inequalities problem solving questions

Solving Inequalities

Sometimes we need to solve Inequalities like these:

Our aim is to have x (or whatever the variable is) on its own on the left of the inequality sign:

We call that "solved".

Example: x + 2 > 12

Subtract 2 from both sides:

x + 2 − 2 > 12 − 2

x > 10

How to Solve

Solving inequalities is very like solving equations , we do most of the same things ...

... but we must also pay attention to the direction of the inequality .

Some things can change the direction !

< becomes >

> becomes <

≤ becomes ≥

≥ becomes ≤

Safe Things To Do

These things do not affect the direction of the inequality:

• Add (or subtract) a number from both sides
• Multiply (or divide) both sides by a positive number
• Simplify a side

Example: 3x < 7+3

We can simplify 7+3 without affecting the inequality:

But these things do change the direction of the inequality ("<" becomes ">" for example):

• Multiply (or divide) both sides by a negative number
• Swapping left and right hand sides

Example: 2y+7 < 12

When we swap the left and right hand sides, we must also change the direction of the inequality :

12 > 2y+7

Here are the details:

Adding or Subtracting a Value

We can often solve inequalities by adding (or subtracting) a number from both sides (just as in Introduction to Algebra ), like this:

Example: x + 3 < 7

If we subtract 3 from both sides, we get:

x + 3 − 3 < 7 − 3    

And that is our solution: x < 4

In other words, x can be any value less than 4.

What did we do?

And that works well for adding and subtracting , because if we add (or subtract) the same amount from both sides, it does not affect the inequality

Example: Alex has more coins than Billy. If both Alex and Billy get three more coins each, Alex will still have more coins than Billy.

What If I Solve It, But "x" Is On The Right?

No matter, just swap sides, but reverse the sign so it still "points at" the correct value!

Example: 12 < x + 5

If we subtract 5 from both sides, we get:

12 − 5 < x + 5 − 5    

That is a solution!

But it is normal to put "x" on the left hand side ...

... so let us flip sides (and the inequality sign!):

Do you see how the inequality sign still "points at" the smaller value (7) ?

And that is our solution: x > 7

Note: "x" can be on the right, but people usually like to see it on the left hand side.

Multiplying or Dividing by a Value

Another thing we do is multiply or divide both sides by a value (just as in Algebra - Multiplying ).

But we need to be a bit more careful (as you will see).

Positive Values

Everything is fine if we want to multiply or divide by a positive number :

Example: 3y < 15

If we divide both sides by 3 we get:

3y /3 < 15 /3

And that is our solution: y < 5

Negative Values

Well, just look at the number line!

For example, from 3 to 7 is an increase , but from −3 to −7 is a decrease.

See how the inequality sign reverses (from < to >) ?

Let us try an example:

Example: −2y < −8

Let us divide both sides by −2 ... and reverse the inequality !

−2y < −8

−2y /−2 > −8 /−2

And that is the correct solution: y > 4

(Note that I reversed the inequality on the same line I divided by the negative number.)

So, just remember:

When multiplying or dividing by a negative number, reverse the inequality

Multiplying or Dividing by Variables

Here is another (tricky!) example:

Example: bx < 3b

It seems easy just to divide both sides by b , which gives us:

... but wait ... if b is negative we need to reverse the inequality like this:

But we don't know if b is positive or negative, so we can't answer this one !

To help you understand, imagine replacing b with 1 or −1 in the example of bx < 3b :

• if b is 1 , then the answer is x < 3
• but if b is −1 , then we are solving −x < −3 , and the answer is x > 3

The answer could be x < 3 or x > 3 and we can't choose because we don't know b .

Do not try dividing by a variable to solve an inequality (unless you know the variable is always positive, or always negative).

A Bigger Example

Example: x−3 2 < −5.

First, let us clear out the "/2" by multiplying both sides by 2.

Because we are multiplying by a positive number, the inequalities will not change.

x−3 2 ×2 < −5  ×2  

x−3 < −10

Now add 3 to both sides:

x−3 + 3 < −10 + 3    

And that is our solution: x < −7

Two Inequalities At Once!

