Question: 1/10 and 1/2

Comment on 1/10 and 1/2

We can also solve for 1) as

We can also solve for 1) as follow, instead of using all values and make calculation
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.

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

Great video. Question: on test day, would we have to go through each of the integers from 3-9 to figure something like this out? Or is there a way we can identify a shortcut when we see something like this? If not, I guess just knowing this question type beforehand will save us time so we can get right to it. 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!

Brent,

Brent,

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,

Brent,

Exercise 224 (Problem Solving - OG 2017)
Could you explain me in a different way from OG 2017 answer solution?

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,

Hey Brent, on statement 2, when I got (x+6)(x-1), I concluded that x must be -6 or 1, and then made the mistake of checking for extraneous roots, which took extra time. How do you know when to check for extraneous roots? 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

To avoid testing all the values for statement 1, I just converted it into an inequality 10>4x+2>2 , then solve it to get 2>x>0, given x is integer, answer must be 1. That works too. Nice job!

That works too. Nice job!

Hi Brent, for statement 1,

Hi Brent, for statement 1, would it be okay to split the inequality into two separate ones.

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,

In the video question above, why doesn't 0 count as an integer?
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

Hi. I tried to actually solve for x. But I wanted to multiply all sides by the reciprocal in the middle. But when I take 1/(4x+2) *(4x +2)/1 then the whole thing turns into 1 and I no longer have an x at all. What is the problem with this method? 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

Hi Brent! Could you tell me if the following approach is valid?

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

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