challenge problem

[namit]

If k is a positive constant and y = |x - k| - |x + k|, what is the maximum value of y?

(1) x < 0

(2) k = 3

how to solve this . I wanted to knw how to solve the inequality??

[RA]

Is the answer C?

Here is how I solved it:

Statement A: Number substitution shows that a lower value of k results in a lower value of y. Since no information is given for k statement (A) alone is not enough.
x=-2; k=3; I-2-3I - I-2+3I = 5 - 1 = 4
x=-2; k=-1; I-2+1I - I-2-1I = 1 - 3 = -2

Statement B: Number substitution would shows that a higher value of x results in a lower value of y. Since no information is given for k statement (B) alone is not enough
x=-2; k=3; I-2-3I - I-2+3I = 5 - 1 = 4
x=-1; k=3: I-1-3I - I-1+3I = 4 - 2 = 2

Statement (A) and (B) together give a definite value for k (=3) and also give us the max value x can have (=-1)
x=-1; k=3; I-1-3I - I-1+3I = 4 - 2 = 2

[RA]

Ignore my previous post, I misread the question.

[RA]

I re-read the question and I am still coming up with (C) as the answer. 6 is the max value Y can have.

Please confirm if this is the correct answer.

[vin7de]

Assuming k=3
Case A
If x>=3 then y=x-3 -(x+3)=-6
Case B
If 3>x>=-3 then y=-(x-3) -(x+3)=-2x
Case C
If -3=>x then y=-(x-3) +(x+3)=6

So, II is not sufficient.
Inuititively, the inequality will depend on the value of k. Hence, the best shot at solving the inequality is knowing both I and II

If x<0, there are two outcomes, Case B and Case C
e.g. If x=-1, y=2; and when x=-4; y=6
However, the maximum value that y can take for any value negative x and k=3 is 6

So the answer is C

[esledge]

It sounds like you all worked this out together, so I'll just throw in another method for you to try.

Absolute value problems can often be done visually on a number line, as absolute value is a "distance." (e.g. |x| = 6 means that x is 6 "away from" zero on the number line, in either direction).

I didn't complete it, but for this one I envision 3 cases drawn on number lines, as we know k > 0, but nothing about x:
A. 0 < k < x
B. 0 < x < k
C. x < 0 < k

You could visually determine the possible y values (or at least whether they could vary). The statements just eliminate cases (e.g. (1) eliminates cases A and B) or lock-in the exact location of k.

[vishalsahdev03]

It sounds like you all worked this out together, so I'll just throw in another method for you to try.

Absolute value problems can often be done visually on a number line, as absolute value is a "distance." (e.g. |x| = 6 means that x is 6 "away from" zero on the number line, in either direction).

I didn't complete it, but for this one I envision 3 cases drawn on number lines, as we know k > 0, but nothing about x:
A. 0 < k < x
B. 0 < x < k
C. x < 0 < k

You could visually determine the possible y values (or at least whether they could vary). The statements just eliminate cases (e.g. (1) eliminates cases A and B) or lock-in the exact location of k.

Could you please elaborate the explanation give, I figure out the answer to be B.
Please verify and explain !
Thanks in advance.

[helloriteshranjan]

Could you please elaborate the explanation give, I figure out the answer to be B.
Please verify and explain !
Thanks in advance.

since statement 1 doesn't say anything about k, its insufficient

statement II: sufficient
we can consider all possible ranges of x, go like this

x :expression
x>3 :-6
x=3 :-6
-3<x<3 :-2x [you have to put algebraic value of x in this, like -1 or -2.3 etc]
x=-3 :6
x<-3 :6

do we need statement I anymore? no i believe
so in my opinion, answer is B

[jerly_vivek]

One golden rule that i always follow in inequality problem is try removing the unknown variables. In the above question II gives us the opportunity to reduce unknown K to known 3. So lets start with II first:
II: y = |x-3|-|x+3|
=>Three conditions arises: x>0,x=0,x<0
a)x>0
=>y=(x-3)-(x+3)=x-3-x-3 = -6.

b)x=0
=>y=(0-3)-(0+3)=-6

c)x<0
=>y=(-x-3)-(-x+3)=-x-3+x-3=-6

As we see, we always get y=-6, we do not need to know value of x.
So, B is the answer.

