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

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

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22 Jan 2020, 23:11
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If a and b are integers with each being greater than 1, is ab > 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 ab > 20?  [#permalink]

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22 Jan 2020, 23:28
3
If a and b are integers with each being greater than 1, is ab > 20?

Possible values of a or b = 2, 3, 4, 5, . . . . . . .

(1) a + b = 9
--> Possible values of (a, b):
1. (a, b) = (2, 7) or (7, 2) --> Product = 2*7 = 14
2. (a, b) = (3, 6) or (6, 3) --> Product = 3*6 = 18
3. (a, b) = (4, 5) or (5, 4) --> Product = 4*5 = 20

So, the product is NEVER greater than 20 --> Sufficient

(2) (a − 4)^2 + (b − 5)^2 = 0
--> Only possible when (a - 4)^2 = 0 & (b - 5)^2 = 0
--> a = 4 & b = 5
--> Product = a*b = 4*5 = 20
Not greater than 20 --> Sufficient

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

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Updated on: 23 Jan 2020, 03:52
1
Integers a>1 and b>1. Is ab > 20?

(1) a + b = 9
a=1, b=8, ab<=20
a=4, b=5, ab<=20
SUFFICIENT

(2) (a − 4)^2 + (b − 5)^2 = 0
The only possible solution is (a,b) =(4,5). ab<=20
SUFFICIENT.

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Originally posted by chondro48 on 22 Jan 2020, 23:31.
Last edited by chondro48 on 23 Jan 2020, 03:52, edited 1 time in total.
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Re: If a and b are integers with each being greater than 1, is ab > 20?  [#permalink]

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22 Jan 2020, 23:54
1
Quote:
If a and b are integers with each being greater than 1, is ab > 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

Question: is ab > 20?

Statement 1: a + b = 9

CONCEPT: For given addition of two (or more) values the product of those values is maximum when they are equally distributed. E.g. for a+b=10, $$(a*b)_{max}= 5*5 = 25$$

Similarly for $$(a*b)_{max}$$, a=4 and b = 5 (cause they are integers)
i.e. $$(a*b)_{max}= 5*4 = 20$$

i.e. a*b can NOT be greater than 20 (Unique and consistent answer) hence

SUFFICIENT

Statement 2: $$(a − 4)^2 + (b − 5)^2 = 0$$

Since square of any value can't be less than 0 therefore each (a-4)^2 and (b-5)^2 must be zero
i.e. a=4 and b=5
i.e. $$(a*b)_{max}= 5*4 = 20$$

i.e. a*b can NOT be greater than 20 (Unique and consistent answer) hence

SUFFICIENT

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

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23 Jan 2020, 01:05
1
(1) a + b = 9
Because
a = 2, b = 7, and ab = 14
a = 3, b = 6, and ab = 18
a = 4, b = 5, and ab = 20
...
If ab > 20? The answer is definitive No. SUFFICIENT.

(2) (a − 4)^2 + (b − 5)^2 = 0
a = 4, b = 5, and ab =20.
If ab > 20? The answer is definitive No. SUFFICIENT.

So, we choose "D".

(I have to admit that I made the mistake of choosing B at first. )
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Re: If a and b are integers with each being greater than 1, is ab > 20?  [#permalink]

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23 Jan 2020, 01:39
#1
a + b = 9
a,b = 4,5 or any other value 7,2; 8,1 ; 3,6 ;
ab>20 not possible
sufficient
#2
a,b = 4,5
sufficient
IMO D
(a − 4)^2 + (b − 5)^2 = 0
possible w

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

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23 Jan 2020, 02:57
1
If a and b are integers with each being greater than 1, is ab > 20?

Constraint: a=Integer >1 ~ +ve int.
b=Integer >1 ~ +ve int.
Asked: Is ab > 20 Yep! Or ab < = 20 Nope!

(1) a + b = 9
When a = 4 ,b=5
a•b =20 so Nope! ab = 20
When a= 3,b= 6
a•b = 18 so Nope! ab < 20
We keep getting Nope! So
(Sufficient)

(2) (a − 4)^2 + (b − 5)^2 = 0
When a= 4 ,b= 5
(4-4)^2 + (5-5)^2 = 0
Any other value of a and b won’t satisfy the equation
Here a•b = 20 so Nope a•b ain’t greater than 20
(Sufficient)

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

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23 Jan 2020, 03:33
Max value of a + b = 9 is when a equals to b....4.5 and 4.5 therefore ( 4.5 * 4.5) is 20.25 i.e greater than 20 but a and b can be 1 and 8 also so less than 20....

So insufficient....

Whereas in 2nd statement ..... A sum of square is zero when a = 4 and b= 5..... Therefore a*b is 20 ....so we get the answer...

B is sufficient

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

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23 Jan 2020, 03:54
1
From the question statement we understand that both a and b belong to the following set:
{2,3,4,5,6,…..}. We are to find out if the product of ab is greater than 20. Since this is a YES-NO type of DS question, we need to obtain a deifinite YES or a definite NO as an answer.

