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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!

## Add a comment