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Re: In the x-y plane, there is line K, (x/a)+(y/b)=1. What is the y-inter
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24 Jan 2017, 09:45

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Top Contributor

MathRevolution wrote:

In the x-y plane, there is line K, (x/a) + (y/b) = 1. What is the y-intercept of line K?

1) a = b 2) b = 5

I like this question a lot!

Target question:What is the y-intercept of line K? Given: The equation of line K is (x/a) + (y/b) = 1

This is a great candidate for rephrasing the target question. If we can rewrite the equation in slope y-intercept form, then this will help us identify the y-intercept of the line.

Slope y-intercept form looks like this y = mx + q, where m is the slope of the line, and q is the y-intercept. So, let's take the equation for line L and rewrite it in slope y-intercept form

Given: (x/a) + (y/b) = 1 Eliminate the fractions by multiplying both sides by ab to get: bx + ay = ab Subtract bx from both sides; ay = -bx + ab Divide both sides by a to get: y = (-b/a)x + b So, the slope of like K will by -b/a, and the y-intercept of line K will be b. Since we're trying to determine the y-intercept, we can rephrase the target question... REPHRASED target question:What is the value of b?

Now onto the statements....

Statement 1: a = b This is not enough information to determine the EXACT value of b. Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: b = 5 PERFECT! This is exactly the information we are looking for! (we now know that 5 is the y-intercept of like K) Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Re: In the x-y plane, there is line K, (x/a)+(y/b)=1. What is the y-inter
[#permalink]

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26 Jan 2017, 00:55

1

==> If you modify the original condition and the question, the y-intercept is the value of y when x=0, so if you substitute x=0, from y/b=1, you get y=b, so you only need to know b. According to con 2), it is unique and sufficient.

Therefore, the answer is B. Answer: B
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Re: In the x-y plane, there is line K, (x/a)+(y/b)=1. What is the y-inter
[#permalink]

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25 Jun 2018, 10:51

Hello from the GMAT Club BumpBot!

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