If \(k\) and \(m\) are numbers such that \(k+m=20\) and

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by Brent@GMATPrepNow » Fri Dec 27, 2019 7:20 am
AAPL wrote:GMAT Prep

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

OA A
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

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by Scott@TargetTestPrep » Fri Jan 10, 2020 12:58 pm
AAPL wrote:GMAT Prep

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

OA A
Squaring the first equation, we have:

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

Substituting, we have:

289 + 2km = 400

2km = 111

km = 55.5

Answer: A

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