DS-1

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DS-1

by bubbliiiiiiii » Mon May 02, 2011 3:52 am
Lines n and p lie on the xy plane. Is the slope of line n less than slope of line p
(1) Lines n and p intersect at (5, 1)
(2) The y-intercept of line n is greater than the y intercept of p

Though this one is discussed earlier .. can someone give a better approach and confirm the answer?

Some say C and some say E.

This one appears to be a GMAT prep 700+ problem.

Experts guidance please.
Regards,

Pranay
Source: — Data Sufficiency |

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by kinji@BTG » Mon May 02, 2011 8:30 am
bubbliiiiiiii wrote:Lines n and p lie on the xy plane. Is the slope of line n less than slope of line p
(1) Lines n and p intersect at (5, 1)
(2) The y-intercept of line n is greater than the y intercept of p

Though this one is discussed earlier .. can someone give a better approach and confirm the answer?

Some say C and some say E.

This one appears to be a GMAT prep 700+ problem.

Experts guidance please.
I am no expert, but I guess the answer should be C.

Both the statements, I can't determine the slope.

Combining the two, as per the diagram,
For the 1st Diagram:
if Y intercept of n > Y intercept of p, then slope of P is more in second quadrant all Tan values are -ve.
For the 2nd Diagram:

if Y intercept of n > Y intercept of p, then slope of P is again is more.
Hence C.

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by Whitney Garner » Mon May 02, 2011 8:35 am
bubbliiiiiiii wrote:Lines n and p lie on the xy plane. Is the slope of line n less than slope of line p
(1) Lines n and p intersect at (5, 1)
(2) The y-intercept of line n is greater than the y intercept of p

Though this one is discussed earlier .. can someone give a better approach and confirm the answer?

Some say C and some say E.

This one appears to be a GMAT prep 700+ problem.

Experts guidance please.
Hi bubbliiiiiiii!

I'm assuming that you understood that each statement alone was insufficient (because you eliminated A, B and D from your question) - BUT I will discuss the overall approach for each statement here as well.

With regard to approach,

Step One: if you read coordinate geometry, immediately consider sketching
Step Two: this is a Yes/No question so our test for insufficiency will need to try to find BOTH a Yes and a No scenario.

Statement (1): Sketch a coordinate plane and put at point at (5,1). Now draw ANY two lines through the point (5,1) and label them N and P. Is the slope of N greater than that of P? Now swap the labels for your line (name the n line with a p now and the p line with an n). The answer just changed - Insufficient!

Statement (2): Same idea, sketch a coordinate plane and put 2 dots on the Y-axis (anywhere you want). Label the higher one N and the lower one P. Now draw 2 lines any way you want. Is the slope of your N line greater than your P line? If you answer Yes, then rotate the P line until the answer is No (and vice versa). We were able to answer BOTH Yes and No - Insufficient.

Statements (Together): Again, the idea is to use our sketch to find AN answer (either yes or no, doesn't matter). And then try to re-sketch the drawing to get the OTHER answer. So, sketch a quick coordinate plane, plot the point (5,1) and then put 2 dots on the Y-axis (anywhere you want) and label the higher one N. Draw your lines - is the slope of N less than P? Now using your 2 fingers for lines, try to keep moving the pair of lines up and down the Y-axis or further apart/closer together on the axis. Can you find a way to get N to have a GREATER slope than P? You're start to see that whatever happens, N will always be at least slightly less than P (or more negative if you have negative slopes).

So, strategy: Geometry (particularly coordinate geo) make a sketch. Yes/No DS problems, work to prove insufficiency - if you cannot, then bets are that it is sufficient.

:)
Whit
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by skspark » Mon May 02, 2011 9:09 am
good qs

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by clock60 » Mon May 02, 2011 9:29 am
hi guys
almost all is said about this problem
but i also see pure numerical approach
n=y=k1*x+b1
p=y=k2*x+b2
we need to estimate k1 vs k2
as all agree neither of the st alone are insuff, so both
1=k1*5+b1
1=k2*5+b2, equate them
k2*5+b2=k1*5+b1
5*k2-5*k1=b1-b2,
5(k2-k1)=b1-b2,
if we provide that b1-b2>0 we prove that k2-k1 also greater then 0, thus k2>k1,
2 st give us that b1>b2, and so k2>k1
C is the answer

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by bubbliiiiiiii » Mon May 02, 2011 9:12 pm
Thanks all and a special thanks to whitney for the beautiful explanation.
Regards,

Pranay