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
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cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx ncert 11th trigonometry, exercise 3.3 11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x]...
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Before taking the driving test you have to get a learner's license.For that you had to to take an objective test on the computer in the ...