It is currently 23 Mar 2018, 11:56

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# (PLEASE SOLVE THIS PROBLEM, ANYONE!) PS Problem #5. Thx!!

Author Message
Intern
Joined: 23 Jun 2009
Posts: 47

### Show Tags

26 Jul 2009, 20:01
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

In the figure above, P and O lie on the circle with center O. What is the value of s?

A) 1/2
B) 1
C) sqrt(2)
D) sqrt(3)
E) sqrt(2)/2

It comes from GMATPrep. Thanks!

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

Attachments

Problem 5.jpg [ 9.04 KiB | Viewed 1112 times ]

Manager
Joined: 10 Jul 2009
Posts: 150

### Show Tags

26 Jul 2009, 20:14
This problem was solved earlier. The link is what-is-the-value-of-t-81370.html
Intern
Joined: 22 Mar 2008
Posts: 49

### Show Tags

29 Jul 2009, 19:56
consider the centre of the circle is at O(0,0).

Since OP and OQ are the radius of the circle
therefore OP=OQ
or, (\sqrt{3})^2+ (1)2 = s^2 + t^2
therefore s^2+t^2 = 4 ...................(1)

again , since POQ is 90 degrees. OP and OQ are perpendicular. hence product of their slopes =-1.
slope of Q = t/s
slope of P = -1/\sqrt{3}
therefore, (t/s)* [-1/\sqrt{3} ] = -1
so, t/s = \sqrt{3}
or, t^2 = 3s^2 ....................(2)

solving (1) & (2) we get s = +/- (1). But , since Q is in first quadrant, s = 1 , hence t = \sqrt{3}
SVP
Joined: 30 Apr 2008
Posts: 1850
Location: Oklahoma City
Schools: Hard Knocks

### Show Tags

29 Jul 2009, 20:04
I'm not sure how the other forum solved it, but here is another way to approach the problem.

If you look at the slope of a line and realize that because we're told the two lines are 90 degrees from each other, you know the lines are perpendicular.

so what ist he slope of a line? $$y = \frac{1}{3}x + 0$$ Sure we don't need the + 0, but it helps to see all the parts there.

Now I used 1/3, but that's just a number to show the formula for a line.

If the coordinates we do know were used to determine the slope of the line, what would the equation look like?

$$y = -\frac{1}{\sqrt{3}}x + 0$$

It's a negative because the slope runs from top left to bottom right and those lines are always negative. So if we know the slope, then how would we figure what the slope is of the perpendicular line, with point {s,t}?

Invert and negate the slope. We can use the same exact numbrs because we know the line is the radius of a circle and therefore the same length. By using the same numbers, we know that it will be the same length and be the correct values for {s,t}

so inverted the slope becomes $$\frac{\sqrt{3}}{1}x$$

Now if before we had $$sqrt{3}$$ on bottom, and it was the x value, and now it is on top, it will not be the x value, but the y value, which corresponds to $$t$$.

Hope this helps present a different way of looking at it. Answering the question won't take nearly as long as exlaining it.

uzonwagba wrote:
In the figure above, P and O lie on the circle with center O. What is the value of s?

A) 1/2
B) 1
C) sqrt(2)
D) sqrt(3)
E) sqrt(2)/2

It comes from GMATPrep. Thanks!

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Intern Joined: 15 Oct 2008 Posts: 5 Re: (PLEASE SOLVE THIS PROBLEM, ANYONE!) PS Problem #5. Thx!! [#permalink] ### Show Tags 31 Jul 2009, 11:06 all you need to determine is that a 30-60-90 triangle can be fitted with OQ as hypotenuse SVP Joined: 30 Apr 2008 Posts: 1850 Location: Oklahoma City Schools: Hard Knocks Re: (PLEASE SOLVE THIS PROBLEM, ANYONE!) PS Problem #5. Thx!! [#permalink] ### Show Tags 31 Jul 2009, 11:21 what part of the picture provided lets you determine the angle measurements? I realize you can see that the readius is 2 from the numbers provided in the first point, but you still have to address how the 90 degree angle between the lines figures intot he problem. travern wrote: all you need to determine is that a 30-60-90 triangle can be fitted with OQ as hypotenuse --== Message from GMAT Club Team ==-- This is not a quality discussion. It has been retired. If you would like to discuss this question please re-post it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Re: (PLEASE SOLVE THIS PROBLEM, ANYONE!) PS Problem #5. Thx!!   [#permalink] 31 Jul 2009, 11:21
Display posts from previous: Sort by