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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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Intern
Joined: 24 Jan 2017
Posts: 3
If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then  [#permalink]

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Updated on: 15 Feb 2017, 12:14
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Difficulty:

45% (medium)

Question Stats:

65% (01:25) correct 35% (01:53) wrong based on 111 sessions

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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

Originally posted by lgarces19 on 15 Feb 2017, 12:08.
Last edited by Bunuel on 15 Feb 2017, 12:14, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Math Expert
Joined: 02 Sep 2009
Posts: 50004
Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then  [#permalink]

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15 Feb 2017, 12:17
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.

P.S. Is this question really from GMAT Prep?
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then  [#permalink]

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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.

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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Joined: 12 Sep 2015
Posts: 3020
Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then  [#permalink]

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16 Feb 2017, 10:31
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Top Contributor
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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Brent Hanneson – GMATPrepNow.com

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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then  [#permalink]

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16 Feb 2017, 11:57
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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Abhishek....

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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 = 289, then &nbs [#permalink] 16 Feb 2017, 11:57
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