Using A : magnitude of x is either 10 or 2.5 [insufficient]
using B : magnitude of x can be 4,5 or more [insufficeitn]

Combining:
magnitude of x is 10

IMO C

from 1st we get to know that x<5
from second we know that x>3 and x<-3

when we combine we get to know the range of x i.e -5<x<-3 and 3<x<5
thus NO
imo c

is this correct?

It should be C.

can you please explain how is it c?

HSPA wrote:using B : magnitude of x can be 4,5 or more [insufficeitn]

Just to add to it, x can also be -4, -5, -6 ..

@Arjunshn, HSPA has describe it nice.

Hope it helps.

what abt the first option..is my explanation right?

I am not sure how you concluded x<5 from option 1.

My approach was to square on both sides to obtain a qudratic equation of x.

Then solve for x to get 5/2 and -10.

i took three intervals for the 1st equation x<-5, -5<x<5 and x>5
when u put the values on the number line only first two satisfy thus x<5

arjunshn wrote:from 1st we get to know that x<5
from second we know that x>3 and x<-3

when we combine we get to know the range of x i.e -5<x<-3 and 3<x<5
thus NO
imo c

is this correct?

The two scenario approach is useful for equations using absolute values.
The equation has two solutions, which can be found using the two scenarios:

scenario 1: Copy the equation as is, without the absolute value. Solve.
X -5 = 3(X + 5)
x-5 = 3x+15
-5-15 = 3x-x
-20 = 2x
-10=x

Scenario 2: copy the equation without the absolute value sign, but put a (-) minus sign on one the right side.
X -5 = -3(X + 5)
x-5 = -3x-15
2x = -10
x=-2.5

As a final step for the two scenario approach, check your work by plugging in both values into the original equation |X -5| = 3 |X + 5| to see that the equation is indeed satisfied for both values:

x=-10 |-10 -5| = 3 |-10 + 5|
|-15| = 3|-5|
15 is indeed equal to 3*5.

x=-2.5 |-2.5 -5| = 3 |-2.5 + 5|
|-7.5| = 3|-2.5|
7.5 is indeed equal to 3*2.5.

So both of these values satisfy the equation, and are thus the solutions of x.

Alone, this is insufficient to answer the question: For x^2 to be greater than 5^2, x must be either >5 or smaller than -5.
-10 is smaller than -5, so the answer here is YES.
-2.5 is greater than -5, so the answer is NO.

Stat. (2) is insufficient for the same reasoning: it is possible to find examples that are both within and outside the required ranges.

combined, only x=-10 satisfies Stat. (2)'s requirements, so the answer to the question is a definite YES. Answer is C.

thank you

arjunshn wrote:Is X^2 > 5^2 ?
1) |X -5| = 3 |X + 5|
2) |X| > 3

Taking into account that |a-b| = dist(a,b) for any two real numbers a and b (where dist means the distance in the real number line), please note that:

The question may be rephrased as: |x| > 5, that is, is dist(x, origin) greater than 5?

(1) dist(x,5) = 3*dist(x, -5)

Please note (diagram follows) that there are ONLY TWO points (real numbers) satisfying this statement, one of them certainly answering the question in the affirmative, the other one in the negative.

From sttm (2) we must "take out" one of the (only two) possibilities allowed by sttm (1), therefore we are done.

Regards,
Fabio.

Its a great way to solve it faster..basically a shortcut or basic logic.
thanx alot for sharing it..
thank you

arjunshn wrote:Its a great way to solve it faster..basically a shortcut or basic logic.
thanx alot for sharing it..
thank you

I´m glad you like it, arjunshn!
Cheers, Fabio.