If k is an integer such that |2k - 7| \le 11 and |k + 2| > 4, how many possible values of … — GMAT

If $k$ is an integer such that $|2k - 7| \le 11$ and $|k + 2| > 4$ , how many possible values of $k$ exist?

Cevap: 7

The total number of possible integer values of

k

is 7.

Unfolding

|2k - 7| \le 11

gives

-2 \le k \le 9

, representing 12 possible integers. Unfolding

|k + 2| > 4

gives

k > 2

k < -6

. Intersecting these two regions for real integers eliminates

k < -6

(since

k \ge -2

) and restricts the set to

3 \le k \le 9

. The integers in this range are

3, 4, 5, 6, 7, 8,

and

9

, yielding a total of 7 valid values.

Solve the first absolute value inequality

|2k - 7| \le 11

-2 \le k \le 9

Expanding the absolute value gives

-11 \le 2k - 7 \le 11

. Adding 7 to all parts yields

-4 \le 2k \le 18

, and dividing by 2 results in

-2 \le k \le 9

Solve the second absolute value inequality

|k + 2| > 4

k > 2

k < -6

Expanding the strict absolute value inequality gives two cases:

k + 2 > 4 \implies k > 2

, or

k + 2 < -4 \implies k < -6

Find the intersection of the two solution sets for integer values of

k

3 \le k \le 9

Since the first inequality requires

k \ge -2

, no integer can satisfy both

k \ge -2

and

k < -6

. Thus,

k

must satisfy

-2 \le k \le 9

and

k > 2

, which reduces to

3 \le k \le 9

Count the number of integers in the range

3 \le k \le 9

The valid integers in this range are

3, 4, 5, 6, 7, 8,

and

9

. The total count is

9 - 3 + 1 = 7

Solving compound absolute value inequalities for integer solution counts

Soru