Problem Solving

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

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.

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.)

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.

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 ...)

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