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## Solving Word Problems in Algebra Inequality Word Problems

How are you with solving word problems in Algebra? Are you ready to dive into the "real world" of inequalities? I know that solving word problems in Algebra is probably not your favorite, but there's no point in learning the skill if you don't apply it.

I promise to make this as easy as possible. Pay close attention to the key words given below, as this will help you to write the inequality. Once the inequality is written, you can solve the inequality using the skills you learned in our past lessons.

I've tried to provide you with examples that could pertain to your life and come in handy one day. Think about others ways you might use inequalities in real world problems. I'd love to hear about them if you do!

Before we look at the examples let's go over some of the rules and key words for solving word problems in Algebra (or any math class).

## Word Problem Solving Strategies

- Read through the entire problem.
- Highlight the important information and key words that you need to solve the problem.
- Identify your variables.
- Write the equation or inequality.
- Write your answer in a complete sentence.
- Check or justify your answer.

I know it always helps too, if you have key words that help you to write the equation or inequality. Here are a few key words that we associate with inequalities! Keep these handy as a reference.

## Inequality Key Words

- at least - means greater than or equal to
- no more than - means less than or equal to
- more than - means greater than
- less than - means less than

Ok... let's put it into action and look at our examples.

## Example 1: Inequality Word Problems

Keith has $500 in a savings account at the beginning of the summer. He wants to have at least $200 in the account by the end of the summer. He withdraws $25 each week for food, clothes, and movie tickets.

- Write an inequality that represents Keith's situation.
- How many weeks can Keith withdraw money from his account? Justify your answer.

Step 1: Highlight the important information in this problem.

Note: At least is a key word that notes that this problem must be written as an inequality.

Step 2 : Identify your variable. What don't you know? The question verifies that you don't know how many weeks.

Let w = the number of weeks

Step 3: Write your inequality.

500 - 25w > 200

I know you are saying, "How did you get that inequality?"

I know the "at least" part is tricky. You would probably think that at least means less than.

But... he wants the amount in his account to be at least $200 which means $200 or greater. So, we must use the greater than or equal to symbol.

Step 4 : Solve the inequality.

The number of weeks that Keith can withdraw money from his account is 12 weeks or less.

Step 5: Justify (prove your answer mathematically).

I'm going to prove that the largest number of weeks is 12 by substituting 12 into the inequality for w. You could also substitute any number less than 12.

Since 200 is equal to 200, my answer is correct. Any more than 12 weeks and his account balance would be less than $200. Any number of weeks less than 12 and his account would stay above $200.

That wasn't too bad, was it? Let's take a look at another example.

## Example 2: More Inequality Word Problems

Yellow Cab Taxi charges a $1.75 flat rat e in addition to $0.65 per mile . Katie has no more than $10 to spend on a ride.

- Write an inequality that represents Katie's situation.
- How many miles can Katie travel without exceeding her budget? Justify your answer.

Note: No more than are key words that note that this problem must be written as an inequality.

Step 2 : Identify your variable. What don't you know? The question verifies that you don't know the number of miles Katie can travel.

Let m = the number of miles

Step 3: Write the inequality.

0.65m + 1.75 < 10

Are you thinking, "How did you write that inequality?"

The "no more than" can also be tricky. "No more than" means that you can't have more than something, so that means you must have less than!

Step 4: Solve the inequality.

Since this is a real world problem and taxi's usually charge by the mile, we can say that Katie can travel 12 miles or less before reaching her limit of $10.

Are you ready to try some on your own now? Yes... of course you are! Click here to move onto the word problem practice problems.

## Take a look at the questions that other students have submitted:

Quite a complicated problem about perimeter and area of a rectangle

This is a toughie.... a compound inequalities word problem

- Inequalities
- Inequality Word Problems

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## Solving Inequalities Word Problems - Examples & Practice - Expii

Solving inequalities word problems - examples & practice, explanations (3), (videos) solving word problems involving inequalities.

by PatrickJMT

This video by Patrick JMT works through an inequality word problem. There is another video with a second example later in the explanation.

