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--- eagleeye wrote: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
---
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.karthikpandian19 wrote:Is it true that |b - 2| + |b + 8| = 10?
1. b ≤ 2
2. b ≥ -8
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