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If k and m are numbers such that k + m = 20 and k^2 + m^2 =
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26 May 2020, 04:57
26 May 2020, 04:57
Expert Reply
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Question Stats:
69% (01:00) correct
30% (01:10) wrong based on 46 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
(A) between 20 and 60
(B) between 60 and 100
(C) between 100 and 150
(D) between 150 and 200
(E) greater than 200
Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 =
[#permalink]
26 May 2020, 04:57
26 May 2020, 04:57
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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Re: If k and m are numbers such that k + m = 20 and k^2 + m^2 =
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26 May 2020, 05:12
26 May 2020, 05:12
2
Carcass 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
(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