The first example is:

A widget factory has a fixed operating cost of $3,600 per day plus costs $1.40 per widget produced. If a widget sells for $4.20, what is the least number of widgets that must be sold per day to make a profit?

To make a profit, the revenue received needs to be greater than the cost needed to make it. \text{Cost}">Revenue>Cost

Now, we want to create an expression for each of these. Let's let x be the number of widgets produced.

For cost, we know there's a fixed cost of $3600 and then $1.40 for each widget. 3600+1.4x">Revenue>3600+1.4x

For revenue, we receive $4.20 for each widget. 3600+1.4x">4.2x>3600+1.4x

Finally, we solve by getting x to one side of the inequality, similar to how you'd do for an equation . 3600+1.4x\\ 4.2\color{#8925AE}{-1.4x}&>3600+1.4x\color{#8925AE}{-1.4x}\\ 2.8x&>3600\\ x&>1285.71 \end{align}">4.2x>3600+1.4x4.2−1.4x>3600+1.4x−1.4x2.8x>3600x>1285.71

So, in order to make a profit, you'd need to sell over 1285 widgets.

Here's another example.

Jason makes and sells fishing poles. If he has costs totaling $12,000 per year plus a cost of $4 per pole, how many fishing poles must he produce to make a profit of $48,000 in one year, if he sells the fishing poles for $28 each?

We want the profit to be greater than or equal to $48,000. Profit≥48000

We can also rewrite profit as the amount you make minus the amount you spend. amount made−amount spent≥48000

If he sells the poles for $28 each, then that is the amount he makes. We can let x be the number of poles. 28x−amount spent≥48000

We also know there's a fixed cost to make the poles for $12000. Also, each pole costs $4 to make. 28x−(12000+4x)≥48000

Finally, we can solve the inequality by getting x alone on one side. 28x−(12000+4x)≥4800028x−12000−4x≥4800024x−12000≥4800024x−12000+12000≥48000+1200024x≥60000x≥2500

To get here, you may want to review the distributive properties with inequalities and solving inequalities via grouping .

Jason will need to sell at least 2,500 poles in order to make a profit of $48,000.

## Related Lessons

Inequality word problems.

The key to word problems is translating the given information into math. In this case, we need to translate word problems into statements of inequality .

Inequalities are denoted by words like "greater than" and "less than". In general, a comparison is made in problems with inequalities.

Let's look at two examples:

Image source: by Hannah Bonville

We need to convert this information into an inequality. We are trying to figure out how many ducks Chris can by while staying at or under budget. In other words, we want to keep the cost less than or equal to his budget. Sounds like a good place to use a ≤ sign, doesn't it?

What inequality translates the information in this word problem? Let x be the number of ducks Chris can buy.

6.95x+9.99≤80

6.95+9.99x≤80

## Solving Word Problems Involving Inequalities

Inequality word problems will always involve comparisons . In real life, you might use inequalities to decide which of two options is better. Look for comparison words to figure out which kind of symbol you should use. Look for rates to write out an equation .

Test Yourself: What are the comparison words in the following problem, and what inequality symbol do they relate to?

Billie earns a salary of $18,000 per year plus a 5% commission on all her sales. How much must her sales be if her annual income is to be no less than $20,000?

"Plus" ">>

"No less than" ≥

"No less than" ≤

"How much" <

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- Yes, I reside in South Africa
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## 2.9 Word problems

2.9 word problems (embfx).

Solving word problems requires using mathematical language to describe real-life contexts. Problem-solving strategies are often used in the natural sciences and engineering disciplines (such as physics, biology, and electrical engineering) but also in the social sciences (such as economics, sociology and political science). To solve word problems we need to write a set of equations that describes the problem mathematically.

Examples of real-world problem solving applications are:

- modelling population growth;
- modelling effects of air pollution;
- modelling effects of global warming;
- computer games;
- in the sciences, to understand how the natural world works;
- simulators that are used to train people in certain jobs, such as pilots, doctors and soldiers;
- in medicine, to track the progress of a disease.

## Problem solving strategy (EMBFY)

Read the problem carefully.

What is the question and what do we need to solve for?

Assign variables to the unknown quantities, for example, \(x\) and \(y\).

