maths problem solving questions and answers grade 9

Grade 9 Algebra Word Problems

Related Pages Grade 8 Algebra Word Problems Algebra Word Problems Solving Equations More Algebra Lessons

These lessons cover grade 9 algebra word problems involving age, distance, rate, time and coins with examples and step-by-step solutions. It includes various examples and solutions for algebra word problems that you will commonly encounter in 9th grade.

Age Word Problems

Age Problems with two unknowns or variables

Example: Taylor is five times as old as Spenser. The sum of their ages is eighteen. How old are Taylor and Spencer?

Solution: Let x represent Spenser’s age Therefore, Taylor’s age is 5x x + 5x = 18 6x = 18 x = 3 Therefore, Spenser is 3 years old and Taylor is 15 years old.

Grade 9 Algebra Word Problems - Age

Example 1: A mother is three times as old as her daughter. Six years ago, the mother’s age was six tines that of her daughter. How old are they now?

Solution: Let x represent the daughter’s age. Therefore, 3x is the mother’s age. 6(x - 6) = 3x - 6 6x - 6 = 3x - 6 3x = 30 x = 10 Therefore, the daughter’s is 10 years old and the mother is 30 years old.

Example 2: A father is now three times as old as his son. Eight years ago, the father was five times as old as his son. How old are they now?

Rate, Distance, Time Word Problems

Grade 9 Algebra Word Problems - Rate, Distance, Time

Example: A bus leaves the terminal and averages 40 km/hr. One hour late, a second bus leaves the same terminal and averages 50 km/hr. In how many hours will the second bus overtake the first?

Grade 9 Rate, Distance, Time Word Problems

Example 1: One motorist travels 5 km.hr faster than another. They leave from the same place and travel in opposite directions. What is the rate of each if they are 195 km apart after 3 hours?

Example 2: A pilot flew from airport A to airport B at a rate of 100 km/hr and flew back from airport B to airport A at 120 km/hr. The total time it took was 11 hours. How far is it from airport A to airport B?

Coin Word Problems

Grade 9 Algebra Word Problems - Coins

Example: A coin collection amounting to $25 consists of nickels and dimes. There are 3 times as many nickels and dimes. There are 3 times as many nickels as dimes. How many coins of each kind are there?

Solution: Let x = number of dimes 3x = number of nickels 10x + 5(3x) = 2500 25x = 2500 x = 100 Therefore, there are 100 dimes and 300 nickels.

Grade 9 Coin Algebra Word Problems

Example: Mr. Rogers has $4.62. He has 3 times as many dimes as nickels and 6 more pennies than dimes. How many coins of each kind does he have?

Coin Algebra Word Problems - Grade 9

Example: Bob has in his pocket a number of pennies, 5 times as many nickels as pennies and 5 more quarters than pennies. The coins amount to $2.27. Find the number of each kind of coin.

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maths problem solving questions and answers grade 9

Grade 9 Math Practice Test Questions

Grade 9 math practice test questions on algebra, geometry, trigonometry, and problem solving are presented along with their solutions on videos.

Solve the following equations. a) $ \quad (x-1)(x+8) = 0 $ b) $ \quad x(x-1) = 1 $ c) $ \quad 2x^2 + 6 x = 8 $ Solution on video at Solve Quadratic Equations, question 8
Solve the following equations. a) $ \quad \sqrt x = 2 $ b) $ \quad \sqrt{x - 2} = 4 $ c) $ \displaystyle \quad \sqrt {\frac{ x}{10}} = 2 $ Solution on video at Solve Equations with Square Roots, question 9
Solve the following inequalities and represent the solution set using intervals, graphs on a number line, and inequality symbols. a) $ \quad 3(x - 2) + 2x \lt -(x+2) $ b) $ \displaystyle \quad \frac{x + 5}{3} \ge \frac{-x+2}{2} $ c) $ \displaystyle \quad - (2x - 2) \le 3x + 12 $ Solution on video at Solve Inequalities with Brackets and Fractions , question 10
For what value of the parameter $ a $ is the point with coordinates $ (a,3) $ on the line whose equation is given by $ 2x - 3y = 4 $ ? Solution on video at Find the x coordinate of a Point on a Line , question 11
What is the point of intersection of the lines given by the equations $ 2 x + y = 5 $ and $ 3x - 2y = 4$ ? Solution on video at Find the point of intersection of two lines , question 12
What is the slope of each of the lines given by the equations? a) $ \quad 4x = -2 y + 4 $ b) $ \quad 3y = -9 $ c) $ \quad - 5x = 10 $ Solution on video at Find the slopes of lines , question 13
Find the x and y intercepts, if possible, of the lines with equations a) $ \quad - 3x = 2 y + 6 $ b) $ \quad \displaystyle 2 y = 8 $ c) $ \quad \displaystyle -3x = 6 $ Solution on video at Find the x and y intercepts of lines , question 14
a) Find the value of the parameter $ s $ so that the slope of the line through the points $ (-2 , s) $ and $ (-4,5) $ is equal to $ -1 $. b) Find the equation of the line Solution on video at Find the y coordinate of a point and the equation of a line , question 15
Find the equation of the line through the point $ (-2,4) $ and perpendicular to the line whose equation is given by $ -2y + 4x = -2 $ ? Solution on video at find the equation of a line through a point and perpendicular to another line , question 16
Which of the following lines whose equations are given below has a negative slope? a) $ \quad 2x - 2y = 0 $ b) $ 3x + 6y = 9 $ c) $ - y = 9 $ d) $ - y = - x + 3 $ Solution on video at find out if the slope of a line is negative , question 18
The population of a small town increased by $ 600 $ in a year. This represents a $ 5\% $ increase over the year before. What was the population of the town the year before the increase? Solution on video at Absolute increase given percentage increase , question 20
Find the area of a regular hexagon whose perimeter is 18 cm. Solution on video at Area of a regular hexagon , question 22
Among 100 students, 70 study math, among which some study physics as well, and 50 study physics, among which some study math as well. a) How many students are enrolled in both math and physics? b) How many are enrolled in math only? c) How many are enrolled in physics only?
One liter of water weighs 1 kilogram and 1 cubic decimeter is equal to 1 liter. What is the weight of water contained in a right rectangular prism container with dimensions of 10 cm, 20 cm, and 200 cm?
Linda traveled at an average speed of 65 miles per hour going to a city and at an average speed of 50 miles per hour coming back using the same road. She drove a total of 6 hours away and back. What is the distance from Linda's house to the city she visited? (round your answer to the nearest mile).
Which real number has its square equal to the sum of its half and its third?
A total of 1100 students are registered at a school. The ratio of boys to girls in this school is equal to 6:5. What is the ratio of boys to the total number of students?
There are 200 balls in a container. The balls have one of three colors: red, blue, and green. The ratio of blue balls to red balls is 3:10, and the ratio of green to blue is 7:3. How many balls of each color are there?

9th Grade Math Worksheets: FREE & Printable

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9th grade math concepts.

Translate Phrases into an Algebraic Statement
Order of Operations
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Quadratic Equations
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Adding and Subtracting Radical Expressions
Domain and Range of Radical Functions
Solving Radical Equations

Rational Expressions

Simplifying and Graphing Rational Expressions
Adding and Subtracting Rational Expressions
Multiplying and Dividing Rational Expressions
Solving Rational Equations and Complex Fractions

Statistics and Probabilities

Probability Problems
Combinations and Permutations

Statistics and Probabilities

9th grade math practice questions, fractions and decimals, real numbers and integers, proportions and ratios, algebraic expressions, equations and inequalities, polynomials, exponents and radicals, solid figures.

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Logical Reasoning

Grade 9 - Logical Reasoning

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Ninth Grade (Grade 9) Quadratic Equations and Expressions Questions

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[math]x^2 - 1x + 30[/math]
[math]x^2 +11 x - 30[/math]
[math]x^2 + x - 30[/math]
[math]x^2 - 11x -30[/math]
[math] 2a^2 - 8a[/math]
[math] 2a^2 + 2a[/math]
[math] 4[/math]
[math]12a^2 - 8a [/math]
[math]( x - 8) (x - 2)[/math]
[math](x -4 ) (x - 4)[/math]
[math](x - 8) (x +2)[/math]
[math](x + 8 ) (x - 2)[/math]
[math]-2 +- sqrt 2[/math]
[math]-2 +- 2 sqrt 2[/math]
[math]2 +- sqrt 2[/math]
[math]2 +- 2 sqrt 2[/math]

[math]y = x[/math]
[math]y = x^2[/math]
[math]y = - x[/math]
[math]y = - x^2[/math]
x=-7 and x=5
x=-7 and x=-5
x=7 and x=-5
x=7 and x=5

[math]y = - x/2[/math]
[math]y = x/2[/math]
[math]y = -(x^2)/2[/math]
[math]y = (x^2)/2[/math]
[math]3x^2-x+5=0[/math]
[math]x^2-3x=5[/math]
[math]x^2-5=-3x[/math]
[math]x^2-5=3x[/math]
[math]9/2, 1/2[/math]
[math]3,2[/math]
[math](-1)/2, (-9)/2[/math]
[math](5 +- sqrt17)/2[/math]

[math]y = 1/2x[/math]
[math]y =-1/2x[/math]
[math]y = 1/2x^2[/math]
[math]y=-1/2x^2[/math]
[math]x^2 - 4 = 0[/math]
[math]-9 + x^2 = 0[/math]
[math]-7x + 12 = 0[/math]
[math]-2 + 9x + x^2 = 0[/math]
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Chapter 1: Whole numbers

In this chapter you will engage with different kinds of numbers that are used for counting, measuring, solving equations and many other purposes.

