Last modified on April 17th, 2024

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 â†’ â€˜greater than or equal toâ€™
- More than â†’ â€˜greater thanâ€™
- No more than or at most â†’ â€˜less than or equal toâ€™
- Less than â†’ â€˜less thanâ€™

**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.**

Solution:

Let â€˜xâ€™ be the total number of groups Mr. Adams can make.

Since each group has 9 students, the total number of students in â€˜xâ€™ 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 isÂ

x < 6

Thus, Mr. Adams can make a maximum of 6 full groups.

**Â 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.**

Solution:

Let â€˜xâ€™ be the number of months Bruce needs to work.

As we know,Â

The amount already saved by Bruce is \$121

He earns \$44 per month

The cost of the tablet is at least \$561

After â€˜xâ€™ months of work, Bruce will have \$(121 + 44x)

Now, the inequality representing the situation is: 121 + 44x â‰¥ 561

On subtracting 121 from both sides,

121 + 44x – 121 â‰¥ 561 – 121

â‡’ 44x â‰¥ 440

On dividing both sides by 44,

x â‰¥ 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?**

Solution:

Let â€˜xâ€™ be the original price of the clothes Ava chooses.

As we know, the original price range is 199 â‰¤ x â‰¤ 299, and the discount is \$26

Now, Ava pays \$(x – 26) after the discount.

Thus, the inequality is: 199 – 26 â‰¤ x – 26 â‰¤ 299 – 26

â‡’ 173 â‰¤ x – 26 â‰¤ 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?**

Solution:

Let â€˜Pâ€™ be the profit, â€˜pâ€™ be the profit per plant, and â€˜nâ€™ be the number of plants.Â

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 â‰¥ 4225 and p = 6.25 â€¦..(i)

Also, P = p Ã— n

Substituting the values of (i), we get

6.25 Ã— n â‰¥ 4225

On dividing both sides by 6.25, we get

${n\geq \dfrac{4225}{6\cdot 25}}$

â‡’ ${n\geq 676}$

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?**

Solution:

As we know, Daniel had \$1200 in his savings account at the start of the year, but he withdrew \$60 for transportation each month.Â

Thus, after â€˜nâ€™ months, he will have \$(1200âˆ’60n) left in his account.

Also, Daniel wants to have at least \$300 in the account at the end of the year.Â

Here, the inequality is: 1200 – 60n â‰¥ 300

â‡’ 1200 – 60n – 1200 â‰¥ 300 – 1200 (by subtraction property)

â‡’ -60n â‰¥ -900

â‡’ 60n â‰¤ 900 (by inversion property)

â‡’ n â‰¤ ${\dfrac{900}{60}}$

â‡’ n â‰¤ 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?**

Solution:

Let â€˜tâ€™ be the number of weeks to weigh less than 155 lb.

As we know, Anne initially weighs 165 lb

After â€˜tâ€™ weeks of cutting 2 lb per week, her weight will be 165 – 2t

Now, Anneâ€™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

â‡’ – 2t < -10

On dividing by -2, the inequality sign is reversed.

${\dfrac{-2t}{-2} >\dfrac{-10}{-2}}$

â‡’ 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?**

Solution:

Let Roryâ€™s score be â€˜r,â€™ and Cinderâ€™s score be â€˜c.â€™

As we know, Rory scored 5 points more than Cinder.

Thus, Roryâ€™s score is r = c + 5 â€¦..(i)

Also, their scores sum up to less than 19 points.

Thus, the inequality is: r + c < 19 â€¦..(ii)

Substituting (i) in (ii), we get

(c + 5) + c < 19

â‡’ 2c + 5 < 19

On subtracting 5 from both sides, we get

2c + 5 – 5 < 19 – 5

â‡’ 2c < 14

On dividing both sides by 2, we get

${\dfrac{2c}{2} >\dfrac{14}{2}}$

â‡’ c < 7 means Cinderâ€™s score is less than 7 points.

Now, from (i), r = c + 5

â‡’ c = r – 5

Thus, c < 7

â‡’ r – 5 < 7

On adding 5 to both sides, we get

r – 5 + 5 < 7 + 5

â‡’ r < 12 means Roryâ€™s score is less than 12 points.

Hence, Roryâ€™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.**

Solution:

Let â€˜câ€™ 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| â‰¤ 4

â‡’ -4 â‰¤ c – 74 â‰¤ 4

On adding 74 to each side, we get

-4 + 74 â‰¤ c – 74 + 74 â‰¤ 4 + 74

â‡’ 70 â‰¤ c â‰¤ 78

Hence, the minimum number of cans in a carton is 70, and the maximum number is 78.

**Â 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?**

Solution:

Let â€˜pâ€™ be Laylaâ€™s total hours of practice in a week, and â€˜sâ€™ 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:

${\dfrac{3}{4}s >p-3}$ â€¦..(i)

As we know, Layla rehearses for at least 12 hours, which means p â‰¥ 12 â€¦..(ii)

From (i), ${\dfrac{3}{4}s >p-3}$

â‡’ ${s >\dfrac{4}{3}\left( p-3\right)}$

From (ii), substituting the value p = 12 in (i), we get

${s >\dfrac{4}{3}\left( 12-3\right)}$

â‡’ ${s >\dfrac{4}{3}\cdot 9}$

â‡’ s > 12

Thus, Layla must complete more than 12 sessions to exceed her weekly rehearsal goal.

Last modified on April 17th, 2024