Translate the words into algebraic expressions by rewriting the given information in terms of the variables.

Set up a system of equations.

Solve for the variables using substitution.

Check the solution.

Write the final answer.

## Simple word problems

Write an equation that describes the following real-world situations mathematically:

Mohato and Lindiwe both have colds. Mohato sneezes twice for each sneeze of Lindiwe's. If Lindiwe sneezes \(x\) times, write an equation describing how many times they both sneezed.

The difference of two numbers is \(\text{10}\) and the sum of their squares is \(\text{50}\). Find the two numbers.

Liboko builds a rectangular storeroom. If the diagonal of the room is \(\sqrt{ \text{1 312}}\) \(\text{m}\) and the perimeter is \(\text{80}\) \(\text{m}\), determine the dimensions of the room.

It rains half as much in July as it does in December. If it rains \(y\) mm in July, write an expression relating the rainfall in July and December.

Zane can paint a room in \(\text{4}\) hours. Tlali can paint a room in \(\text{2}\) hours. How long will it take both of them to paint a room together?

\(\text{25}\) years ago, Arthur was \(\text{5}\) years more than a third of Bongani's age. Today, Bongani is \(\text{26}\) years less than twice Arthur's age. How old is Bongani?

The product of two integers is \(\text{95}\). Find the integers if their total is \(\text{24}\).

## Worked example 24: Gym membership

The annual gym subscription for a single member is \(\text{R}\,\text{1 000}\), while an annual family membership is \(\text{R}\,\text{1 500}\). The gym is considering increasing all membership fees by the same amount. If this is done then a single membership would cost \(\dfrac{5}{7}\) of a family membership. Determine the amount of the proposed increase.

## Identify the unknown quantity and assign a variable

Let the amount of the proposed increase be \(x\).

## Use the given information to complete a table

Set up an equation, solve for \(x\), write the final answer.

The proposed increase is \(\text{R}\,\text{250}\).

## Worked example 25: Corner coffee house

Erica has decided to treat her friends to coffee at the Corner Coffee House. Erica paid \(\text{R}\,\text{54,00}\) for four cups of cappuccino and three cups of filter coffee. If a cup of cappuccino costs \(\text{R}\,\text{3,00}\) more than a cup of filter coffee, calculate how much a cup of each type of coffee costs?

## Method 1: identify the unknown quantities and assign two variables

Let the cost of a cappuccino be \(x\) and the cost of a filter coffee be \(y\).

## Use the given information to set up a system of equations

Solve the equations by substituting the second equation into the first equation.

\begin{align*} 4(y+3) + 3y &= 54 \\ 4y+12 + 3y &= 54 \\ 7y &= 42 \\ y &= 6 \end{align*}

If \(y=6\), then using the second equation we have \begin{align*} x &= y + 3 \\ &= 6 + 3 \\ &= 9 \end{align*}

## Check that the solution satisfies both original equations

A cup of cappuccino costs \(\text{R}\,\text{9}\) and a cup of filter coffee costs \(\text{R}\,\text{6}\).

## Method 2: identify the unknown quantities and assign one variable

Let the cost of a cappuccino be \(x\) and the cost of a filter coffee be \(x-3\).

## Use the given information to set up an equation

\[4x + 3(x-3) = 54\]

\begin{align*} 4x + 3(x-3) &= 54 \\ 4x + 3x-9 &= 54 \\ 7x &= 63 \\ x &= 9 \end{align*}

## Worked example 26: Taps filling a container

Two taps, one more powerful than the other, are used to fill a container. Working on its own, the less powerful tap takes \(\text{2}\) hours longer than the other tap to fill the container. If both taps are opened, it takes \(\text{1}\) hour, \(\text{52}\) minutes and \(\text{30}\) seconds to fill the container. Determine how long it takes the less powerful tap to fill the container on its own.

## Identify the unknown quantities and assign variables

Let the time taken for the less powerful tap to fill the container be \(x\) and let the time taken for the more powerful tap be \(x - 2\).

