PSAT Math : Inequalities

Study concepts, example questions & explanations for PSAT Math

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Example Questions

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

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|0.5

w/2

3w/2

|w|

w2

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.

Example Question #2 : How To Find The Solution To An Inequality With Addition

Solve for .

Possible Answers:

Correct answer:

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.

Example Question #3 : How To Find The Solution To An Inequality With Addition

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

Possible Answers:

Correct answer:

Explanation:

To solve this problem, add the two equations together:

x+1<4

y-2<-1

x+1+y-2<4-1

x+y-1<3

x+y<4

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

Example Question #4 : How To Find The Solution To An Inequality With Addition

If , which of the following could be a value of ?

Possible Answers:

-

Correct answer:

Explanation:

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

Therefore, the only option that solves the inequality is

 

Example Question #5 : How To Find The Solution To An Inequality With Addition

What values of  make the statement  true?

Possible Answers:

Correct answer:

Explanation:

First, solve the inequality :

Since we are dealing with absolute value,  must also be true; therefore:

Example Question #4 : How To Find The Solution To An Inequality With Multiplication

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

Possible Answers:

|n2 - 1| > 1

n2 < 2n

16n2 - 1 = 0

(n-1)2 > n

n2 < n

Correct answer:

|n2 - 1| > 1

Explanation:

N_part_1

N_part_2

N_part_3

N_part_4

N_part_5

Example Question #5 : How To Find The Solution To An Inequality With Multiplication

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

Possible Answers:

-1

All of the answers choices are valid.

-2

-3

-4

Correct answer:

-1

Explanation:

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

Example Question #6 : How To Find The Solution To An Inequality With Multiplication

Solve for x

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

 

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

\small 3x \geq -2x-3

\small 5x \geq -3

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

Example Question #7 : How To Find The Solution To An Inequality With Multiplication

We have , find the solution set for this inequality. 

Possible Answers:

Correct answer:

Explanation:

Example Question #12 : Inequalities

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

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

 

Possible Answers:

None of the other answers are correct.

The rational expression is undefined.

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

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

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

Correct answer:

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

Explanation:

(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)

So we can cancel out as follows:

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

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

 

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