Properties of numbers

Different types of numbers, the natural numbers.

The numbers that we use to count are called natural numbers :

$1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 7 \quad 8 \quad 9 \quad 10 \quad 11 \quad 12 \quad 13 \quad 14$

Natural numbers have the following properties:

When you add two or more natural numbers, you get a natural number again.

When you multiply two or more natural numbers, you get a natural number again.

Mathematicians describe this by saying: The system of natural numbers is closed under addition and multiplication .

However, when a natural number is subtracted from another natural number the answer is not always a natural number again. For example, there is no natural number that provides the answer to $5 - 20$.

Similarly, when a natural number is divided by another natural number the answer is not always a natural number again. For example, there is no natural number that provides the answer to $10 \div 3$.

When subtraction or division is done with natural numbers, the answers are not always natural numbers.

The system of natural numbers is not closed under subtraction or division .

Is there a smallest natural number, that means a natural number that is smaller than all other natural numbers? If so, what is it?

Is there a largest natural number, in other words, a natural number that is larger than all other natural numbers? If so, what is it?

In each of the following cases, say whether the answer is a natural number or not.

$100 + 400$
$100 - 400$
$100 \times 400$
$100 \div 400$

The whole numbers

Although we don't use 0 for counting, we need it to write numbers. Without 0, we would need a special symbol for 10, all multiples of 10 and some other numbers. For example, all the numbers that belong in the yellow cells below would need a special symbol.

The natural numbers combined with 0 is called the system of whole numbers .

If you are working with natural numbers and you add two numbers, the cd will always be different from any of the two numbers added. For example: $21 + 25 = 46$ and $24 + 1 = 25$. If you are working with whole numbers, in other words including 0, this is not the case. When 0 is added to a number the answer is just the number you start with: $24 + 0 = 24$.

For this reason, 0 is called the identity element for addition. In the set of natural numbers there is no identity element for addition.

Is there an identity element for multiplication in the whole numbers? Explain your answer.

What is the smallest natural number?

What is the smallest whole number?

The integers

In the set of whole numbers, no answer is available when you subtract a number from a number smaller than itself. For example there is no whole number that is the answer for $5 - 8$. But there is an answer to this subtraction in the system of integers.

$5 - 8 = -3$. The number $-3$ is read as "negative 3" or "minus 3".

The whole numbers start with 0 and extend in one direction:

The integers extend in both directions:

All whole numbers are also integers. The set of whole numbers forms part of the set of integers. For each whole number, there is a negative number that corresponds with it. The negative number -5 corresponds to the whole number 5 and the negative number -120 corresponds to the whole number 120.

Within the set of integers, the sum of two numbers can be 0.

For example $20 + (-20) = 0 $ and $135 + (-135) = 0$.

20 and -20 are called additive inverses of each other.

Calculate the following without using a calculator.

$100-165$
$300-700$

You may use a calculator to calculate the following:

$123-765$
$385-723$

The rational numbers

Five people share 12 slabs of chocolate equally among them.

Will each person get more or less than two full slabs of chocolate?

Can each person get another half of a slab?

How much more than two full slabs can each person get, if the two remaining slabs are divided as shown here?

Will each person get $2$, $4$ or 2$\frac{2}{5}$ slab?

The system of integers does not provide an answer for all possible division questions. For example, as we see above, the answer for 12 $\div$ 5 is not an integer.

To have answers for all possible division questions, we have to extend the number system to include fractions and negative fractions, in other words, numbers of the form $\frac{\text{integer}}{\text{integer}}$. This system of numbers is called the rational numbers . We can represent rational numbers as common fractions or as decimal numbers.

Express the answers for each of the following division problems in two ways: using the common fraction notation and using the decimal notation for fractions.

$23 \div 10$
$23 \div 5$
$230 \div 100$
$8 \div 10$

Answer the statement by writing 'yes' or 'no' in the appropriate cell.

Irrational numbers

Rational numbers do not provide for all situations that may occur in mathematics. For example, there is no rational number which will produce the answer 2 when it is multiplied by itself.

$2 \times 2 = 4$ and $1 \times 1 = 1$, so clearly, this number must be between 1 and 2.

But there is no number which can be expressed as a fraction, in either the common fraction or the decimal notation, which will solve this problem. Numbers like these are called irrational numbers .

Here are some more examples of irrational numbers:

The rational and the irrational numbers together are called the real numbers .

Calculations with whole numbers

Do not use a calculator at all in Section 1.2.

Estimating, rounding off and compensating

A shop owner wants to buy chickens from a farmer. The farmer wants R38 for each chicken. Answer the following questions without doing written calculations.

If the shop owner has R10 000 to buy chickens, do you think he can buy more than 500 chickens?

Do you think he can buy more than 200 chickens?

Do you think he can buy more than 250 chickens?

What you were trying to do in question 1 is called estimation . To estimate, when working with numbers, means to try to get close to an answer without actually doing the calculations. However, you can do other, simpler calculations to estimate.

When the goal is not to get an accurate answer, numbers may be rounded off. For example, the cost of 51 chickens at R38 each may be approximated by calculating $50 \times 40$. This is clearly much easier than calculating $51 \times \text{R }38$.

To approximate something means to try to find out more or less how much it is, without measuring or calculating it precisely.

How much is $5 \times 4$?
How much is $5 \times 40$?
How much is $50 \times 40$?

The cost of 51 chickens at R38 each is approximately R2 000.

This approximation was obtained by rounding both 51 and 38 off to the nearest multiple of 10, and then calculating with the multiples of 10.

In each case, estimate the cost by rounding off to calculate the approximate cost, without using a calculator. In each case make two estimates. First make a rough estimate by rounding the numbers off to the nearest 100 before calculating. Then make a better estimate by rounding the numbers off to the nearest 10 before calculating.

83 goats are sold for R243 each.

121 chairs are sold for R258 each.

R5 673 is added to R3 277.

R874 is subtracted from R1 234.

Suppose you have to calculate $\text{R }823 - \text{R }273$.

An estimate can be made by rounding the numbers off to the nearest 100:

$\text{R }800 - \text{R }300 = \text{R }500$.

By working with R800 instead of R823, an error was introduced into your answer. How can this error be corrected: by adding R23 to the R500, or by subtracting it from R500?

Correct the error to get a better estimate.

Now also correct the error that was made by subtracting R300 instead of R273.

What you did in question 4 is called compensating for errors .

Estimate each of the following by rounding off the numbers to the nearest 100.

$812 - 342$
$2 342 - 1 876$
$812 + 342$
$2 342 + 1 876$
$9 + 278$
$3 231 - 1 769$
$8 234 - 2 776$
$5 213 - 3 768$

Find the exact answer for each of the calculations in question 5, by working out the errors caused by rounding, and compensating for them.

Adding in columns

Write $8 000 + 1 100 + 130 + 14$ as a single number:

Write $3 000 + 700 + 50 + 8$ as a single number:

Write $5 486$ in expanded notation, as shown in 1(b).

You can calculate $3 758 + 5 486$ as shown on the left below.

Explain how the numbers in each of steps 1 to 4 are obtained.

It is only possible to use the shorter method if you add the units first, then add the tens, then the hundreds and finally, the thousands. You can then do what you did in question 1(a), without writing the separate terms of the expanded form.

Calculate each of the following without using a calculator.

$3 878 + 3 784$
$298 + 8 594$
$10 921 + 2 472$
$1 298 + 18 782$

A farmer buys a truck for R645 840, a tractor for R783 356, a plough for R83 999 and a bakkie for R435 690.

Estimate to the nearest R100 000 how much these items will cost altogether.

Use a calculator to calculate the total cost.

An investor makes R543 682 in one day on the stock market and then loses R264 359 on the same day.

Estimate to the nearest R100 000 how much money she has made in total on that day.

Use a calculator to determine how much money she has made.