## Convert all units of time to be the same

First we must convert \(\text{1}\) hour, \(\text{52}\) minutes and \(\text{30}\) seconds to hours: \[1 + \frac{52}{60} + \frac{30}{(60)^2} = \text{1,875}\text{ hours}\]

Write an equation describing the two taps working together to fill the container: \begin{align*} \frac{1}{x} + \frac{1}{x-2} &= \frac{1}{\text{1,875}} \end{align*}

## Multiply the equation through by the lowest common denominator and simplify

\begin{align*} \text{1,875}(x-2) + \text{1,875}x &= x(x-2) \\ \text{1,875}x - \text{3,75} + \text{1,875}x &= x^2 - 2x \\ 0 &= x^2 - \text{5,75}x + \text{3,75} \end{align*}

Multiply the equation through by \(\text{4}\) to make it easier to factorise (or use the quadratic formula) \begin{align*} 0 &= 4x^2 - 23x + 15 \\ 0 &= (4x-3)(x-5) \end{align*} Therefore \(x = \frac{3}{4}\) or \(x = 5\).

We have calculated that the less powerful tap takes \(\frac{3}{4}\) hours or \(\text{5}\) hours to fill the container, but we know that when both taps are opened it takes \(\text{1,875}\) hours. We can therefore discard the first solution \(x = \dfrac{3}{4}\) hours.

So the less powerful tap fills the container in \(\text{5}\) hours and the more powerful tap takes \(\text{3}\) hours.

## Check that the solution satisfies the original equation

The less powerful tap fills the container in \(\text{5}\) hours and the more powerful tap takes \(\text{3}\) hours.

Mr. Tsilatsila builds a fence around his rectangular vegetable garden of \(\text{8}\) \(\text{m$^{2}$}\). If the length is twice the breadth, determine the dimensions of Mr. Tsilatsila's vegetable garden.

We let the length be \(l\) and the breadth be \(b\). The area of the garden is \(\text{8}\) \(\text{m$^{2}$}\) and is given by \(A = l \times b\).

Since the length is twice the breadth we can express the length in terms of the breadth: \(l = 2b\). We now have the following:

Therefore the breadth is \(\text{2}\) \(\text{m}\) and the length is twice this, \(\text{4}\) \(\text{m}\). Note that the breadth cannot be a negative number and so we do not consider this solution.

Kevin has played a few games of ten-pin bowling. In the third game, Kevin scored \(\text{80}\) more than in the second game. In the first game Kevin scored \(\text{110}\) less than the third game. His total score for the first two games was \(\text{208}\). If he wants an average score of \(\text{146}\), what must he score on the fourth game?

We let the score for the first game be \(a\), the score for the second game be \(b\), the score for the third game be \(c\) and the score for the fourth game be \(d\).

Now we note the following:

We make the \(c\) the subject of the first two equations:

And then we use \(a = 208 - b\) to solve for \(b\):

Now we can find \(a\):

And we can find \(c\):

Finally we can find \(d\):

Kevin must score \(\text{187}\) in the fourth game.

When an object is dropped or thrown downward, the distance, \(d\), that it falls in time, \(t\), is described by the following equation:

In this equation, \({v}_{0}\) is the initial velocity, in \(\text{m·s$^{-1}$}\). Distance is measured in meters and time is measured in seconds. Use the equation to find how long it takes a tennis ball to reach the ground if it is thrown downward from a hot-air balloon that is \(\text{500}\) \(\text{m}\) high. The tennis ball is thrown at an initial velocity of \(\text{5}\) \(\text{m·s$^{-1}$}\).

We are given the distance that the ball falls and the initial velocity so we can solve for \(t\):

Since time cannot be negative the only solution is \(t = \dfrac{-1 + \sqrt{401}}{2} \approx \text{9,5}\text{ s}\).

The table below lists the times that Sheila takes to walk the given distances.

Plot the points.

Find the equation that describes the relationship between time and distance. Then use the equation to answer the following questions:

How long will it take Sheila to walk \(\text{21}\) \(\text{km}\)?

How far will Sheila walk in \(\text{7}\) minutes?

If Sheila were to walk half as fast as she is currently walking, what would the graph of her distances and times look like?

The equation is \(t = 5d\).

