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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
61% (02:24) correct
39%
(02:32)
wrong
based on 1110
sessions
History
Date
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For how many integer values of x, is \(|3x-3|+|2x+8|<15\)?
A. 2
B. 3
C. 4
D. 5
E. 6
Untitled.jpg [ 61.01 KiB | Viewed 17373 times ]
A. 2
B. 3
C. 4
D. 5
E. 6
Attachment:
Untitled.jpg [ 61.01 KiB | Viewed 17373 times ]
ziyuenlau
\(|3x-3|+|2x+8|<15\)
\(3*|x-1|+2*|x+4|<15\)
We want the values of x such that the sum of "thrice their distance from 1" and "twice their distance from -4" is less than 15.
Let's try to find the point where this distance is equal to 15.
........................ (-4) ...................................... (0) ........... (1) ..........................
The distance between -4 and 1 is 5. Thrice this distance is 15. So at the point x = -4, the sum will be 15. As we move to the right of -4, the sum will reduce (since the twice component will keep increasing). At x = 1, the sum becomes 0 + 2*5 = 10.
What happens when you go to the right of 1? Now the sum starts increasing since the thrice components increasing now.
At x = 2, the sum becomes 3*1 + 2*6 = 15.
To the right of 2, the sum will keep increasing.
So the sum will be less than 15 between -4 and 2. This gives us 5 integer values (-3, -2, -1, 0, 1).
Answer (D)
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\(3x-3 = 0\) then \(x = 1\)
\(2x+8 = 0\) then \(x = -4\)
there can be 3 range
\(x\leq{-4}\) | \(-4 < x < 1\) | \(x\geq{1}\)
1. When \(x\leq{-4}\), then both \(|3x-3|\) and \(|2x+8|\) will be negative.
\(-3x+3 -2x-8 < 15\)
\(-5x-5 < 15\)
\(x > -4\) (this is opposite to \(x\leq{-4}\). )
Hence \(x\leq{-4}\) is not a possibility.
2. When \(-4 < x < 1\), then \(|3x-3|\) would still be negative but \(|2x+8|\) will be positive. hence
\(-3x+3 +2x+8 < 15\)
-x+11 < 15
\(-x < 4\) i.e. \(x > -4\) (this lies within \(-4 < x < 1\)) so correct range.
3. When \(x\geq{1}\), then both \(|3x-3|\) and \(|2x+8|\) will be positive.
\(3x-3+2x+8 < 15\)
\(5x+5 < 15\)
\(x+1 < 3\)
\(x < 2\) this range is also fine because \(x\geq{1}\). so x must be 1.
So our range is \(-4 <\)\(x\geq{1}\), so howmany integers within this range?
-3, -2, -1, 0 and 1
Count is 5
Answer D.
\(2x+8 = 0\) then \(x = -4\)
there can be 3 range
\(x\leq{-4}\) | \(-4 < x < 1\) | \(x\geq{1}\)
1. When \(x\leq{-4}\), then both \(|3x-3|\) and \(|2x+8|\) will be negative.
\(-3x+3 -2x-8 < 15\)
\(-5x-5 < 15\)
\(x > -4\) (this is opposite to \(x\leq{-4}\). )
Hence \(x\leq{-4}\) is not a possibility.
2. When \(-4 < x < 1\), then \(|3x-3|\) would still be negative but \(|2x+8|\) will be positive. hence
\(-3x+3 +2x+8 < 15\)
-x+11 < 15
\(-x < 4\) i.e. \(x > -4\) (this lies within \(-4 < x < 1\)) so correct range.
3. When \(x\geq{1}\), then both \(|3x-3|\) and \(|2x+8|\) will be positive.
\(3x-3+2x+8 < 15\)
\(5x+5 < 15\)
\(x+1 < 3\)
\(x < 2\) this range is also fine because \(x\geq{1}\). so x must be 1.
So our range is \(-4 <\)\(x\geq{1}\), so howmany integers within this range?
-3, -2, -1, 0 and 1
Count is 5
Answer D.
General Discussion
hazelnut
\(positive:|3x-3|≥0…3x≥3…x≥1…negative:x<1\)
\(positive:|2x+8|≥0…2x≥-8…x≥-4…negative:x<-4\)
\(range:--(neg)--(-4)---(pos,neg)--(1)--(pos)---\)
\(x≥1:|3x-3|+|2x+8|<15…3x-3+2x+8<15…5x<10…x<2:1≤x<2=[1]\)
\(4≤x<1:…-3x+3+2x+8<15…-x<4…x>-4:-4<x<1=[-3,-2,-1,0]\)
\(x<-4:…-3x+3-2x-8<15…-5x<20…x>-4:invalid=x<-4\)
\(x=[-3,-2,-1,0,1]=5\)
Ans (D)
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-15 < 3x-3+2x+8 < 15
-15 < 5x + 5 < 15
-20 < 5x < 10
-4 < x < 2
X can be -3, -2, -1, 0 , 1, 2 . ( D )
-15 < 5x + 5 < 15
-20 < 5x < 10
-4 < x < 2
X can be -3, -2, -1, 0 , 1, 2 . ( D )
One can always break the modulus and then proceed with the usual process as mentioned is other solutions but I find solving such problems by plotting the graph of the inequality. The visual is far better to interpret and avoids missing any values even if there are multiple such modulus.
Below is a descriptive solution for each step.
Posted from my mobile device
Below is a descriptive solution for each step.
Posted from my mobile device
Attachments
IMG_20240712_054515695~2.jpg [ 1.27 MiB | Viewed 6578 times ]
Posts: 478
The thing to learn from this question is the value is not given as (x-1) + (x+2) < 15
but there is a coefficient of x involved
the way HKD1710 has solved is exactly how I would approach
but there is a coefficient of x involved
the way HKD1710 has solved is exactly how I would approach
Posts: 1
Hi,
Thanks for such a concise solution.
Do you happen to have a video explanation on how to apply this approach?
Thanks for such a concise solution.
Do you happen to have a video explanation on how to apply this approach?
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