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a + b = 8. Is ab > 8? (1) a^2 - b^2 = 32 (2) a^2 + b^2 = 40

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a + b = 8. Is ab > 8? (1) a^2 - b^2 = 32 (2) a^2 + b^2 = 40  [#permalink]

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New post 09 Mar 2017, 17:15
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Question Stats:

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a + b = 8. Is ab > 8?


(1) a^2 - b^2 = 32

(2) a^2 + b^2 = 40
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Re: a + b = 8. Is ab > 8? (1) a^2 - b^2 = 32 (2) a^2 + b^2 = 40  [#permalink]

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New post 09 Mar 2017, 19:09
2
Cez005 wrote:
a + b = 8. Is ab > 8?

(1) a2 - b2 = 32

(2) a2 + b2 = 40

A. Statement (1) by itself is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) by itself is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) taken together are sufficient to answer the question, even though neither statement by itself is sufficient.
D. Either statement by itself is sufficient to answer the question.
E. Statements (1) and (2) taken together are not sufficient to answer the question.


Kudos if you like the q! Note a2 is equal to a squared and b2 is equal to b squared.



(1) a^2-b^2 = (a+b)(a-b)=32
putting a+b=8----------(x)
we get a-b= 32/8=4-------(y)
solving (x) & (y) we get definite values of a& b
thus we can find value ab
suff

(2) given a+b= 8
sq. both sides
a^2+b^2+2ab=64----(z)
but per option (2) a^2+b^2=40
substituting in eq(z) we get
2ab=64-40=24
thus ab=12
suff

Ans D
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Re: a + b = 8. Is ab > 8? (1) a^2 - b^2 = 32 (2) a^2 + b^2 = 40  [#permalink]

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New post 09 Mar 2017, 23:31
rohit8865 wrote:
Cez005 wrote:
a + b = 8. Is ab > 8?

(1) a2 - b2 = 32

(2) a2 + b2 = 40

A. Statement (1) by itself is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) by itself is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) taken together are sufficient to answer the question, even though neither statement by itself is sufficient.
D. Either statement by itself is sufficient to answer the question.
E. Statements (1) and (2) taken together are not sufficient to answer the question.


Kudos if you like the q! Note a2 is equal to a squared and b2 is equal to b squared.



(1) a^2-b^2 = (a+b)(a-b)=32
putting a+b=8----------(x)
we get a-b= 32/8=4-------(y)
solving (x) & (y) we get definite values of a& b
thus we can find value ab
suff

(2) given a+b= 8
sq. both sides
a^2+b^2+2ab=64----(z)
but per option (2) a^2+b^2=40
substituting in eq(z) we get
2ab=64-40=24
thus ab=12
suff

Ans D



For 2) how about this approach?

a2 = (8-b)2 = 64-16b+b2, so 64-16b+b2+b2=40. Subbing into a2+b2=40:
24-16b+2b2=0
b2-8b+12=0
(b-2)(b-6)=0
b=2 or 6, both of which give answers of 6 and 2 for a respectively, therefore ab>8.
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Re: a + b = 8. Is ab > 8? (1) a^2 - b^2 = 32 (2) a^2 + b^2 = 40  [#permalink]

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New post 10 Mar 2017, 01:21
1
a+b = 8
St 1: a^2 -b^2 = 32
or (a+b)(a-b) = 32
or 8(a-b) = 32
or a-b = 4 and a+b = 8. therefore a=6 and b=2. hence ab = 12 which is greater than 8.ANSWER

St 2: a^2+b^2 = 40
or(a+b)^2 - 2ab =40
or8^2 -2ab =40
or 2ab = 44 or ab =22 which is greater than 8. ANSWER

Option D
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Re: a + b = 8. Is ab > 8? (1) a^2 - b^2 = 32 (2) a^2 + b^2 = 40  [#permalink]

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New post 10 Apr 2019, 20:53
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Re: a + b = 8. Is ab > 8? (1) a^2 - b^2 = 32 (2) a^2 + b^2 = 40   [#permalink] 10 Apr 2019, 20:53
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