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We have a condition that a < b < c. What is the solution of a^2 + b^2

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We have a condition that a < b < c. What is the solution of a^2 + b^2  [#permalink]

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New post 11 Oct 2017, 06:43
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A
B
C
D
E

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We have a condition that a < b < c. What is the solution of a^2 + b^2 + c^2 = 14?

1) a, b and c are integers.
2) a, b and c are positive.

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We have a condition that a < b < c. What is the solution of a^2 + b^2  [#permalink]

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New post Updated on: 11 Oct 2017, 10:06
MathRevolution wrote:
We have a condition that a < b < c. What is the solution of a^2 + b^2 + c^2 = 14?

1) a, b and c are integers.
2) a, b and c are positive.


the wording of the question is not at all clear. i am assuming that the question is asking to find the value of \(a\), \(b\) & \(c\). Bunuel correct me if I am wrong here.

Statement 1
: we know all are integers and distinct. as \(4^2=16\), hence \(c<4\) or \(c>-4\), so we get

\(a=1\) or \(-3\), \(b=2\) or \(-2\) and \(c=3\) or \(-1\). in either case \(a^2 + b^2 + c^2 = 14\) but there is no unique value of \(a\), \(b\) or \(c\). Hence Insufficient

Statement 2: \(a\), \(b\) & \(c\) can be any positive number (including decimals) less than \(4\). Hence no unique solution. Insufficient

Combining 1 & 2, we know that only possible solution is \(a=1\), \(b=2\) & \(c=3\). Sufficient

Option C

Originally posted by niks18 on 11 Oct 2017, 07:42.
Last edited by niks18 on 11 Oct 2017, 10:06, edited 1 time in total.
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Re: We have a condition that a < b < c. What is the solution of a^2 + b^2  [#permalink]

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New post 11 Oct 2017, 09:59
I agree the questions is somewhat unclear, but it seems to be asking for a, b, and c.

Statement 1: I also believe that from the first statement, we know that the solution contains either 1, 2 and 3 or -1, -2, and -3 with a < b < c. Therefore either a = 1, b = 2 and c = 3 or a = -3, b = -2 and c = -3. I suppose you could even say something like a = -2, b = 1 and c = 3 which could be a solution.

Statement 2: We know now we are dealing with all positive numbers but do not know if they are all integers.

Combining the two, we get that the solution must be a= 1, b = 2 and c = 3.

Choice C is correct.
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We have a condition that a < b < c. What is the solution of a^2 + b^2  [#permalink]

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New post 15 Oct 2017, 06:10
Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution.

There are 3 variables and 0 equation. E is most likely to be the answer.

Condition 1) & 2)
Assume a = 1 and b = 2. Then, we have 1 + 4 +c^2 = 14 or c^2 = 9.
Thus, we have a solution a =1, b = 2 and c =3, since a, b and c are positive integers.

Assume a = 2 and b = 3. Then, we have 4 + 9 + c2 = 14 or c2 = 1.
However, c2 must be greater than 9.
Thus, we can realize that a = 1, b = 2 and c = 3 is a unique solution.

Condition 1)
Both of a = 1, b = 2, c = 3 and a = -1, b = 2, c = 3, are solutions.
Thus, we don’t have a unique solution.
This is not sufficient.

Condition 2)
Both of a = 1, b = 2, c = 3 and a = 1/2, b = 1, c = √51/2, are solutions.
Thus, we don’t have a unique solution.
This is not sufficient either.
Therefore, C is the answer.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both con 1) and con 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Then, E is most likely to be the answer using con 1) and con 2) together according to DS definition. Obviously, there may be cases where the answer is A, B, C or D. .
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We have a condition that a < b < c. What is the solution of a^2 + b^2 &nbs [#permalink] 15 Oct 2017, 06:10
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