. If there are no real numbers that satisfy the inequality 3x – k(x +

Question

If there are no real numbers that satisfy the inequality  3x – k(x + 2) ≥ 10 – 3(1 + x) , where k is a constant, what is the value of k ?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

nick1816 · Answer

3x – k(x + 2) ≥ 10 – 3(1 + x)

3x-kx-2k ≥ 10-3-3x

(6-k)x ≥ 7+2k

x ≥ \frac{7+2k}{(6-k)}

(6-k) can't be equal to 0 or k can't be equal to 6.

E

GMATGuruNY · Answer

Let x=10.
Plugging x=10 into  3x – k(x + 2) ≥ 10 – 3(1 + x) , we get:
30 - k(12) ≥ 10 - 3(11)
-12k ≥ -53
k ≤ 53/12

Of the five answer choices, only E is not less than 53/12.
Implication:
If k=0, k=1, k=2, or k=4, then the inequality has at least one real solution (x=10).
Eliminate A, B, C and D.

E

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=_nLKi_24ZYM

Given that no real number satsfies 3x – k(x + 2) ≥ 10 – 3(1 + x), where k is a constant and we need to find the value

3x – k(x + 2) ≥ 10 – 3(1 + x)
=> 3x - kx - 2k ≥ 10 - 3 - 3x
=> 3x - kx - 2k ≥ 7 - 3x
=> 3x - kx + 3x ≥ 7 + 2k
=> 6x - kx ≥ 7 + 2k
=> x*(6 - k) ≥ 7 + 2k
   Now, there can be two cases depending on the value of 6 - k

-   Case 1  : 6 - k ≥ 0 
=> When we divided both the sides by 6 - k then sign of inequality will not change
=> x ≥  \frac{7 + 2k}{6 - k}

Now, fraction will be undefined if denominator = 0
=> 6 - k = 0
=>  k = 6

-   Case 2  : 6 - k ≤ 0 
=> When we divided both the sides by 6 - k then sign of inequality will reverse
=> x ≤  \frac{7 + 2k}{6 - k}

Now, fraction will be undefined if denominator = 0
=> 6 - k = 0
=>  k = 6

So,  Answer will be E 
Hope it helps!

Watch the following video to MASTER Inequalities

https://www.youtube.com/watch?v=1THrnWJtOqc

shawrochis · Answer

". If there are no real numbers that satisfy the inequality .." What is the significance of this part?

BrushMyQuant · Answer

It means that no real numbers from very big negative number to 0 to very big positive number (including decimal points) can satisfy the inequality.

Example x^2 = -1 does not have any real solutions as only value of x which will satisfy this will be x =  \sqrt{-1}  = i (iota) which is NOT a real number.

. If there are no real numbers that satisfy the inequality 3x – k(x +

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Now, there can be two cases depending on the value of 6 - k
-Case 1 : 6 - k ≥ 0 => When we divided both the sides by 6 - k then sign of inequality will not change => x ≥ \(\frac{7 + 2k}{6 - k}\) Now, fraction will be undefined if denominator = 0 => 6 - k = 0 => k = 6	-Case 2 : 6 - k ≤ 0 => When we divided both the sides by 6 - k then sign of inequality will reverse => x ≤ \(\frac{7 + 2k}{6 - k}\) Now, fraction will be undefined if denominator = 0 => 6 - k = 0 => k = 6

GMAT Problem Solving (PS) Questions

. If there are no real numbers that satisfy the inequality 3x – k(x +

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards