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If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then

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Bunuel
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  15 Feb 2017, 12:17
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lgarces19 wrote:
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200


Square k + m = 20;

(k + m)^2 = 20^2;

k^2 + 2km + m^2 = 400.

Since given that k^2 + m^2 = 289, then 289 + 2km = 400.

km = 111/2 = 55.5.

Answer: A.

P.S. Is this question really from GMAT Prep?
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BrentGMATPrepNow
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  16 Feb 2017, 10:31
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lgarces19 wrote:
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200


Bunuel's solution is perfect.
I just want to point out that there's no shortage of questions that test the concepts tested here.

Important expansions: (x + y)² = x² + 2xy + y² and (x - y)² = x² - 2xy + y²
So, if you see questions that include information about x² + y² and 2xy (or just xy), then you should consider their relationship with the above expansions.

Cheers,
Brent
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  15 Feb 2017, 14:47
Bunuel wrote:
lgarces19 wrote:
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200


Square k + m = 20;

(k + m)^2 = 20^2;

k^2 + 2km + m^2 = 400.

Since given that k^2 + m^2 = 289, then 289 + 2km = 400.

km = 111/2 = 55.5.

Answer: A.

P.S. Is this question really from GMAT Prep?



clearly, I need to brush up my quant skills...I took into consideration that k and m are integers...damn!
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Abhishek009
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  16 Feb 2017, 11:57
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lgarces19 wrote:
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200


\(( k + m ) ^2= k^2 + m^2 + 2km\)

So, \(( 20 ) ^2= 289 + 2km\)

Or, \(400 = 289 + 2km\)

Or, \(2km = 111\)

Or, \(km = 55.5\)

Thus, km must be between 20 and 60, answer must be option (A) between 20 and 60
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  24 Sep 2019, 01:31
mvictor wrote:
Bunuel wrote:
lgarces19 wrote:
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200


Square k + m = 20;

(k + m)^2 = 20^2;

k^2 + 2km + m^2 = 400.

Since given that k^2 + m^2 = 289, then 289 + 2km = 400.

km = 111/2 = 55.5.

Answer: A.

P.S. Is this question really from GMAT Prep?



clearly, I need to brush up my quant skills...I took into consideration that k and m are integers...damn!


Different year - same problem.

It's funny how the test makers can correctly measure the probability of test takers to fall into certain traps.
Because after you see it the light bulb goes on, but when you actually in the "zone" and calculating you don't even look left or right and watch for signs.

I did not even consider that the numbers might not be integers. Beginner mistake!
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BrentGMATPrepNow
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  27 Dec 2019, 08:19
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lgarces19 wrote:
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200


GIVEN: k + m = 20 and k² + m² = 289

Take: k + m = 20
Square both sides to get: (k + m)² = 20²
Expand and simplify the left side: k² + 2km + m² = 400
Rewrite as: (k² + m²) + 2km= 400
Substitute to get: (289) + 2km= 400
Subtract 289 from both sides to get: 2km = 111
Divide both sides by 2 to get: km = 55.5

Answer: A

Cheers,
Brent
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  15 Aug 2020, 04:32
Expert Reply
lgarces19 wrote:
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200


Solution:

Since k + m = 20, we have:

(k + m)^2 = 20^2

k^2 + m^2 + 2km = 400

Since k^2 + m^2 = 289, we have:

289 + 2km = 400

2km = 111

km = 55.5

Answer: A
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  15 Aug 2020, 08:22
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(A) between 20 and 60 --> correct: \(k^2 + m^2 = 289\) & k + m = 20 => \( (k + m)^2 = (20)^2 => k^2+m^2+2km = 400 => 289+2km=400=> 2km=111 => km = 55.5\), 55.5 is between 20 and 60, so km is between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  15 Aug 2020, 09:16
Expert Reply
k + m = 20

On squaring both the sides, we get

\((k + m )^2\) = 400

On expanding, \(k^2\) + \(m^2\) + 2km = 400

Given: \(k^2\) + \(m^2\) = 289

=> 289 + 2km = 400

=> 2km = 400-289 = 111

=> km = \(\frac{111 }{ 2}\) = 55.5

It lies between 20 and 60.

Answer A
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  15 Aug 2020, 09:19
Asked: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then the value of the product km is

(k + m)^2 = 400 = k^2 + m^2 + 2km = 289 + 2km
2km = 400 - 289 = 111
km = 55.55

IMO A
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If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  23 Sep 2020, 09:21
How come this is not solvable by this manner?
(20-M)^2 + M^2 = 289
...
2M^2-40M+111 = 0
Factoring this into form (2M... ) (M....)

Or this method?

Plugging this into Quadratic Equation ----> 2M^2-40M+111 = 0

The crucial step of converting k + m = 20 into this --> (k + m)^2 = 20^2; never occurred to me. How would I know next time NOT to take the two steps I took above and instead identify this key step which makes the numbers work out nicely? Perhaps there was a clue in the problem I should have identified? Thank you!
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  23 Sep 2020, 11:48
k+m = 20
(k+m)^2 = 400
k^2 + m^2 + 2km = 400
2km = 111
km = 55.5
20<55.5<60
So, A
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  24 Oct 2020, 13:23
Given: k + m = 20 and k^2 + m^2 = 289, what is km?

Notice 289 is the square of 17; thus, it's not possible for k and m to be integers.

Square k + m = 20 to get k^2 + 2km + m^2 = 400.

We're given k^2 + m^2 is 289:

400 - 289 = 111

2km = 111
km = 55.5

Answer is A.
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink] New post  26 Mar 2023, 06:05
k and m are numbers such that k + m = 20 and k^2 + m^2 = 289

Helpful to know - 289 is basically 17^2

k + m = 20
square both sides
k^2 + m^2 + 2km = 20^2
or
17^2 + 2km = 20^2
or
KM = (0.5)(20^2 - 17^2) = (0.5)(37)(3)
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then [#permalink]
26 Mar 2023, 06:05
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