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Comment on 1/10 and 1/2
We can also solve for 1) as
we can rewrite expression in 1) as
10>4x+2>2, this means, 4x+2 can be 3,4,5,6,7,8 or 9
if we solve above, we get
8>4x>0
divide by 4
2>x>0,
that leave us option: x=1.
Can we do following?
1/x<1/y<1/z then x>y>z, i think so.
Fantastic!
Fantastic!
Your suggested rule (if 1/x <1/y < 1/z then x > y > z) is true AS LONG AS x, y, and z are all positive or all negative.
For example, 1/(-2) < 1/(-3) < 1/2, however we can't say that -2 > -3 > 2
Thank you so much.
I really liked your videos. The way that you solves and explains little facts are really amazing as these little facts can trick in simple questions.
Cheers
Atul
Great video. Question: on
Once we know that 4x+2 equals
Once we know that 4x+2 equals one of the integers from 3 to 9, we can automatically eliminate all of the ODD integers, since 4x will be EVEN and when we add 2, the result is EVEN.
This leaves us with 4, 6 or 8 top test.
Thank you! 3-9 I think you
Brent,
Please help me, I solved the first stated as below.
1/10 < 1/4x+2 < 1/2
0.1 < 1/4x+2 < 0.5
Since there is no integer that lies between 0.1 & 0.5, I declared this statement to be insufficient.
Be careful. We are told that
Be careful. We are told that x is an integer. We are not told that 1/(4x+2) is an integer.
For example, ix x = 1, then 1/(4x+2) = 1/6 = 0.166...
Brent,
Exercise 224 (Problem Solving - OG 2017)
Could you explain me in a different way from OG 2017 answer solution?
Thank you in advance.
Cheers,
Pedro
At the risk of appearing to
At the risk of appearing to "pass the buck," I'd like to direct you to a very rich discussion of this question here: http://www.beatthegmat.com/og-13-ps-218-is-this-prob-really-doable-withi...
In particular, Mitch (GMATGuruNY) provides a concise solution.
Also, Ceilidh (from Manhattan Prep) explains why this might be a great candidate for guessing and moving on.
If you have any questions about those various solutions, I'm happy to respond. I just don't think I'd do much better than Mitch's solution.
Hey Brent, on statement 2,
You only need check for
You only need check for extraneous roots when dealing with equations with square roots and equations with absolute value.
Cheers,
Brent
To avoid testing all the
That works too. Nice job!
That works too. Nice job!
Hi Brent, for statement 1,
Such that 1/10 < 1/4x+2 < 1/2 becomes:
1) 1/10 < 1/4x+2 and 2) 1/4x+2 < 1/2
This gives us 1) x < 2 and 2) x > 0
If we combine 1) and 2) again to bring back us back to to the original equality, we get:
0<x<2. Since x is an integer, x must be 1.
Just want to know if this approach works for all inequality questions with 2 inequality signs.
Thanks.
That's a perfectly valid
That's a perfectly valid solution. Nice work.
Cheers,
Brent
In the video question above,
x could be 0 - couldn't it?
x is, indeed, an integer.
x is, indeed, an integer.
However, x = 0 is not a solution to either statement.
Let's see why:
Statement 1) 1/10 < 1/(4x + 2) < 1/2
Plug in x = 0 to get: 1/10 < 1/(4(0) + 2) < 1/2
Simplify to get: 1/10 < 1/2 < 1/2
Since it is NOT the case that 1/2 < 1/2, x = 0 is not a possible value of x.
Statement 2) x² + 5x - 6 = 0
Plug in x = 0 to get: 0² + 5(0) - 6 = 0
Simplify to get: 0 + 0 - 6 = 0 (doesn't work)
So, 0 is not a possible value of x.
Does that help?
Cheers,
Brent
Hi. I tried to actually solve
Let's see how that approach
Let's see how that approach will work.
Given: 1/10 < 1/(4x+2) < 1/2
Multiply all sides by (4x+2) to get : (4x+2)/10 < 1 < (4x+2)/2
Multiply all sides by 10 to get : (4x+2) < 10 < 5(4x+2)
Expand to get : 4x + 2 < 10 < 20x + 10
We can treat this a two separate inequalities:
4x + 2 < 10 and 10 < 20x + 10
Let's solve each one.
