On the real number line, the distance between a real number k and -3 is strictly less than… — GRE General Test

On the real number line, the distance between a real number $k$ and $-3$ is strictly less than $7$ , and the distance between $k$ and $5$ is at least $4$ . Which of the following inequalities represents the complete set of all possible values of $k$ ?

$-10 < k \le 1$ Answer
B
$-10 \le k \le 1$
C
$-10 < k < 1$
D
$-4 < k \le 1$
E
$1 \le k < 4$

Answer

-10 < k \le 1

The correct inequality

-10 < k \le 1

properly combines the strict bound from the distance to

-3

(which gives

-10 < k < 4

) with the non-strict bound from the distance to

5

(which gives

k \le 1

k \ge 9

). Taking their intersection gives

-10 < k \le 1

Step-by-Step Solution

Express the first condition using absolute value notation and solve for

k

|k - (-3)| < 7 \implies |k + 3| < 7 \implies -7 < k + 3 < 7 \implies -10 < k < 4

Distance on a number line between

x

and

y

is given by

|x - y|

Express the second condition using absolute value notation and solve for

k

|k - 5| \ge 4 \implies k - 5 \le -4

k - 5 \ge 4 \implies k \le 1

k \ge 9

The phrase 'at least 4' means greater than or equal to 4.

Find the overlap (intersection) of the two solution sets.

(-10 < k < 4) \cap (k \le 1 \text{ or } k \ge 9) = -10 < k \le 1

Since

k

must be less than

4

, the region

k \ge 9

contains no valid solutions, leaving only

-10 < k \le 1

Key Concept

Absolute value as distance on the number line and solving compound absolute value inequalities.

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

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