How do we solve something with two inequalities at once?

Example: −2 < 6−2x 3 < 4

First, let us clear out the "/3" by multiplying each part by 3.

Because we are multiplying by a positive number, the inequalities don't change:

−6 < 6−2x < 12

−12 < −2x < 6

Now divide each part by 2 (a positive number, so again the inequalities don't change):

−6 < −x < 3

Now multiply each part by −1. Because we are multiplying by a negative number, the inequalities change direction .

6 > x > −3

And that is the solution!

But to be neat it is better to have the smaller number on the left, larger on the right. So let us swap them over (and make sure the inequalities point correctly):

−3 < x < 6

• Many simple inequalities can be solved by adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own.
• Multiplying or dividing both sides by a negative number
• Don't multiply or divide by a variable (unless you know it is always positive or always negative)

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Examples of Solving Harder Linear Inequalities

Intro & Formatting Worked Examples Harder Examples & Word Prob's

Once you'd learned how to solve one-variable linear equations, you were then given word problems. To solve these problems, you'd have to figure out a linear equation that modelled the situation, and then you'd have to solve that equation to find the answer to the word problem.

Content Continues Below

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Solving Inequalities

So, now that you know how to solve linear inequalities — you guessed it! — they give you word problems.

• The velocity of an object fired directly upward is given by V = 80 − 32 t , where the time t is measured in seconds. When will the velocity be between 32 and 64 feet per second (inclusive)?

This question is asking when the velocity, V , will be between two given values. So I'll take the expression for the velocity,, and put it between the two values they've given me. They've specified that the interval of velocities is inclusive, which means that the interval endpoints are included. Mathematically, this means that the inequality for this model will be an "or equal to" inequality. Because the solution is a bracket (that is, the solution is within an interval), I'll need to set up a three-part (that is, a compound) inequality.

I will set up the compound inequality, and then solve for the range of times t :

32 ≤ 80 − 32 t ≤ 64

32 − 80 ≤ 80 − 80 − 32 t ≤ 64 − 80

−48 ≤ −32 t ≤ −16

−48 / −32 ≥ −32 t / −32 ≥ −16 / −32

1.5 ≥ t ≥ 0.5

Note that, since I had to divide through by a negative, I had to flip the inequality signs.

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Note also that you might (as I do) find the above answer to be more easily understood if it's written the other way around, with "less than" inequalities.

And, because this is a (sort of) real world problem, my working should show the fractions, but my answer should probably be converted to decimal form, because it's more natural to say "one and a half seconds" than it is to say "three-halves seconds". So I convert the last line above to the following:

0.5 ≤ t ≤ 1.5

Looking back at the original question, it did not ask for the value of the variable " t ", but asked for the times when the velocity was between certain values. So the actual answer is:

The velocity will be between 32 and 64 feet per second between 0.5 seconds after launch and 1.5 seconds after launch.

Okay; my answer above was *extremely* verbose and "complete"; you don't likely need to be so extreme. You can probably safely get away with saying something simpler like, "between 0.5 seconds and 1.5 seconds". Just make sure that you do indeed include the approprioate units (in this case, "seconds").

Always remember when doing word problems, that, once you've found the value for the variable, you need to go back and re-read the problem to make sure that you're answering the actual question. The inequality 0.5 ≤  t  ≤ 1.5 did not answer the actual question regarding time. I had to interpret the inequality and express the values in terms of the original question.

• Solve 5 x + 7 < 3( x  + 1) .

First I'll multiply through on the right-hand side, and then solve as usual:

5 x + 7 < 3( x + 1)

5 x + 7 < 3 x + 3

2 x + 7 < 3

2 x < −4

x < −2

In solving this inequality, I divided through by a positive 2 to get the final answer; as a result (that is, because I did *not* divide through by a minus), I didn't have to flip the inequality sign.

