How much greater is the square of the sum of three different positive

Question

How much greater is the square of the sum of three different positive integers than the sum of their squares?

(1) The sum of the products of all possible pairs of two different integers out of the original set of three is 61.

(2) The largest of the three integers, 7, is equal to the sum of the two smaller integers.

Kudos for a correct  solution .

King407 · Accepted Answer

(a+b+c)^2 = {a^2}+b^2+c^2 + 2ab+2bc+2ac = {a^2}+b^2+c^2 + 2(ab+bc+ac)

Now from A, ab + bc + ca = 61 => (a+b+c)^2 = {a^2}+b^2+c^2 + 61  - Sufficient

From B, c=7=a+b, for this a,b could be 1,6; 2,5 or 3,4 etc. since there is no single solution - Insufficient

Hence, answer is A

funkyleaf · Answer

To provide sufficient data to answer the question posed in the stem we need to be able to know the value of each of the three integers.

1) If we use the three integers 3,4,5 we get the pairs 3-4,3-5, and 4-5. the sum of the products of these pairs is only 47. Next I guessed 3,4,6 and got 54. Then I tried 3,4,7 and ended up with 61. I could not find any other combination of integers that gave this value to be 61 so we have only one set of three integers, 3,4,7 that can be used to determine the answer for the initial question posed. This is sufficient. Eliminate choices B,C,E.

2) From the information here we could have the pairs 1,6,7 or 2,5,7 or 3,4,7 giving us no clear way to chose a pair of three integers distinctly. This is not sufficient to answer the initial question posed. Eliminate choice D.

Select answer choice A.

I think this reasoning is correct if I am wrong please let me know!

Happy Friday everyone!!

sudh · Answer

Let the three different positives be x, y, z
 (x + y + z)^2 - (x^2 + y^2 + z^2) =\,? 
 x^2 + y^2 + z^2 + 2xy + 2yz + 2zx - x^2 - y^2 - z^2\,=\,? 
 2(xy + yz + zx)\,=\,?

Statement (1):
The sum of the products of all possible pairs of two different integers out of the original set of three is 61.
 xy + yz + zx = 61 
then  2(xy + yz + zx) = 122 
Sufficient

Statement (2):
The largest of the three integers, 7, is equal to the sum of the two smaller integers.
if C = 7, then A = 1 or 2 and B = 6 or 5
Not Sufficient

Answer A

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

Translate this word problem into algebra. First, name the three integers x, y, and z (making x the smallest and z the largest integer, since they’re all different and it might be handy to keep track of their relative sizes).

The question refers to the square of the sum of the integers, or (x + y + z)^2. The question also refers to the sum of the squares of these integers, or x^2+y^2+z^2. Finally, you are asked “how much greater” the first quantity is than the second quantity. In other words, what is their difference:

(x + y + z)^2 – (x^2+y^2+z^2) = ?

Before going further, try to simplify this expression. Expand (x + y + z)^2:

(x + y + z)^2
= (x + y + z)(x + y + z)
= x^2+ xy + xz + yx + y^2 + yz + zx + zy + z^2
= x^2+ y^2+z^2+ 2xy + 2xz + 2yz

So the difference we’re looking for is just 2xy + 2xz + 2yz, or 2(xy + xz + yz). Ultimately, we can simplify the question to this:

xy + xz + yz = ?

Statement (1): SUFFICIENT. This statement tells you that the sum of every possible pair of different integers out of the set of x, y, and z is 61. In algebraic terms, you are told that xy + xz + yz = 61.

Statement (2): INSUFFICIENT: You know that z = 7 = x + y, so you can get rid of two variables in the target expression, as shown:

xy + xz + yz = x(7 – x) + 7x + 7(7 – x) = 7x – x^2 + 7x + 49 – 7x = 7x – x^2 + 49. You do know that x can be no smaller than 1 and no larger than 5 (to allow y to be 6, different from either x or z and between them both). However, these constraints are not enough to determine a single value for the difference you’re looking for.

The correct answer is A.

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How much greater is the square of the sum of three different positive

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