From statement I alone, a+b = 9. Since neither a nor b can be equal to 1, the only combinations that satisfy the above equation will be (5,4), (6,3) and (7,2).

For each of these combinations, the product is either equal to 20 or less than 20.
Is ab>20? Clearly, NO. Statement I alone is sufficient.

Answer options B, C and E can be eliminated. Possible answer options are A or D.

From statement II alone, $$(a-4)^2$$ + $$(b-5)^2$$ = 0.
Remember that the square of any number is always a NON-NEGATIVE value. This means that the square of a number can be either ZERO or POSITIVE, it will never be negative.

In the equation given, we have two squares being added to give us ZERO. This can happen only if both the squares are ZERO themselves i.e. $$(a-4)^2$$ = 0 and $$(b-5)^2$$ = 0. As per this, a = 4 and b = 5.

Using these values of a and b, ab = 20. Is ab>20? Clearly, the answer is a NO again.
Statement II alone is sufficient. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
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Re: If a and b are integers with each being greater than 1, is ab > 20?  [#permalink]

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23 Jan 2020, 05:28
1
Quote:

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

(1) a + b = 9

(2) (a − 4)^2 + (b − 5)^2 = 0

(a,b)>1=ie.(2,2)

(1) a + b = 9 sufic

a+b=9; max(a,b)=(4,5)=20, min=()=9, ans=no

(2) (a − 4)^2 + (b − 5)^2 = 0 sufic

$$(a-4)^2…and…(b-5)^2≥0…(a,b)=(4,5)≤20$$

Ans (D)
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Re: If a and b are integers with each being greater than 1, is ab > 20?  [#permalink]

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23 Jan 2020, 10:26
1
I think D shoukd be answer, as both are sufficient to answer the question.
First statement, highest product can go upto 20.
Second statement, also has the only product of 20.

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

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23 Jan 2020, 15:29
1
a,b --integers (a >1, b>1)
is ab >20 ???

(Statement1): a + b = 9
--> the maximum value of (a,b) pairs from (1,8),(2,7),(3,6),(4,5) is ( 4,5 )
ab =20 --> (Cannot be greater than 20-NO)
Sufficient

(Statement2): $$(a − 4)^2 + (b − 5)^2 = 0$$
--> In order $$(a − 4)^2 + (b − 5)^2$$ to be equal to zero, $$(a − 4)^2=0$$ and $$(b − 5)^2=0$$ at the same time.
--> a=4, b =5
ab= 4*5=20
(Cannot be greater than 20-NO)
Sufficient

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

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23 Jan 2020, 19:01
1
given a and b each greater than 1

1) a+b= 9
now product is maximun when numbers are equal
hence nearest integers will be
4 and 5
their prod = 20
thus product will be <= 20
sufficient

(2) (a − 4)^2 + (b − 5)^2 = 0

or (a − 4)^2 = 0
thus a = 4
(b − 5)^2 = 0

or b = 5
product = 20
not greater
thus sufficient

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

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23 Jan 2020, 19:24
1
(1) a + b = 9

Consider all possible pairs (2,7) (3,6) (4,5)

ab is never greater than 20

1 is sufficient

(2) (a − 4)^2 + (b − 5)^2 = 0

Sum of two perfect squares is 0. Since a square cannot be negative, the only possibility is that a=4 and b=5 and so ab=20

2 is sufficient

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

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23 Jan 2020, 19:45
1
We know that a and b are integers greater than 1. We are to determine if ab>20.

Statement 1: a+b=9
possible pairs are a=2, b=7 or b=2 and a=7, ab=14 !> 20 No.
a=3, b=6, or b=3, and a=6, ab=18 !> 20 No.
a=4, b=5 or a=5 and b=4, ab=20 !>0 No.
Statement 1 is sufficient.

Statement 2: (a-4)^2 + (b-5)^2 = 0
The square of a number is always a positive number or zero, hence the only way to get (a-4)^2 + (b-5)^2 = 0 is when a-4=0 and b-5=0, implying a=4 and b=5.
So ab=20 and 20 !>20 so statement 2 is also sufficient.

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

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23 Jan 2020, 21:25
1
Both are sufficient individually

a) all the below possibilities satisfies the condition
1,8
2,7
3,6
4,5
Sufficient

b)only possible solution is a=4,b=5...... sufficient

OA:D

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

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23 Jan 2020, 22:38
1
Quote:
If a and b are integers with each being greater than 1, is ab > 20?

(1) a + b = 9

(2) $$(a − 4)^2 + (b − 5)^2 = 0$$

Statement (1)
2*7=14
3*6 = 18
4*5 = 20
=> $$ab\leq{20}$$
=> Suff
Statement (2)
$$(a − 4)^2 + (b − 5)^2 = 0$$
=> $$=> a = 4; b=5 => ab =20$$
=> Suff
=> Choice D
Re: If a and b are integers with each being greater than 1, is ab > 20?   [#permalink] 23 Jan 2020, 22:38
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