Inequalities - PS - Set 31 - Q15

Hi,
You have tested the below two:
for A,B if we take x as -2 it fails the condition
for C ,D if we take x as 4 it fails the condition

However, note that the given boundary condition is: –2 < x < 4
i.e., X is not equal to -2 or +4; hence your test conditions would always fail.
For such problems, I would suggest using the below approach:
If |A| < B is given, then it is equivalent to: -B < A < +B
Now solving the given conditions:
(A) | x – 2 | < 4: -2 < X < 6
(B) | x – 1 | < 3: -2 < X < 4
(C) | x + 1 | < 3: -4 < X < 2
(D) | x + 2 | < 4: -6 < X < 2

Hence answer is B.
Let me know in case of any doubts about this solution.

gmatrant,

PS - Set 31 - Q15
B option is | x – 1 | < 3

when u remove the absolute you can get either a positive value or a negative value.

so the expression B becomes:
| x – 1 | < 3
= (x - 1) < 3
= x < 3 +1
= x < 4 ...... I

simillarly,
| x – 1 | < 3
= - (x - 1) < 3
= - x < 3 -1
= - x < 2
= x > -2 ...... II

joining I and II u get
–2 < x < 4

Guest,

PS - Set 37 Q 14

E Option is | x + 1 | &#8804; 4

| x + 1 | &#8804; 4
= (x + 1) &#8804; = 4
= X &#8804; 3 ......... I

simillarly,

| x + 1 | &#8804; 4
= - (x + 1) &#8804; 4
= -x <= 5
= x >= -5 ...... II

combining I and II u get
-5<x<3

Suyog wrote:gmatrant,

PS - Set 31 - Q15
B option is | x – 1 | < 3

when u remove the absolute you can get either a positive value or a negative value.

so the expression B becomes:
| x – 1 | < 3
= (x - 1) < 3
= x < 3 +1
= x < 4 ...... I

simillarly,
| x – 1 | < 3
= - (x - 1) < 3
= - x < 3 -1
= - x < 2
= x > -2 ...... II

joining I and II u get
–2 < x < 4

Guest,

PS - Set 37 Q 14

E Option is | x + 1 | &#8804; 4

| x + 1 | &#8804; 4
= (x + 1) &#8804; = 4
= X &#8804; 3 ......... I

simillarly,

| x + 1 | &#8804; 4
= - (x + 1) &#8804; 4
= -x <= 5
= x >= -5 ...... II

combining I and II u get
-5<x<3

Inequalities seems to be my weak area, I tried doing a similar problem using your approach but got stuck without an answer.

Find the range of all real values of x if |7-x|<3-5x?
A) -1<X<5/3 B)X>1 C)X>5/3 D)X<-1 E)X>-1
Answer : D

Case 1: x>0
7-x < 3-5x
4x<-4
x<-1
But x has to be greater than 0, so can take this as x value

Case 2: x<0
-7+x<3-5x
4x<10
x<5/3

I did not know how to proceed after this.
Can you tell me how to solve this and also point me to any tutorials on inequalities.

think u solved it rite ..is'nt the answer D

gauravp wrote:think u solved it rite ..is'nt the answer D

Case 2: x<0
-7+x<3-5x
4x<10
x<5/3

Yes D is the answer

But x has to be lesser than 0, but I came to a point where x < 5/3.This is were I got confused because the case 2 condition i.e x<0 does not get fully satisfied for x< 5/3, because x can be 1.12 or 1.13 or any positive value lesser than 5/3


What if one of the answer choices were x<-2?