Multiplying in columns

Write 3 489 in expanded notation:
Write an expression without brackets that is equivalent to $7 \times (3 000 + 400 + 80 + 9)$:

$7 \times 3 489$ may be calculated as shown on the left below.

Explain how the numbers in each of steps 1 to 4 on the above left are obtained.

$47 \times 3 489$ may be calculated as shown on the left below.

Explain how the numbers in each of steps 5 to 8 on the above left are obtained.
Explain how the number 139 560 that appears in the shorter form on the above right is obtained.

Subtracting in columns

$8 000 + 400 + 30 + 2$
$7 000 + 1 300 + 120 +12$
$3000 + 900 + 50 + 7$

If you worked correctly you should have obtained the same answers for questions 1(a) and 1(b). If this was not the case, redo your work.

The expression $7 000 + 1 300 + 120 + 12$ was formed from $8 000 + 400 + 30 + 2$ by

taking 1 000 away from 8 000 and adding it to the hundreds term to get 1 400,
taking 100 away from 1 400 and adding it to the tens term to get 130, and
taking 10 away from 130 and adding it to the units term to get 12.

Form an expression like the expression in 1(b) for each of the following:

$8 000 + 200 + 100 + 4$
$3 000 + 400 + 30 + 1$

Write expressions like in question 1(b) for the numbers below.

$8 432 - 3 957$ can be calculated as shown below.

To do the subtraction in each column, you need to think of $8 432$ as $8 000 + 400 + 30 + 2$, in fact you have to think of it as $7 000 + 1 300 + 120 + 12$.

In step 1, the 7 of 3 957 is subtracted from 12.

How is the 70 in step 2 obtained?
How is the 400 in step 3 obtained?
How is the 4 000 in step 4 obtained?
How is the 4 475 in step 5 obtained?

Because of the zeros obtained in steps 2, 3 and 4, the answers need not be written separately as shown above. The work can actually be shown in the short way below.

$9 123 - 3 784$
$8 284 - 3 547$

Use a calculator to check your answers. If your answers are wrong, try again.

$7 243 - 3 182 $
$6 221 - 1 888$

You may use a calculator to do the questions below.

Bettina has R87 456 in her savings account. She withdraws R44 800 to buy a car. How much money is left in her savings account?

Liesbet starts a savings account by making a deposit of R40 000. Over a period of time she does the following transactions on the savings account:

a withdrawal of R4 000

a withdrawal of R2 780

a deposit of R1 200

a deposit of R7 550

a withdrawal of R5 230

a deposit of R8 990

a deposit of R1 234

How much money does she have in her savings account now?

$\text{R }34 537 - \text{R }13 267$
$\text{R }135 349 - \text{R }78 239$

Long division

Study this method for calculating $13 254 \div 56$:

So $13 254 \div 56 = 236$ remainder 38, or $13 254 \div 56 = 236 \frac{38}{56} = 236\frac{19}{28} $, which can also be written as 236,68 (correct to two decimal figures).

The work can also be set out as follows:

Mlungisi's work to do a certain calculation is shown on the right. What is the question that Mlungisi tries to answer?

Where does the number 31 200 in step 1 come from? How did Mlungisi obtain it, and for what purpose did he calculate it?

Explain step 2 in the same way as you explained step 1.

Explain step 3.

$33 030 \div 63$

$18 450 \div 27$

Use a calculator to check your answers to question 2. If your answers are wrong, try again. It is important that you learn to do long division correctly.

$ 76 287 \div 287$

$ 65 309 \div 44$

Use your calculator to do questions 5 and 6 below. A municipality has budgeted R85 000 for putting up new street name boards. The street name boards cost R72 each. How many new street name boards can be put up, and how much money will be left in the budget?

A furniture dealer quoted R840 000 for supplying 3 450 school desks. A school supply company quoted R760 000 for supplying 2 250 of the same desks. Which provider is cheapest, and what do the two providers actually charge for one school desk?

Multiples and factors

Lowest common multiples and prime factorisation.

Consecutive multiples of 6, starting at 6 itself, are shown in the table below.

The table below also shows multiples of a number. What is the number?

Draw rough circles around all the numbers that occur in both the above tables.

What is the smallest number that occurs in both tables?

90 is a multiple of 6, it is also a multiple of 15.

90 is called a common multiple of 6 and 15, it is a multiple of both.

The smallest number that is a multiple of both 6 and 15 is the number 30.

30 is called the lowest common multiple or LCM of 6 and 15.

Calculate, without using a calculator.

$2 \times 3 \times 5 \times 7 \times 11$

$2 \times 2 \times 5 \times 7 \times 13$

$2 \times 3 \times 3 \times 3 \times 5 \times 13$

$3 \times 5 \times 5 \times17$

Check your answers by using a calculator or by comparing with some classmates.

2 is a factor of each of the numbers 2 310, 1 820 and 3 510.

Another way of saying this is: 2 is a common factor of 2 310, 1 820 and 3 510.

Is $2 \times 3$, in other words, 6, a common factor of 2 310 and 3 510?

Is $2 \times 3 \times 5$, in other words, 30, a common factor of 2 310 and 3 510?

Is there any bigger number than 30 that is a common factor of 2 310 and 3 510?

30 is called the highest common factor or HCF of 2 310 and 3 510.

In question 2 you can see the list of prime factors of the numbers 2 310, 1 820, 3 510 and 1 275.

The LCM of two numbers can be found by multiplying all the prime factors of both numbers, without repeating (except where a number is repeated as a factor in one of the numbers). The HCF of two numbers can be found by multiplying the factors that are common to the two numbers, i.e. in the list of prime factors of both numbers

In each case find the HCF and LCM of the numbers.

1 820 and 3 510

2 310 and 1 275

1 820 and 3 510 and 1 275

2 310 and 1 275 and 1 820

780 and 7 700

360 and 1 360

Ratio and rate problems

You may use a calculator in this section.

Moeneba collects apples in the orchard. She picks about 5 apples each minute. Approximately how many apples will Moeneba pick in each of the following periods of time?

In the situation described in question 1, Moeneba picks apples at a rate of about 5 apples per minute .

Garth and Kate also collect apples in the orchard, but they both work faster than Moeneba. Garth collects at a rate of about 12 apples per minute, and Kate collects at a rate of about 15 apples per minute. Complete the following table to show approximately how many apples they will each collect in different periods of time.

In this situation, the number of apples picked is directly proportional to the time taken.

If you filled the table in correctly, you will notice that during any period of time, Kate collected 3 times as many apples as Moeneba. We can say that during any time interval, the ratio between the numbers of apples collected by Moeneba and Kate is 3 to 1 , which can be written as 3:1 . For any period of time, the ratio between the numbers of apples collected by Garth and Moeneba is 12:5.

What is the ratio between the numbers of apples collected by Kate and Garth during a period of time?

Would it be correct to also say that the ratio between the numbers of apples collected by Kate and Garth is 5:4? Explain your answer.

To make biscuits of a certain kind, 5 parts of flour has to be mixed with 2 parts of oatmeal, and 1 part of cocoa powder. How much oatmeal and how much cocoa powder must be used if 500 g of flour is used?

A motorist covers a distance of 360 km in exactly 4 hours.

Approximately how far did the motorist drive in 1 hour?

Do you think the motorist covered exactly 90 km in each of the 4 hours? Explain your answer briefly.

Approximately how far will the motorist drive in 7 hours?

Approximately how long will the motorist need to travel 900 km?

Some people use these formulae to do calculations like those in question 5.

average speed = $\dfrac{\textbf{distance}}{\textbf{time}}$ ,which here means distance $\div $ time

distance = ${\textbf{average speed}} \times {\textbf{time}}$

time = $\dfrac{\textbf{distance}}{\textbf{average speed}}$ , which here means distance $\div $ average speed

For each of questions 5(c) and 5(d), state which formula will produce the correct answer.

A motorist completes a journey in three sections, making two long stops to eat and relax between sections. During section A he covers 440 km in 4 hours. During section B he covers 540 km in 6 hours. During section C he covers 280 km in 4 hours.

Calculate his average speed over each of the three sections.

Calculate his average speed for the journey as a whole.

On the next day, the motorist has to travel 874 km. How much time (stops excluded) will he need to do this? Justify your answer with calculations.

Different vehicles travel at different average speeds. A large transport truck with a heavy load travels much slower than a passenger car. A small bakkie is also slower than a passenger car. In the following table, some average speeds and the times needed are given for different vehicles that all have to be driven for the same distance of 720 km. Complete the table.

Look at the table you have just completed.

What happens to the time needed if the average speed increases?

What happens to the average speed if the time is reduced?

What can you say about the product average speed $\times$ time, for the numbers in the table?

In the situation above, the average speed is said to be indirectly proportional to the time needed for the journey.