• You want to invest $30,000 . Part of this will be invested in a stable 5% -simple-interest account. The remainder will be "invested" in your father's business, and he says that he'll pay you back with 7% interest. Your father knows that you're making these investments in order to pay your child's college tuition with the interest income. What is the least you can "invest" with your father, and still (assuming he really pays you back) get at least $1900 in interest?

First, I have to set up equations for this. The interest formula for simple interest is I = Prt , where I is the interest, P is the beginning principal, r is the interest rate expressed as a decimal, and t is the time in years.

Since no time-frame is specified for this problem, I'll assume that t  = 1 ; that is, I'll assume (hope) that he's promising to pay me at the end of one year. I'll let x be the amount that I'm going to "invest" with my father. Then the rest of my money, being however much is left after whatever I give to him, will be represented by "the total, less what I've already given him", so 30000 −  x will be left to invest in the safe account.

Then the interest on the business investment, assuming that I get paid back, will be:

I = ( x )(0.07)(1) = 0.07 x

The interest on the safe investment will be:

(30 000 − x )(0.05)(1) = 1500 − 0.05 x

The total interest is the sum of what is earned on each of the two separate investments, so my expression for the total interest is:

0.07 x + (1500 − 0.05 x ) = 0.02 x + 1500

I need to get at least $1900 ; that is, the sum of the two investments' interest must be greater than, or at least equal to, $1,900 . This allows me to create my inequality:

0.02 x + 1500 ≥ 1900

0.02 x ≥ 400

x ≥ 20 000

That is, I will need to "invest" at least $20,000 with my father in order to get $1,900 in interest income. Since I want to give him as little money as possible, I will give him the minimum amount:

I will invest $20,000 at 7% .

• An alloy needs to contain between 46% copper and 50% copper. Find the least and greatest amounts of a 60% copper alloy that should be mixed with a 40% copper alloy in order to end up with thirty pounds of an alloy containing an allowable percentage of copper.

This is similar to a mixture word problem , except that this will involve inequality symbols rather than "equals" signs. I'll set it up the same way, though, starting with picking a variable for the unknown that I'm seeking. I will use x to stand for the pounds of 60% copper alloy that I need to use. Then 30 −  x will be the number of pounds, out of total of thirty pounds needed, that will come from the 40% alloy.

Of course, I'll remember to convert the percentages to decimal form for doing the algebra.

How did I get those values in the bottom right-hand box? I multiplied the total number of pounds in the mixture ( 30 ) by the minimum and maximum percentages ( 46% and 50% , respectively). That is, I multiplied across the bottom row, just as I did in the " 60% alloy" row and the " 40% alloy" row, to get the right-hand column's value.

The total amount of copper in the mixture will be the sum of the copper contributed by each of the two alloys that are being put into the mixture. So I'll add the expressions for the amount of copper from each of the alloys, and place the expression for the total amount of copper in the mixture as being between the minimum and the maximum allowable amounts of copper:

13.8 ≤ 0.6 x + (12 − 0.4 x ) ≤ 15

13.8 ≤ 0.2 x + 12 ≤ 15

1.8 ≤ 0.2 x ≤ 3

9 ≤ x ≤ 15

Checking back to my set-up, I see that I chose my variable to stand for the number of pounds that I need to use of the 60% copper alloy. And they'd only asked me for this amount, so I can ignore the other alloy in my answer.

I will need to use between 9 and 15 pounds of the 60% alloy.

Per yoozh, I'm verbose in my answer. You can answer simply as " between 9 and 15 pounds ".

• Solve 3( x − 2) + 4 ≥ 2(2 x − 3)

First I'll multiply through and simplify; then I'll solve:

3( x − 2) + 4 ≥ 2(2 x − 3)

3 x − 6 + 4 ≥ 4 x − 6

3 x − 2 ≥ 4 x − 6

−2 ≥ x − 6            (*)

Why did I move the 3 x over to the right-hand side (to get to the line marked with a star), instead of moving the 4 x to the left-hand side? Because by moving the smaller term, I was able to avoid having a negative coefficient on the variable, and therefore I was able to avoid having to remember to flip the inequality when I divided through by that negative coefficient. I find it simpler to work this way; I make fewer errors. But it's just a matter of taste; you do what works for you.

