I dont want to confuse others with wrong solutions
GMATPrep DS Obstacles
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Source: Beat The GMAT — Data Sufficiency |
Careful. GMATPrep says the answer is A. I assume this first step is troublesome
"as r2+2 is a positive number so question reduces to - ((r2+2)/p)<r"
Thanks for the effort anyways. I am sure its something simple I am overseeing, just can't pinpoint it.
[quote="kul512"][size=18][b]Question3----[/b][/size]
Question asks whether-(1/p)<(r/(r2+2))
as r2+2 is a positive number so question reduces to -
((r2+2)/p)<r
((r2+2-pr)/p)<0
A. putting p=r
(r2+2-r2)/p<0
(2/p)<0
but as we don't know the sign of p we cant say anything.
NOT SUFFICIENT
B. r>0
only sign of r only cant say anything about the inequality as sign of P is also required.
NOT SUFFICIENT
Now combining the two information-
as r=p and r>0 so p>0 so (2/p)>0 , SUFFICIENT
[spoiler]Answer (C)[/spoiler][/quote]
"as r2+2 is a positive number so question reduces to - ((r2+2)/p)<r"
Thanks for the effort anyways. I am sure its something simple I am overseeing, just can't pinpoint it.
[quote="kul512"][size=18][b]Question3----[/b][/size]
Question asks whether-(1/p)<(r/(r2+2))
as r2+2 is a positive number so question reduces to -
((r2+2)/p)<r
((r2+2-pr)/p)<0
A. putting p=r
(r2+2-r2)/p<0
(2/p)<0
but as we don't know the sign of p we cant say anything.
NOT SUFFICIENT
B. r>0
only sign of r only cant say anything about the inequality as sign of P is also required.
NOT SUFFICIENT
Now combining the two information-
as r=p and r>0 so p>0 so (2/p)>0 , SUFFICIENT
[spoiler]Answer (C)[/spoiler][/quote]
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krusta80
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Question 1
4x + 4y <= 10?
Part (1)
x < 1
Let's plug in 1 for x in both formulas to see what happens:
Given: y <= 5/3
y <= 3/2 ?
Clearly this is TRUE.
Now, plugging in 0 for x in both forumulas to see what happens:
y <= 10/3
y <= 10/4?
Not always TRUE... INSUFFICIENT!
Part (2)
11x <= 10
x <= 10/11
Plugging the min value for x into the given equation...
x = 0 -> y <= 10/3
4x + 4y <= 10?
y <= (5-2x)/2
Plugging in 0 for x again...
y <= 5/2 ? NOT NECESSARILY NOT SUFFICIENT
Combining both doesn't help the case when x = 0, which still allows for either case.
Answer is E.
Given: 5x + 3y <= 10myfish wrote:I have the right answers from GMATPrep but no idea how they got there, even after 10 minPlease take a look at the screenshots and take a crack at it. Thanks a lot!!!!!
4x + 4y <= 10?
Part (1)
x < 1
Let's plug in 1 for x in both formulas to see what happens:
Given: y <= 5/3
y <= 3/2 ?
Clearly this is TRUE.
Now, plugging in 0 for x in both forumulas to see what happens:
y <= 10/3
y <= 10/4?
Not always TRUE... INSUFFICIENT!
Part (2)
11x <= 10
x <= 10/11
Plugging the min value for x into the given equation...
x = 0 -> y <= 10/3
4x + 4y <= 10?
y <= (5-2x)/2
Plugging in 0 for x again...
y <= 5/2 ? NOT NECESSARILY NOT SUFFICIENT
Combining both doesn't help the case when x = 0, which still allows for either case.
Answer is E.
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krusta80
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Question 2
2x - 3y < x^2
Part (1)
2x - 3y = -2
Since x^2 can never be less than 0, we know that the main inequality will always hold true.
SUFFICIENT
Part (2)
x > 2 and y > 0
The y > 0 tells us that we will always be subtracting from the 2x, which will be greater than 4. On the right-hand-side, we have x^2, which will also always be greater than 4. Since even at x = 2 the RHS is greater than the left, we know that it ALWAYS will be greater for all values of x > 2, because the RHS will always rise more quickly.
SUFFICIENT
D
2x - 3y < x^2
Part (1)
2x - 3y = -2
Since x^2 can never be less than 0, we know that the main inequality will always hold true.
SUFFICIENT
Part (2)
x > 2 and y > 0
The y > 0 tells us that we will always be subtracting from the 2x, which will be greater than 4. On the right-hand-side, we have x^2, which will also always be greater than 4. Since even at x = 2 the RHS is greater than the left, we know that it ALWAYS will be greater for all values of x > 2, because the RHS will always rise more quickly.