Solving problems in financial contexts

You may use a calculator in Section 1.5.

Discount, profit and loss

R12 800 is divided equally between 100 people.

How much money does each person get?

How much money do eight of the people together get?

Another word for hundredths is percent .

Instead of $\frac{5}{100}$ we can write 5%. The symbol % means exactly the same as $\frac{}{100}$ .

In question 1(a) you calculated $\frac{1}{100}$ or 1% of R12 800, and in question 1(b) you calculated $\frac{8}{100}$ or 8% of R12 800.

The amount that a dealer pays for an article is called the cost price . The price marked on the article is called the marked price and the price of the article after discount is the selling price .

The marked prices of some articles are given below. A discount of 15% is offered to customers who pay cash. In each case calculate how much a customer who pays cash will actually pay.

Lina bought a couch at a sale. It was marked R3 500 but she paid only R2 800.

She was given a discount of R700.

What percentage discount was given to Lina?

This question means:

How many hundredths of the marked price were taken off?

To answer the question we need to know how much $\frac{1}{100}$ (one hundredth) of the marked price is.

How much is $\frac{1}{100}$ of R3 500?

How many hundredths of R3 500 is the same as R700?

What percentage discount was given to Lisa: 10% or 20%?

The cost price, marked price and selling price of some articles are listed below.

Article A: Cost price = R240, marked price = R360, selling price = R324.

Article B: Cost price = R540, marked price = R700, selling price = R560.

Article C: Cost price = R1 200, marked price = R2 000, selling price = R1 700.

The profit is the difference between the cost price and the selling price.

For each of the above articles, calculate the percentage discount and profit.

Remey decided to work from home and bought herself a sewing machine for R750. She planned to make 40 covers for scatter cushions. The fabric and other items she needed cost her R3 600. Remey planned to sell the covers at R150 each.

How much profit could Remey make if she sold all 40 covers at this price?

Remey managed to sell only 25 of the covers and decided to sell the rest at R100 each. Calculate her percentage profit.

Zadie bakes and sells pies to earn some extra income. The cost of the ingredients for her chicken pies came to about R68. She sold the pies for R60. Did she make a profit or a loss? Calculate the percentage loss or profit.

Hire purchase

Sometimes you need an item but do not have enough money to pay the full amount immediately. One option is to buy the item on hire purchase (HP) . You will have to pay a deposit and sign an agreement in which you undertake to pay monthly instalments until you have paid the full amount. Therefore:

HP price = deposit + total of instalments

The difference between the HP price and the cash price is the interest that the dealer charges you for allowing you to pay off the item over a period of time.

Calculate the total HP price.
How much interest does she pay?

Susie buys a car on hire purchase. The car costs R130 000. She pays a 10% deposit on the cash price and will have to pay monthly instalments of R4 600 for a period of three years. David buys the same car, but chooses another option where he has to pay a 35% deposit on the cash price and monthly instalments of R3 950 for two years.

Calculate the HP price for both options.

Calculate the difference between the total price paid by Susie and by David.

Calculate the interest that Susie and David have to pay as a percentage of the cash price.

Simple interest

When interest is calculated for a number of years on an amount (i.e. a fixed deposit) without the interest being added to the amount each year for the purpose of later interest calculations, it is referred to as simple interest. If the amount is invested for part of a year, the time must be written as a fraction of a year.

Interest rates are normally expressed as percentages. This makes it easier to compare rates. Express each of the following as a percentage:

A rate of R5 for every R100

A rate of R7,50 for every R50

A rate of R20 for every R200

A rate of x rands for every a rands

Annie deposits R8 345 into a savings account at Bonus Bank. The interest rate is 9% per annum.

Per annum means "per year".

How much interest will she have earned at the end of the first year?

Annie decides to leave the deposit of R8 345 with the bank for an indefinite period, and to withdraw only the interest at the end of every year. How much interest does she receive over a period of five years?

Maxi invested R3 500 at an interest rate of 5% per annum. Her total interest was R875. For what period did she invest the amount?

Money is invested for 1 year at an interest rate of 8% per annum. Complete the table of equivalent rates.

Interest on overdue accounts is charged at a rate of 20% per annum. Calculate the interest due on an account that is 10 days overdue if the amount owing is R260. (Give your answer to the nearest cent.)

A sum of money invested in the bank at 5% per annum, simple interest, amounted to R6 250 after 5 years. This final amount includes the interest. Thuli figured out that the final amount is (1 + 0,05 $\times$ 5 ) $\times$ amount invested.

Explain Thuli's thinking.

Calculate the amount that was invested.

Compound interest

When the interest earned each year is added to the original amount, and the interest for the following year is calculated on this new amount, the result is known as compound interest .

R2 000 is invested at 10% per annum compound interest:

End of 1st year: Amount = R2 000 + R200 interest = R2 200

End of 2nd year: Amount = R2 200 + R220 interest = R2 420

End of 3rd year: Amount = R2 420 + R242 interest = R2 662

An amount of R20 000 is invested at 5% per annum compound interest.

What is the total value of the investment after 1 year?

What is the total value of the investment after 2 years?

What is the total value of the investment after 3 years?

Bonus Bank is offering an investment scheme over two years at a compound interest rate of 15% per annum. Mr Pillay wishes to invest R800 in this way.

How much money will be due to him at the end of the two-year period?

How much interest will have been earned during the two years?

Andrew and Zinzi are arguing about interest on money that they have been given for Christmas. They each received R750. Andrew wants to invest his money in ABC Building Society for 2 years at a compound interest rate of 14% per annum, while Zinzi claims that she will do better at Bonus Bank, earning 15% simple interest per annum over 2 years. Who is correct?

Mr Martin invests R12 750 for 3 years compounded quarterly (i.e. four times a year) at 5,3%.

How many conversion periods will his investment have altogether?

How much is his investment worth after 3 years?

Calculate the total interest that he earns on his initial investment.

Calculate the interest generated by an investment (P) of R5 000 at 10% (r) compound interest over a period (n) of 3 years. A is the final amount. Use the formula $A = P(1 + \frac{r}{100})^n$ to calculate the interest.

Exchange rate and commission

Tim bought Â£650 at the foreign exchange desk at Gatwick Airport in the UK at a rate of R15,66 per Â£1. The desk also charged 2,5% commission on the transaction. How much did Tim spend to buy the pounds?

What was the value of R1 in British pounds on that day?

Mandy wants to order a book from the internet. The price of the book is $25,86. What is the price of the book in rands? Take the exchange rate as R9,95 for $1.

Bongani is a car salesperson. He earns a commission of 3% on the sale of a car with the value of R220 000. Calculate how much commission he earned.

Free Printable Quadratic Worksheets for 9th Grade

Quadratic Math Worksheets: Discover a comprehensive collection of free printable resources tailored for Grade 9 students and teachers to explore, practice, and master quadratic equations and functions.

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Explore printable Quadratic worksheets for 9th Grade

Quadratic worksheets for Grade 9 are an essential resource for teachers looking to enhance their students' understanding of quadratic equations and functions. These worksheets provide a variety of problems that challenge students to apply their knowledge of quadratic equations, factoring, graphing, and solving word problems. By incorporating these worksheets into their lesson plans, teachers can ensure that their Grade 9 students have a strong foundation in math and are well-prepared for more advanced mathematical concepts. Furthermore, these worksheets can be easily tailored to suit individual students' needs, making them an invaluable tool for both classroom instruction and personalized learning.

Quizizz offers a wide range of resources for teachers, including Quadratic worksheets for Grade 9, as well as other math-related materials. This platform allows teachers to create engaging quizzes and interactive lessons that can be accessed by students on any device. With Quizizz, teachers can track their students' progress and identify areas where they may need additional support. In addition to Quadratic worksheets for Grade 9, Quizizz also offers resources for other math topics, such as algebra, geometry, and trigonometry, making it a one-stop-shop for all of your math teaching needs. By incorporating Quizizz into your classroom, you can provide your Grade 9 students with a fun, interactive, and effective way to learn and practice math skills.

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What can QuickMath do?

QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students.

The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and cancelling common factors within a fraction.
The equations section lets you solve an equation or system of equations. You can usually find the exact answer or, if necessary, a numerical answer to almost any accuracy you require.
The inequalities section lets you solve an inequality or a system of inequalities for a single variable. You can also plot inequalities in two variables.
The calculus section will carry out differentiation as well as definite and indefinite integration.
The matrices section contains commands for the arithmetic manipulation of matrices.
The graphs section contains commands for plotting equations and inequalities.
The numbers section has a percentages command for explaining the most common types of percentage problems and a section for dealing with scientific notation.

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NCERT Solutions
NCERT Class 9
NCERT 9 Maths

NCERT Solutions for Class 9 Maths

Ncert solutions for class 9 maths – free pdf updated for 2023-24 session.

