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Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b

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Bunuel
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Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b [#permalink] New post   14 Apr 2020, 03:13
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Is \(b^2≤a^2\)


(1) \(b ≤ a\)

(2) \(a+b = a−b\)


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Re: Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b [#permalink] New post   14 Apr 2020, 03:33
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Bunuel wrote:
Is \(b^2≤a^2\)


(1) \(b ≤ a\)

(2) \(a+b = a−b\)



Please find below the video explanation of the question

Answer: Option B

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Re: Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b [#permalink] New post   14 Apr 2020, 03:41
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We are to determine if \(b^2≤a^2\)

Statement 1: \(b≤a\)
Statement 1 is insufficient. When \(a=3\) and \(b=2\), then the answer is No. However, when \(b=-2\) and \(a=-1\) satisfying statement 1, \(b^2=4\) while \(a^2=1\) and the answer is Yes.

Statement 2: \(a+b=a-b\)
Sufficient since we know that \(b=0\) from statement 2. Irrespective of the value of \(a\), \(b^2≤a^2.\)

The answer is therefore B.
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Re: Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b [#permalink] New post   14 Apr 2020, 03:51
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Bunuel wrote:
Is \(b^2≤a^2\)


(1) \(b ≤ a\)

(2) \(a+b = a−b\)


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Solution


Step 1: Analyse Question Stem


Is \(b^2 ≤ a^2\) ?

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1:\( b ≤ a\)
    • Case I: If a= 2 and b= 1,\( a^2 = 4\) and \(b^2 = 1\)
      o Then, \(b^2 ≤ a^2\)
    • Case II: If a = 2 and b = -3, then \(a^2 = 4\) and \(b^2 = 9\)
      o Then \(b^2 > a^2\)
    • We are getting different results.
Hence, statement 1 is not sufficient, we can eliminate the answer options A and D.
Statement 2: \(a + b = a- b\)
    • \(a -a = -b -b\)
    • \(0 = -2b\)
      o \(b = 0\)
      o Case 1: If a =0, then \(a^2 = b^2 \)
      o Case 2: If a < 0, then \(a^2\) will always be a positive number
        o \(0 < a^2, b^2 < a^2\)
      o Case 3: If a > 0, then \(a^2\) will always be a positive number
        o \(0 < a^2, b^2 < a^2\)
      o We can say that, \(b^2 ≤ a^2 \)
Hence, statement 2 is also sufficient, the correct answer is Option B
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Re: Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b [#permalink] New post   17 Apr 2020, 21:53
Bunuel wrote:
Is \(b^2≤a^2\)


(1) \(b ≤ a\)

(2) \(a+b = a−b\)


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Asked: Is \(b^2≤a^2\)


(1) \(b ≤ a\)
If b=-2; a=1; b<a but b^2 = 4>a^2 = 1
But if b=1; a = 2; b<a and b^2 = 1<a^2 = 4
NOT SUFFICIENT

(2) \(a+b = a−b\)
b = 0
b^2 = 0 <= a^2 for any value of a
SUFFICIENT

IMO B
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Re: Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b [#permalink] New post   18 Apr 2020, 05:14
Is \(b^2 ≤ a^2\) ?
--> \((a-b)(a+b) ≤0\) ???

(Statement1): \(b ≤ a\)
--> \(b-a ≤0\)
Since nothing tells us about what \(a\) and \(b\) are, \((b+a)\) could be positive or negative.
Insufficient

(Statement2): \(a+b =a−b\)
--> \(2b =0 \) --> \(b=0\)
\(a\) could be negative, zero or positive number

Case1: a=-5, b =0
---> \(0^{2} ≤ (-5)^{2}\) (YES)

Case2: a=0, b =0
---> \(0^{2} ≤ 0^{2}\) (YES)

Case3: a=5, b =0
---> \(0^{2} ≤ (5)^{2}\) (YES)

Sufficient

Answer (B).
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Re: Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b [#permalink] New post   11 Sep 2021, 04:19
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Re: Is b^2 ≤ a^2 (1) b ≤ a (2) a + b = a − b [#permalink]
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