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PSAT Math : Inequalities

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Example Questions

Example Question #131 : Equations / Inequalities

If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?

Possible Answers:

|w|

w²

w/2

3w/2

|w|^0.5

Correct answer:

3w/2

Explanation:

3w/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.

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Example Question #21 : Inequalities

Solve for $\displaystyle z$.

$\left | z-3 \right |\geq 5$ $\displaystyle \left | z-3 \right |\geq 5$

Possible Answers:

$z\leq -2\ \text{or}\ z\geq 8$ $\displaystyle z\leq -2\ \text{or}\ z\geq 8$

$z\geq 8$ $\displaystyle z\geq 8$

$-3\leq z\leq 5$ $\displaystyle -3\leq z\leq 5$

$z\leq -3\ \text{or}\ z\geq 5$ $\displaystyle z\leq -3\ \text{or}\ z\geq 5$

$-2\leq z\leq 8$ $\displaystyle -2\leq z\leq 8$

Correct answer:

$z\leq -2\ \text{or}\ z\geq 8$ $\displaystyle z\leq -2\ \text{or}\ z\geq 8$

Explanation:

Absolute value problems always have two sides: one positive and one negative.

First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.

Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).

We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.

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Example Question #256 : New Sat

If $x+1< 4$ $\displaystyle x+1< 4$ and $y-2<-1$ $\displaystyle y-2< -1$ , then which of the following could be the value of x+y $\displaystyle x+y$?

Possible Answers:

$\displaystyle 16$

$\displaystyle 12$

$\displaystyle 4$

$\displaystyle 0$

$\displaystyle 8$

Correct answer:

$\displaystyle 0$

Explanation:

To solve this problem, add the two equations together:

$x+1<4$ $\displaystyle x+1< 4$

$y-2<-1$ $\displaystyle y-2< -1$

$x+1+y-2<4-1$ $\displaystyle x+1+y-2< 4-1$

$x+y-1<3$ $\displaystyle x+y-1< 3$

$x+y<4$ $\displaystyle x+y< 4$

The only answer choice that satisfies this equation is 0, because 0 is less than 4.

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Example Question #1 : How To Find The Solution To An Inequality With Addition

If 17-x>19 $\displaystyle 17-x>19$, which of the following could be a value of $\displaystyle x$?

Possible Answers:

-$\displaystyle -2$

$\displaystyle 4$

$\displaystyle 2$

$\displaystyle 0$

$\displaystyle -4$

Correct answer:

$\displaystyle -4$

Explanation:

In order to solve this inequality, you must isolate $\displaystyle x$ on one side of the equation.

17-x>19 $\displaystyle 17-x>19$

$\displaystyle 17 > 19 +x$

-2 > x $\displaystyle -2 > x$

Therefore, the only option that solves the inequality is $\displaystyle -4$.

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Example Question #22 : Inequalities

What values of $\displaystyle x$ make the statement $\left |5x-9 \right |\geq6$ $\displaystyle \left |5x-9 \right |\geq6$ true?

Possible Answers:

$x\geq5,x\leq \frac{1}{5}$ $\displaystyle x\geq5,x\leq \frac{1}{5}$

$x\geq15,x\leq \frac{2}{5}$ $\displaystyle x\geq15,x\leq \frac{2}{5}$

$x\geq3, x\leq \frac{3}{5}$ $\displaystyle x\geq3, x\leq \frac{3}{5}$

$x\geq6,x\leq \frac{1}{3}$ $\displaystyle x\geq6,x\leq \frac{1}{3}$

$x\geq4,x\leq -\frac{1}{2}$ $\displaystyle x\geq4,x\leq -\frac{1}{2}$

Correct answer:

$x\geq3, x\leq \frac{3}{5}$ $\displaystyle x\geq3, x\leq \frac{3}{5}$

Explanation:

First, solve the inequality $5x-9 \geq6$ $\displaystyle 5x-9 \geq6$:

$5x-9 \geq6$ $\displaystyle 5x-9 \geq6$

$5x\geq15$ $\displaystyle 5x\geq15$

$x\geq3$ $\displaystyle x\geq3$

Since we are dealing with absolute value, $5x-9\leq-6$ $\displaystyle 5x-9\leq-6$ must also be true; therefore:

$5x-9\leq-6$ $\displaystyle 5x-9\leq-6$

$5x\leq3$ $\displaystyle 5x\leq3$

$x\leq \frac{3}{5}$ $\displaystyle x\leq \frac{3}{5}$

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Example Question #151 : Algebra

If –1 < n < 1, all of the following could be true EXCEPT:

Possible Answers:

(n-1)² > n

16n² - 1 = 0

n² < n

n² < 2n

|n² - 1| > 1

Correct answer:

|n² - 1| > 1

Explanation:

N_part_1

N_part_2

N_part_3

N_part_4

N_part_5

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Example Question #166 : Equations / Inequalities

(√(8) / -x ) < 2. Which of the following values could be x?