NCERT Solutions for Class 9 Maths includes solutions to all the questions given in the NCERT textbook for Class 9. The students can download PDFs of chapter-wise solutions to these problems from the links provided below on this page. These NCERT Solutions for Class 9 cover all the topics included in the NCERT textbook, like Number System, Coordinate Geometry, Polynomials, Euclid’s Geometry, Quadrilaterals, Triangles, Circles, Constructions, Surface Areas and Volumes, Statistics, Probability, etc.

With the help of these Solutions of NCERT Books for Class 9 Maths , students can practise all types of questions from the chapters. The CBSE Class 9 Maths Solutions have been designed by our experts in a well-structured format to provide several possible methods of answering the problems and ensure a proper understanding of concepts. The students are suggested to practise all these solutions thoroughly for their exams. It will also help them in building a foundation for higher-level classes.

NCERT Solutions Class 9 Maths Chapters

Chapter 1 Number System
Chapter 2 Polynomials
Chapter 3 Coordinate Geometry
Chapter 4 Linear Equations in Two Variables
Chapter 5 Introduction to Euclids Geometry
Chapter 6 Lines and Angles
Chapter 7 Triangles
Chapter 8 Quadrilaterals
Chapter 9 Circles
Chapter 10 Heron’s Formula
Chapter 11 Surface Areas and Volumes
Chapter 12 Statistics

The following chapters have been removed from the NCERT Class 9 Maths textbook 2023-24.

Areas of Parallelograms and Triangles
Constructions
Probability

Along with NCERT Solutions , students are also provided with other online learning materials available at BYJU’S, such as notes, books, question papers, exemplar problems, worksheets etc. These materials are designed with respect to the CBSE syllabus and NCERT curriculum. Students are also advised to practise CBSE Class 9 Sample Papers to get an idea of the question pattern in the final exam.

15 chapters are present in the NCERT Solutions for Class 9 Maths . These chapters given in the NCERT Class 9 Maths lay a foundation for the chapters present in Class 10. The PDFs given at BYJU’S are accessible to everyone and can be downloaded by the students for free.

NCERT Solutions of Class 9 Maths Book Chapter Brief:

Students having trouble solving tough Math problems can refer to these CBSE Maths Class 9 Solutions of NCERT for better guidance and for quick review. Solving these exercises in each chapter will ensure positive results.

NCERT Solutions Class 9 Maths Chapter 1 Number System

This chapter discusses different topics, including rational numbers and irrational numbers. Students will also be learning the extended version of the number line and how to represent numbers (integers, rational and irrational) on it. A total of 6 exercises are present in this chapter that contains the problems based on all the topics asked in the chapter. This chapter also teaches students the representation of terminating/non-terminating recurring decimals (and successive magnification method) as well as the presentation of square roots of 2, 3 and other non-rational numbers on the number line. The chapter also deals with the laws of integral powers and rational exponents with positive real bases in the Number System.

Topics Covered in Class 9 Maths Chapter 1 Number System

Review of representation of natural numbers, integers, and rational numbers on the number line. Representation of terminating/non-terminating recurring decimals on the number line through successive magnification. Rational numbers as recurring/terminating decimals.

Examples of nonrecurring/non-terminating decimals such as √2, √3, √5 etc. Existence of non-rational numbers (irrational numbers) such as √2, √3 and their representation on the number line. Explaining that every real number is represented by a unique point on the number line, and conversely, every point on the number line represents a unique real number.

Existence of √x for a given positive real number x (visual proof to be emphasised). Definition of n th root of a real number.

Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing the learner to arrive at the general laws).

Rationalisation (with precise meaning) of real numbers of the type (and their combinations) $\begin{array}{l}\frac{1}{a+b\sqrt{x}}\: and\: \frac{1}{\sqrt{x}+\sqrt{\sqrt{y}}}\end{array} $ where x and y are natural numbers, and a and b are integers.

Important Formulas –

Operations on Real Numbers

√(a/b) = √a/√b

√ab = √a √b

(√a + √b) (√a – √b) = a – b

(a + √b) (a – √b) = a 2 – b

(√a + √b) (√c + √d) = √ac + √ad + √bc + √bd

(√a + √b) 2 = a + 2√ab + b

Laws of Exponents for Real Numbers

a m . a n = a m + n

(a m ) n = a mn

a m /a n = a m – n , m > n

a m b m = (ab) m

Also, access the following resources for Class 9 Chapter 1 Number System, at BYJU’S:

NCERT Number System Class 9 Notes
Number System Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Number Systems
RD Sharma Solutions Number Systems Class 9

NCERT Solutions Class 9 Maths Chapter 2 Polynomials

This chapter discusses a particular type of algebraic expression called polynomial and the terminology related to it. Polynomial is an expression that consists of variables and coefficients involving the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. The chapter also deals with the Remainder Theorem and the Factor Theorem with the uses of these theorems in the factorisation of polynomials. Students will be taught several examples as well as the definition of different terms like polynomials, degrees, coefficients, zeros and terms of a polynomial. A total of 5 exercises are present in this chapter, which includes problems related to all the topics mentioned in the chapter.

Topics Covered in Class 9 Maths Chapter 2 Polynomials

Definition of a polynomial in one variable, its coefficients, with examples and counter examples, its terms, zero polynomial. Degree of a polynomial. Constant, linear, quadratic, cubic polynomials; monomials, binomials, trinomials. Factors and multiples. Zeros/roots of a polynomial/equation. State and motivate the Remainder Theorem with examples and analogy to integers. Statement and proof of the Factor Theorem. Factorisation of ax 2 + bx + c, a ≠ 0 where a, b, c are real numbers, and of cubic polynomials using the Factor Theorem.

Recall algebraic expressions and identities. Further identities of the type:

(x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx

(x ± y) 3 = x 3 ± y 3 ± 3xy (x ± y)

x 3 + y 3 + z 3 – 3xyz = (x + y + z) (x 2 + y 2 + z 2 – xy – yz – zx)

and their use in the factorisation of polynomials. Simple expressions are reducible to these polynomials.

Important Formulas –

(x + y) 2 = x 2 + 2xy + y 2

(x – y) 2 = x 2 – 2xy + y 2

x 2 – y 2 = (x + y) (x – y)

(x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx

(x + y) 3 = x 3 + y 3 + 3xy(x + y)

(x – y) 3 = x 3 – y 3 – 3xy(x – y)

x 3 + y 3 + z 3 – 3xyz = (x + y + z) (x 2 + y 2 + z 2 – xy – yz – zx)

Dividend = (Divisor × Quotient) + Remainder

Remainder Theorem

Let p(x) be any polynomial of degree greater than or equal to one, and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

Factor Theorem

If p(x) is a polynomial of degree n > 1 and a is any real number, then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).

Also, access the following resources for Class 9 Chapter 2 Polynomials, at BYJU’S:

NCERT Polynomials Class 9 Notes
Polynomials Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Polynomials
RD Sharma Solutions factorisation of Polynomials Class 9

NCERT Solutions Class 9 Maths Chapter 3 Coordinate Geometry

The chapter Coordinate Geometry includes the concepts of the cartesian plane, coordinates of a point in xy–plane, terms, and notations associated with the coordinate plane, including the x-axis, y-axis, x- coordinate, y-coordinate, origin, quadrants and more. Students, in this chapter, will also be studying the concepts of abscissa and ordinates of a point as well as plotting and naming a point in xy–plane. There are 3 exercises in this chapter that contain questions revolving around the topics mentioned in the chapter, helping the students get thorough with the concepts.

Topics Covered in Class 9 Maths Chapter 3 Coordinate Geometry

The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations, plotting points in the plane, graph of linear equations as examples; focus on linear equations of the type ax + by + c = 0 by writing it as y =mx + c and linking with the chapter on linear equations in two variables.

Important Points –

To locate the position of an object or a point in a plane, we require two perpendicular lines. One of them is horizontal, and the other is vertical.
The plane is called the Cartesian, or coordinate plane, and the lines are called the coordinate axes.
The horizontal line is called the x-axis, and the vertical line is called the y-axis.
The coordinate axes divide the plane into four parts called quadrants.
The point of intersection of the axes is called the origin.
The distance of a point from the y-axis is called its x-coordinate, or abscissa, and the distance of the point from the x-axis is called its y-coordinate, or ordinate.

Also, access the following resources for Class 9 Chapter 3 Coordinate Geometry, at BYJU’S:

NCERT Coordinate Geometry Class 9 Notes
Coordinate Geometry Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Coordinate Geometry
RD Sharma Solutions Coordinate Geometry Class 9

NCERT Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables

Along with recalling the knowledge of linear equations in one variable, this chapter will introduce the students to the linear equation in two variables, i.e. ax + by + c = 0. Students will also learn to plot the graph of a linear equation in two variables. There are 4 exercises in this chapter that consist of questions related to finding the solutions of a linear equation, plotting a linear equation on the graph and other topics discussed in the chapter.

