that at most two surgeries are successful

At most two successes means 0 or 1 or 2 successes

Probability that at most two surgeries are successful

= 1 − P(all three successful)

= 1 - P[ SSS]

= 1 - [0.9 × 0.9 × 0.9]

= 1 - 0.729

= 0.271

= 271/1000

see this video for more explanation

basic probability , independent events , cbse 12th maths old board exam question paper 20252026

Tuesday, June 30, 2026

From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

Cylinder and conical cavity have same radius and height

r = 2.1 cm, h = 2.8 cm

.Volume of remaining solid

Volume of cylinder − Volume of cone

Volume of cylinder = πr²h = π × (2.1)² × 2.8 = π × 4.41 × 2.8 = 12.348π cm³

Volume of cone = (1/3)πr²h = (1/3) × 12.348π = 4.116π cm³

Remaining volume = 12.348π − 4.116π = 8.232π cm³

Using π = 22/7

Remaining volume = 8.232 × 22/7 = 25.872 cm³ ≈ 25.87 cm³

. Total surface area of remaining solid

Surfaces left after hollowing:

Bottom circular base of cylinder = πr²

Curved surface of cylinder = 2πrh

Curved surface of cone = πrl

Slant height of cone: l = √(r² + h²) = √(2.1² + 2.8²) = √(4.41 + 7.84) = √12.25 = 3.5 cm

Required surface area = πr² + 2πrh + πrl

= π(2.1)² + 2π(2.1)(2.8) + π(2.1)(3.5)

= 4.41π + 11.76π + 7.35π = 23.52π cm² Using π = 22/7:

= 23.52 × 22/7 = 73.92 cm²

for more details watch this video

mensuration, surface area and volume of solids , cbse th maths old board exam question paper 2025 2026

Thursday, June 25, 2026

cbse 10th maths questions

cbse 10th maths questions

Linear equations in two unknowns

The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

solution

Quadratic equations

Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

solution

Coordinate Geometry

Find a relation between x and y such that the point P(x, y) is equidistant from the

points A(5, 3) and B(1, 7)

solution

The coordinates of the centre of a circle are (x − 7, 2x). Find the value(s) of ‘x’, if the circle passes through the point (−9, 11) and has radius 5√2 units.

solution

Trigonometry

If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

solution

Mensuration surface area and volume of solids

From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

solution

Statistics

Find the mean of the following distribution :

Class 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80

Frequency 6 13 8 12 11

solution

Probability

Based on the above, answer the following questions:

(a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket.

(b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball

solution

While playing badminton Ravi has set the barrier chain hung between two posts at the edge of the walkway of a street. It is hung in the shape of a parabola. Based on the above information answer the following questions : (a) Which type of the polynomial (linear, quadratic, cubic etc.) is graphically represented by a parabola ? (b) If the polynomial represented by a parabola, intersects the x-axis at −2 and 3 and y-axis at −3, then write the zeroes of the parabola. (c) Find the expression for the above polynomial. OR (c) If the zeroes of the polynomial are −5 and 3, find its expression.

While playing badminton Ravi has set the barrier chain hung between two posts at the edge of the walkway of a street. It is hung in the shape of a parabola.

Based on the above information answer the following questions :

(a) Which type of the polynomial (linear, quadratic, cubic etc.) is graphically represented by a parabola ?

(b) If the polynomial represented by a parabola, intersects the x-axis at −2 and 3 and y-axis at −3, then write the zeroes of the parabola.

(d) If the zeroes of the polynomial are −5 and 3, find its expression.

[a] A parabola is the graph of a quadratic polynomial

Given

parabola, intersects the x-axis at −2 and 3

therefore zeroes are [-2] and 3

Polynomial with roots α and β is given by

f(x) = k(x − α)(x − β)

f(x) = k[x² − (α + β)x + αβ]

where k ≠ 0 is any real number.

given that zeroes are [-2] and 3

Take them as

α = [-2]

β = 3

f(x) = k(x + 2)(x − 3)

use distribution law

f(x) =k(x² − x − 6)

to find k use the condition that parabola cuts y-axis at −3

which means it passes through (0,-3)

so Use f(0) = −3

k[0-0-6] = [-3]

we get k =(1/2)

and f(x) = (1/2)(x² − x − 6)

in the last part the zeroes are given to be −5 and 3

so g(x) = k(x + 5)(x − 3) =kx² + 2x − 15

choose k=1

g(x) = x² + 2x − 15

see this video for more details

cbse 10th maths old board exam question papers 2025 2026 to find the equation of a polynomial whose zeroes are given

Wednesday, June 24, 2026

If (sin x)ʸ = yᶜᵒˢˣ, then find dy/dx

If (sin x)ʸ = yᶜᵒˢˣ, then find dy/dx.

