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Mathematics Past Questions and Answers

Topic: Change of subject of formula

Jamb Mathematics Questions - Change of subject of formula

Question 16:
Make y the subject of the formula Z = x\(^2\) + \(\frac{1}{y^3}\)
  • A Y = \(\frac{1}{(Z - x^2)^3}\)
  • B Y = \(\frac{1}{(Z + x^2)^{\frac{1}{3}}}\)
  • C Y = \(\frac{1}{(Z - x^2)^{\frac{1}{3}}}\)
  • D Y = \(\frac{1}{\sqrt[3]{Z} - \sqrt[3]{x^2}}\)
View Answer & Explanation
Question 17:
Find p in terms of q if \(\log_{3} p + 3\log_{3} q = 3\)
  • A (\(\frac{3}{q}\))<sup style='font-size: smaller;'>3</sup>
  • B (\(\frac{q}{3}\))<sup style='font-size: smaller;'>\(\frac{1}{3}\)</sup>
  • C (\(\frac{q}{3}\))<sup style='font-size: smaller;'>3</sup>
  • D (\(\frac{3}{q}\))<sup style='font-size: smaller;'>\(\frac{1}{3}\)</sup>
View Answer & Explanation
Question 18:
Make t the subject of formula S = ut + \(\frac{1}{2} at^2\)
  • A \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))
  • B \(\frac{1}{a}\) {u \(\pm\) (U<sup style='font-size: smaller;'>2</sup> - 2as)}
  • C \(\frac{1}{a}\) {u \(\pm\) \(\sqrt{2as}\)}
  • D \(\frac{1}{a}\) {-u + \(\sqrt{( 2as)}\)}
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Question 19:
Find T in terms of K, Q and S if S = 2r(\(\piQT + K)
  • A \(\frac{S^2}{2 \pi r^2Q} - \frac{k}{Q}\)
  • B \(\frac{S^2}{2 \pi r^2Q}\) - k
  • C \(\frac{S^2}{4 \pi r^2Q} - \frac{k}{Q}\)
  • D \(\frac{s^2}{4 \pi r^2Q}\)
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Question 20:
Two variables x and y are such that \(\frac{dy}{dx}\) = 4x - 3 and y = 5 when x = 2. Find y in terms of x
  • A 2x<sup style='font-size: smaller;'>2</sup> - 3x + 5
  • B 2x<sup style='font-size: smaller;'>2</sup> - 3x + 3
  • C 2x<sup style='font-size: smaller;'>2</sup> - 3x
  • D 4
View Answer & Explanation