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modulus DS question

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Source: — Data Sufficiency |
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by pemdas » Thu Mar 01, 2012 10:32 am
pappueshwar wrote:Is |a^2 -b^2|<6?

1) |a+b|<2
2) |a-b|<3

please xplain how to solve? i am heavily getting confused when solving modulus problems
|a^2-b^2| falls into a^2-b^2 and b^2-a^2 as modulus may get either negative sign of non-negative sign
|a^2-b^2|=a^2-b^2 when the value inside mod is positive and |a^2-b^2|=-(a^2-b^2)=b^2-a^2 when the value inside mod is negative
Prompt: |a^2-b^2|<6? asks if a^2-b^2<6 and b^2-a^2<6
st(1) |a+b|<2 implies either a+b<2 and -(a+b)<2. The latter could be written as a+b>-2 (flip the signs when multiplying both sides of inequality by negative value)

Thus, -2<a+b<2. The new inequality expression is Not Sufficient, as we don't know about a-b and b-a; a^2-b^2 and b^2-a^2 are equivalent to (a-b)(a+b) and (b-a)(b+a)

st(2) according to the same logic as with st(1) it's Not Sufficient
-3<a-b<3

Combining st(1&2): Sufficient as the limits set less than 6

c
Last edited by pemdas on Thu Mar 01, 2012 4:31 pm, edited 1 time in total.
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by krusta80 » Thu Mar 01, 2012 12:42 pm
pappueshwar wrote:Is |a^2 -b^2|<6?

1) |a+b|<2
2) |a-b|<3

please xplain how to solve? i am heavily getting confused when solving modulus problems

Step 1: Can the main statement be proven independently of parts (1) and (2)?

No, since there are no restrictions on a and b. ie. a = b = 0 works and a = 3 | b = 0 does NOT.


Step 2: Let's look at parts (1) and (2)

Always keep your eye out for common themes. In this case, a^2 - b^2 = (a-b)(a+b)

|(a-b)(a+b)| < 6

(1) |a+b| < 2

If you picture a number line, |a+b| < 2 allows for two possibilities:
i.) |a|+|b| < 2
ii.) The distance between a and 0 is within 2 of the distance between b and 0 and the two have opposite signs, assuming |a|+|b| => 2

Since the second possiblity allows for (a-b) to be largely positive (if a is largely positive and b is largely negative for example), |(a-b)(a+b)| can certainly be much greater than 6, but since the first possibility allows for (a-b) to be very small (a and b are both very close to 0), |(a-b)(a+b)| can be less than 6. INSUFFICIENT


(2) |a-b| < 3

Again, picturing a number line, |a-b| represents the distance between a and b. Since a and b can be moved in unison to any area of the number line, clearly this allows for (a+b) to be as large as we want (or as negatively small)...or of course it can be 0 if we want. INSUFFICIENT


(1) and (2)

Since (a-b)(a+b) is within an absolute value sign, we can ignore negatives and simply try to make a+b and a-b each as big as possible. We already know that setting a = b = 0 works for all constraints, so let's try to violate the main statement.

|a+b| < 2, which can be rewritten as
-2 < a+b < 2

So, a+b cannot exceed 2.

|a-b| < 3, which can be rewritten as
-3 < a-b < 3

So a-b cannot exceed 3.

Therefore, |(a-b)(a+b)| < 6 is ALWAYS true.

Answer is C.
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by GMATGuruNY » Fri Mar 02, 2012 7:22 am
pappueshwar wrote:Is |a^2 -b^2|<6?

1) |a+b|<2
2) |a-b|<3

please xplain how to solve? i am heavily getting confused when solving modulus problems
Question rephrased: Is -6 < (a+b)(a-b) < 6?

Statement 1: |a+b|<2.
-2 < a+b < 2.
No restrictions on a-b.

Plugging a+b = 1 and a-b = 1 into -6 < (a+b)(a-b) < 6, we get:
-6 < 1*1 < 6
-6 < 1 < 6.
YES.

Plugging a+b = 1 and a-b = 100 into -6 < (a+b)(a-b) < 6, we get:
-6 < 1*100 < 6
-6 < 100 < 6.
NO.

Since in the first case the answer is YES, and in the second case the answer is NO, INSUFFICIENT.

Statement 2: |a-b|<3
-3 < a-b < 3.
No restrictions on a+b.

Plugging a-b = 1 and a+b = 1 into -6 < (a+b)(a-b) < 6, we get:
-6 < 1*1 < 6
-6 < 1 < 6.
YES.

Plugging a-b = 1 and a+b = 100 into -6 < (a+b)(a-b) < 6, we get:
-6 < 1*100 < 6
-6 < 100 < 6.
NO.

Since in the first case the answer is YES, and in the second case the answer is NO, INSUFFICIENT.

Statements 1 and 2 combined:
Statement 1: -2 < a+b < 2
Statement 2: -3 < a-b < 3.

To determine the range when two inequalities are MULTIPLIED -- in this case, the range of (a+b)(a-b) -- calculate the product of EVERY COMBINATION OF ENDPOINTS.
(lower limit of a+b) * (lower limit of a-b) = (-2)(-3) = 6.
(lower limit of a+b) * (upper limit of a-b) = (-2)(3) = -6.
(upper limit of a+b) * (lower limit of a-b) = (2)(-3) = -6.
(upper limit of a+b) * (upper limit of a-b) = (2)(3) = 6.

The resulting products indicate the LOWER LIMIT and the UPPER LIMIT of (a+b)(a-b).
Since the smallest product is -6 and the greatest product is 6:
-6 < (a+b)(a-b) < 6.
SUFFICIENT.

The correct answer is C.
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by pappueshwar » Fri Mar 02, 2012 7:44 am
ultimate explanation mitch...

thanks a ton
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