Can anyone explain how to solve / deal with a double equality problem?
Double Inequality
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- karthikpandian19
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- eagleeye
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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
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
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@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
Basically you are letting me to know that we need to use Plugging In values over here along with the Boldface equations.
Nice explanation
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
Last edited by karthikpandian19 on Sun May 27, 2012 10:36 pm, edited 1 time in total.
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON
---If you find my post useful, click "Thank" ---
---Never stop until cracking GMAT---
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON
---If you find my post useful, click "Thank" ---
---Never stop until cracking GMAT---
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- Anurag@Gurome
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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
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.
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