differential calculus topic of logarthimic differentiation

given

(sin x)ʸ = yᶜᵒˢˣ [power contains variables]

Take ln both sides

ln[(sin x)ʸ] = ln[yᶜᵒˢˣ] [use properties of logarithms]

y. ln(sin x) = cos x . ln y

Differentiate w.r.t. x

[use product rule and chain rule]

[dy/dx] ln(sin x) + y cot x = −sin x · ln y + cos x · (1/y) ·[ dy/dx]

[dy/dx ] ln(sin x) + y cot x = −sin x ln y + (cos x / y) [dy/dx]

Collect dy/dx terms

[dy/dx] ln(sin x) − (cos x / y)[ dy/dx ]= −sin x ln y − y cot x

[ dy/dx] [ln(sin x) − cos x / y] = −[sin x ln y + y cot x]

dy/dx = −[sin x ln y + y cot x] / [ln(sin x) − cos x / y]

Multiply numerator and denominator by y:

dy/dx = −y[sin x ln y + y cot x] / [y ln(sin x) − cos x]

see this video for more explanation

cbse 12th maths old board exam question paper

2025 2026 differentiation

logarthmic differentiation

Tuesday, June 23, 2026

Find a particular solution of the differential equation (x + 1) dy/dx = 2 e⁻ʸ − 1, given that y = 0 when x = 0

Find a particular solution of the differential equation

(x + 1) dy/dx = 2 e⁻ʸ − 1,

given that y = 0 when x = 0

variable separable differential equation with initial condition

given

(x + 1) dy/dx = 2 e⁻ʸ − 1,

separating

dy / (2 e⁻ʸ − 1) = dx / (x + 1)

multiply numerator and denominator of LHS with eʸ

eʸ dy / (2 − eʸ) = dx / (x + 1)

integrating

∫ eʸ / (2 − eʸ) dy = ∫ 1 / (x + 1) dx

for the LHS use the substitution

u = 2 − eʸ

⇒ du = −eʸ dy

eʸ dy =-du

we get

∫ −du / u = ∫ 1 / (x + 1) dx

−ln|u| = ln|x + 1| + C

re substitute the value of u = 2 − eʸ

−ln|2 − eʸ| = ln|x + 1| + C

ln|2 − eʸ| +ln|x + 1| = -C

using properties of logarithms,

ln|(2 − eʸ)(x + 1)| = -C

using definition of logarithm

(x + 1)(2 − eʸ) = A-----(1)

, where A is another arbitrary constant

use the initial condition y = 0 when x = 0 to solve for A

we get A =1

(1)⇒ (x + 1)(2 − eʸ) =1

2 − eʸ = 1 / (x + 1)

eʸ = 2 − [ 1 / (x + 1)]

eʸ = (2x + 1) / (x + 1)

using definition of log

y = ln[(2x + 1) / (x + 1)]

for more details watch this video

cbse 12th past year question papers 2025 2026 differential equations

Monday, June 22, 2026

variable separable differential equation dy/dx = (2 − y) / (x + 1)

Solve the following variable separable differential equation

dy/dx = (2 − y) / (x + 1)

clearly it is variable separable type

dy / (2 − y) = dx / (x + 1)

integrating both sides

∫ dy / (2 − y) = ∫ dx / (x + 1)

− ln|2 − y| = ln|x + 1| + lnC [ note the ( -1), coefficient of y when integrating]

0 = ln|2 − y| + ln|x + 1| + lnC

use the property of logarithms

0=ln [ |2 − y| |x + 1| C ]

[ |2 − y| |x + 1| C ] = 1

for more details refer to this video

cbse 12th applied mathematics 2025 2026 old board exam question papers applied mathematics

variable sepaprable differential equation

Wednesday, June 17, 2026

Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

Let the three consecutive positive integers be n, n+1, n+2

Given that

that sum of square of the first and the product of the other two is 67

n² + (n+1)(n+2) = 67

n² + (n² + 2n + n + 2) = 67

n² + n² + 3n + 2 = 67

2n² + 3n + 2 = 67

2n² + 3n + 2 - 67 = 0

2n² + 3n − 65 = 0

Solve by quadratic formula or factorisation

Factorisation method

Find two numbers that gives a product of 2 × (−65) = −130 and add to 3.

by trial and error the numbers13 and −10 satisfy the condition

use this to split the middle term of

2n² + 3n − 65 = 0

.2n² + 13n − 10n − 65 = 0

n(2n + 13) − 5(2n + 13) = 0

(n − 5)(2n + 13) = 0

solving

n − 5 = 0 ⇒ n = 5

2n + 13 = 0 ⇒ n = −13/2 (rejecct)

required positive integers are

n=5

n+1 =6

n+2=7

for a video of the solution

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quadratic equation word problem formation and solution by splitting the middle term