It will take Sheila \(t = 5(21) = 105\) minutes to walk \(\text{21}\) \(\text{km}\).

Sheila will walk \(d = \frac{7}{5} = \text{1,4}\) kilometres in \(\text{7}\) minutes.

The gradient of the graph will be twice the gradient of the first graph. The graph will be steeper and lie closer to the \(y\)-axis.

The power \(P\) (in watts) supplied to a circuit by a \(\text{12}\) volt battery is given by the formula \(P = 12I - \text{0,5}I^2\) where \(I\) is the current in amperes.

Since both power and current must be greater than \(\text{0}\), find the limits of the current that can be drawn by the circuit.

Draw a graph of \(P = 12I - \text{0,5}I^2\) and use your answer to the first question to define the extent of the graph.

What is the maximum current that can be drawn?

From your graph, read off how much power is supplied to the circuit when the current is \(\text{10}\) A. Use the equation to confirm your answer.

At what value of current will the power supplied be a maximum?

We set the equation equal to \(\text{0}\) to find the limits:

The maximum current that can be drawn is \(\text{24}\) \(\text{A}\).

The power is \(\text{70}\) \(\text{W}\).

\(\text{12}\) \(\text{A}\). This is the turning point of the parabola.

A wooden block is made as shown in the diagram. The ends are right-angled triangles having sides \(3x\), \(4x\) and \(5x\). The length of the block is \(y\). The total surface area of the block is \(\text{3 600}\) \(\text{cm$^{2}$}\).

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## Chapter 1: Sets

- Representation of a Set
- Types Of Sets
- Universal Sets
- Venn Diagram
- Operations on Sets
- Union of Sets

## Chapter 2: Relations & Functions

- Cartesian Product of Sets
- Relation and Function
- Introduction to Domain and Range
- Piecewise Function
- Range of a Function

## Chapter 3: Trigonometric Functions

- Measuring Angles
- Trigonometric Functions
- Trigonometric Functions of Sum and Difference of Two Angles

## Chapter 4: Principle of Mathematical Induction

- Principle of Mathematical Induction

## Chapter 5: Complex Numbers and Quadratic Equations

- Complex Numbers
- Algebra of Real Functions
- Algebraic Operations on Complex Numbers | Class 11 Maths
- Polar Representation of Complex Numbers
- Absolute Value of a Complex Number
- Conjugate of Complex Numbers
- Imaginary Numbers

## Chapter 6: Linear Inequalities

- Compound Inequalities
- Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation - Linear Inequalities | Class 11 Maths
- Graphical Solution of Linear Inequalities in Two Variables

## Solving Linear Inequalities Word Problems

Chapter 7: permutations and combinations.

- Fundamental Principle of Counting
- Permutation
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## Chapter 8: Binomial Theorem

- Binomial Theorem
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## Chapter 9: Sequences and Series

- Sequences and Series
- General and Middle Terms - Binomial Theorem - Class 11 Maths
- Arithmetic Series
- Arithmetic Sequence
- Geometric Progression
- Geometric Series
- Arithmetic Progression and Geometric Progression
- Special Series - Sequences and Series | Class 11 Maths

## Chapter 10: Straight Lines

- Slope of a Line
- Introduction to Two-Variable Linear Equations in Straight Lines
- Forms of Two-Variable Linear Equations - Straight Lines | Class 11 Maths
- Point-slope Form - Straight Lines | Class 11 Maths
- Slope Intercept Form
- Writing Slope-Intercept Equations
- Standard Form of a Straight Line
- X and Y Intercept
- Graphing slope-intercept equations - Straight Lines | Class 11 Maths

## Chapter 11: Conic Sections

- Conic Sections
- Equation of a Circle
- Focus and Directrix of a Parabola
- Identifying Conic Sections from their Equation

