GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 May 2019, 04:47

### 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

# In the figure above, V represents an observation point at one end of

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55265
In the figure above, V represents an observation point at one end of  [#permalink]

### Show Tags

01 Sep 2017, 07:41
00:00

Difficulty:

25% (medium)

Question Stats:

92% (01:45) correct 8% (01:22) wrong based on 19 sessions

### HideShow timer Statistics

In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object.

(A) 10 – 5√3
(B) 10 – 4√3
(C) 5
(D) 10 – 5√2
(E) 6

Attachment:

2017-09-01_1836_002.png [ 4.89 KiB | Viewed 698 times ]

_________________
Math Expert
Joined: 02 Aug 2009
Posts: 7684
Re: In the figure above, V represents an observation point at one end of  [#permalink]

### Show Tags

01 Sep 2017, 09:42
Bunuel wrote:

In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object.

(A) 10 – 5√3
(B) 10 – 4√3
(C) 5
(D) 10 – 5√2
(E) 6

Attachment:
2017-09-01_1836_002.png

Let the point of right angle be A..
So VRA is a right angle triangle where VR = hypotenuse=10
Perpendicular=VA=5, so $$AR^2=10^2-5^2=100-25=75....AR=5√3$$...

But AS=10, so RS=AS-AR=10-5√3

A
_________________
Senior SC Moderator
Joined: 22 May 2016
Posts: 2755
In the figure above, V represents an observation point at one end of  [#permalink]

### Show Tags

01 Sep 2017, 10:13
Bunuel wrote:

In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object.

(A) 10 – 5√3
(B) 10 – 4√3
(C) 5
(D) 10 – 5√2
(E) 6

Attachment:
2017-09-01_1836_002.png

To find perceived distance, use right triangle properties. VR is the hypotenuse of a right triangle.

Let the right angle in the diagram be X.

XR = perceived distance, from R to X, of object

For ∆ XRV, find length of leg XR using Pythagorean theorem

Length of hypotenuse VR, given, = 10
Length of leg XV, given, = 5

$$5^2 + (XR)^2 = 10^2$$
$$(XR)^2 = 75$$

$$XR$$ = $$\sqrt{75}$$, so

$$XR$$ = 5$$\sqrt{3}$$

Length of XS = 10 (actual distance of object from S to A)

Length of XR = 5$$\sqrt{3}$$ (perceived distance of object from R to A)

Difference between actual and perceived is XS - XR =

10 - 5$$\sqrt{3}$$

_________________
Listen, are you breathing just a little, and calling it a life?
-- Mary Oliver

For practice SC questions go to SC Butler, here.

In the figure above, V represents an observation point at one end of   [#permalink] 01 Sep 2017, 10:13
Display posts from previous: Sort by