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Quote:
If a^2 + b^2 = 144 and ab ≠ 0, then the greatest possible value for b is between

(A) 16 and 13
(B) 12 and 3
(C) 11 and 4
(D) 10 and 5
(E) 9 and 6

a^2 + b^2 = 144, a^2 + b^2 = 12^2
a=0: 0^2+b^2=12^2, b=12; but ab≠0 so, b<12

Ans (B) assuming the range in the answer choices are NOT INCLUSIVE.
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If a^2 + b^2 = 144 and ab ≠ 0, then -12<b<0 and 0<b<12. b can be an integer or a rational number.

The greatest possible value for b is between:

(A) 16 and 13 --> NO! b cannot be greater than or equal to 12.
(B) 12 and 3 --> YES! Correct answer. b should range as follows: -12<b<0 and 0<b<12

We stop at B because we have found the right answer.

FINAL ANSWER IS (B)

Posted from my mobile device
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\(144 = 12^2\)
\( a*b \neq{0} => b^2 < 12^2\)
\(=> b<12 \)
=> Choice B
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a^2+b^2=144, where ab≠0.
ab≠0 means that a≠0, b≠0, a≠12, and b≠12. However, a and b can be small fractions.
The word between to modify the range suggests that the limits are not part of the set of numbers considered. The greatest possible value of b is, therefore, between 12 to 3. This is because a can be a small fraction and b can be any number close to 12 but not 12 itself. b can be 11.99 with a being the complementary fraction that makes a^2+b^2=144.

B is the answer.
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144-b^2= + value use plugin technique value of a must be +ve
IMO B 12 and 3




If a^2 + b^2 = 144 and ab ≠ 0, then the greatest possible value for b is between

(A) 16 and 13
(B) 12 and 3
(C) 11 and 4
(D) 10 and 5
(E) 9 and 6

Posted from my mobile device
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If \(a^2 + b^2 = 144\) and ab ≠ 0, then the greatest possible value for b is between -???
--> \(a^2\) -always positive --> \(b^2< 144\)
(b+12)(b-12) <0
-12 <b< 12
Only B satisfies this inequality

The answer is B.
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If \(a^2 + b^2 = 144\) and ab ≠ 0, then the greatest possible value for b is between

(A) 16 and 13
(B) 12 and 3
(C) 11 and 4
(D) 10 and 5
(E) 9 and 6

\(a^2 + b^2 = 144\)
\((a+b)^2 - 2ab = 144\)
The value of \(a^2\) should be as small as possible so that b takes the maximum value. Hence 2ab is almost negligible.

\((a+b)^2 = 144 + 2ab\)
\((a+b)^2 > 12^2\)
a + b > 12
b > 12 - a (such that b < 12 to satisfy \(a^2 + b^2 = 144\))
b < 12

IMO Answer B.
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Bunuel

Competition Mode Question



If a^2 + b^2 = 144 and ab ≠ 0, then the greatest possible value for b is between

(A) 16 and 13
(B) 12 and 3
(C) 11 and 4
(D) 10 and 5
(E) 9 and 6

If a = 0, then the greatest possible value of b is 12. However, since a can’t be 0 (but it still can be very close to 0), the greatest possible value of b can be just slightly less than 12 (for example, if a = 1, then b = √143 ≈ 11.96).

Answer: B
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can this be a right approcah?
a^2+b^2-144=0
2ab=144
2*12*6 = 144
therefore highest value is 12 and the other value can be 6 but not below 3.
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