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Joined: 23 Jan 2021
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Concentration: , International Business
If 6 ·| 4-k/3|> 12, which of the following could be the value of k? I
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11 Nov 2021, 00:20
11 Nov 2021, 00:20
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If 6 ·| 4-k/3|> 12, which of the following could be the value of k?
Indicate all such values.
A. -15
B. -10
C. -5
D. 0
E. 5
Indicate all such values.
A. -15
B. -10
C. -5
D. 0
E. 5
ShowHide Answer
Official Answer
A,B,C,D,E
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Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
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Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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Re: If 6 ·| 4-k/3|> 12, which of the following could be the value of k? I
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12 Nov 2021, 05:45
12 Nov 2021, 05:45
1
Absolute Value problem
We are given that 6 ·| 4-k/3|> 12 and we need to indicate which all can be the possible values for k
Let's solve it using two methods
Method 1: Substitution
Let's take each answer choice and plug it into the answer choices, before doing that let's divide both the sides by +6 to simply the equation
We get 6 x | 4-k/3| / 6> 12/6 => | 4-k/3| > 2
A. k = -15. Substituting k = -15 in | 4-k/3|> 2 we get, | 4-(-15)/3|> 2 => | 4+5 |> 2 => |9|> 2
Since 9 is positive so |9| = 9 => 9> 2 , which is true => A is possible
B. k = -10. Substituting k = -10 in | 4-k/3|> 2 we get, | 4-(-10)/3|> 2 => | 4+10/3 |> 2 => |22/3|> 2
Since 22/3 is positive so |22/3| = 22/3 => 22/3> 2 ,which is true => B is possible
C. k = -5. Substituting k = -5 in | 4-k/3|> 2 we get, | 4-(-5)/3|> 2 => | 4+5/3 |> 2 => |17/3|> 2
Since 17/3 is positive so |17/3| = 17/3 => 17/3> 2 , which is true => C is possible
D. k = 0. Substituting k = 0 in | 4-k/3|> 2 we get, | 4-(0)/3|> 2 => | 4+0 |> 2 => |4|> 2
Since 4 is positive so |4| = 4 => 4 > 2 , which is true => D is possible
E. k = 5. Substituting k = 5 in | 4-k/3|> 2 we get, | 4-5/3|> 2 => |7/3|> 2
Since 7/3 is positive so |7/3| = 7/3 => 7/3 > 2 , which is true => E is possible
So, all A to E are possible
Method 2: Algebra
6 x | 4-k/3|> 12
Divide both the sides by 6 we get (As above)
| 4-k/3| > 2
=> | (12-k)/3| > 2
=> | 12-k | > 2*|3|
=> |12-k | > 2*3
=> |12-k| > 6
=> 12-k > 6 or 12-k < -6
=> k < 12-6 or k > 12+6
=> k < 6 or k > 18
So, any value < 6 or > 18 will satisfy the equation
All values given in the option choice are < 6 so all values are true.
Hope it helps!
Watch the following video to learn the Basics of Absolute Values
We are given that 6 ·| 4-k/3|> 12 and we need to indicate which all can be the possible values for k
Let's solve it using two methods
Method 1: Substitution
Let's take each answer choice and plug it into the answer choices, before doing that let's divide both the sides by +6 to simply the equation
We get 6 x | 4-k/3| / 6> 12/6 => | 4-k/3| > 2
A. k = -15. Substituting k = -15 in | 4-k/3|> 2 we get, | 4-(-15)/3|> 2 => | 4+5 |> 2 => |9|> 2
Since 9 is positive so |9| = 9 => 9> 2 , which is true => A is possible
B. k = -10. Substituting k = -10 in | 4-k/3|> 2 we get, | 4-(-10)/3|> 2 => | 4+10/3 |> 2 => |22/3|> 2
Since 22/3 is positive so |22/3| = 22/3 => 22/3> 2 ,which is true => B is possible
C. k = -5. Substituting k = -5 in | 4-k/3|> 2 we get, | 4-(-5)/3|> 2 => | 4+5/3 |> 2 => |17/3|> 2
Since 17/3 is positive so |17/3| = 17/3 => 17/3> 2 , which is true => C is possible
D. k = 0. Substituting k = 0 in | 4-k/3|> 2 we get, | 4-(0)/3|> 2 => | 4+0 |> 2 => |4|> 2
Since 4 is positive so |4| = 4 => 4 > 2 , which is true => D is possible
E. k = 5. Substituting k = 5 in | 4-k/3|> 2 we get, | 4-5/3|> 2 => |7/3|> 2
Since 7/3 is positive so |7/3| = 7/3 => 7/3 > 2 , which is true => E is possible
So, all A to E are possible
Method 2: Algebra
6 x | 4-k/3|> 12
Divide both the sides by 6 we get (As above)
| 4-k/3| > 2
=> | (12-k)/3| > 2
=> | 12-k | > 2*|3|
=> |12-k | > 2*3
=> |12-k| > 6
=> 12-k > 6 or 12-k < -6
=> k < 12-6 or k > 12+6
=> k < 6 or k > 18
So, any value < 6 or > 18 will satisfy the equation
All values given in the option choice are < 6 so all values are true.
Hope it helps!
Watch the following video to learn the Basics of Absolute Values
gmatclubot
Re: If 6 ·| 4-k/3|> 12, which of the following could be the value of k? I [#permalink]
12 Nov 2021, 05:45
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