SUFFICIENT
D
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krusta80
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Question 3
1/p > r/(r^2+2) ?
Part (1)
p = r
If p and r are positive, then we can see than 1/p will always be greater, because r/r^2 equals 1/p but the extra +2 in the denominator will always make the RHS a touch smaller!
If p and r are negative, however, the fractions become negative, which means that the inverse is true: the RHS will be slightly larger.
INSUFFICIENT
Part (2)
r > 0
This is not enough, because it tells us nothing of p.
Combined, parts (1) and (2) are SUFFICIENT, since now we know that r and p are positive, which always makes the main inequality true.
C
1/p > r/(r^2+2) ?
Part (1)
p = r
If p and r are positive, then we can see than 1/p will always be greater, because r/r^2 equals 1/p but the extra +2 in the denominator will always make the RHS a touch smaller!
If p and r are negative, however, the fractions become negative, which means that the inverse is true: the RHS will be slightly larger.
INSUFFICIENT
Part (2)
r > 0
This is not enough, because it tells us nothing of p.
Combined, parts (1) and (2) are SUFFICIENT, since now we know that r and p are positive, which always makes the main inequality true.
C
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ranjeet75
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[quote="krusta80"][size=18]Question 2[/size]
2x - 3y < x^2
Part (1)
2x - 3y = -2
Since x^2 can never be less than 0, we know that the main inequality will always hold true.
SUFFICIENT
Part (2)
x > 2 and y > 0
The y > 0 tells us that we will always be subtracting from the 2x, which will be greater than 4. On the right-hand-side, we have x^2, which will also always be greater than 4. Since even at x = 2 the RHS is greater than the left, we know that it ALWAYS will be greater for all values of x > 2, because the RHS will always rise more quickly.
SUFFICIENT
D[/quote]
Please clarify Part 2 as I could not understand
2x - 3y < x^2
Part (1)
2x - 3y = -2
Since x^2 can never be less than 0, we know that the main inequality will always hold true.
SUFFICIENT
Part (2)
x > 2 and y > 0
The y > 0 tells us that we will always be subtracting from the 2x, which will be greater than 4. On the right-hand-side, we have x^2, which will also always be greater than 4. Since even at x = 2 the RHS is greater than the left, we know that it ALWAYS will be greater for all values of x > 2, because the RHS will always rise more quickly.
SUFFICIENT
D[/quote]
Please clarify Part 2 as I could not understand
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krusta80
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Sure...so, we are are posed with the question: Is 2x - 3y < x^2?ranjeet75 wrote:Please clarify Part 2 as I could not understandkrusta80 wrote:Question 2
2x - 3y < x^2
Part (1)
2x - 3y = -2
Since x^2 can never be less than 0, we know that the main inequality will always hold true.
SUFFICIENT
Part (2)
x > 2 and y > 0
The y > 0 tells us that we will always be subtracting from the 2x, which will be greater than 4. On the right-hand-side, we have x^2, which will also always be greater than 4. Since even at x = 2 the RHS is greater than the left, we know that it ALWAYS will be greater for all values of x > 2, because the RHS will always rise more quickly.
SUFFICIENT
D
In part 2, we are told that x > 2 and y > 0.
Let's pretend for the moment that y = 0, which we know is less than the smallest value we can make y in this part but it helps to make my point.
If y = 0, then the question simplifies to:
2x < x^2?
Without getting into heavy algebra, it is pretty simple to see that 2x will always be less than x^2 if x is greater than 2, since 2x = x^2 at x = 2 and since x^2 grows much more quickly than 2x (imagine it drawn out on a graph...or see below).
So if y = 0, then 3y = 0, which means that we take away NOTHING from the 2x on the left hand side of the original equation and we STILL can say that the equation holds true for x > 2. But what happens when y > 0? Well, then 3y > 0, and we end up lessening the value of the left hand side. So clearly, this will NOT change the outcome of the question. It is still always true when y > 0 and x > 2.
WOW! With real MATLAB graphic! Thanks so much!!!!
[quote="krusta80"][quote="ranjeet75"][quote="krusta80"][size=18]Question 2[/size]
2x - 3y < x^2
Part (1)
2x - 3y = -2
Since x^2 can never be less than 0, we know that the main inequality will always hold true.
SUFFICIENT
Part (2)
x > 2 and y > 0
The y > 0 tells us that we will always be subtracting from the 2x, which will be greater than 4. On the right-hand-side, we have x^2, which will also always be greater than 4. Since even at x = 2 the RHS is greater than the left, we know that it ALWAYS will be greater for all values of x > 2, because the RHS will always rise more quickly.
