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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

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by BTGModeratorVI » Sat Aug 08, 2020 7:07 am

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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

Answer: A
Source: GMAT prep
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Source: — Problem Solving |

BTGModeratorVI wrote:
Sat Aug 08, 2020 7:07 am
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

Answer: A
Source: GMAT prep
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
Brent Hanneson - Creator of GMATPrepNow.com
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BTGModeratorVI wrote:
Sat Aug 08, 2020 7:07 am
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

Answer: A
Source: GMAT prep
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

Scott Woodbury-Stewart
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