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Inequality problems

by greatchap » Sat Jun 21, 2008 8:30 pm
Hi Guys,

Inequality is a topic that is bothering me. I do know the formulas but still I get stuck in a few ques -

Q-1 If z^2 - 4z > 5 then which of the following is always true

A) z > -5
B) z < 5
C) z > -1
D) z < 1
E) z < -1

my solving method -
(z-5) (z+1) > 0
z-5> 0 or z>5
OR z+1>0 pr z>-1

here answer is z<-1 but why I get otherwise???

Q-2 What is the solution set of inequality 4 / (x-2) < 2 ?

(a) x > 4
(b) x < 4
(c) x < 4 union x > 4
(d) x < 2 union x > 4
(e) infinity

OA is (d)
can anyone explain ??

Q-3 Solve x^2 + 8x + 7 < 0

A-3 I do - (x+7) (x+1) < 0
x+7 < 0 or x < -7
x+1 < 0 or x<-1
so -7<x<-1 Is this OK ?

If that was OK then if an equation is
(x-7) (x-1) < 0 then would ans be 1<x<7 becoz
as per formula (x-a)(x-b) < 0 then a<x<b ..(minus sign is query here)

Will be glad if anyone can help me out here.

Thanks,
GR
Last edited by greatchap on Sun Jun 22, 2008 8:06 pm, edited 1 time in total.
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Re: Inequality problems

by Ian Stewart » Sun Jun 22, 2008 6:23 am
I'll just look at the first of your examples, and hopefully you can then apply the logic to the others. Factoring is helpful here, as you've done, and up to this point:

(z-5)(z+1) > 0

all is fine. How do we analyze this inequality? Well, we're multiplying (z-5) by (z+1) and getting a postive result. If you multiply two things and get a positive answer, both things must have the same sign:

either: z-5 > 0 *and* z+1 > 0
or: z-5 < 0 *and* z+1 < 0

In the first case, z > 5 and z > -1 must be true. For both of these to be true, z must be greater than 5.

In the second case, both z < 5 and z < -1 must be true. For both of these to be true, z must be less than -1.
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Re: Inequality problems

by beeparoo » Sun Jun 22, 2008 10:27 am
greatchap wrote:Q-2 What is the solution set of inequality 4 / (x-2) <2> 4?
I don't understand how you've typed out Q2... Is there something missing in the latter part of the inquality expression?

2 > 4 doesn't make sense....
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Re: Inequality problems

by greatchap » Sun Jun 22, 2008 8:06 pm
beeparoo wrote:
greatchap wrote:Q-2 What is the solution set of inequality 4 / (x-2) <2> 4?
I don't understand how you've typed out Q2... Is there something missing in the latter part of the inquality expression?

2 > 4 doesn't make sense....
Question 2 has been fixed. Thanks for identifying the typing error.
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by g_beatthegmat » Sun Jun 22, 2008 10:31 pm
Question # 2.

4 / (x - 2) Less than 2

=> 2 / (x - 2) less than 1 ... (i)

Now, inequality (i) can be possible in two ways -

1) If the denominator (which is x-2) is greater than the numerator (2) then the left hand side would be less than 1.

Putting the words into an equation -
x -2 > 2
=> x > 4 ... (ii)

2) The other way of left hand side (LHS) to be less than 1 is if the denominator is negative. That ways, LHS would be negative and hence less than 1.

Putting the words into an equation -
x - 2 less than 0
=>x less than 2


Thus, the solution is x less than 2 & x > 4
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by g_beatthegmat » Sun Jun 22, 2008 11:04 pm
For Question # 3 -

Q-3 Solve x^2 + 8x + 7 less than 0

A-3 I do - (x+7) (x+1) less than 0
x+7 less than 0 or x less than -7
x+1 less than 0 or x less than -1
so -7 less than x less than -1 Is this OK ?
If you see above closely, you got x less than -7 and x less than -1, x less than both -1 and -7, so how did you derive -7 less than x less than -1?

Here's what the problem is with the above solution -

We get 0 > (x+7)(x+1)
Or in other words, 0 > A*B
So, for LHS to be less than 0, either A has to be negative or B has to be negative, but both A & B cannot be negative - else that would make LHS positive.

So we get two options -
i) 0>A and B>0
ii) A>0 and 0>B

Solving (i), we get x>-1 and -7>x. On the number line, this is not possible because there's nothing common between x>-1 and -7>x. So this is useless information for us.