Take: 4x + 2 < 10
Subtract 2 from both sides to get: 4x < 8
Divide both sides by 4 to get: x < 2
Take: 10 < 20x + 10
Subtract 10 from both sides to get: 0 < 20x
Divide both sides by 20 to get: 0 < x
If we combine x < 2 and 0 < x, we get: 0 < x < 2
Since x is an INTEGER, x must equal 1
Does that help?
Cheers,
Brent
Hi Brent! Could you tell me
In respect of statement 1 can we assume 2 situations i.e. when x is positive and when x is negative.
When x is positive the equation becomes 10<4x+2<2 which is not possible and hence can be discarded.
When x is negative the equation becomes 10>4x+2>2 which simplified gives us 2>x>0 and the only integer value possible in this case is x=1.
Is this a valid approach?
That's a good idea, but that
That's a good idea, but that's not quite how the inequality works.
You write: "When x is positive the equation becomes 10 < 4x+2 < 2 which is not possible and hence can be discarded"
If x is positive, then it must be the case that 2 < 4x+2 < 10 (not 10 < 4x+2 < 2)
Since, x is an integer, we can be certain that x = 1.
So, we can't discard that scenario.
---------------------------------
Conversely, the statement "When x is negative the equation becomes 10>4x+2>2 which simplified gives us 2>x>0 and the only integer value possible in this case is x=1" is contradictory.
If x is negative, then we can't then conclude that x = 1 (a positive number)
Does that help?
Cheers,
Brent
Hi Brent, I was able to solve
The main property at play
The main property at play here is as follows:
If a/b and c/d are positive fractions such that a/b < c/d, we can conclude that d/c < b/a.
To understand why, we need to apply some inequality rules (covered here: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...)
Take: a/b < c/d
Multiply both sides by b to get: a < cb/d
Multiply both sides by d to get: ad < cb
Divide both sides by a to get: d < cb/a
Divide both sides by c to get: d/c < b/a
Voila!
Hi Brent. Please help me to
1/10 < 1/(4x+2) < 1/2
I really struggled with x being in the denominator so I tried the following:
10 > 4x +2 > 2.
Then follows, 8 > 4x > 0 , and finally 2 > x > 0. As the only possible integer between these values is 1. x equals 1.
Is that correct or should I stop doing such things?
Cheers.
Alex
That approach is perfectly
That approach is perfectly valid.
In fact, it's pretty much the same as my approach :-)
You concluded that 10 > 4x +2 > 2
I concluded that 4x + 2 = 3, 4, 5, 6, 7, 8, or 9 (since x must be an integer)
Cheers,
Brent
Hi Brent. Looking at
Hi tan27,
Hi tan27,
You're making a very common mistake. When we analyze statement 2, we must be sure to IGNORE information from statement 1.
The following video addressed that mistake: https://www.gmatprepnow.com/module/gmat-data-sufficiency/video/1097
Cheers,
Brent
Hi there,
For working out part 1 I did:
1/10 < 1/4x+2 < 1/2
10 > 4x+2 > 2
8 > 4x > 0
2 > x > 0
Therefore, from reciprocal-ing everything and inverting the signs I was able to determine x = 1
Is that a correct approach?
That approach will work AS
That approach will work AS LONG AS the fractions are ALL positive or ALL negative.
For example, it is true that -1/2 < 1/4 < 1/3.
However, we can't then conclude that -2 > 4 > 3
In the above question, we can see that 1/10 and 1/2 are both positive. Since 1/4x+2 lies BETWEEN 1/10 and 1/2, we can be certain that 1/4x+2 is also positive. So, your strategy works perfectly!
Cheers,
Brent
Hi Brent,
Just want your opinion about my approach for statement 1.
1/10 < 1/(4x+2) < 1/2
What I did is, to power them to -1 and change the direction of the inequality signs giving me
1/1/10 > 1/1/(4x+2) > 1/1/2
I then simplified each fraction giving me 10 > 4x+2 > 2. After that I solved for x. Is it a correct approach?
That approach works perfectly
That approach works perfectly. Nice work!
https://gmatclub.com/forum/is
for this question I factored out statement 2 and solve for statement 2 so I didn't get it right, I have ruled our statement 1 but incorrectly chose C, so my approach isn't quiet right...
Question link: https:/
Question link: https://gmatclub.com/forum/is-x-5-1-x-2-5-2-x-2-x-243300.html
How did you factor statement 2?
THANKS Brent! the only one i
I determined that Statement A is sufficient, ruled our BCE
Then I chose D and correct!