Region R is defined as the region in the first quadrant satisfying the

Question

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

GMATWhizTeam · Answer

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) 
 https://www.dropbox.com/s/5si46zl3zj05ojc/Probability_Region.png?dl=1 
• 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}  
Thus, the correct answer is  Option C.

CrackverbalGMAT · Answer

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

Fdambro294 · Answer

Given 3x + 4y < 12

In Slope Intercept Form: y < -(3/4)x + 3

The Region represented by this inequality is all the Area UNDERNEATH the downward sloping line BUT ONLY in quadrant 1.

This region described above underneath the Line 3x + 4y = 12 ends up forming a Right Triangle with:

Y Axis from (0 , 0) up to (0 , 3) as One Leg

and

X Axis from (0 , 0) up to (4 , 0) as Second Leg

The Region in which the X-Coordinate of Point P (r) is > 2 is defined by the Region to the RIGHT of the Vertical Line X = 2 but INSIDE the Right Triangle.

Since this Vertical Line divides the Whole Right Triangle into 2 Similar Triangles:

Probability = (Area of Smaller Right Triangle) / (Area of ENTIRE Right Triangle)

Corresponding Side of Smaller Triangle / Corresponding Side of ENTIRE Triangle = 2/4 = 1/2

Therefore, because the triangles are Similar Triangles:

(Area of Smaller Triangle) / ( Area of ENTIRE Triangle) = (1)^2 / (2)^2 = 1 /4

-Answer-

Probability = 1/4

-C-

Posted from my mobile device

BusySnipe · Answer

Hi,
Is it possible to get this type of questions in the new gmat (without geometrics) ? 
Thanks !

Bunuel · Answer

Probably not. You can ignore this question.

Region R is defined as the region in the first quadrant satisfying the

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