What values of X will be closer to the point A = -5 than the

Question

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

Bunuel · Accepted Answer

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<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).

GoLeafsGo · Answer

From a logic perspective, you see the question is asking about distances from -5 and 7. The distance between -5 and 7 is (7- (-5)) = 12. One half of 12 is 6, so 6 + (-5) is +1 and 7 - 6 = +1 as well, which is the mid-point between -5 and 7. Therefore, we are trying to find if the points in the given answers are < +1. The only answer that fulfills this requirement for all possible numbers is choice III.

AlexanderSupertramp · Answer

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.

WoundedTiger · Answer

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

KanishkM · Answer

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

satyam43298 · Answer

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.

lewdtoucan · Answer

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 · Answer

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

luisdicampo · Answer

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

What values of X will be closer to the point A = -5 than the

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What values of X will be closer to the point A = -5 than the

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