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Formula used:
Distance between points K(x1,y1) and G(x2,y2) = \(\sqrt{(x1 - x2)^2 + (y1 - y2)^2}\)


On analyzing statement 1 independently, we get A = 6,-1
When A=6, Points K(6,0) and G(16,\(\sqrt{21}\)) have a distance of 11(a prime number)
But if A=-1, Points K(-1,0) and G(2,\(\sqrt{7}\)) have a distance of 4(not a prime number)
Insufficient.

Similarly on analyzing statement 2 independently, using A = 3,6
Since we have got the distance, when A=6 to be prime.
Let analyze if A=3
Points K(3,0) and G(10,\(\sqrt{15}\)) have a distance of 8(not a prime number)
Insufficient.

But, on combining both the statements, we can clearly say A=6, which has a prime number as the distance between K and G.
Sufficient(Option C)
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In order to solve this question we need a restriction on the possibilities for X- statement 1 and 2 individual fail to provide criteria that allow us to narrow down a discrete value of "a" but statement 1 and 2 together allow us to factor the equation in one and asses one value of a.

Posted from my mobile device
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TheKingInTheNorth
Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\)). Is the distance between point K and G prime?

(1) A^2 – 5A – 6 = 0
(2) A > 2

Distance between K and G = \(\sqrt{((2A + 9)^{1/2} - 0)^2 + (2A + 4 - A)^2}\)
= \(\sqrt{(2A + 9 + (A + 4)^2)}\)
= \(\sqrt{(2A + 9 + A^2 + 8A + 16)}\)
= \(\sqrt{(A^2 + 10A + 25)}\)
= \(\sqrt{(A + 5)^2}\)
= \(A + 5\)

(1) \(A^2 – 5A – 6 = 0\)
--> \(A^2 – 6A + A – 6 = 0\)
--> \(A(A - 6) + 1(A – 6) = 0\)
--> \((A - 6)(A + 1) = 0\)
--> A = 6 or -1
--> Two values are possible --> Insufficient

(2) \(A > 2\)
--> Infinite values are possible --> Insufficient

Combining (1) & (2),
--> \(A = 6 or -1\) & \(A > 2\)
--> \(A = 6\) only

--> Distance = \(A + 5 = 6 + 5 = 11\) --> Sufficient

IMO Option C
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TheKingInTheNorth
Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\)). Is the distance between point K and G prime?

(1) A^2 – 5A – 6 = 0
(2) A > 2

I suggest my students to use statements when you do not know how to proceed and this is one such scenario...

If you do not get hang of it or are short of time, --
There is one variable-A and if we know A, we can answer for sure.
Statement I gives two values of A as it is a quadratic equation. It may be possible that both lead to prime or non-prime, but in most cases it will not.
Statement II gives a range, so insuff
Combined, I says -1 and 6, II says>2, so a unique value C


Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\))..
Two ways..

(I) Don't waste time finding an equation in variable here
We know that to know the distance here, we will require to know the value of A, so let us check the statements
(1) \(A^2 – 5A – 6 = 0...(A-6)(A+1)=0\) A =-1 or A=6..
A is -1.. \(K=(-1,0)\) and G\(=(2,\sqrt{7}\)...Distance = \(\sqrt{9+7}=\sqrt{16}=4\)..NO
A is 6.. \(K=(6,0)\) and G\(=(16,\sqrt{21}\)...Distance = \(\sqrt{100+21}=\sqrt{121}=11\).. Yes
Insuff

(2) A > 2
Clearly insuff. There may be cases where it is true and places where it may not be true.

Combined..ans is 11 and YES
Suff

C

(II) Simplify the question stem
Distance = \(\sqrt{(2A+4-A)^2+\sqrt{2A+9}^2}=\sqrt{A^2+10A+25}=\sqrt{(A+5)}^2=A+5\)
So, if A=2, 7 etc ans is YES and if A=3, 4 etc ans is NO
(1) A^2 – 5A – 6 = 0
A= -1 or 6
Insuff

(2) A > 2
Insuff

Combined A=6
Suff

C
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TheKingInTheNorth
Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\)). Is the distance between point K and G prime?

(1) A^2 – 5A – 6 = 0
(2) A > 2

I suggest my students to use statements when you do not know how to proceed and this is one such scenario...

If you do not get hang of it or are short of time, --
There is one variable-A and if we know A, we can answer for sure.
Statement I gives two values of A as it is a quadratic equation. It may be possible that both lead to prime or non-prime, but in most cases it will not.
Statement II gives a range, so insuff
Combined, I says 2 and 3, II says>2, so a unique value C


Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\))..
Two ways..