Topics Covered in Class 9 Maths Chapter 4 Linear Equations in Two Variables:

Recall of linear equations in one variable. Introduction to the equation in two variables. Prove that a linear equation in two variables has infinitely many solutions, and justify their being written as ordered pairs of real numbers, plotting them and showing that they seem to lie on a line. Examples, problems from real life, including problems on Ratio and Proportion and with algebraic and graphical solutions being done simultaneously.

Important Points –

An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables.
A linear equation in two variables has infinitely many solutions.
The graph of every linear equation in two variables is a straight line.
x = 0 is the equation of the y-axis, and y = 0 is the equation of the x-axis.
The graph of x = a is a straight line parallel to the y-axis.
The graph of y = a is a straight line parallel to the x-axis.
An equation of the type y = mx represents a line passing through the origin.

Also, access the following resources for Class 9 Chapter 4 Linear Equations in Two Variables, at BYJU’S:

NCERT Linear Equations in Two Variables Class 9 Notes
Linear Equations in Two Variables Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Linear Equations in Two Variables
RD Sharma Solutions Linear Equations in Two Variables Class 9

NCERT Solutions Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry

This chapter discusses Euclid’s approach to geometry and tries to link it with present-day geometry. Introduction to Euclid’s Geometry provides the students with a method of defining common geometrical shapes and terms. Students will be taken deeper into the topic of axioms, postulates and theorems with the two exercises present in the chapter.

Topics Covered in Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry:

History – Euclid and geometry in India. Euclid’s method of formalising observed phenomenon into rigorous mathematics with definitions, common/obvious notions, axioms/postulates, and theorems. The five postulates of Euclid. Equivalent versions of the fifth postulate. Showing the relationship between axiom and theorem. 1. Given two distinct points, there exists one and only one line through them. 2. (Prove) Two distinct lines cannot have more than one point in common.

Important Axioms and Postulates –

Some of Euclid’s axioms are:

(1) Things which are equal to the same thing are equal to one another.

(2) If equals are added to equals, the wholes are equal.

(3) If equals are subtracted from equals, the remainders are equal.

(4) Things which coincide with one another are equal to one another.

(5) The whole is greater than the part.

(6) Things which are double of the same things are equal to one another.

(7) Things which are halves of the same thing are equal to one another.

Euclid’s Five Postulates

Postulate 1 – A straight line may be drawn from any one point to any other point.

Postulate 2 – A terminated line can be produced indefinitely.

Postulate 3 – A circle can be drawn with any centre and any radius.

Postulate 4 – All right angles are equal to one another.

Postulate 5 – If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

Also, access the following resources for Class 9 Chapter 5 Introduction to Euclid’s Geometry, at BYJU’S:

NCERT Introduction to Euclid’s Geometry Class 9 Notes
Introduction to Euclid’s Geometry Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Introduction to Euclid’s Geometry
RD Sharma Solutions Introduction to Euclid’s Geometry Class 9

NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles

This chapter revolves around the theorems present in the topics of Lines and Angles. Students might often be asked to prove the statements given in the questions. There are 3 exercises in the chapter, solving which students would be able to understand the concepts covered in the chapter thoroughly. There are four axioms and eight theorems covered in the chapter.

Topics Covered in Class 9 Maths Chapter 6 Lines and Angles:

1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and the converse. 2. (Prove) If two lines intersect, the vertically opposite angles are equal. 3. (Motivate) Results on corresponding angles, alternate angles, and interior angles when a transversal intersects two parallel lines. 4. (Motivate) Lines, which are parallel to a given line, are parallel. 5. (Prove) The sum of the angles of a triangle is 180°. 6. (Motivate) If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice versa. This property is called the Linear pair axiom.
If two lines intersect each other, then the vertically opposite angles are equal.
If a transversal intersects two parallel lines, then

(i) each pair of corresponding angles is equal,

(ii) each pair of alternate interior angles is equal,

(iii) each pair of interior angles on the same side of the transversal is supplementary.

If a transversal intersects two lines such that, either

(i) any one pair of corresponding angles is equal, or

(ii) any one pair of alternate interior angles is equal, or

(iii) any one pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel.

Also, access the following resources for Class 9 Chapter 6 Lines and Angles, at BYJU’S:

NCERT Lines and Angles Class 9 Notes
Lines and Angles Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Lines and Angles
RD Sharma Solutions Lines and Angles Class 9

NCERT Solutions Class 9 Maths Chapter 7 Triangles

In this chapter, students will study in detail the congruence of triangles, rules of congruence, some properties of triangles and the inequalities in triangles. The chapter has a total of 5 exercises, in which the students are asked “to-prove” as well as application-level problems. With this chapter, students will also learn to prove the properties that they learnt in earlier classes. The chapter also teaches the students to apply the various congruence rules while solving the problems. There are about eight theorems covered in this chapter.

Topics Covered in Class 9 Maths Chapter 7 Triangles:

1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence). 2. (Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence). 3. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to the three sides of the other triangle (SSS Congruence). 4. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. 5. (Prove) The angles opposite to equal sides of a triangle are equal. 6. (Motivate) The sides opposite to equal angles of a triangle are equal. 7. (Motivate) Triangle inequalities and the relation between ‘angle and facing side’; inequalities in a triangle.

Important Axioms and Theorems –

Axiom 7.1 (SAS congruence rule) – Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

Theorem 7.1 (ASA congruence rule) – Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.

Theorem 7.2 – Angles opposite to equal sides of an isosceles triangle are equal.

Theorem 7.3 – The sides opposite to equal angles of a triangle are equal.

Theorem 7.4 (SSS congruence rule) – If the three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Theorem 7.5 (RHS congruence rule) – If in two right triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Theorem 7.6 – If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).

Theorem 7.7 – In any triangle, the side opposite to the larger (greater) angle is longer.

Theorem 7.8 – The sum of any two sides of a triangle is greater than the third side.

Also, access the following resources for Class 9 Chapter 7 Triangles, at BYJU’S:

NCERT Triangles Class 9 Notes
Triangles Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Triangles
RD Sharma Solutions Triangles Class 9

NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals

A figure obtained by joining four points in order is called a quadrilateral. This chapter takes the students to the depth of the topics of Quadrilaterals. The chapter contains 2 exercises that contain only one theorem to prove. However, there are a total of nine theorems that can be used to solve the application or conceptual-level questions asked. The angle sum property of a quadrilateral, types of quadrilaterals, properties of a parallelogram, and the mid-point theorem are explained in this chapter to help the students in learning the concepts thoroughly.

Topics Covered in Class 9 Maths Chapter 8 Quadrilaterals:

1. (Prove) The diagonal divides a parallelogram into two congruent triangles. 2. (Motivate) In a parallelogram, opposite sides are equal and conversely. 3. (Motivate) In a parallelogram, opposite angles are equal and conversely. 4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides are parallel and equal. 5. (Motivate) In a parallelogram, the diagonals bisect each other conversely. 6. (Motivate) In a triangle, the line segment joining the midpoints of any two sides is parallel to the third side and (motivate) its converse.

Important Theorems –

A quadrilateral has four sides, four angles and four vertices.

Angle Sum Property of a Quadrilateral – The sum of the angles of a quadrilateral is 360 o .

Theorem 8.1 – A diagonal of a parallelogram divides it into two congruent triangles

Theorem 8.2 – In a parallelogram, opposite sides are equal.

Theorem 8.3 – If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Theorem 8.4 – In a parallelogram, opposite angles are equal.

Theorem 8.5 – If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Theorem 8.6 – The diagonals of a parallelogram bisect each other.

Theorem 8.7 – If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Theorem 8.8 – A quadrilateral is a parallelogram if a pair of opposite sides are equal and parallel.

Theorem 8.9 – The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Theorem 8.10 – The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

Also, access the following resources for Class 9 Chapter 8 Quadrilaterals, at BYJU’S:

NCERT Quadrilaterals Class 9 Notes
Quadrilaterals Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Quadrilaterals
RD Sharma Solutions Quadrilaterals Class 9

NCERT Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

In this chapter, an attempt has been made to consolidate the knowledge about the formulae to find the areas of different figures by studying relationships between the areas of geometric figures, provided they lie on the same base and between the same parallels. This study will also be useful in understanding some results on the ‘similarity of triangles’. The chapter contains 4 exercises, of which most of the questions ask the students to prove the statements given.

Topics Covered in Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles:

Review the concept of area and recall the area of a rectangle. 1. (Prove) Parallelograms on the same base and between the same parallels have the same area. 2. (Motivate) Triangles on the same base and between the same parallels are equal in area and its converse.