Monday, June 15, 2026

∫ cos x / [(2 + sin x)(4 + sin x)] dx

find ∫ cos x / [(2 + sin x)(4 + sin x)] dx

integration by substitution and partial fractions

Substitution

Let t = sin x

Then dt = cos x dx

∫ cos x / [(2 + sin x)(4 + sin x)] dx = ∫ 1 / [(2 + t)(4 + t)] dt

Method of partial fractions

1 / [(t + 2(t + 4)] = A/(t + 2) + B/(t + 4)

multiply [t+2][t+4]

1 = A(t + 4) + B(t + 2)

Put t = −2

1 = A(2)

⇒ A = ½

Put t = −4:

1 = B(−2)

⇒ B = −½

1 / [(t + 2)(t + 4)] = ½[1/(t + 2) − 1/(t + 4)]

integrate

I = ½ ∫ [1/(t + 2) − 1/(t + 4)] dt

= ½[ln|t + 2| − ln|t + 4|] + C

= ½ ln|(t + 2)/(t + 4)| + C use t = sin x

=½ ln[(2 + sin x)/(4 + sin x)] + C

for explanation watch this video

Thursday, June 11, 2026

The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

Let numerator = x, denominator = y. Fraction = x⁄y

Given

sum of numerator and denominator of a fraction is 4 less than twice the denominator

x + y = 2y − 4

⇒ x = y − 4 --- (1)

Given

If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃

(x − 1)⁄(y − 1) = ¹⁄₃

⇒ 3(x − 1) = y − 1

⇒ 3x − 3 = y − 1

⇒ 3x = y + 2 --- (2)

Solve using substitution method

Substitute (1) into (2):

3(y − 4) = y + 2

3y − 12 = y + 2

2y = 14

y = 7

Then

x = y − 4

x = 7 -4

x= 3

x =3

y=7

The fraction is ³⁄₇

linear equations in two unknowns

formation of linear equations and solution by substitution method

cbse 10th maths previous years question papers

2025 2026

for a detailed video of the solution using elimination method

\refer

Wednesday, June 10, 2026

If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

given

cos A + sin A = (√2) cos A

sin A = (√2) cos A - cos A

sin A = (√2 - 1 ) cos A

sin A/ (√2 - 1 ) = cos A

rationalising

(√2 + 1 ) sin A/ [ (√2 - 1 ) (√2 + 1 ) ] = cos A

(√2 + 1 ) sin A / [2-1] = cosA

(√2 + 1 ) sin A / [1] = cosA

(√2 + 1 ) sin A = cosA

apply distribution law

(√2) sin A + 1 sinA = cosA

(√2) sin A + sin A = cosA

(√2) sin A = cosA - sin A

cos A − sin A = (√2) sin A

cbse 10th maths old board exam question paper 2025 2026 trigonometry

video of an alternate method

Tuesday, June 9, 2026

AB is a chord of length 24 cm of a circle of radius 15 cm. The tangents at A and B intersect at a point P. Find the length PA.

Let O be the centre of the circle and Q the midpoint of chord /AB

Given: OA = OB = 15 cm, AB = 24 cm

Q is midpoint of AB

AQ = QB = 24/2 = 12 cm,

also OQ ⊥ AB

In right triangle ΔOQA:

OA² = OQ² + AQ²

15² = OQ² + 12²

225 = OQ² + 144

OQ² = 81

OQ = 9 cm

∠OAP = 90° since radius ⊥ tangent

so ΔOAQ ~ ΔOPA

using corresponding sides

OA/OP = OQ/OA

OA² = OQ × OP

15² = 9 × OP

225 = 9 × OP

OP = 25 cm

In right triangle ΔOPA:

PA² = OP² − OA²

PA² = 25² − 15² = 625 − 225 = 400

PA = 20 cm

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circles tangents to circles

a detailed video on an alternative method to solve the problem

Monday, June 8, 2026

Find a relation between x and y such that the point P(x, y) is equidistant from the points A(5, 3) and B(1, 7).

Find a relation between x and y such that the point P(x, y) is equidistant from the

points A(5, 3) and B(1, 7)

Given

P(x, y) is equidistant from A and B,

PA = PB

squaring both sides

PA² = PB²

Using distance formula to find PA and PB

(x - 5)² + (y - 3)² = (x - 1)² + (y - 7

Expand both sides using identity

x² - 10x + 25 + y² - 6y + 9 = x² - 2x + 1 + y² - 14y + 49

-10x - 6y + 34 = -2x - 14y + 50

-10x + 2x - 6y + 14y = 50 - 34

-8x + 8y = 16

-x + y = 2

y = x + 2

watch this video for more details

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