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Extreme Values - Inequalities

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by MBA_Ziggy » Sat Jul 03, 2010 7:12 pm
3. If 2h+k<8, g+3h=15, k=4, what is the possible range of values for g?
[spoiler]
Plugging in 4....
2h+4 <8
2h<4
h<2

LT= Less than
GT = Greater than

g+3h=15
g+3(LT2)=15
g+LT6=15
g=15-LT6 <-----why did this change to GT?...I can't visualize what is going on
g=GT(9)
g>9 [/spoiler]

Thanks.
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Source: — Problem Solving |

by jeremy8 » Sat Jul 03, 2010 8:55 pm
The way I see it, g>9 and there's no upper limit.

If k=4, then 2h+k<8 leads to:

2h+4<8
2h<4
h<2

This leads us to g+3(LT2)=15, where LT stands for "Less Than". (This is the technique taught in the MGMAT books)

This means that g+LT6=15. When you subtract LT6 from both sides, you end up with g=GT9 (Greater Than).
This makes intuitive sense, because if you take away less than 6 from 15, you necessarily end up with more than 9.
Similarly, if you multiply less than 2 by 3, you necessarily end up with less than 6.

g + something less than 6 = 15. The minimum value for g is going to be 9 point something.
There is no upper limit because all we know is that h<2.
h could be -198,479,263 for all we know, in which case g would have to be 198,479,278 to make g+h=15.
Or h could be 5.9999, in which case g would have to be 9.0001
So g>9 is all we know.

I'm really not a math buff by any standards and I just started studying for all that stuff, so this could be wrong, but I think it's probably ok. I'm actually curious to see what other ways people come up with.

Good luck with your studying.
Last edited by jeremy8 on Sat Jul 03, 2010 9:35 pm, edited 1 time in total.
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by kvcpk » Sat Jul 03, 2010 9:22 pm
MBA_Ziggy wrote:3. If 2h+k<8, g+3h=15, k=4, what is the possible range of values for g?
Plugging in 4....
2h+4 <8
2h<4
h<2

LT= Less than
GT = Greater than

g+3h=15
g+3(LT2)=15
g+LT6=15
g=15-LT6 <-----why did this change to GT?...I can't visualize what is going on
g=GT(9)
g>9

Thanks.
Let me explain:
I hope you understood till h<2
so 3h<6
when you multiply by -1 on both sides, the inequality gets reversed.
for example: 3>2. but -3<-2 [not -3>-2]

So -3h>-6
add 15 to both sides
15-3h>15-6
15-3h>9
we know that 15-3h = g
so g>9

Hope this helps!!
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by MBA_Ziggy » Mon Jul 05, 2010 10:11 pm
Thanks everyone, excellent explanations!
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by rsvaishu » Tue Jul 06, 2010 7:13 am
we know K= 4 and 2h+K <8

As for now let we assume 2h+K = 8 for simplification. And it comes h = 2 Therefore h<2 since 2h+k<8

Next step g+3h=15. We have h=2, which gives g = 9

so definitely g<9 range
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by rsvaishu » Tue Jul 06, 2010 7:17 am
rsvaishu wrote:we know K= 4 and 2h+K < 8

As for now let we assume 2h+K = 8 for simplification. And it comes h = 2 Therefore h< 2 since 2h+k<8

Next step g+3h=15. We have h<2, and put h = 2 as for now which gives g = 9

so definitely g >9 range
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by Minh » Sun Nov 15, 2015 8:54 am
hey I came reviewing for the gre and came across the same confusion on inequality and extreme, and i found a thought process that work for me:

if you have
g = 15 -LT6 (i know that less than LT6 can mean it can go to negative infinity)
so i see it as g = 15 - (-inf)
since 15 minus negative infinity is 15 + infinity it will be the GT

g = 15 + GT6
i apply the same thought process here (GT6 can go up to positive infinity)
so g = 15 + positive infinity result in GT
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