let r = x and t = y for simplicity
q: whether 2x + 3y <=6 that is represented by area under the blue line.
ST1. its represented by ALL POINTS ONLY ON THE RED LINE.
Since some part of red line lies within the blue curve and some lies outside of it, not sufficient
St1: orange shaded area.
Again as some area lies within the blue line region and others dont, not sufficient.
Taking 1 and 2 together: Blue portion on the line satifies St1 and St2 and lies within the required region but BLACK portion of the line satifies both ST1 and St2 but DOES nOT LIE within the required region
Hence E
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OG Data sufficiency prob no.129
Source: Beat The GMAT — Data Sufficiency |
While solving coordinate geometry problems, specifically DS problems, try to draw the scenario and solve by visualizing possible situations. In this case, the following diagram will help us to solve the problem:

Statement 1: (r, s) may or may not lie in region R.
Not sufficient
Statement 2: If we take r = 3 and s = 2, then the point (3, 2) does not lie in region R.
r ≤ 3 and s ≤ 2 implies we can also take negative values for r and s. If r = -2, s = -3, then (-2, -3) lies in region R.
Not sufficient
1 & 2 Together: If r = 2, s = 0 then (2, 0) lies in region R. But if r = 2/3 and s = 2 then (2/3, 2) lies above the line 2x + 3y = 6, which means (2/3, 2) does not lie in region R. Combining also doesn't give a unique answer.
The correct answer is E.

Statement 1: (r, s) may or may not lie in region R.
Not sufficient
Statement 2: If we take r = 3 and s = 2, then the point (3, 2) does not lie in region R.
r ≤ 3 and s ≤ 2 implies we can also take negative values for r and s. If r = -2, s = -3, then (-2, -3) lies in region R.
Not sufficient
1 & 2 Together: If r = 2, s = 0 then (2, 0) lies in region R. But if r = 2/3 and s = 2 then (2/3, 2) lies above the line 2x + 3y = 6, which means (2/3, 2) does not lie in region R. Combining also doesn't give a unique answer.
The correct answer is E.
Anju Agarwal
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Region R is composed of all the points on or below y=(-2/3)x + 2.In the XY plane, region R consists of all the points (x,y) such that 2x+3y≤6. Is the point (r,s) in region R?
1. 3r+2s=6
2. r≤3 and s≤2
Statement 1: s = (-3/2)r + 3.

The figure above shows that some points on s=(-3/2)r + 3 lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.
Statement 2: r≤3 and s≤2.

Inside the green box are points such that r≤3 and s≤2.
Some of the points inside the green box lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.
Statements 1 and 2 combined:

Inside the green box are points on s=(-3/2)r + 3 such that r≤3 and s≤2.
Some of these points lie BELOW y=(-2/3)x + 2, while others lie ABOVE y=(-2/3)x + 2.
INSUFFICIENT.
The correct answer is E.
An alternate way to combine the two statements is to treat this as MAX/MIN problem.
R MAXIMIZED:
In statement 2, the maximum possible value of r is 3.
If r=3 and s=0, both statements 1 and 2 are satisfied.
Check whether (3,0) is within the region defined by y ≤ (-2/3)x + 2:
0 = (-2/3)(3) + 2
0 ≤ 0.
YES.
S MAXIMIZED:
In statement 2, the maximum possible value of s is 2.
If s=2 and r=(2/3), both statements 1 and 2 are satisfied.
Check whether (2/3, 2) is within the region defined by y ≤ (-2/3)x + 2:
2 ≤ (-2/3)(2/3) + 2
2 ≤ 2/3
NO.
Since in the first case (r,s) is within the required region, but in the second case (r,s) is not within the required region, the two statements combined are INSUFFICIENT.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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