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SAT Math : Inequalities

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Example Questions

Example Question #1 : Inequalities

|12x + 3y| < 15

What is the range of values for y, expressed in terms of x?

Possible Answers:

5 + 4x < y < 5 – 4x

–5 – 4x < y < 5 – 4x

y < 5 – 4x

5 – 4x < y < 5 + 4x

y > 15 – 12x

Correct answer:

–5 – 4x < y < 5 – 4x

Explanation:

Recall that with absolute values and "less than" inequalities, we have to hold the following:

12x + 3y < 15

AND

12x + 3y > –15

Otherwise written, this is:

–15 < 12x + 3y < 15

In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:

–15 – 12x < 3y < 15 – 12x

Now, we have to divide each element by 3:

(–15 – 12x)/3 < y < (15 – 12x)/3

This simplifies to:

–5 – 4x < y < 5 – 4x

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Example Question #2 : Inequalities

|4x + 14| > 30

What is a possible valid value of x?

Possible Answers:

–11

–3

Correct answer:

Explanation:

This inequality could be rewritten as:

4x + 14 > 30 OR 4x + 14 < –30

Solve each for x:

4x + 14 > 30; 4x > 16; x > 4

4x + 14 < –30; 4x < –44; x < –11

Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.

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Example Question #3 : Inequalities

Given the inequality, |2x – 2| > 20,

what is a possible value for x?

Possible Answers:

–10

–8

Correct answer:

–10

Explanation:

For this problem, we must take into account the absolute value.

First, we solve for 2x – 2 > 20. But we must also solve for 2x – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).

First step:

2x – 2 > 20

2x > 22

x > 11

Second step:

2x – 2 < –20

2x < –18

x < –9

Therefore, x > 11 and x < –9.

A possible value for x would be –10 since that is less than –9.

Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.

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Example Question #4 : Inequalities

Solve for $x$ .

$-2x+5\leq 10$

Possible Answers:

$None\ of\ the\ above$

$x\geq -\frac{5}{2}$

$x\leq 5$

$x\geq \frac{5}{2}$

$x\leq \frac{5}{2}$

Correct answer:

$x\geq -\frac{5}{2}$

Explanation:

Move +5 using subtraction rule which will give you $-2x\leq 5$ .

Divide both sides by 2 (using division rule) and you will get $-x\leq \frac{5}{2}$ which is the same as $x\geq \frac{5}{2}$

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Example Question #4 : Inequalities

If 2 more than is a negative integer and if 5 more than is a positive integer, which of the following could be the value of ?

Possible Answers:

-7

Correct answer:

Explanation:

m+2 < 0 and m+5 > 0 , so m < -2 and m > -5 . The only integers between and are and .

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Example Question #5 : Inequalities

If $\frac{a}{5}+5> 6$ , which of the following MUST be true?

I. $a> 2$

II. $a> 10$

III. $a< 6$

Possible Answers:

I only

III only

II only

I and II only

I, II, and III

Correct answer:

I only

Explanation:

Subtract 5 from both sides of the inequality:

$\frac{a}{5}> 1$

Multiply both sides by 5:

$a> 5$

Therefore only I must be true.

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Example Question #6 : Inequalities

Which of the following is equivalent to $\left | x-3 \right |<2$ ?

Possible Answers:

1 < x < 5

x<5

x>1

x>-1

-3 < x < 2

Correct answer:

1 < x < 5

Explanation:

Solve for both x – 3 < 2 and –(x – 3) < 2.

x – 3 < 2 and –x + 3 < 2

x < 2 + 3 and –x < 2 – 3

x < 5 and –x < –1

x < 5 and x > 1

The results are x < 5 and x > 1.

Combine the two inequalities to get 1 < x < 5

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Example Question #6 : Inequalities

Given $\left | 7x+9\right |>30$ , what is a possible value of ?

Possible Answers:

$-\frac{39}{7}$

Correct answer:

Explanation:

In order to find the range of possible values for $\left | 7x+9\right |>30$ , we must first consider that the absolute value applied to this inequality results in two separate equations, both of which we must solve:

7x+9>30 and 7x+9< -30 .

Starting with the first inequality:

7x+9>30

7x>21

x>3

Then, our second inequality tells us that

7x+9< -30

7x< -39

$x<-\frac{39}{7}(or -5\frac{4}{7})$

Therefore, is the correct answer, as it is the only value above for which x>3 (NOT greater than or equal to) or $x<-\frac{39}{7}$ (NOT less than or equal to).

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Example Question #9 : Inequalities

The cost, in cents, of manufacturing $\dpi{100} \small x$ pencils is $\dpi{100} \small 1200+20x$ , where 1200 is the number of cents required to run the factory regardless of the number of pencils made, and 20 represents the per-unit cost, in cents, of making each pencil. The pencils sell for 50 cents each. What number of pencils would need to be sold so that the revenue received is at least equal to the manufacturing cost?

Possible Answers:

$\dpi{100} \small 27$

$\dpi{100} \small 40$

$\dpi{100} \small 30$

$\dpi{100} \small 33$

$\dpi{100} \small 36$

Correct answer:

$\dpi{100} \small 40$

Explanation:

If each pencil sells at 50 cents, $\dpi{100} \small x$ pencils will sell at $\dpi{100} \small 50x$ . The smallest value of $\dpi{100} \small x$ such that

$\dpi{100} \small 50x\geq 1200+20x$

$\dpi{100} \small x\geq 40$

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Example Question #5 : Inequalities

Solve for :

-3x+5<101

Possible Answers:

x<-32

x>-32

x>32

x<32

x=32

Correct answer:

x>-32

Explanation:

The correct method to solve this problem is to substract 5 from both sides. This gives -3x<96 .

Then divide both sides by negative 3. When dividing by a negative it is important to remember to change the inequality sign. In this case the sign goes from a less than to a greater than sign.

This gives the answer x>-32 .

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SAT Math : Inequalities

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1 : Inequalities

Example Question #2 : Inequalities

Example Question #3 : Inequalities

Example Question #4 : Inequalities

Example Question #4 : Inequalities

Example Question #5 : Inequalities

Example Question #6 : Inequalities

Example Question #6 : Inequalities

Example Question #9 : Inequalities

Example Question #5 : Inequalities

All SAT Math Resources

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