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Intern Joined: 24 May 2016
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In the figure above, what are the values of X and Y?

A. $$x= \frac{ak}{b}; y = b - k$$

B. $$x = a; y= b$$

C. $$x = \frac{a}{b}; y = b - k$$

D. $$x = \frac{a}{bk}; y=b$$

E. $$x = \frac{ak}{b}; y = b + k$$
[Reveal] Spoiler: OA
Manager Joined: 27 Feb 2017
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Re: In the figure above [#permalink]
Anyone have an answer to this question? and possibly an explanation too please? Retired Moderator Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Re: In the figure above [#permalink]
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Expert's post
These types of questions are prime candidate for elemination methods.

From the figure we can see that the point $$(x,y)$$ lies on the line which starts from (a,0) and ends at (0,b). The y coordinate of the point (x,y) is shown to be k units below point (0,b).

So $$y = k-b$$.

Eleminate all other options apart from A and C.

Now if you cant intuitively figure out that option A is most likely. Just write the equation of the line.

$$y= mx + c$$ where m is the slope and c is the intercept. So points (a,0) and (0, b) must both satisfy the equation.

So substitute (a,0) $$0= m \times a + c$$ and substitute (0,b) $$b= m \times 0 + c$$. Solve for m and c

$$c=b$$ and $$m= \frac{-b}{a}$$.

So equation of line is $$y=\frac{-b}{a} \times x + b$$.

Put value of y as b-k to solve for x $$b-k=\frac{-b}{a} \times x +b$$ so $$x=\frac{ak}{b}$$.

Hence option A.

Note there is an easier form of the equation of line if the line if the points of intersection with x and y axis are known. That form is

$$\frac{x}{a}+ \frac{y}{b}=1$$.
_________________

Sandy
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Try our free Online GRE Test Re: In the figure above   [#permalink] 05 May 2018, 12:24
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