Possible Answers:

-1

-3

-2

-4

All of the answers choices are valid.

Correct answer:

-1

Explanation:

The equation simplifies to x > -1.41. -1 is the answer.

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Example Question #31 : Inequalities

Solve for x

$\small 3x+7 \geq -2x+4$ $\displaystyle \small 3x+7 \geq -2x+4$

Possible Answers:

$\small x \geq -\frac{3}{5}$ $\displaystyle \small x \geq -\frac{3}{5}$

$\small x \leq \frac{3}{5}$ $\displaystyle \small x \leq \frac{3}{5}$

$\small x \geq \frac{3}{5}$ $\displaystyle \small x \geq \frac{3}{5}$

$\small x \leq -\frac{3}{5}$ $\displaystyle \small x \leq -\frac{3}{5}$

Correct answer:

$\small x \geq -\frac{3}{5}$ $\displaystyle \small x \geq -\frac{3}{5}$

Explanation:

$\small 3x+7 \geq -2x+4$ $\displaystyle \small 3x+7 \geq -2x+4$

$\small 3x \geq -2x-3$ $\displaystyle \small 3x \geq -2x-3$

$\small 5x \geq -3$ $\displaystyle \small 5x \geq -3$

$\small x\geq -\frac{3}{5}$ $\displaystyle \small x\geq -\frac{3}{5}$

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Example Question #151 : Algebra

We have x^2-4<0 $\displaystyle x^2-4< 0$, find the solution set for this inequality.

Possible Answers:

x>-2 $\displaystyle x>-2$

x<2 $\displaystyle x< 2$

-2<x<2 $\displaystyle -2< x< 2$

x=0 $\displaystyle x=0$

x>2, x<-2 $\displaystyle x>2, x< -2$

Correct answer:

-2<x<2 $\displaystyle -2< x< 2$

Explanation:

$x^2-4<0 \Rightarrow x^2<4 \Rightarrow -2<x<2$ $\displaystyle x^2-4< 0 \Rightarrow x^2< 4 \Rightarrow -2< x< 2$

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Example Question #32 : Inequalities

Fill in the circle with either $<$ $\displaystyle < $, $>$ $\displaystyle >$, or $=$ $\displaystyle =$ symbols:

$(x-3)\circ\frac{x^2-9}{x+3}$ $\displaystyle (x-3)\circ\frac{x^2-9}{x+3}$ for $x\geq 3$ $\displaystyle x\geq 3$.

Possible Answers:

The rational expression is undefined.

$(x-3)=\frac{x^2-9}{x+3}$ $\displaystyle (x-3)=\frac{x^2-9}{x+3}$

None of the other answers are correct.

$(x-3)> \frac{x^2-9}{x+3}$ $\displaystyle (x-3)> \frac{x^2-9}{x+3}$

$(x-3)< \frac{x^2-9}{x+3}$ $\displaystyle (x-3)< \frac{x^2-9}{x+3}$

Correct answer:

$(x-3)=\frac{x^2-9}{x+3}$ $\displaystyle (x-3)=\frac{x^2-9}{x+3}$

Explanation:

$(x-3)\circ\frac{x^2-9}{x+3}$ $\displaystyle (x-3)\circ\frac{x^2-9}{x+3}$

Let us simplify the second expression. We know that:

$(x^2-9)=(x+3)(x-3)$ $\displaystyle (x^2-9)=(x+3)(x-3)$

So we can cancel out as follows:

$\frac{x^2-9}{x+3}=\frac{(x+3)(x-3)}{(x+3)}=x-3$ $\displaystyle \frac{x^2-9}{x+3}=\frac{(x+3)(x-3)}{(x+3)}=x-3$

$(x-3)=\frac{x^2-9}{x+3}$ $\displaystyle (x-3)=\frac{x^2-9}{x+3}$

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PSAT Math : Inequalities

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #131 : Equations / Inequalities

Example Question #21 : Inequalities

Example Question #256 : New Sat

Example Question #1 : How To Find The Solution To An Inequality With Addition

Example Question #22 : Inequalities

Example Question #151 : Algebra

Example Question #166 : Equations / Inequalities

Example Question #31 : Inequalities

Example Question #151 : Algebra

Example Question #32 : Inequalities

All PSAT Math Resources

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