## Chapter 12: Introduction to Three-dimensional Geometry

- Coordinate Axes and Coordinate planes in 3D space
- 3D Distance Formula

## Chapter 13: Limits and Derivatives

- Formal Definition of Limits
- Strategy in Finding Limits
- Determining Limits using Algebraic Manipulation
- Limits of Trigonometric Functions
- Properties of Limits
- Limits by Direct Substitution
- Estimating Limits from Graphs
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- Sandwich Theorem
- Derivatives
- Average and Instantaneous Rate of Change
- Algebra of Derivative of Functions
- Product Rule in Derivatives
- Quotient Rule
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- Application of Derivatives
- Applications of Power Rule

## Chapter 14: Mathematical Reasoning

- Statements - Mathematical Reasoning
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## Chapter 15: Statistics

- Measures of spread - Range, Variance, and Standard Deviation
- Mean Absolute Deviation
- Measures of Central Tendency in Statistics
- Difference Between Mean, Median, and Mode with Examples
- Frequency Distribution
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## Chapter 16: Probability

- Random Experiment - Probability
- Types of Events in Probability
- Events in Probability
- Axiomatic Approach to Probability
- Measure of Dispersion
- CBSE Class 11 Maths Formulas
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We are well versed with equations in multiple variables. Linear Equations represent a point in a single dimension, a line in a two-dimensional, and a plane in a three-dimensional world. Solutions to linear inequalities represent a region of the Cartesian plane. It becomes essential for us to know how to translate real-life problems into linear inequalities.

## Linear Inequalities

Before defining the linear inequalities formally, let’s see them through a real-life situation and observe why their need arises in the first place. Let’s say Albert went to buy some novels for himself at the book fair. He has a total of Rs 200 with him. The book fair has a special sale policy which offers any book at Rs 70. Now he knows that he may not be able to spend the full amount on the books. Let’s say x is the number of books he bought. This situation can be represented mathematically by the following equation,

70x < 200

Since he can’t spend all the amount on books, and also the amount spent by him will always be less than Rs 200. The present situation can only be represented by the equation given above. Now let’s study the linear inequalities with a formal description,

Two real numbers or two algebraic expressions which are related by symbols such as ‘>’, ‘<‘, ‘≥’ and’≤’ form the inequalities. Linear inequalities are formed by linear equations which are connected with these symbols. These inequalities can be further classified into two parts: Strict Inequalities: The inequalities with the symbols such as ‘>’ or ‘<‘. Slack Inequalities: The inequalities with the symbols such as ‘≥’ or ‘≤’.

Rules of Solving Linear Inequalities:

There are certain rules which we should keep in mind while solving linear inequalities.

- Equal numbers can be added or subtracted from both sides of the inequality without affecting its sign.
- Both sides of Inequality can be divided or multiplied by any positive number but when they are multiplied or divided by a negative number, the sign of the linear inequality is reversed.

Now with this brief introduction to linear inequalities, let’s see some word problems on this concept.

## Sample Problems

Question 1: Considering the problem given in the beginning. Albert went to buy some novels for himself at the book fair. He has a total of Rs 200 with him. The book fair has a special sale policy which offers any book at Rs 70. Now he knows that he may not be able to spend the full amount on the books. Let’s say x is the number of books he bought. Represent this situation mathematically and graphically.

Solution:

We know that Albert cannot buy books for all the money he has. So, let’s say the number of books he buys is “x”. Then, 70x < 200 ⇒ x < To plot the graph of this inequality, put x = 0. 0 < Thus, x = 0 satisfies the inequality. So, the graph for the following inequality will look like,

Question 2: Consider the performance of the strikers of the football club Real Madrid in the last 3 matches. Ronaldo and Benzema together scored less than 9 goals in the last three matches. It is also known that Ronaldo scored three more goals than Benzema. What can be the possible number of goals Ronaldo scored?

Let’s say the number of goals scored by Benzema and Ronaldo are y and x respectively. x = y + 3 …..(1) x + y < 9 …..(2) Substituting the value of x from equation (1) in equation (2). y + 3 + y < 9 ⇒2y < 6 ⇒y < 3 Possible values of y: 0,1,2 Possible values of x: 3,4,5

Question 3: A classroom can fit at least 9 tables with an area of a one-meter square. We know that the perimeter of the classroom is 12m. Find the bounds on the length and breadth of the classroom.

