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I solved this by drawing a number a line and then getting the minimum and maximum distance from both points A and B for each option of X.

I) 0 < x < 2 - (Min,Max) distances from A: (5,7) and B:(5,7)
II) 1 < x < 1.5 -(Min,Max) distances from A: (6,6.5) and B:(6,5.5)
III) x < -10 -(Min,Max) distances from A: (5,undefined) and B:(17,undefined)

So we see that only option 3 satisfies X being closer to A than B.Hence, the answer is C.
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Narenn
What values of X will be closer to the point A = -5 than the point B = 7 on the number line?

I) 0 < x < 2
II) 1 < x < 1.5
III) x < -10

A) Only I
B) Only II
C) Only III
D) Both I and III
E) Both II and III

Sol: Consider Range in option I

If x =0.9 then x is close to pt A but if x= 1.9 then pt x is close to pt B so option A and D can be ruled out
Consider range in option II. All the points in this range is close to Pt B. So option B and E is ruled out

Ans is III only or Option C only

Alternatively the question can be done by drawing a no. line. We can clearly see that that values to the left of Pt A will be closer to it than Pt B. Points to the right may or may not be close to Pt A in the given ranges in option I and II. Ans is C
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Narenn
What values of X will be closer to the point A = -5 than the point B = 7 on the number line?

I) 0 < x < 2
II) 1 < x < 1.5
III) x < -10

A) Only I
B) Only II
C) Only III
D) Both I and III
E) Both II and III

So when we plot the given lines on the umber line, we can actually see that 1 and 2 will be giving values that will be far from A.

Only 3, will give you a value which will be nearer to point A

C
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I think statement 1 can be correct too if we take the value of "x" in statement 1 as 0.1, the statement says the value is between "0 < x < 2", so if we take the value as 0.1 then 0.1 is much closer to -5 than it is to +7, please comment on this.
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Hello!

If two absolute value terms are expressed in an inequality, is squaring both sides always a valid way to solve? Can you please point me to the theory behind this approach?

Bunuel
Narenn
What values of X will be closer to the point A = -5 than the point B = 7 on the number line?

I) 0 < x < 2
II) 1 < x < 1.5
III) x < -10

A) Only I
B) Only II
C) Only III
D) Both I and III
E) Both II and III

x is closer to -5 than to 7 means \(|x-(-5)|<|x-7|\) --> \(|x+5|<|x-7|\).

Both sides are non-negative, thus we can safely square: \(x^2+10x+25<x^2-14x+49\) --> \(x<1\). So, we have that for x to be closer to -5 than it is to 7, x must be less than 1. Only the third option gives the values of x all of which are less than 1.

Answer: C (III only).
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lewdtoucan
Hello!

If two absolute value terms are expressed in an inequality, is squaring both sides always a valid way to solve? Can you please point me to the theory behind this approach?

Bunuel
Narenn
What values of X will be closer to the point A = -5 than the point B = 7 on the number line?

I) 0 < x < 2
II) 1 < x < 1.5
III) x < -10

A) Only I
B) Only II
C) Only III
D) Both I and III
E) Both II and III

x is closer to -5 than to 7 means \(|x-(-5)|<|x-7|\) --> \(|x+5|<|x-7|\).

Both sides are non-negative, thus we can safely square: \(x^2+10x+25<x^2-14x+49\) --> \(x<1\). So, we have that for x to be closer to -5 than it is to 7, x must be less than 1. Only the third option gives the values of x all of which are less than 1.

Answer: C (III only).

You can square an inequality when both sides are non-negative. Since absolute values are always non-negative, squaring both sides of |expression| < |other expression| is always valid.


Inequalities Made Easy!
Inequalities: Tips and hints
Arithmetic with Inequalities
Wavy Line Method Application - Complex Algebraic Inequalities
Solving Quadratic Inequalities: Graphic Approach
Inequalities - Quadratic Inequalities
Graphic approach to problems with inequalities
Inequalities trick
INEQUATIONS(Inequalities) Part 1
INEQUATIONS(Inequalities) Part 2
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Deconstructing the Question

We need values of x such that the distance from -5 is less than the distance from 7.

This translates to:
\(|x + 5| < |x - 7|\)

Key idea: numbers closer to -5 than 7 lie to the left of the midpoint.

Midpoint:
\((-5 + 7)/2 = 1\)

So the solution is:
\(x < 1\)

Step-by-step

Check each statement.

I) \(0 < x < 2\)

This interval includes values both less than 1 and greater than 1.
Not all values satisfy \(x < 1\).
So this is not always true.

II) \(1 < x < 1.5\)

All values are greater than 1.
These are closer to 7, not -5.
So this is false.

III) \(x < -10\)

All values are less than 1.
So all are closer to -5.
This is valid.

Answer C
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