Linear Inequalities

Help Questions

PSAT Math › Linear Inequalities

Questions 1 - 10
1

A student can spend no more than $$\18$$ on snacks. Chips cost $$\2$$ and drinks cost $$\3$$. If the student buys 4 bags of chips and $d$ drinks, which inequality shows the possible values of $d$?

$8+3d\ge18$

$8+3d\le18$

$3d-8\le18$

$2(4+d)\le18$

Explanation

The question requires modeling spending no more than $18 on 4 bags of chips at $2 each and $d$ drinks at $3 each. The inequality is $4 \times 2 + 3d \le 18$, or $8 + 3d \le 18$. This uses $\le$ for 'no more than.' A common error is using the wrong coefficients. Another pitfall is adding extras. Define costs clearly in budgeting problems.

2

Solve $10-4x\le2x+1$. What is the solution?

$x\ge-\dfrac{3}{2}$

$x\ge\dfrac{3}{2}$

$x\le-\dfrac{3}{2}$

$x\le\dfrac{3}{2}$

Explanation

The question asks to solve 10 - 4x ≤ 2x + 1. Add 4x to both sides to 10 ≤ 6x + 1, subtract 1 to 9 ≤ 6x, divide by 6 to 9/6 ≤ x, or x ≥ 3/2. Positive division keeps the direction. A common error is not flipping when needed, but here it's positive. Another is sign errors in adding terms. Test with x = 3/2.

3

For a fundraiser, a team must sell at least 80 items. They have already sold 27 items and plan to sell the same number each day for $x$ days, with 6 items per day. Which inequality represents meeting the goal?

$6x-27\ge 80$

$27-6x\ge 80$

$6x+27\ge 80$

$6x+27\le 80$

Explanation

This question requires an inequality for selling at least 80 items total, with 27 already sold and 6 per day for x days. Total items: 6x + 27, and 'at least' means ≥80. So, 6x + 27 ≥ 80. Choice B uses ≤, which is the opposite, and D rearranges terms incorrectly. A key error is using subtraction instead of addition for already sold items. Another pitfall is confusing 'at least' with 'at most.' For goal-oriented inequalities, ensure the symbol matches 'at least' (≥) and test with x=0 to check baseline.

4

Solve the inequality $-3(2x-5)\ge 9$ and select the solution set written in simplest form.

$x\le 1$

$x\ge -1$

$x\le -1$

$x\ge 1$

Explanation

This question requires solving the inequality -3(2x - 5) ≥ 9 and selecting the simplest solution set. First, distribute the -3: -6x + 15 ≥ 9. Subtract 15 from both sides: -6x ≥ -6. Divide by -6, and since dividing by a negative number, reverse the inequality: x ≤ 1. A key error is forgetting to reverse the inequality when dividing by -6, which would give x ≥ 1 incorrectly. Another pitfall is mishandling distribution, like treating -3(2x - 5) as positive. For inequalities involving negatives, always double-check the reversal step to avoid direction errors.

5

A gym requires members to be at least 16 years old and younger than 65 to use the weight room without supervision. Let $x$ be a member’s age. Which compound inequality represents this policy?

$16\le x<65$

$16<x\le 65$

$16<x<65$

$16\le x\le 65$

Explanation

This question requires a compound inequality for ages at least 16 and younger than 65. 'At least 16' means ≥16, and 'younger than 65' means <65. Combining gives 16 ≤ x < 65. Choice A uses >16, excluding 16, and C uses ≤65, including 65. A key error is mixing strict and non-strict symbols, like using ≤ for 'younger than.' Always translate words precisely: 'at least' is ≥, 'younger than' is <. For compound inequalities, ensure both parts are correctly joined with 'and.'

6

Solve the compound inequality $x-2\ge1$ OR $x+5<0$. Which solution set is correct?

$x\le3\ \text{or}\ x>-5$

$x\ge3\ \text{and}\ x<-5$

$x\ge3\ \text{or}\ x<-5$

$-5\le x\le3$

Explanation

This question asks for the solution set to the compound inequality x - 2 ≥ 1 or x + 5 < 0. Solve each: x ≥ 3 or x < -5. Union of solutions. A common error is using 'and' instead of 'or', leading to no solution. Another mistake is solving inequalities wrongly. Remember 'or' compounds mean union of sets.

7

A theater must sell at least 120 tickets to break even. It has already sold 45 tickets. Let $t$ be the additional tickets sold. Which inequality shows how many more tickets must be sold?

$t-45\ge120$

$t+45\ge120$

$45-t\ge120$

$t+45\le120$

Explanation

This question asks for the inequality showing how many additional tickets t must be sold to reach at least 120 total, with 45 already sold. The total tickets are 45 + t, and it must be at least 120, so 45 + t ≥ 120. No negative multiplication occurs, so the direction stays the same. A common error is using ≤ instead of ≥, as if it were at most. Another mistake is subtracting 45 incorrectly, like in choice B or D. Carefully match words like 'at least' to ≥ in modeling problems.

8

A student needs an average of at least 85 on 4 tests. The first three scores are 82, 90, and 84. Let $x$ be the fourth test score. Which inequality shows the requirement?

$82+90+84+x\ge340$

$82+90+84+x\le85$

$\dfrac{82+90+84+x}{4}\ge85$

$\dfrac{82+90+84+x}{4}\le85$

Explanation

This question asks for the inequality showing the fourth test score x needed for an average of at least 85 on four tests, with prior scores 82, 90, 84. The average is (82 + 90 + 84 + x)/4 ≥ 85, which simplifies to (256 + x)/4 ≥ 85. No negatives involved in solving. A key error is using ≤ instead of ≥, or summing without dividing. Another pitfall is incorrect total for ≥ 340 without averaging. Ensure the model matches 'at least' with ≥ and includes all elements.

9

Solve $7x-9\ge5x+11$. What is the solution?

$x\le-10$

$x\ge-10$

$x\ge10$

$x\le10$

Explanation

This question asks for the solution to 7x - 9 ≥ 5x + 11. Subtract 5x from both sides to get 2x - 9 ≥ 11, add 9 to get 2x ≥ 20, divide by 2 (positive, no reversal) to get x ≥ 10. The steps maintain the inequality direction. A common error is subtracting incorrectly, leading to x ≤ 10. Another mistake is reversing unnecessarily. Combine like terms carefully before isolating the variable.

10

A school bus can carry at most 52 students. There are already 18 students on the bus. Let $s$ be the number of additional students. Which inequality represents the situation?

$52-s\le18$

$18s\le52$

$s\le34$

$s\ge34$

Explanation

The question requires an inequality for additional students s on a bus with at most 52 total and 18 already aboard. The situation is 18 + s ≤ 52, which simplifies to s ≤ 34 by subtracting 18. This uses ≤ because 'at most' includes equality. A key error is using greater than for capacity limits. Another mistake is adding instead of subtracting the existing students. In modeling problems, define variables clearly and choose the correct inequality direction based on limits.

Page 1 of 7