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29 Dec 2021, 05:30
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:41) correct
65%
(02:59)
wrong
based on 43
sessions
History
Date
Time
Result
Not Attempted Yet
In the x-y plane, region S is defined by x ≤ 0, y ≥ 0 and 3y -5x ≤ 15. Does the point (a, b) lie in the defined region S, given a and b are integers?
(1) a*b = -4
(2) b – a = 5
(1) a*b = -4
(2) b – a = 5
29 Dec 2021, 14:00
Kudos
Bookmarks
Bunuel
Breaking Down the Info:
A key detail given is a point in region S must lie in the 2nd quadrant. We may use that to eliminate answers first. The line given can be expressed as \(y ≤ \frac{5}{3}x + 5\). This line goes through Quadrant 2, and thus region S is triangular.
Statement 1 Alone:
\(a*b = -4\) has two curves in the 2nd and 4th quadrant. It is possible that the curve in the 2nd quadrant does not intersect with region S at all. To check if the curve intersects \(3y -5x ≤ 15\) or not, we do the following:
Plug in y = b = \(-\frac{4}{a}\) and \(x = a\) to get \(-\frac{12}{a} - 5a ≤ 15\) and \(5a^2 + 15a + 12 ≤ 0 \) (recall \(a\) has to be negative if we want a point in the region, so we pretend \(a\) is negative). By finding the discriminant, we can see \(b^2 - 4ac = 225 - 240 < 0\). This tells us \(5a^2 + 15a + 12\) must be positive, and the curve \(a*b = -4\) is too far away from \(3y -5x ≤ 15\). Thus there is no intersection, and no point in \(a*b = -4\) is within the region. Then statement 1 alone is sufficient.
Statement 2 Alone:
We can rewrite this as \(b = a + 5\). The left portion of this is above the line \(y = \frac{5}{3}x + 5\), thus none of \(b = a + 5\) lies in the triangular region except (0, 5). Then this is still insufficient, but there is only one possible point that would lie within the region.
Answer: A
05 Jan 2022, 09:38
Kudos
Bookmarks
Bunuel
Decoding the stem:
Attachment:
graph 3.9.png [ 37.04 KiB | Viewed 1665 times ]
Above is the defined region:
\(x \leq 0 \)
\(y \geq 0 \)
\(3y \leq 15 +5x\)
max value of \(x =0\) as it is given \(x \leq 0\)
Hence max value of \(y \leq 5\)
when \(x=0\) then \(y=5\)
Least value of \(x=-3\) as lesser than that \(y < 0 \) but we need \(y \geq 0\)
When \(x=-3\) then \(y =0 \)
Hence we need to know whether point (a,b) lies in the given green region.
\(-3 \leq x \leq 0 \)
\(\hspace{4mm} \)\(0 \leq y \leq 5\)
(1) a*b = -4
if \(a =1\) and \(b = -4 \) then outside the region
if \(a =-1 \)and \(b=4 \) then inside the region
INSUFF.
(2) b – a = 5
if \(b=6 \) and \(a=1 \) outside the region
if \(b=3 \) and \(a=-2 \) then inside the region
INSUFF.
1+2
\(a*b = -4\) (i)
\(b – a = 5\) (ii)
\(b + \frac{4}{b}=5 \) \(\hspace{2mm}\) :putting value of a from (i) in (ii)
\(b^2-5b+4=0\)
\((b-4)(b-1)=0 \)
\(b=4 \) or \(b=1 \)
if \(b=4\) then \(a=-1\) inside the green region
if \(b=1\) then \(a=-4\) outside the green region
INSUFF.
IMO E
