BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Hard 800 problem

Expert replies
by lime777 » Mon Feb 04, 2013 10:47 pm
Is there a positive integer x such that the fraction (8x+3)/(6x+2) is not in lowest terms? Why or why not?

I will post answer keys later. This is out of the intense prep. course, that's why it's in a bit different format, but wanted to see if anybody can crack it.
Join the discussion
Source: — Problem Solving |

by Tommy Wallach » Tue Feb 05, 2013 12:36 pm
Hey Lime,

Questions like this are a little dangerous on the GMAT, because they encourage you to worry about things that, if you worried about them on the real test, would be an enormous waste of time. If you were doing this on the GMAT, you'd simply try the first few x values, and extrapolate out:

x = 1 --> 11/8; x = 2 --> 19/14; x = 3 --> 27/20; x = 4 --> 35/26; x = 5 --> 43/32

You can immediately see that nothing's ever going to reduce. If you wanted to know the reason, it's nothing particularly interesting, merely that the primes don't ever line up. The numerator is always odd, and the denominator even, so it'll never reduce by any even number. And 6x + 2 can never be a multiple of 3, because 6x is a multiple of 3 and you're adding 2 to it.

Is there something more interesting that one should extrapolate from this? If so, I don't know what it is! : )

-t
Tommy Wallach, Company Expert
ManhattanGMAT

If you found this posting mega-helpful, feel free to thank and/or follow me!
Join the discussion

by lime777 » Tue Feb 05, 2013 8:51 pm
Thanks!

It's a prep problem, not a real one, but your answer is correct!
Join the discussion

by Anju@Gurome » Tue Mar 19, 2013 4:56 am
lime777 wrote:Is there a positive integer x such that the fraction (8x+3)/(6x+2) is not in lowest terms? Why or why not?
Let us assume that the fraction is not in lowest term, i.e. (8x + 3) is a multiple of (6x + 2) or (8x + 3) = k*(6x + 2), where k is some positive integer definitely greater than 1.

So, (8x + 3) = (6kx + 2k)
--> x = (2k - 3)/(8 - 6k)

As k > 1, (2k - 3) > 0 but (8 - 6k) < 0
This indicates x cannot be positive --> A contradiction

Hence, our initial assumption is definitely not true, i.e. the fraction is in fact in lowest term.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
Join the discussion