Theorem 9.1 – Parallelograms on the same base and between the same parallels are equal in area.

Theorem 9.2 – Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Theorem 9.3 – Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Also, access the following resources for Class 9 Chapter 9 Areas of Parallelograms and Triangles, at BYJU’S:

NCERT Areas of Parallelograms and Triangles Class 9 Notes
Areas of Parallelograms and Triangles Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Areas of Parallelograms and Triangles
RD Sharma Solutions Areas of Parallelograms and Triangles Class 9

NCERT Solutions Class 9 Maths Chapter 10 Circles

A circle can be defined as a collection of all the points in a plane at a fixed distance from a fixed point in the plane. Topics like Angle Subtended by a Chord at a Point, Equal Chords and their respective distances from the Centre, the Angle Subtended by an Arc of a Circle, Cyclic Quadrilaterals and other terms related to circles are covered in this chapter. A total of twelve theorems are present in this chapter, learning which the students will get a clearer idea of the concepts taught. There are 6 exercises in this chapter which consist of questions from all the concepts present in the chapter.

Topics Covered in Class 9 Maths Chapter 10 Circles:

Through examples, arrive at definitions of circle-related concepts, radius, circumference, diameter, chord, arc, and subtended angle. 1. (Prove) Equal chords of a circle subtend equal angles at the centre and (motivate) its converse. 2. (Motivate) The perpendicular from the centre of a circle to a chord bisects the chord, and conversely, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. 3. (Motivate) There is one and only one circle passing through three given non-collinear points. 4. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the centre(s) and conversely. 5. (Prove) The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. 6. (Motivate) Angles in the same segment of a circle are equal. 7. (Motivate) If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, the four points lie on a circle. 8. (Motivate) The sum of either pair of opposite angles of a cyclic quadrilateral is 180 0 and its converse.

Theorem 10.1 – Equal chords of a circle subtend equal angles at the centre.

Theorem 10.2 – If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Theorem 10.3 – The perpendicular from the centre of a circle to a chord bisects the chord.

Theorem 10.4 – The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Theorem 10.5 – There is one and only one circle passing through three given non-collinear points.

Theorem 10.6 – Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Theorem 10.7 – Chords equidistant from the centre of a circle are equal in length.

Theorem 10.8 – The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Theorem 10.9 – Angles in the same segment of a circle are equal.

Theorem 10.10 – If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

Theorem 10.11 – The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.

Theorem 10.12 – If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

Also, access the following resources for Class 9 Chapter 10 Circles, at BYJU’S:

NCERT Circles Class 9 Notes
Circles Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Circles
RD Sharma Solutions Circles Class 9

NCERT Solutions Class 9 Maths Chapter 11 Constructions

In this chapter, students will learn some basic constructions. The method learnt will then be used to construct certain kinds of triangles. There are 2 exercises present in this chapter, of which the first exercise deals with the construction of a certain angle or the bisector of a given angle. On the other hand, the second exercise deals with the constructions of triangles when different parameters are given.

Topics Covered in Class 9 Maths Chapter 11 Constructions:

1. Construction of bisectors of a line segment and angle, 60°, 90°, 45° angles etc., equilateral triangles. 2. Construction of a triangle given its base, sum/difference of the other two sides and one base angle. 3. Construction of a triangle of given perimeter and base angles.

Construction 11.1 – To construct the bisector of a given angle.

Construction 11.2 – To construct the perpendicular bisector of a given line segment.

Construction 11.3 – To construct an angle of 600 at the initial point of a given ray.

Construction 11.4 – To construct a triangle, given its base, a base angle and the sum of the other two sides.

Construction 11.5 – To construct a triangle given its base, a base angle and the difference between the other two sides.

Construction 11.6 – To construct a triangle, given its perimeter and its two base angles.

Also, access the following resources for Class 9 Chapter 11 Constructions, at BYJU’S:

NCERT Constructions Class 9 Notes
Constructions Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Constructions
RD Sharma Solutions Constructions Class 9

NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula

The chapter discusses Heron’s formula, which can be used to calculate the area of a triangle when the length of all three sides is given. In this method, there is no need to calculate the angles or other distances in the triangle. This formula can be used not only to find the area of triangles but also to find the areas of quadrilaterals and other polygons by dividing them into triangles. There are 2 exercises in this chapter which help the student understand the method of solving the problems based on Heron’s formula.

Topics Covered in Class 9 Maths Chapter 12 Heron’s Formula:

Area of a triangle using Heron’s formula (without proof) and its application in finding the area of a quadrilateral.

Area of a triangle = 1/2 × base × height

Area of a triangle by Heron’s Formula = $\begin{array}{l}\sqrt{s(s-a)(s-b)(s-c)}\end{array} $

Where a, b and c are the sides of a triangle

s is the semi-perimeter, i.e. half of the perimeter of a triangle = (a + b + c)/2

Also, access the following resources for Class 9 Chapter 12 Heron’s Formula, at BYJU’S:

NCERT Heron’s Formula Class 9 Notes
Heron’s Formula Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Heron’s Formula
RD Sharma Solutions Heron’s Formula Class 9

NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes

In this chapter, students shall learn to find the surface areas and volumes of cuboids and cylinders in detail and broaden the study to some other solids, such as cones and spheres. This chapter is just an extended version of the chapter mensuration, in which the students learnt about the surface areas and volumes in earlier classes. There are 8 exercises in this chapter, and these exercises contain problems that are based on surface areas and volumes of different solids such as cubes, cuboids, spheres, cylinders, cones, and hemispheres.

Topics Covered in Class 9 Maths Chapter 13 Surface Areas and Volumes :

Surface areas and volumes of cubes, cuboids, spheres (including hemispheres) and right circular cylinders/cones.

Surface Area of a Cuboid = 2(lb + bh + hl)

where l, b and h are, respectively, the three edges of the cuboid

Surface Area of a Cube = 6a 2

where a is the edge of the cube.

Curved Surface Area of a Cylinder = 2πrh

where r is the radius of the base of the cylinder and h is the height of the cylinder

Total Surface Area of a Cylinder = 2πr(r + h)

where h is the height of the cylinder and r is its radius

Curved Surface Area of a Cone = 1/2 × l × 2πr = πrl

where r is its base radius, and l is its slant height

Total Surface Area of a Cone = πrl + πr 2 = πr(l + r)

Surface Area of a Sphere = 4 π r 2

where r is the radius of the sphere.

Curved Surface Area of a Hemisphere = 2πr 2

where r is the radius of the sphere of which the hemisphere is a part.

Total Surface Area of a Hemisphere = 3πr 2

Volume of a Cuboid = base area × height = length × breadth × height

Volume of a Cube = edge × edge × edge = a 3

Volume of a Cylinder = πr 2 h

where r is the base radius, and h is the height of the cylinder.

Volume of a Cone = 1/3 πr 2 h

where r is the base radius, and h is the height of the cone.

Volume of a Sphere = 4/3 πr 3

Volume of a Hemisphere = 2/3 πr 3

where r is the radius of the hemisphere.

Also, access the following resources for Class 9 Chapter 13 Surface Areas and Volumes, at BYJU’S:

NCERT Surface Areas and Volumes Class 9 Notes
Surface Areas and Volumes Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Surface Areas and Volumes
RD Sharma Solutions Surface Areas and Volumes Class 9

NCERT Solutions Class 9 Maths Chapter 14 Statistics

The branch of Mathematics in which the extraction of meaningful information is studied is known as Statistics. It can also be defined as the collection of data on different aspects of the life of people useful to the State. The chapter teaches about the different presentations of the data, including the frequency distribution as well. The chapter also helps the students learn the graphical representation of data using different graphs such as Bar graphs, Histograms, Frequency polygons, etc. The chapter also lets the students learn the measure of central tendency mean, median and mode of the raw data. A total of 4 exercises are present in the chapter that includes problems related to all these concepts.

Topics Covered in Class 9 Maths Chapter 14 Statistics:

Introduction to Statistics: Collection of data, presentation of data – tabular form, ungrouped/grouped, bar graphs, histograms (with varying base lengths), frequency polygons, qualitative analysis of data to choose the correct form of presentation for the collected data. Mean, median, and mode of ungrouped data.

Statistics deals with the collection, presentation, and analysis of data, as well as the drawing of meaningful conclusions from the data.

Mean – It is found by adding all the values of the observations and dividing it by the total number of observations. It is denoted by $\begin{array}{l}\overline{x}\end{array} $ .

So, $\begin{array}{l}\overline{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}\end{array} $

For an ungrouped frequency distribution, it is $\begin{array}{l}\overline{x}=\frac{\sum_{i=1}^{n}f_{i}x_{i}}{\sum_{i=1}^{n}f_{i}}\end{array} $

Median – It is the value of the middle-most observation (s).

