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Difficulty:
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65% (02:27) correct
35%
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How many distinct integer values of n satisfy the inequality ||n-3| + 4| ≤ 12 ?
A. 15
B. 16
C. 17
D. 18
E. 19
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A. 15
B. 16
C. 17
D. 18
E. 19
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GMATinsight
METHOD-1
||n-3| + 4| ≤ 12 can be re-written as
i.e. -12 ≤ |n-3| + 4 ≤ 12
i.e. -12-4 ≤ |n-3| ≤ 12-4
i.e. -16 ≤ |n-3| ≤ 8
Here the corrective action needs to be taken.
-16 ≤ |n-3| in the above Inequation is REDUNTANT because we know that absolute value of any expression can't be less than zero. So an absolute value is greater than -16 doesn't contribute to our information at all
i.e. The Inequation can be sufficiently defined by the expression |n-3| ≤ 8
i.e. -8 ≤ n-3 ≤ 8
i.e. -8+3 ≤ n ≤ 8+3
i.e. -5 ≤ n ≤ 11
The Number of Integer values from -5 to 11 (Inclusive) is 17
Answer: Option C
METHOD-2
The primary expression is ||n-3| + 4| ≤ 12
i.e. The Absolute value of the expression on Left Hand Side (LHS) can go upto 12
i.e. The value of |n-3| can go upto 8
i.e. the Extreme absolute Value of n-3 can be +8
i.e. when n-3 = +8, The extreme value of n = +11
and when n-3 = -8, The extreme value of n = -5
i.e. The Value of n can range from -5 to +11 i.e. 17 Integer values
Answer: Option C
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General Discussion
GMATinsight
can anyone tell me my approach....
considering LHS IIn-3I+4I is always +ve since In-3I and 4 both r +ve.
so i can write this as In-3I+4<=12
or In-3I<=8
so n can have -5 to 11 = 17 true values....
rohit8865
Yes that's right...
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Posts: 17
the best way to solve is by solving the outer modules and the inner one.
||n-3|+4|=12
-12<=|n-3|+4<=12
-12-4<=|n-3|<=12-4 (subtract 4 from both sides)
-16<=|n-3|<=8
As the value of |n-3| can never be negative, so ignoring the -16 value.
-8<=|n-3|<=8
-8+3<=n<=8+3
-5<=n<=11
resulting in 17 distinct values.
||n-3|+4|=12
-12<=|n-3|+4<=12
-12-4<=|n-3|<=12-4 (subtract 4 from both sides)
-16<=|n-3|<=8
As the value of |n-3| can never be negative, so ignoring the -16 value.
-8<=|n-3|<=8
-8+3<=n<=8+3
-5<=n<=11
resulting in 17 distinct values.
Posts: 198
GMATinsight
hi
I have solved some problems of this kind previously, without any corrective action taken as such. I think the double absolute modulus has made this difference. Do you think, however, we are better off turning the double modulus into one, before setting the final inequality....? For example ...
|n-3| + 4| ≤ 12
| n-3 | <= 8
now we can set the inequality as under:
-8 <| n-3 | <= 8
OR, if there is any guiding principle to follow when solving questions as such, please say to me ...
thanks in advance ...
gmatcracker2017
Only problem in this expression -8 <| n-3 | <= 8 is that |n-3| can be between -8 and 0 as per the given expression while in real it's impossible because absolute value of expression can't be negative
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GMATinsight
nakulmk - I would use the graphical approach here:
Attachment:
Screenshot 2019-06-18 at 13.42.14.png [ 16.64 KiB | Viewed 9860 times ]
This is what the graph of ||n - 3| + 4| will look like. Since |n - 3| + 4 has only positive values, ||n - 3| + 4| has the same graph.
(discussed in the posts below:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... h-to-mods/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/1 ... solutions/ )
If y <= 12, then x will take values from -5 to 11 (since slope of the lines are 1 and -1). So it will give us 17 values. Once you understand your graphs well, you can get the answer in under 1 min by just imagining it.
Answer (C)
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Are these double modulus questions even tested on the GMAT?? I see a number of questions on GC based on these. But I have never come across an official questions which tests these.. does it make sense to practice these kind of questions?
Sappy
I agree - not a lot of sense in practicing them!
Here is the thing about GMAT - the concepts tested are basic but there are no boundaries on the form in which they could be tested. If the topic being tested is "absolute values", what says that multiple modulus terms are fine but nested are not? Both lie at the far right of the difficulty scale - the nested modulus just a bit farther off. You can use the same concepts to solve these questions too. Hence, only if one is aiming for 50/51 does one need to spend time on these to cover all ground (as one would do in all other relevant topics too). Else, one can ignore.
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TheNightKing
Joined: 18 Dec 2017
Last visit: 20 Mar 2024
Posts: 1,119
Location: United States (KS)
GMAT 1: 600 Q46 V27
GMAT 1: 600 Q46 V27
Posts: 1,119
GMATinsight
I did not do any fancy approach. Started with n=0 went upto n=11. All of these 12 satisfy.
Lets start with n=-1 you can go upto n=-5 all of these 5 satisfy. Hence 17 values.
P.S: I did not write n=-5, missed it and marked B. Haha. Last mile delivery is tough (just like for Amazon). Hahaha!
\(\left|\right|n-3\left|+4\right|\ \le\ 12\)
In the given inequality. \(\left|n-3\right|\) is always positive.
Considering \(\left|n-3\right| \)as x which is always non negative.
Hence we have : \(\left|x+4\right|\ \le\ 12\)
x takes values from 0 to 8. A total of 9 values.
When x = 0, we have\( \left|n-3\right|\) is equal to zero and hence n = 3.
Similarly when x = 1, we have :\( \left|n-3\right|\ =\ 1\ or\ n-3\ =\ 1\ or\ -1\)
Similarly for every value of x from 1..............8 we have 2 values of n.
When x = 0 we only have one value. =8*2 + 1 = 17
A total of 17 values exist.
In the given inequality. \(\left|n-3\right|\) is always positive.
Considering \(\left|n-3\right| \)as x which is always non negative.
Hence we have : \(\left|x+4\right|\ \le\ 12\)
x takes values from 0 to 8. A total of 9 values.
When x = 0, we have\( \left|n-3\right|\) is equal to zero and hence n = 3.
Similarly when x = 1, we have :\( \left|n-3\right|\ =\ 1\ or\ n-3\ =\ 1\ or\ -1\)
Similarly for every value of x from 1..............8 we have 2 values of n.
When x = 0 we only have one value. =8*2 + 1 = 17
A total of 17 values exist.
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