Why did I switch the inequality in the last line and put the variable on the left? Because I'm more comfortable with inequalities when the answers are formatted this way. Again, it's only a matter of taste. The form of the answer in the previous line, 4 ≥ x , is perfectly acceptable.

As long as you remember to flip the inequality sign when you multiply or divide through by a negative, you shouldn't have any trouble with solving linear inequalities.

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Linear Inequalities Questions

Linear inequalities questions and solutions help the students to solve various problems on inequalities in mathematics. In this article, you will get solved questions on linear inequalities, and additional questions for practice. You will find detailed explanations to the solutions of linear inequalities in one variable and two variables.

What are Linear inequalities?

In maths, linear inequalities are the equations in which both sides of the equation are separated by the inequality symbol, such as <, >, ≤ and ≥. That means LHS and RHS are not equal. For example, 2x – 4 < 12, x – y > 7, 3x – 4 ≤ 0, and so on.

Also, check: Linear inequalities

Linear Inequalities Questions and Answers

1. Solve for x:

3(x – 1) ≤ 2 (x – 3)

The above inequality can be written as,

3x – 3 ≤ 2x – 6

Adding 3 to both the sides, we get;

3x – 3+ 3 ≤ 2x – 6+ 3

3x ≤ 2x – 3

Subtracting 2x from both the sides,

3x – 2x ≤ 2x – 3 – 2x

Therefore, the solutions to the given inequality are defined by all the real numbers less than or equal to -3.

Hence, the required solution set for x is (-∞, -3].

2. Solve the inequality: (x – 2)/(x + 5) > 2

(x – 2)/(x + 5) > 2

Subtracting 2 from both sides, we get;

(x – 2)/(x + 5) – 2 > 0

[(x – 2) – 2(x + 5)]/ (x + 5) > 0

(x – 2 – 2x – 10)/(x + 5) > 0

-(x + 12)/(x + 5) > 0

Multiplying -1 on both sides, we get;

(x + 12)/(x + 5) < 0

⇒ x + 12 < 0 and x + 5 > 0 (or) x + 12 > 0 and x + 5 < 0

⇒ x < -12 and x > -5 (or) x > -12 and x < -5

⇒ -12 < x < -5

Therefore, x ∈ (-12, -5).

3. Solve for x from the following:

1/(|x| – 3) ≤ ½

Subtracting ½ from both sides, we get;

[1/ (|x| – 3)] – (½) ≤ 0

(2 – |x| + 3)/ 2(|x| – 3) ≤ 0

(5 – |x|)/(|x| – 3) ≤ 0

⇒ 5 – |x| ≤ 0 and |x| – 3 > 0 or 5 – |x| ≥ 0 and |x| – 3 < 0

⇒ |x| ≥ 5 and |x| > 3 or |x| ≤ 5 and |x| < 3

⇒ |x| ≥ 5 or |x| < 3

⇒ x ∈ (- ∞ , – 5] or [5, ∞) or x ∈ ( -3 , 3)

⇒ x ∈ (- ∞ , – 5] ∪ ( -3 , 3) ∪ [5, ∞)

4. Solve for x: |x + 1| + |x| > 3

|x + 1| + |x| > 3

In LHS, we have two terms with modulus.

So, let’s equate the given expression within the modulus to 0.

Then, we get critical points x = -1, 0.

Thus, we can write the real line intervals for these critical points as: (-∞, -1), [-1, 0), [0, ∞)

When – ∞ < x < – 1

⇒ – x – 1 – x > 3

⇒ x < – 2.

When – 1 ≤ x < 0,

⇒ x + 1 – x > 3

⇒ 1 > 3 (not possible)

When 0 ≤ x < ∞,

⇒ x + 1 + x > 3

⇒ x > 1.

