|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
05 May 2020, 03:15
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
52% (02:17) correct
48%
(02:29)
wrong
based on 205
sessions
History
Date
Time
Result
Not Attempted Yet
Region R is defined as the region in the first quadrant satisfying the condition 3x + 4y < 12. Given that a point P with coordinates (r, s) lies within the region R, what is the probability that r > 2?
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
Are You Up For the Challenge: 700 Level Questions
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
Are You Up For the Challenge: 700 Level Questions
05 May 2020, 06:38
Kudos
Bookmarks
Bunuel
Solution
- • If we draw the line \(3x + 4y = 12\) in co-ordinate plane, we can see that, it passes through Points A(0,3) and B(4,0).
- o So, region R is nothing but triangle, AOB (as shown in the below diagram)
• Now, all the points inside region R whose x co-ordinate is greater than 2 must lies inside the shaded region, as shown in the above diagram. (We can find the co-ordinates of point D by substituting \(x = 2\) in the line \(3x + 4y = 12\)).
• We can observe that the area of triangle AOB represents all the points inside the region R and area of triangle CDB represents the points inside region R whose x- co-ordinate is greater than 2.
• Thus, the required probability = area of triangle CDB / area of the triangle AOB \(= \frac{0.5*CB*DC}{0.5*AO*OB} = \frac{0.5*2*1.5}{0.5*3*4} = \frac{1.5 }{ 6} = \frac{1}{4}\)
02 Oct 2020, 04:43
Kudos
Bookmarks
This is a pretty decent question combining Co -ordinate geometry and probability.
We are given that 3x + 4y < 12.
The x intercept (when y = 0) is (4, 0) and y intercept (when x = 0) is (0, 3)
The region R is a right angled triangle (Triangle OAB) who's area is 6 units formed by the Origin (0,0), Point A (0,3) and Point B (4,0)
Now to find point (r,s), where r is > 2. The value of s can be found by putting r = 2
6 + 4y = 12
4y = 6
y = 1.5
Point (r, s) as shown by point C is (2,1.5).
This point along with (2,0) shown as Point D and (4,0) forms another right angled triangle (Triangle CDB)
Area = (1/2) * (4 - 2) * 1.5 = 1.5 units
\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)
Favorable outcomes = Point lies in the area of 1.5 units
Total Outcomes = Area of R = 6 units.
Therefore the required probability = \(\frac{1.5}{6} = \frac{1}{4}\)
Option C
Arun Kumar
We are given that 3x + 4y < 12.
The x intercept (when y = 0) is (4, 0) and y intercept (when x = 0) is (0, 3)
The region R is a right angled triangle (Triangle OAB) who's area is 6 units formed by the Origin (0,0), Point A (0,3) and Point B (4,0)
Now to find point (r,s), where r is > 2. The value of s can be found by putting r = 2
6 + 4y = 12
4y = 6
y = 1.5
Point (r, s) as shown by point C is (2,1.5).
This point along with (2,0) shown as Point D and (4,0) forms another right angled triangle (Triangle CDB)
Area = (1/2) * (4 - 2) * 1.5 = 1.5 units
\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)
Favorable outcomes = Point lies in the area of 1.5 units
Total Outcomes = Area of R = 6 units.
Therefore the required probability = \(\frac{1.5}{6} = \frac{1}{4}\)
Option C
Arun Kumar
Attachments
Probability.jpg [ 569.63 KiB | Viewed 9655 times ]
