Data Sufficiency

Mo2men wrote:If k does not equal -1, 0 or 1, does the point of intersection of line y = kx+b and line x = ky+b have a negative x-coordinate?

1. kb > 0
2. k > 1

Can we solve this question be drawing lines???

Mitch has beautifully explained the graphical solution, here is the Algebraic one.

We wish to know the nature of x-coordinate.

We have two intersecting line: y = kx+b and x = ky+b;

By plugging in the value of y from the first line into the second, we get,

x = k(kx+b)+b

=> x=k^2x+bk+b

=> x(1-k^2) = b(k+1)

=> x(1+k)(1-k)=b(k+1)

=> x=b/(1-k)

Signs of b and k will determine whether x is negative/positive.

S1: kb > 0

kb > 0 implies that kb is positive. Thus, either both are positive or both are negative.

Case 1: b and k are positive.

x=b/(1-k) may or may not be negative.

(a) If 0 < k < 1, x = b / |(1-k)| => x is not negative.

(b) If k > 1, x = b / (-|(1-k)|) => x is negative. No unique answer.

There is no need to discuss case 2 (b and k are negative.) as we already concluded that S1 itself in not sufficient.

S2: k > 1

There is no information about the sign of b. Insufficient.

S1 and S2:

From S1, we know that b and k are of the same sign and from S2, we know that k is positive thus, b and k each is positive.

We saw in the Case 1 analysis that even if b and k each is positive, there is no unique answer. However, if k > 1, we have do not have Case 1.a. Thus, x is negative. Sufficient.

The correct answer: C

Relevant book: Manhattan Review GMAT Coordinate Geometry Guide

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: New York | Frankfurt | Hong Kong | Zurich | and many more...

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Mo2men wrote:If k does not equal -1, 0 or 1, does the point of intersection of line y = kx+b and line x = ky+b have a negative x-coordinate?

1. kb > 0
2. k > 1

Can we solve this question be drawing lines???

Mitch has beautifully explained the graphical solution, here is the Algebraic one.

We wish to know the nature of x-coordinate.

We have two intersecting line: y = kx+b and x = ky+b;

By plugging in the value of y from the first line into the second, we get,

x = k(kx+b)+b

=> x=k^2x+bk+b

=> x(1-k^2) = b(k+1)

=> x(1+k)(1-k)=b(k+1)

=> x=b/(1-k)

Signs of b and k will determine whether x is negative/positive.

S1: kb > 0

kb > 0 implies that kb is positive. Thus, either both are positive or both are negative.

Case 1: b and k are positive.

x=b/(1-k) may or may not be negative.

(a) If 0 < k < 1, x = b / |(1-k)| => x is not negative.

(b) If k > 1, x = b / (-|(1-k)|) => x is negative. No unique answer.

There is no need to discuss case 2 (b and k are negative.) as we already concluded that S1 itself in not sufficient.

S2: k > 1

There is no information about the sign of b. Insufficient.

S1 and S2:

From S1, we know that b and k are of the same sign and from S2, we know that k is positive thus, b and k each is positive.

We saw in the Case 1 analysis that even if b and k each is positive, there is no unique answer. However, if k > 1, we have do not have Case 1.a. Thus, x is negative. Sufficient.

The correct answer: C

Relevant book: Manhattan Review GMAT Coordinate Geometry Guide

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: New York | Frankfurt | Hong Kong | Zurich | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

Mo2men wrote:

Jay@ManhattanReview wrote:
Case 1: b and k are positive.

x=b/(1-k) may or may not be negative.

(a) If 0 < k < 1, x = b / |(1-k)| => x is not negative.

(b) If k > 1, x = b / (-|(1-k)|) => x is negative. No unique answer.

Thanks Jay.

Why did you add modulus sign for (1-k). Is absolute value?

Hi Mo2men,

A modulus is a great tool to demonstrate the outcome of a sign of an expression.

Let us bring the above.

Case 1: b and k are positive.

x=b/(1-k) may or may not be negative.

(a) If 0 < k < 1, x = b / |(1-k)| => x is not negative.

Putting (1-k) as |(1-k)| ensures that the dominator is positive. Since (1-k) is not positive for all k > 0, we have to express unambiguously. x = Positive / Positive. I did not write b as |b| since there is no ambiguity with b. It's a positive quantity. Though there is no harm in writing b as |b|.

(b) If k > 1, x = b / (-|(1-k)|) => x is negative. No unique answer.

Same here. Since for k > 1, (1-k) is negative, I fixed the negative sign with (1-k). So, within -|(1-k)| whatever is happening is already taken care. Thus, x = Positive / Negative = Negative.

Hope this is clear.

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
_________________
Manhattan Review GMAT Prep

Locations: New York | Frankfurt | Hong Kong | Zurich | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.