Solving (ii), we get -1>x and x>-7. This is helpful. And this should be the answer as well -

i.e. -7 less than x less than -1

Hope it helps.
-gr

ps: I'm not able to use the "less than" sign properly on the posts, it some how dis-aligns the complete post. Does someone know how to fix it?
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Re: Inequality problems

by AleksandrM » Mon Jun 23, 2008 9:31 am
Ian Stewart wrote:both things must have the same sign:

either: z-5 > 0 *and* z+1 > 0
or: z-5 < 0 *and* z+1 <0> 5 and z > -1 must be true. For both of these to be true, z must be greater than 5.

In the second case, both z < 5 and z < -1 must be true. For both of these to be true, z must be less than -1.
Ian,

I don't understand your explanation. I get the part where we subtract 5 from both sides and foil and consider two signs. But I don't understand how we then conclude from all of this that the answer is z<-1.
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Re: Inequality problems

by Ian Stewart » Mon Jun 23, 2008 9:51 am
AleksandrM wrote:
Ian Stewart wrote:both things must have the same sign:

either: z-5 > 0 *and* z+1 > 0
or: z-5 < 0 *and* z+1 <0> 5 and z > -1 must be true. For both of these to be true, z must be greater than 5.

In the second case, both z < 5 and z < -1 must be true. For both of these to be true, z must be less than -1.
Ian,

I don't understand your explanation. I get the part where we subtract 5 from both sides and foil and consider two signs. But I don't understand how we then conclude from all of this that the answer is z<-1.
Editing, because forgot to disable html...

We concluded at the end that either z > 5, or z < -1.

We had two cases- either the two factors are both positive, or the two factors are both negative. In each case we arrive at two conditions. For example, in the second case, for both factors to be negative we learned that z must be less than 5 and z must be less than -1. If both of these must be true, then z must be less than -1. This is what must be true for both factors to be negative. If both factors will be positive, then we concluded that z must be larger than 5, by similar reasoning. In each case, (z-5)(z+1) will be greater than 0.
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by aj5105 » Wed Apr 29, 2009 1:33 am
Ian,

Below is an excerpt from a post.

Some fundamental rules of Inequality

Case I

(x-a)(x-b)>0

where a>b

so,

x>a and x<b

Case II

(x-a)(x-b) < 0

so b<x< a

provided a>b


Question 1: Are the above rules right?

Question 2: If I apply the above fundamentals, I get Z>5 AND Z<-1.
Z>5 is not in the answer choices. Z<-1 is there. I picked E. Am i right here?
Last edited by aj5105 on Sun May 03, 2009 11:22 am, edited 1 time in total.
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by maihuna » Wed Apr 29, 2009 9:52 am
Q-2 What is the solution set of inequality 4 / (x-2) < 2 ?

(a) x > 4
(b) x < 4
(c) x < 4 union x > 4
(d) x < 2 union x > 4
(e) infinity

(x-2) /4 > 1/2
x-2 > 2
x >4

Choose A

Q-3 Solve x^2 + 8x + 7 < 0

(x+1)(x+7) < 0

they will of opposite sign:
x>-1 or x<-7
x<-1 or x > -7
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Re: Inequality problems

by kanha81 » Thu Apr 30, 2009 9:27 am
greatchap wrote:Hi Guys,

Inequality is a topic that is bothering me. I do know the formulas but still I get stuck in a few ques -

Q-1 If z^2 - 4z > 5 then which of the following is always true

A) z > -5
B) z < 5
C) z > -1
D) z < 1
E) z < -1

my solving method -
(z-5) (z+1) > 0
z-5> 0 or z>5
OR z+1>0 pr z>-1

here answer is z<-1 but why I get otherwise???
Hey greatchap,
your approach is sound; only tweaking that is required is plugging in the values back into the equation to make sure the equation- z^2-4z-5 > 0 proves right. Here's what I mean:

case-i: z>5 & z>-1
z=0: 0-0-5 = -5 > 0 ? No. Discard this values

case-ii: z<5 & z<-1
z=4: 16-16-5 = -5 > 0? No.
z=-2: 4+8-5 = 7 > 0? Yes. Keep it

Hence [spoiler][E][/spoiler]
greatchap wrote: Q-2 What is the solution set of inequality 4 / (x-2) < 2 ?