From the three cases written above, we get x ∈ (– ∞ , – 2) ∪ (1, ∞).

5. The longest side of a triangle is 3 times the shortest side, and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Let x cm be the length of the shortest side of the triangle.

∴ According to the question, length of the longest side = 3x cm

Length of the third side = (3x – 2) cm

The least perimeter of the triangle = 61 cm (given)

Thus, x + 3x + (3x – 2) cm ≥ 61 cm

= 7x – 2 ≥ 61

Dividing by 7 on both sides, we get;

= 7x/7 ≥ 63/7

Hence, the minimum length of the shortest side will be 9 cm.

6. Solve (3x – 4)/2 ≥ (x + 1)/4 – 1. Show the graph of the solutions on the number line.

(3x – 4)/2 ≥ (x + 1)/4 – 1

(3x – 4)/2 ≥ [(x + 1) – 4]/4

(3x – 4)/2 ≥ (x – 3)/4

Multiplying by 4 on both sides,

(3x – 4) ≥ (x – 3)/2

2(3x – 4) ≥ (x – 3)

6x – 8 ≥ x – 3

Adding 8 on both sides,

Subtracting x from both sides, we get;

i.e., x ≥ 1

Therefore, the graphical representation of this solution (on the number line) is given as:

7. Solve the following system of linear inequalities graphically.

x + y ≥ 5….(i)

x – y ≤ 3….(ii)

From (i), x + y ≥ 5

Let x + y = 5

y = 5 – x

From (ii), x – y ≤ 3

Let x – y = 3

y = x – 3

Now, let’s draw the graph for the above two linear equations, i.e., x + y = 5 and x – y = 3.

Now, the solution of inequality (i) can be represented by shading the region towards the right of the line x + y = 5, including the points on the line.

On the same set of axes, we draw the graph of the equation x – y = 3. The inequality (ii) represents the shaded region towards the left side of the line x – y = 3, including the points on the line.

The double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

8. A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that the acid content in the resulting mixture will be more than 15% but less than 18%?

Let x litres of 30% acid solution be required to be added.

Total mixture = (x + 600) litres

Thus, 30% x + 12% of 600 > 15% of (x + 600)

30% x + 12% of 600 < 18% of (x + 600)

⇒ (30x/100) + (12/100) × (600) > (15/100) (x + 600)

(30x/100) + (12/100) × (600) < (18/100) (x + 600)

⇒ 30x + 7200 > 15x + 9000 and 30x + 7200 < 18x + 10800

⇒ 15x > 1800 and 12x < 3600

⇒ x > 120 and x < 300,

i.e., 120 < x < 300

Hence, the number of litres of the 30% acid solution will have to be more than 120 litres but less than 300 litres.

9. Solve the following system of inequalities graphically.

x + y ≥ 4, 2x – y < 0

x + y ≥ 4….(i)

2x – y < 0….(ii)

From (i), x + y ≥ 4

Let x + y = 4

y = 4 – x

When x = 0, y = 4

When x = 4, y = 0

From (ii), 2x – y < 0

Let 2x – y = 0

When x = 0, y = 0

When x = 1, y = 2

From the above computation we can draw the graph of linear equations x + y = 4 and 2x – y = 0.

Now, shade the region for x + y ≥ 4, on the right of the line of x + y = 4.

For (ii), the shaded region will be on the left side of the line 2x – y = 0.

These can be represented graphically as:

Hence, the combined shaded region will be the solution region for the given system of inequalities.

10. Solve the following inequalities and represent the solution set on the number line.

5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

5(2x – 7) – 3(2x + 3) ≤ 0….(i)

2x + 19 ≤ 6x + 47….(ii)

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ x ≤ 11 ……(iii)

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ x ≥ -7 ……….(iv)

From equations (iii) and (iv), the solution of the given inequalities is (-7, 11).

This can be shown on the number line as:

Practice Questions on Linear Inequalities

• The cost and revenue functions of a product are given by C(x) = 20 x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold. How many items must be sold to realise some profit?
• Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
• Solve the following inequalities and represent the solution set graphically on the number line.

