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Solve; \(\frac{P}{2} + \frac{k}{3}\) = 5 and 2p = k = 6 simultaneously

Further Mathematics
WAEC 2019

Solve; \(\frac{P}{2} + \frac{k}{3}\) = 5 and 2p = k = 6 simultaneously

  • A. p = -6, k = -6
  • B. p = -6, k = 6
  • C. p = 6, k = 6
  • D. p = 6, k = -6
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Correct Answer: Option C
Explanation

\(\frac{P}{2} + \frac{k}{3}\) = 5

\(\frac{3p + 2k}{6}\) = 5

2p + 3k = 30

- 2p - k = 6

\(\overline{\frac{4k}{4} = \frac{24}{4}}\) 

k = 6

 

From 2p - k = 6

2p - 6 = 6

\(\frac{2p}{2} = \frac{12}{2}\)

p = 6


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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995