If n is an odd number, the median = value of the [(n + 1)/2] th observation

If n is an even number, median = Mean of the values of the (n/2) th and [(n/2) + 1] th observations.

Mode – The mode is the most frequently occurring observation.

Also, access the following resources for Class 9 Chapter 14 Statistics, at BYJU’S:

NCERT Statistics Class 9 Notes
Statistics Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Statistics
RD Sharma Solutions Statistics Class 9

NCERT Solutions Class 9 Maths Chapter 15 Probability

The collection of some outcomes of an experiment is known as an event of an experiment. The chances of occurrence of an event are known as probability. In this chapter, students will learn to measure the chance of occurrence of a particular outcome in an experiment. This chapter contains only 1 exercise. The problems covered in this exercise are based on real-life incidents, enhancing the interest of the students in solving the questions.

Topics Covered in Class 9 Maths Chapter 15 Probability:

History, Repeated experiments and observed frequency approach to probability. Focus is on empirical probability. (A large amount of time to be devoted to group and to individual activities to motivate the concept; the experiments to be drawn from real-life situations and from examples used in the chapter on statistics).

The empirical probability P(E) of an event E happening is given by

P (E) = Number of trials in which the event happened/The total number of trials

Also, access the following resources for Class 9 Chapter 15 Probability, at BYJU’S:

NCERT Probability Class 9 Notes
Probability Class 9 Important Questions
NCERT Exemplar Solutions Class 9 Probability
RD Sharma Solutions Probability Class 9

Benefits of BYJU’S NCERT Maths Solutions for Class 9

The solutions are given in easy steps so that students can easily understand them.
Students can access these solutions for free and clear any doubts instantly.
Exercise-wise PDFs are also provided for all chapters of CBSE 9th Class Maths.
Detailed explanations also help students to develop problem-solving skills and help them to learn concepts more effectively.
The questions are solved using the easiest method.
Diagrams are also given to help students visualise the questions and solutions.

CBSE Class 9 Maths Unit-wise Weightage for 2023-24

Internal assessment class 9.

During the exam preparation, students should first complete the units carrying more weightage and then move on to the lower weightage. This helps students to first cover the important concepts and improves their confidence level to perform well in the annual exam.

The students are advised to analyse all the concepts covered in the Syllabus of Class 9 Maths and prepare a proper preparation strategy. It’s merely impossible to score well in Maths by simply reading and memorising the concepts. Students should always try to understand the concept and logic given in any particular topic. The students are suggested to practise these NCERT Solutions Class 9 materials on a regular basis to develop a strong understanding of basic mathematics concepts.

Along with the NCERT Solutions, we have also provided a brief summary of important formulas and different methods of solving similar problems to help students understand the concepts thoroughly. Keep visiting BYJU’S to get more updated learning materials, and download the BYJU’S app for a better and personalised learning experience with engaging video lessons.

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Check answers. 1) x = 0 , its square is 0 2 = 0 . Hence x and its square are equal. 2) x = 1 , its square is 1 2 = 1 . Hence x and its square are equal. Which number (s) is (are) equal to half its (their) square? Let x be the number to find. Half its square is (1/2) x 2. First write with right side equal to zero.
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Write the ratio of the number of blue marbles to the total number of marbles in terms of r, b and w . The perimeter of a rectangle is equal to 280 meters. The ratio of its length to its width is 5:2. Find the area of the rectangle. The angles of a triangle are in the ratio 1:3:8. Find the measures of the three angles of this triangle.
Grade 9 Algebra Word Problems
Coin Algebra Word Problems - Grade 9. Example: Bob has in his pocket a number of pennies, 5 times as many nickels as pennies and 5 more quarters than pennies. The coins amount to $2.27. Find the number of each kind of coin. Show Video Lesson.
Grade 9 Math Practice Test Questions
Grade 9 math practice test questions on algebra, geometry, trigonometry, and problem solving are presented along with their solutions on videos. Simplify the expressions. Solution on video at Simplify expressions with radicals, question 1. Expand and simplify the polynomials below. Solution on video at Expand and simplify polynomials, question 2.
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Free 9th Grade Math Problems Worksheets. This collection shares over 100 free printable 9th grade math worksheets on topics including order of operations, fractions and decimals, solving equations, and graphing. Plus every worksheet includes a free answer key. Just scroll through the list of topics, select any one that you want, and click on it ...
50+ Math worksheets for 9th Grade on Quizizz
Free Printable Math Worksheets for 9th Grade. Math just got more exciting for Grade 9 students! Discover a vast collection of free printable math worksheets by Quizizz, tailored to enhance learning and understanding. Dive into a world of fun and interactive problem-solving activities that will transform your classroom experience.
9th Grade Math Worksheets: FREE & Printable
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Free Printable algebra Worksheets for 9th Grade
Explore printable algebra worksheets for 9th Grade. Algebra worksheets for Grade 9 are essential resources for teachers who want to provide their students with a comprehensive understanding of algebraic concepts and applications. These worksheets cover a wide range of topics, including linear equations, quadratic functions, systems of equations ...
Algebra 1
The Algebra 1 course, often taught in the 9th grade, covers Linear equations, inequalities, functions, and graphs; Systems of equations and inequalities; Extension of the concept of a function; Exponential models; and Quadratic equations, functions, and graphs. Khan Academy's Algebra 1 course is built to deliver a comprehensive, illuminating, engaging, and Common Core aligned experience!
Grade 9 Math Lessons and Practice
Free grade 9 math lessons and practice: videos, printable notes and worksheets, online quizzes and games. No subscription required.
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Khan Academy's 100,000+ free practice questions give instant feedback, don't need to be graded, and don't require a printer. Math Worksheets. Khan Academy. Math worksheets take forever to hunt down across the internet. Khan Academy is your one-stop-shop for practice from arithmetic to calculus. Math worksheets can vary in quality from ...
Grade 9
Grade 9 - Logical Reasoning problems, online practice, tests, worksheets, quizzes, and teacher assignments. ... Math; Grade 9; Logical Reasoning; Grade 9 - Logical Reasoning ... New Worksheet button, you will get a brand new printable PDF worksheet on Logical Reasoning. You can choose to include answers and step-by-step solutions. New Worksheet
Grade 9 Math Unit 2
Free lessons, worksheets, and video tutorials for students and teachers. Topics in this unit include: solving degree 1 (linear) equations, balance method, inverse operations, solving simple and multi-step equations, solving equations involving fractions, and rearranging formulas. This follows chapter 4 of the principles of math grade 9 McGraw ...
Ninth Grade (Grade 9) Quadratic Equations and Expressions Questions
You are browsing Grade 9 questions. View questions in All Grades. ... You can create printable tests and worksheets from these Grade 9 Quadratic Equations and Expressions questions! Select one or more questions using the checkboxes above each question. ... (-1)/2, (-9)/2[/math] [math](5 +- sqrt17)/2[/math] Grade 9 Quadratic Equations and ...
Solving problems about ratio, rate and proportion
Siyavula's open Mathematics Grade 9 textbook, chapter 1 on Whole numbers covering Solving problems about ratio, rate and proportion. Home Practice. ... If you worked correctly you should have obtained the same answers for questions 1(a) and 1(b). If this was not the case, redo your work. ...
Free Printable Quadratic Worksheets for 9th Grade
Free Printable Quadratic Worksheets for 9th Grade. Quadratic Math Worksheets: Discover a comprehensive collection of free printable resources tailored for Grade 9 students and teachers to explore, practice, and master quadratic equations and functions. grade 9 Quadratic. Solving Quadratics (Factor, SquareRoot & Quadratic Formula) 20 Q. 9th - 11th.
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QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and ...
NCERT Solutions for Class 9 Maths (Updated for 2023-24 Exam)
Students having trouble solving tough Math problems can refer to these CBSE Maths Class 9 Solutions of NCERT for better guidance and for quick review. Solving these exercises in each chapter will ensure positive results. ... ☞ Exercise 12.2 - 9 Questions (4 long answers, 5 very long answers) Also, access the following resources for Class 9 ...
30 Math Puzzles (with Answers) to Test Your Smarts
Which number should replace the question mark? Answer: 9. Go figure. ... Answer: 6—it should be yellow. RELATED: Hard Math Problems That'll Test Your ... Try solving the viral numbers on ...
PDF GRADE 5 Mathematics Practice Assessment
Mathematics Page 8 DIRECTIONS Read each question carefully. Choose the best answer to each question. For open-response questions, determine the best answer to the question. 1 A recipe for one batch of cookies uses 1.25 cups of butter. How many cups of butter are needed to make 6 batches of cookies? A 7.5 cups. B 4.75 cups. C 7.25 cups. D 6.5 ...

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Grade 9 - Logical Reasoning

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