5x + 1 > – 24, 5x – 1 < 24

• Show that the following system of linear inequalities has no solution: x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
• Solve for x: -5 ≤ (5 – 3x)/2 ≤ 8

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Last modified on April 17th, 2024

Inequality Word Problems

Inequalities are common in our everyday life. They help us express relationships between quantities that are unequal. Writing and solving word problems involving them helps develop our problem-solving approach, understanding, logical reasoning, and analytical skills.

Here are the 4 main keywords commonly used to write mathematical expressions involving inequalities.   

• At least →  ‘greater than or equal to’
• More than → ‘greater than’
• No more than or at most → ‘less than or equal to’
• Less than → ‘less than’

To participate in the annual sports day, Mr. Adams would like to have nine students in each group. But fewer than 54 students are in class today, so Mr. Adams is unable to make as many full groups as he wants. How many full groups can Mr. Adams make? Write the inequality that describes the situation.

Let ‘x’ be the total number of groups Mr. Adams can make. Since each group has 9 students, the total number of students in ‘x’ groups is 9x As we know, fewer than 54 students are in a class today. Thus, the inequality that represents the situation is: 9x < 54 On dividing both sides by 9, the maximum number of groups Mr. Adams can make is  x < 6 Thus, Mr. Adams can make a maximum of 6 full groups.

 Bruce needs at least \$561 to buy a new tablet. He has already saved \$121 and earns \$44 per month as a part-timer in a company. Write the inequality and determine how long he has to work to buy the tablet.

Let ‘x’ be the number of months Bruce needs to work. As we know,  The amount already saved by Bruce is \$121 He earns \$44 per month The cost of the tablet is at least \$561 After ‘x’ months of work, Bruce will have \$(121 + 44x) Now, the inequality representing the situation is: 121 + 44x ≥ 561 On subtracting 121 from both sides, 121 + 44x – 121 ≥ 561 – 121 ⇒ 44x ≥ 440 On dividing both sides by 44, x ≥ 10 Thus, Bruce needs to work for at least 10 months to buy the new tablet.

A store is offering a \$26 discount on all women’s clothes. Ava is looking at clothes originally priced between \$199 and \$299. How much can she expect to spend after the discount?

Let ‘x’ be the original price of the clothes Ava chooses. As we know, the original price range is 199 ≤ x ≤ 299, and the discount is \$26 Now, Ava pays \$(x – 26) after the discount. Thus, the inequality is: 199 – 26 ≤ x – 26 ≤ 299 – 26 ⇒ 173 ≤ x – 26 ≤ 273 Thus, she can expect to spend between \$173 and \$273 after the discount.

A florist makes a profit of \$6.25 per plant. If the store wants to profit at least \$4225, how many plants does it need to sell?

Let ‘P’ be the profit, ‘p’ be the profit per plant, and ‘n’ be the number of plants.  As we know, the store wants a profit of at least \$4225, and the florist makes a profit of \$6.25 per plant. Here, P ≥ 4225 and p = 6.25 …..(i) Also, P = p × n Substituting the values of (i), we get 6.25 × n ≥ 4225 On dividing both sides by 6.25, we get ${n\geq \dfrac{4225}{6\cdot 25}}$ ⇒ ${n\geq 676}$ Thus, the store needs to sell at least 676 plants to make a profit of \$4225.

Daniel had \$1200 in his savings account at the start of the year, but he withdraws \$60 each month to spend on transportation. He wants to have at least \$300 in the account at the end of the year. How many months can Daniel withdraw money from the account?

