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The sum of the squares of three numbers is 138, while the

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by DavidG@VeritasPrep » Tue Oct 03, 2017 6:37 am
nkmungila1 wrote:The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

A. 20
B. 30
C. 40
D. 50
E. 60
This is kind of an odd question. You could mix and match perfect squares until you find numbers that work. 8^2 + 7^2 + 5^2 = 64 + 49 + 25 = 138. There's actually no reason to use the second piece of information here (which could have been worded a bit more clearly), but if we took 8, 7, and 5, there are three products we can generate by selecting two of the three numbers and then multiplying them: 8*7, 8*5, and 7*5, or 56, 40, and 35. These do, in fact, sum to 131.

8+7+5 =20. And the answer is A.
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by DavidG@VeritasPrep » Tue Oct 03, 2017 6:41 am
nkmungila1 wrote:The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

A. 20
B. 30
C. 40
D. 50
E. 60
(Worth noting that 11^2 + 4^2 + 1^2 would also get us to 138, but because 11 + 4 + 1 = 16 isn't an option, we can discard this possibility without testing to see if the second condition would be met.)
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by EconomistGMATTutor » Sat Oct 07, 2017 2:52 pm
nkmungila1 wrote:The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

A. 20
B. 30
C. 40
D. 50
E. 60
Hi nkmungila1,
Let's take a look at your question.

This is a very easy algebraic question if we just take a closer look at it.
Let the three numbers be x, y and z.

The question states that,
"The sum of the squares of three numbers is 138."
x^2 + y^2 + z^2 = 138 ---- (i)

"The sum of their products taken two at a time is 131."
xy + yz + zx = 131 --- (ii)

We are asked to find the sum of the numbers.
x + y + z = ?

We will use the following polynomial identity to solve this problem.
(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)
Plugin the known values from (i) and (ii)
(x + y + z)^2 = 138 + 2(131)
(x + y + z)^2 = 138 + 262
(x + y + z)^2 = 400

Taking square root on both sides,
x + y + z = √400
x + y + z = 20

The sum of the numbers is 20.
Therefore, Option A is true.

I am available if you'd like any followup.
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by Matt@VeritasPrep » Thu Oct 12, 2017 9:30 pm
DavidG@VeritasPrep wrote:
nkmungila1 wrote:The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

A. 20
B. 30
C. 40
D. 50
E. 60
(Worth noting that 11^2 + 4^2 + 1^2 would also get us to 138, but because 11 + 4 + 1 = 16 isn't an option, we can discard this possibility without testing to see if the second condition would be met.)
Those were the first squares that popped into my head - I was pretty surprised that there was another set of three squares that sums to 138! (which, apropos of nothing, has been a favorite number of mine ever since I heard that Misfits song about it 20 years ago ...)
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by DavidG@VeritasPrep » Fri Oct 13, 2017 7:15 am
Matt@VeritasPrep wrote:
DavidG@VeritasPrep wrote:
nkmungila1 wrote:The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

A. 20
B. 30
C. 40
D. 50
E. 60
(Worth noting that 11^2 + 4^2 + 1^2 would also get us to 138, but because 11 + 4 + 1 = 16 isn't an option, we can discard this possibility without testing to see if the second condition would be met.)
Those were the first squares that popped into my head - I was pretty surprised that there was another set of three squares that sums to 138! (which, apropos of nothing, has been a favorite number of mine ever since I heard that Misfits song about it 20 years ago ...)
I'm making an official request that we all make a more concerted effort to make more GMAT-Misfit connections on these boards :)
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Re: The sum of the squares of three numbers is 138, while the

by Scott@TargetTestPrep » Sun Feb 02, 2020 10:06 am
nkmungila1 wrote: ↑
Tue Oct 03, 2017 2:44 am
 The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

A. 20
B. 30
C. 40
D. 50
E. 60

Let the three numbers be a, b, and c. We can create equations:

a^2 + b^2 + c^2 = 138

and

ab + ac + bc = 131

Instead of solving the two equations above, let’s expand (a + b + c)^2 since we are asked to determine a + b + c:

(a + b + c)(a + b + c)

a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc

a^2 + b^2 + c^2 + 2(ab + ac + bc)

Substituting 138 for a^2 + b^2 + c^2 and 131 for ab + ac + bc, we have:

(a + b + c)^2 = 138 + 2(131)

(a + b + c)^2 = 400

Taking the square root of both sides, we have:

a + b + c = 20

Answer: A

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