Problem Solving

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

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....

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.

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

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?

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.

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?

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

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!

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 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).