As we know, Daniel had \$1200 in his savings account at the start of the year, but he withdrew \$60 for transportation each month.  Thus, after ‘n’ months, he will have \$(1200−60n) left in his account. Also, Daniel wants to have at least \$300 in the account at the end of the year.  Here, the inequality is: 1200 – 60n ≥ 300 ⇒ 1200 – 60n – 1200 ≥ 300 – 1200 (by subtraction property) ⇒ -60n ≥ -900 ⇒ 60n ≤ 900 (by inversion property) ⇒ n ≤ ${\dfrac{900}{60}}$ ⇒ n ≤ 15 Thus, Daniel can withdraw money from the account for at most 15 months.

Anne is a model trying to lose weight for an upcoming beauty pageant. She currently weighs 165 lb. If she cuts 2 lb per week, how long will it take her to weigh less than 155 lb?

Let ‘t’ be the number of weeks to weigh less than 155 lb. As we know, Anne initially weighs 165 lb After ‘t’ weeks of cutting 2 lb per week, her weight will be 165 – 2t Now, Anne’s weight will be less than 155 lb Here, the inequality from the given word problem is: 165 – 2t < 155 On subtracting 165 from both sides, we get 165 – 2t – 165 < 155 – 165 ⇒ – 2t < -10 On dividing by -2, the inequality sign is reversed. ${\dfrac{-2t}{-2} >\dfrac{-10}{-2}}$ ⇒ t > 5

Rory and Cinder are on the same debate team. In one topic, Rory scored 5 points more than Cinder, but they scored less than 19 together. What are the possible points Rory scored?

Let Rory’s score be ‘r,’ and Cinder’s score be ‘c.’ As we know, Rory scored 5 points more than Cinder. Thus, Rory’s score is r = c + 5 …..(i) Also, their scores sum up to less than 19 points. Thus, the inequality is: r + c < 19 …..(ii) Substituting (i) in (ii), we get (c + 5) + c < 19 ⇒ 2c + 5 < 19 On subtracting 5 from both sides, we get 2c + 5 – 5 < 19 – 5 ⇒ 2c < 14 On dividing both sides by 2, we get ${\dfrac{2c}{2} >\dfrac{14}{2}}$ ⇒ c < 7 means Cinder’s score is less than 7 points. Now, from (i), r = c + 5 ⇒ c = r – 5 Thus, c < 7 ⇒ r – 5 < 7 On adding 5 to both sides, we get r – 5 + 5 < 7 + 5 ⇒ r < 12 means Rory’s score is less than 12 points. Hence, Rory’s scores can be 6, 7, 8, 9, 10, or 11 points.

An average carton of juice cans contains 74 pieces, but the number can vary by 4. Find out the maximum and minimum number of cans that can be present in a carton.

Let ‘c’ be the number of juice cans in a carton. As we know, the average number of cans in a carton is 74, and it varies by 4 cans. Thus, the required inequality is |c – 74| ≤ 4 ⇒ -4 ≤ c – 74 ≤ 4 On adding 74 to each side, we get -4 + 74 ≤ c – 74 + 74 ≤ 4 + 74 ⇒ 70 ≤ c ≤ 78 Hence, the minimum number of cans in a carton is 70, and the maximum number is 78.

 Layla rehearses singing for at least 12 hours per week, for three-fourths of an hour each session. If she has already sung 3 hours this week, how many more sessions remain for her to exceed her weekly practice goal?

Let ‘p’ be Layla’s total hours of practice in a week, and ‘s’ be the number of sessions she needs to complete. As we know, Layla has already rehearsed 3 hours, then her remaining rehearsal time is (p – 3) Each session lasts for three-fourths of an hour. Thus, we have the inequality: ${\dfrac{3}{4}s >p-3}$ …..(i) As we know, Layla rehearses for at least 12 hours, which means p ≥ 12 …..(ii) From (i), ${\dfrac{3}{4}s >p-3}$ ⇒ ${s >\dfrac{4}{3}\left( p-3\right)}$ From (ii), substituting the value p = 12 in (i), we get ${s >\dfrac{4}{3}\left( 12-3\right)}$ ⇒ ${s >\dfrac{4}{3}\cdot 9}$ ⇒ s > 12 Thus, Layla must complete more than 12 sessions to exceed her weekly rehearsal goal.

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