It can fit 9 tables, that means the area of the classroom is atleast 9m 2 . Let’s say the length of the classroom is x and breadth is y meters. 2(x + y) = 12 {Perimeter of the classroom} ⇒ x + y = 6 Area of the rectangle is given by, xy > 9 ⇒x(6 – x) > 9 ⇒6x – x 2 > 9 ⇒ 0 > x 2 – 6x + 9 ⇒ 0 > (x – 3) 2 ⇒ 0 > x – 3 ⇒ x < 3 Thus, length of the classroom must be less than 3 m. So, then the breadth of the classroom will be greater than 3 m.

Question 4: Formulate the linear inequality for the following situation and plot its graph.

Let’s say Aman and Akhil went to a stationery shop. Aman bought 3 notebooks and Akhil bought 4 books. Let’s say cost of each notebook was “x” and each book was “y”. The total expenditure was less than Rs 500.

Cost of each notebook was “x” and for each book, it was “y”. Then the inequality can be described as, 3x + 4y < 500 Putting (x,y) → (0,0) 3(0) + 4(0) < 500 Origin satisfies the inequality. Thus, the graph of its solutions will look like, x.

Question 5: Formulate the linear inequality for the following situation and plot its graph.

A music store sells its guitars at five times their cost price. Find the shopkeeper’s minimum cost price if his profit is more than Rs 3000.

Let’s say the selling price of the guitar is y, and the cost price is x. y – x > 3000 ….(1) It is also given that, y = 5x ….(2) Substituting the value of y from equation (2) to equation (1). 5x – x > 3000 ⇒ 4x > 3000 ⇒ x > ⇒ x> 750 Thus, the cost price must be greater than Rs 750.

Question 6: The length of the rectangle is 4 times its breadth. The perimeter of the rectangle is less than 20. Formulate a linear inequality in two variables for the given situation, plot its graph and calculate the bounds for both length and breadth.

Let’s say the length is “x” and breadth is “y”. Perimeter = 2(x + y) < 20 ….(1) ⇒ x + y < 10 Given : x = 4y Substituting the value of x in equation (1). x + y < 10 ⇒ 5y < 10 ⇒ y < 2 So, x < 8 and y < 2.

Question 7: Rahul and Rinkesh play in the same football team. In the previous game, Rahul scored 2 more goals than Rinkesh, but together they scored less than 8 goals. Solve the linear inequality and plot this on a graph.

The equations obtained from the given information in the question, Suppose Rahul scored x Number of goals and Rinkesh scored y number of goals, Equations obtained will be, x = y+2 ⇢ (1) x+y< 8 ⇢ (2) Solving both the equations, y+2 + y < 8 2y < 6 y< 3 Putting this value in equation (2), x< 5

Question 8: In a class of 100 students, there are more girls than boys, Can it be concluded that how many girls would be there?

Let’s suppose that B is denoted for boys and G is denoted for girls. Now, since Girls present in the class are more than boys, it can be written in equation form as, G > B The total number of students present in class is 100 (given), It can be written as, G+ B= 100 B = 100- G Substitute G> B in the equation formed, G> 100 – G 2G > 100 G> 50 Hence, it is fixed that the number of girls has to be more than 50 in class, it can be 60, 65, etc. Basically any number greater than 50 and less than 100.

Question 9: In the previous question, is it possible for the number of girls to be exactly 50 or exactly 100? If No, then why?

No, It is not possible for the Number of girls to be exactly 50 since while solving, it was obtained that, G> 50 In any case if G= 50 is a possibility, from equation G+ B= 100, B = 50 will be obtained. This simply means that the number of boys is equal to the number of girls which contradicts to what is given in the question. No, it is not possible for G to be exactly 100 as well, as this proves that there are 0 boys in the class.

Question 10: Solve the linear inequality and plot the graph for the same,

7x+ 8y < 30

The linear inequality is given as, 7x+ 8y< 30 At x= 0, y= 30/8= 3.75 At y= 0, x= 30/7= 4.28 These values are the intercepts. The graph for the above shall look like, Putting x= y/2, that is, y= 2x in the linear inequality, 7x + 16x < 30 x = 1.304 y = 2.609

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