Double Inequality

[karthikpandian19]

Can anyone explain how to solve / deal with a double equality problem?

[eagleeye]

hi karthikpandian19:

This is how I solved it
We have |b-2| + |b+8| , the concept to use with |X| questions is simple but important
1)if X<=0; |X| = -X
2)if X>=0; |X| = X

Now let's look at this one
1. b <=2 , then b-2<=0; if let's use b = -100, then |b-2| + |b+8| = |-100-2| + |-100+8| = 102 + 92 = 294, hence not sufficient

2. b>=-8 , then b+8>=0; let b = 100, again we see that |b-2| + |b-8| > 10; hence insufficient.
Let's look at both of them together,
we have b<=2, which implies b-2<=0; therefore |b-2| = -(b-2)
also we have, b>=-8 which implies b+8>=0; therefore |b+8| = (b+8)

Then we get |b-2| + |b-8| = -(b-2) + b+8 = -b + 2 + b + 8 = 10; hence C is the correct answer.

One can always use different cases for |x| questions and use the boldface expressions to solve almost all questions of this type.

1)if X<=0; |X| = -X
2)if X>=0; |X| = X

btw, karthikpandian19, you post really good questions

let me know if this helps

[karthikpandian19]

@eagleeye,

Basically you are letting me to know that we need to use Plugging In values over here along with the Boldface equations.
Nice explanation

hi karthikpandian19:

This is how I solved it
We have |b-2| + |b+8| , the concept to use with |X| questions is simple but important
1)if X<=0; |X| = -X
2)if X>=0; |X| = X

Now let's look at this one
1. b <=2 , then b-2<=0; if let's use b = -100, then |b-2| + |b+8| = |-100-2| + |-100+8| = 102 + 92 = 294, hence not sufficient

2. b>=-8 , then b+8>=0; let b = 100, again we see that |b-2| + |b-8| > 10; hence insufficient.
Let's look at both of them together,
we have b<=2, which implies b-2<=0; therefore |b-2| = -(b-2)
also we have, b>=-8 which implies b+8>=0; therefore |b+8| = (b+8)

Then we get |b-2| + |b-8| = -(b-2) + b+8 = -b + 2 + b + 8 = 10; hence C is the correct answer.

One can always use different cases for |x| questions and use the boldface expressions to solve almost all questions of this type.

1)if X<=0; |X| = -X
2)if X>=0; |X| = X

btw, karthikpandian19, you post really good questions

let me know if this helps

[eagleeye]

yes, karthikpandian19, plugging in alongside algebra works well in these types. Once, we do more practice, the DS answers should become really fast.

Thanks for the kudos

[Anurag@Gurome]

Is it true that |b - 2| + |b + 8| = 10?

1. b ≤ 2
2. b ≥ -8

Let's solve this problem more conceptually than going into algebraic complexities. |x| is essentially the distance of x from zero on the number line. Similarly, |b - 2| and |b + 8| are the distances of b from 2 and -8 on the number line.

Let's draw the number line with 2 and -8 on it.

<----------(-8)--------------0----2----------------->

From the above diagram, it's obvious that sum of |b - 2| and |b + 8|, i.e. sum of the distances of b from 2 and -8 on the number line will be equal to 10 only when b lies between -8 and 2, both inclusive.

Only by taking both statements together, we can conclude that -8 ≤ b ≤ 2.

The correct answer is C.