[cyber_office]

I don't think you can treat the absolute value the way you treat parens. You have to solve the absolute value first. | x -3| is not the same as (x-3).

[terpneil]

I agree with the answer B.
When plugging in numbers, need to make sure you run the full range of numbers.

Therefore, run x<-3, -3, -3<x<0, and so forth.

Also, the question asks for the maximum value of Y, not the value of Y.

[imanemekouar]

Can a gmat instructor evaluate my understanding;
There is 4 possibilities for the 3 absolute value .I onlu consider the cas when the 2 of them have the same sign or a different sign.

1 case :+,+ Y=X-k-x-k=-2k
y=-2k

2cas : different sign for the absolute value.
y=-X+K-X-K
Y=-2x
First answer choice : no sufficient
Second answer choice:K=3 so Y=-6

B is sufficient.
please correct me if i m wrong
s

[esledge]

The last time I posted here, I didn't complete the solution. Here's a way, similar to jerly_vivek's "condition" method. But there are actually 4 conditions to consider, including one I didn't consider before.

Scenario 1: x is positive, and x>k
(i.e. the expressions in both absolute values are positive)
y = |x - k| - |x + k| = (x-k) - (x+k) = -2k
y = -2k, which is negative.

Scenario 2: x is positive, but 0<x<k
(i.e. the expression in the left abs value is negative, the one in the right is positive)
y = |x - k| - |x + k| = -(x-k) - (x+k) = -2x
y = -2x, which is negative.

Scenario 3: x is negative, but not more negative than k is positive (-k<x<0)
(i.e. the expression in the left abs value is negative, the one in the right is positive)
y = |x - k| - |x + k| = -(x-k) - (x+k) = -2x
y = -2x, which is positive.

Scenario 4: x is negative--more negative than k is positive (x<-k<0)
(i.e. the expression in both absolute value signs are negative)
y = |x - k| - |x + k| = -(x-k) - (-1)(x+k) = -x+k+x+k
y = 2k, which is positive.

So clearly the maximum value for y is one of the positive results (Scenario 3 or 4).

Statement (1) is INSUFFICIENT, as it tells us that Scenario 3 or 4 applies, but gives no info on the value of 2k or -2x.

Statement (2) is SUFFICIENT, as we already know that Scenario 3 or 4 yield the max y. In Scenario 3, x>-3, so y<6. In Scenario 4, y = 2*3. The maximum y is 6.

[NanoBotZ44]

If k is a positive constant and y = |x - k| - |x + k|, what is the maximum value of y?

(1) x < 0
(2) k = 3

I tried the approach below, but I am not sure if it is fool-proof. Can you please let me know if there is a flaw.

For MAX value of |x - k| - |x + k|,
|x-k| should be MAXIMUM &
|x+k| should be MINIMUM

|x-k| is MAX when x is negative ( Say Condition 1: x<0)
AND
|x+k| is MIN when x+k < 0 ( Say Condition 2: x < - k)
x < - k

Statement 1 satisfies condition 1, but doesn't satisfy condition 2
Statement 2 satisfies both condition 1 & 2

Hence B.

Regards,
Andy

[RonPurewal]

um.

i may be missing something, but, as far as i can tell, no work at all is necessary to show that statement (2) is sufficient.

if you substitute the actual value k = 3, you get "what is the max value of |x - 3| + |x + 3| ?"
i mean, the problem is asking for the maximum value. that's a single function of x, so it will automatically have a single maximum value.
done.
sufficient.

(the only possible objection would be that the function might become infinite -- in which case it wouldn't have a max value at all -- but that's not a thing, since the gmat doesn't include problems with no solution.)

hmmm