(a) x > 4
(b) x < 4
(c) x < 4 union x > 4
(d) x < 2 union x > 4
(e) infinity

OA is (d)
can anyone explain ??
4 < 2(x-2)
4 < 2x-4
8 < 2x
4 < x => x > 4 ...(i)

This is one part of the solution.

4 / (x-2) < 2
We know that x <> 2, otherwise undefined.
Hence x < 2 ---> The statement above remains valid for all values....(ii)

Hence [spoiler][D][/spoiler]
greatchap wrote: Q-3 Solve x^2 + 8x + 7 < 0

A-3 I do - (x+7) (x+1) < 0
x+7 < 0 or x < -7
x+1 < 0 or x<-1
so -7<x<-1 Is this OK ?

If that was OK then if an equation is
(x-7) (x-1) < 0 then would ans be 1<x<7 becoz
as per formula (x-a)(x-b) < 0 then a<x<b ..(minus sign is query here)

Will be glad if anyone can help me out here.

Thanks,
GR
x^2 + 8x + 7 < 0
(x+7)(x+1) < 0

case-i: x+7 < 0 & x+1 > 0 (opposite signs result in -ve)
x < -7 & x > -1
So, -7 > x > -1

case-ii: x+7 > 0 & x+1 < 0
x > -7 & x < -1
So, -7 < x < -1

Hope it's all clear now!
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Re: Inequality problems

by Uri » Thu Apr 30, 2009 10:48 am
kanha81 wrote: case-i: z>5 & z>-1
z=0: 0-0-5 = -5 > 0 ? No. Discard this values

case-ii: z<5 & z<-1
z=4: 16-16-5 = -5 > 0? No.
z=-2: 4+8-5 = 7 > 0? Yes. Keep it

Hence [spoiler][E][/spoiler]
kanha81, imho, two values for z must satisfy Q-1: z>5 and z<-1. you have tested case-i with z=0 and since it did not make sense, you discarded it. what would you have done if you had tested with z=6? that would give 7>0, which is obviously true.

i feel that since only z<-1 is given in the answer choices, we have to pick that one. and gmat would never give two right answer choices in a question, unless its one of those numbered (I,II,III) questions! please rectify me if i am wrong.
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Re: Inequality problems

by kanha81 » Thu Apr 30, 2009 11:38 am
Uri wrote:
kanha81 wrote: case-i: z>5 & z>-1
z=0: 0-0-5 = -5 > 0 ? No. Discard this values

case-ii: z<5 & z<-1
z=4: 16-16-5 = -5 > 0? No.
z=-2: 4+8-5 = 7 > 0? Yes. Keep it

Hence [spoiler][E][/spoiler]
kanha81, imho, two values for z must satisfy Q-1: z>5 and z<-1. you have tested case-i with z=0 and since it did not make sense, you discarded it. what would you have done if you had tested with z=6? that would give 7>0, which is obviously true.

i feel that since only z<-1 is given in the answer choices, we have to pick that one. and gmat would never give two right answer choices in a question, unless its one of those numbered (I,II,III) questions! please rectify me if i am wrong.
Uri,
I understand what you're saying, BUT the reason for not picking z = 6 or any other value was clearly evident from the answer choices. If you take a look, you'll see that option z > 5 is not present in the answer choice, so there is no point in testing values z>5. Hence I chose to test only with z=0 from case-(i).
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by Ian Stewart » Fri May 01, 2009 12:10 pm
aj5105 wrote:Ian,

Below is an excerpt from a post.

Some fundamental rules of Inequality
Case I

(x-a)(x-b)>0

where a>b

so,

x>a and x<b

Case II

(x-a)(x-b) < 0

so b<x< a
provided a>b


Question 1: Are the above rules right?
Not sure where you got these rules from, but the phrasing in the first case is a bit odd. The conclusion in Case I should read "either x > a OR x < b." Using 'and' here doesn't make any sense, since x can't be both greater than a and less than b. Besides that one issue with the wording, the conclusion is correct (in both cases), provided you keep in mind the condition that a > b.
aj5105 wrote: Question 2: If I apply the above fundamentals, I get Z>5 AND Z<-1.
Z>5 is not in the answer choices. Z<-1 is there. I picked E. Am i right here?
Yes, you've reached the correct conclusion. There is (as I probably should have been more clear about above) clearly a problem with the answer choices to Q1 in greatchap's original post - none of the answer choices 'is always true'. All you can conclude is that either x > 5 or x < -1.
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