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If a and b are integers with each being greater than 1, is a × b > 20?

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If a and b are integers with each being greater than 1, is a × b > 20?  [#permalink]

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New post 29 Dec 2018, 21:41
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If a and b are integers with each being greater than 1, is a × b > 20?

(1) a + b = 9

(2) (a − 4)^2 + (b − 5)^2 = 0
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Re: If a and b are integers with each being greater than 1, is a × b > 20?  [#permalink]

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New post 29 Dec 2018, 21:56
From statement 1:

a+b = 9
As a and b are greater than 1. then a and b can take values (6,3) or (4,5) or (7,2)
6*3 = 18 < 20.
7*2 = 14 < 20
4*5 = 20 and 20 is not greater than 20.
Hence, Insufficient.

From statement 2:

\((a − 4)^2 + (b − 5)^2 = 0\)
Here a and b can only take values 4 and 5. and 20 is not greater than 20.
Hence, a*b can never be greater than 20.
Sufficient.

B is the answer.
akurathi12, is the OA correct?
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Re: If a and b are integers with each being greater than 1, is a × b > 20?  [#permalink]

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New post 29 Dec 2018, 22:10
I think OA is wrong. The correct answer is B.

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Re: If a and b are integers with each being greater than 1, is a × b > 20?  [#permalink]

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New post 30 Dec 2018, 04:08
1
(1) a + b = 9

Combinations possible are (2,7), (3,6) and (4,5)
2*7 < 20
3*6 < 20
4*5 = 20

Is a*b > 20, No it is less than or equal to 20
So, clearly I can say a*b not greater than 20
i.e.,\(a*b <= 20\)
Sufficient

(2) (a − 4)^2 + (b − 5)^2 = 0
for some of squares i.e., positive values to be zero, they should be zero
a = 4 and b = 5 is the only possibility
a*b = 20 which is not greater than
Sufficient

So, option D is correct
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Re: If a and b are integers with each being greater than 1, is a × b > 20?  [#permalink]

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New post 30 Dec 2018, 06:16
Afc0892 wrote:
From statement 1:

a+b = 9
As a and b are greater than 1. then a and b can take values (6,3) or (4,5) or (7,2)
6*3 = 18 < 20.
7*2 = 14 < 20
4*5 = 20 and 20 is not greater than 20.
Hence, Insufficient.

From statement 2:

\((a − 4)^2 + (b − 5)^2 = 0\)
Here a and b can only take values 4 and 5. and 20 is not greater than 20.
Hence, a*b can never be greater than 20.
Sufficient.

B is the answer.
akurathi12, is the OA correct?

I am in a confusion with OA, that is the reason I have posted.
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Re: If a and b are integers with each being greater than 1, is a × b > 20?  [#permalink]

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New post 30 Dec 2018, 06:18
Dare Devil wrote:
(1) a + b = 9

Combinations possible are (2,7), (3,6) and (4,5)
2*7 < 20
3*6 < 20
4*5 = 20

Is a*b > 20, No it is less than or equal to 20
So, clearly I can say a*b not greater than 20
i.e.,\(a*b <= 20\)
Sufficient

(2) (a − 4)^2 + (b − 5)^2 = 0
for some of squares i.e., positive values to be zero, they should be zero
a = 4 and b = 5 is the only possibility
a*b = 20 which is not greater than
Sufficient

So, option D is correct


Thank you for the statement 1 explanation.
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Re: If a and b are integers with each being greater than 1, is a × b > 20?  [#permalink]

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New post 30 Dec 2018, 09:42
akurathi12 wrote:
Dare Devil wrote:
(1) a + b = 9

Combinations possible are (2,7), (3,6) and (4,5)
2*7 < 20
3*6 < 20
4*5 = 20

Is a*b > 20, No it is less than or equal to 20
So, clearly I can say a*b not greater than 20
i.e.,\(a*b <= 20\)
Sufficient

(2) (a − 4)^2 + (b − 5)^2 = 0
for some of squares i.e., positive values to be zero, they should be zero
a = 4 and b = 5 is the only possibility
a*b = 20 which is not greater than
Sufficient

So, option D is correct


Thank you for the statement 1 explanation.


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Re: If a and b are integers with each being greater than 1, is a × b > 20?  [#permalink]

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New post 08 Jan 2019, 12:10
akurathi12 wrote:
If a and b are integers with each being greater than 1, is a × b > 20?

(1) a + b = 9

(2) (a − 4)^2 + (b − 5)^2 = 0


Given
a and b are integers with each being greater than 1

Statement 1
a + b = 9
2+7, is 14>20, No
3+6, is 18>20, No
4+5, is 20>20, No
Consistent, No.
Sufficient.

Statement 2
\((a − 4)^2 + (b − 5)^2\) = 0
Can only be true when a = 4 and b = 5
is 20>20, No
Sufficient.

Answer D
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Re: If a and b are integers with each being greater than 1, is a × b > 20?   [#permalink] 08 Jan 2019, 12:10
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