SUFFICIENT
D[/quote]
Please clarify Part 2 as I could not understand[/quote]
Sure...so, we are are posed with the question: Is 2x - 3y < x^2?
In part 2, we are told that x > 2 and y > 0.
Let's pretend for the moment that y = 0, which we know is less than the smallest value we can make y in this part but it helps to make my point.
If y = 0, then the question simplifies to:
2x < x^2?
Without getting into heavy algebra, it is pretty simple to see that 2x will always be less than x^2 if x is greater than 2, since 2x = x^2 at x = 2 and since x^2 grows much more quickly than 2x (imagine it drawn out on a graph...or see below).
[url=https://postimage.org/image/4g2b9e1vx/][img]https://s14.postimage.org/4g2b9e1vx/graph.jpg[/img][/url]
So if y = 0, then 3y = 0, which means that we take away NOTHING from the 2x on the left hand side of the original equation and we STILL can say that the equation holds true for x > 2. But what happens when y > 0? Well, then 3y > 0, and we end up lessening the value of the left hand side. So clearly, this will NOT change the outcome of the question. It is still always true when y > 0 and x > 2.[/quote]
[quote="krusta80"][quote="ranjeet75"][quote="krusta80"][size=18]Question 2[/size]
2x - 3y < x^2
Part (1)
2x - 3y = -2
Since x^2 can never be less than 0, we know that the main inequality will always hold true.
SUFFICIENT
Part (2)
x > 2 and y > 0
The y > 0 tells us that we will always be subtracting from the 2x, which will be greater than 4. On the right-hand-side, we have x^2, which will also always be greater than 4. Since even at x = 2 the RHS is greater than the left, we know that it ALWAYS will be greater for all values of x > 2, because the RHS will always rise more quickly.
SUFFICIENT
D[/quote]
Please clarify Part 2 as I could not understand[/quote]
Sure...so, we are are posed with the question: Is 2x - 3y < x^2?
In part 2, we are told that x > 2 and y > 0.
Let's pretend for the moment that y = 0, which we know is less than the smallest value we can make y in this part but it helps to make my point.
If y = 0, then the question simplifies to:
2x < x^2?
Without getting into heavy algebra, it is pretty simple to see that 2x will always be less than x^2 if x is greater than 2, since 2x = x^2 at x = 2 and since x^2 grows much more quickly than 2x (imagine it drawn out on a graph...or see below).
[url=https://postimage.org/image/4g2b9e1vx/][img]https://s14.postimage.org/4g2b9e1vx/graph.jpg[/img][/url]
So if y = 0, then 3y = 0, which means that we take away NOTHING from the 2x on the left hand side of the original equation and we STILL can say that the equation holds true for x > 2. But what happens when y > 0? Well, then 3y > 0, and we end up lessening the value of the left hand side. So clearly, this will NOT change the outcome of the question. It is still always true when y > 0 and x > 2.[/quote]
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krusta80
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Not a problem (Excel actually)...the whole point was just to help drive home the point of different rates of growth, etc. After enough exposure, things start to become second nature.myfish wrote:WOW! With real MATLAB graphic! Thanks so much!!!!
krusta80 wrote:Sure...so, we are are posed with the question: Is 2x - 3y < x^2?ranjeet75 wrote:Please clarify Part 2 as I could not understandkrusta80 wrote:Question 2
2x - 3y < x^2
Part (1)
2x - 3y = -2
Since x^2 can never be less than 0, we know that the main inequality will always hold true.
SUFFICIENT
Part (2)
x > 2 and y > 0
The y > 0 tells us that we will always be subtracting from the 2x, which will be greater than 4. On the right-hand-side, we have x^2, which will also always be greater than 4. Since even at x = 2 the RHS is greater than the left, we know that it ALWAYS will be greater for all values of x > 2, because the RHS will always rise more quickly.
SUFFICIENT
D
In part 2, we are told that x > 2 and y > 0.
Let's pretend for the moment that y = 0, which we know is less than the smallest value we can make y in this part but it helps to make my point.
If y = 0, then the question simplifies to:
2x < x^2?
Without getting into heavy algebra, it is pretty simple to see that 2x will always be less than x^2 if x is greater than 2, since 2x = x^2 at x = 2 and since x^2 grows much more quickly than 2x (imagine it drawn out on a graph...or see below).
So if y = 0, then 3y = 0, which means that we take away NOTHING from the 2x on the left hand side of the original equation and we STILL can say that the equation holds true for x > 2. But what happens when y > 0? Well, then 3y > 0, and we end up lessening the value of the left hand side. So clearly, this will NOT change the outcome of the question. It is still always true when y > 0 and x > 2.
