SAT MATH FORMULA SHEET FOR QUICK REFERENCE
ALGEBRA
SO;VING LINEAR EQUATIONS [PART 1]
SOLVING LINEAR EQUATIONS [PART II]
QUADRATIC EQUATIONS [PART I]
QUADRATIC EQUATIONS [PART II]
Tuesday, July 21, 2026
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
A jacket costing ₹3,200 receives a 25% discount. Find the sale price.
Increase ₹850 by 12%.
A phone costs ₹30,000. After a 15% discount, an 8% sales tax is added. Find the final price.
A town with 65,000 people grows by 6% and then by 4% the following year. Find the population after two years.
A shop buys a microwave oven for ₹9,600 and sells it for ₹11,040. Find the profit percentage.
A bicycle is purchased for ₹15,000 and sold for ₹13,800. Find the loss percentage.
Answers
₹2,400
₹952
₹27,540
71,656
15%
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.
Monday, July 20, 2026
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
A museum welcomed 720 visitors on Monday and 900 visitors on Tuesday. Find the percentage increase.
A reservoir contained 4,500 cubic metres of water. After a dry season, it contained 3,960 cubic metres. Find the percentage decrease.
Increase 840 by 18%.
Increase 360 by 12.5%.
A factory produced 2,400 bicycles last year and 2,760 this year. Find the percentage increase.
A theatre sold 640 tickets on Friday and 560 on Saturday. Find the percentage decrease.
Answers
25%
12%
991.2
405
15%
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.
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%
= 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.
28
Step 2
Identify the whole.
40
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
e
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)
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:
9x
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.
If
A × B = 0
then
A = 0
or
B = 0
or both.
This rule is called the Zero Product Property.
Example:
(x + 4)(x − 7) = 0
Either
x + 4 = 0
or
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
or
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
0
Correct.
Now check 5.
25 − 45 + 20
0
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)
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
14
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
11
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:
6
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
10
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
9
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
y
a
b
m
n
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
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
so
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
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