(I) Don't waste time finding an equation in variable here
We know that to know the distance here, we will require to know the value of A, so let us check the statements
(1) \(A^2 – 5A – 6 = 0...(A-3)(A-2)=0\) A =2 or A=3..
A is 2.. \(K=(2,0)\) and G\(=(8,\sqrt{13}\)...Distance = \(\sqrt{36+13}=\sqrt{49}=7\)..Yes
A is 3.. \(K=(3,0)\) and G\(=(10,\sqrt{15}\)...Distance = \(\sqrt{49+15}=\sqrt{64}=8\).. No
Insuff

(2) A > 2
Clearly insuff. There may be cases where it is true and places where it may not be true.

Combined..ans is 3 and NO
Suff

C

(II) Simplify the question stem
Distance = \(\sqrt{(2A+4-A)^2+\sqrt{2A+9}^2}=\sqrt{A^2+10A+25}=\sqrt{(A+5)}^2=A+5\)
So, if A=2, 7 etc ans is YES and if A=3, 4 etc ans is NO
(1) A^2 – 5A – 6 = 0
A= 2 or 3
Insuff

(2) A > 2
Insuff

Combined A=3
Suff

C
chetan2u
factorising is incorrect here. Mistakenly you have calculated (A-3) (A-2)
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TheKingInTheNorth
Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\)). Is the distance between point K and G prime?

(1) A^2 – 5A – 6 = 0
(2) A > 2

I suggest my students to use statements when you do not know how to proceed and this is one such scenario...

If you do not get hang of it or are short of time, --
There is one variable-A and if we know A, we can answer for sure.
Statement I gives two values of A as it is a quadratic equation. It may be possible that both lead to prime or non-prime, but in most cases it will not.
Statement II gives a range, so insuff
Combined, I says 2 and 3, II says>2, so a unique value C


Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\))..
Two ways..

(I) Don't waste time finding an equation in variable here
We know that to know the distance here, we will require to know the value of A, so let us check the statements
(1) \(A^2 – 5A – 6 = 0...(A-3)(A-2)=0\) A =2 or A=3..
A is 2.. \(K=(2,0)\) and G\(=(8,\sqrt{13}\)...Distance = \(\sqrt{36+13}=\sqrt{49}=7\)..Yes
A is 3.. \(K=(3,0)\) and G\(=(10,\sqrt{15}\)...Distance = \(\sqrt{49+15}=\sqrt{64}=8\).. No
Insuff

(2) A > 2
Clearly insuff. There may be cases where it is true and places where it may not be true.

Combined..ans is 3 and NO
Suff

C

(II) Simplify the question stem
Distance = \(\sqrt{(2A+4-A)^2+\sqrt{2A+9}^2}=\sqrt{A^2+10A+25}=\sqrt{(A+5)}^2=A+5\)
So, if A=2, 7 etc ans is YES and if A=3, 4 etc ans is NO
(1) A^2 – 5A – 6 = 0
A= 2 or 3
Insuff

(2) A > 2
Insuff

Combined A=3
Suff

C
chetan2u
factorising is incorrect here. Mistakenly you have calculated (A-3) (A-2)

Thanks .. Yes, I calculated it for A^2-5A+6
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TheKingInTheNorth
Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\)). Is the distance between point K and G prime?

(1) A^2 – 5A – 6 = 0
(2) A > 2

Point K = (A, 0), Point G = (2A + 4, \(\sqrt{2A + 9}\)). Is the distance between point K and G prime?

Distance between K & G = \(\sqrt{(2A+4-A)^2 + (2A+9)} = \sqrt{(A+4)^2 + (2A+9)} = \sqrt{A^2 + 8A + 16 + 2A + 9} = \sqrt{A^2 + 10A + 25} = A+5\)
Q. A+5 is prime?

(1) A^2 – 5A – 6 = 0
(A-5/2)^2 = 6 + 25/4 = 49/4
A = 5/2 +- 7/2
A = 6 or -1
A + 5 = 11 or 4; 11 is prime but 4 is not a prime number
NOT SUFFICIENT

(2) A > 2
If A = 3; A+5 = 8 is not a prime number
If A = 6; A+5 = 11 is a prime number
NOT SUFFICIENT

(1) + (2)
(1) A^2 – 5A – 6 = 0
(A-5/2)^2 = 6 + 25/4 = 49/4
A = 5/2 +- 7/2
A = 6 or -1
A + 5 = 11 or 4; 11 is prime but 4 is not a prime number
(2) A > 2
A = 6; A+5 = 11 is a prime number
SUFFICIENT

IMO C
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\(Distance \ = \sqrt{A^2-6A+25}\)

\(Statement \ 1) \ \ (A-6)(A+1)=0 \implies A_1=6 \ (YES) \ A_2=-1 \ (NO)\)

\(Statement \ 2) \ \ A>2\)


\(Statement \ 1 \& \ 2) A=6 \ (YES)\)
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