WAEC 2012 Mathematics · Question 47

Make p the subject of the relation: q = 3pr+s2\frac{3p}{r} + \frac{s}{2}

Options

  1. A.p = 2q−rs6\frac{2q - rs}{6}
  2. B.p = 2qr - sr - 3
  3. C.p = 2qr−s6\frac{2qr - s}{6}
  4. D.p = 2qr−rs6\frac{2qr - rs}{6}

Answer: D. p = 2qr−rs6\frac{2qr - rs}{6}

Explanation written with AI and checked against the answer key

Analyze the question: We need to isolate p in q = 3p/r + s/2.

Concept: Use inverse operations. First move the term without p to the other side, then clear the fractions, then divide by the coefficient of p.

Solution:

  1. Start with: q = 3p/r + s/2
  1. Subtract s/2 from both sides: q − s/2 = 3p/r
  1. Combine the left side over a common denominator: (2q − s)/2 = 3p/r
  1. Cross-multiply: r(2q − s) = 6p
  1. Divide by 6: p = r(2q − s)/6 = (2qr − rs)/6

Check against the options:

  • A: (2q − rs)/6 is wrong because r was not multiplied through the q term.
  • B: wrong form entirely.
  • C: (2qr − s)/6 is wrong because r was not multiplied through the s term.
  • D: (2qr − rs)/6 matches our result.

Tip: When you clear fractions, multiply every term by the denominator. Here r multiplies both 2q and s. A quick check is to substitute simple numbers, such as p = 2, r = 1, s = 2. Then q = 6 + 1 = 7, and option D gives (14 − 2)/6 = 2, which is correct.

AI can make mistakes. Report anything that looks wrong.

Past questions © the examination body (JAMB/WAEC/NECO). DeeepSolve is not affiliated with them. Rights holders: see /copyright.