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WAEC 2015 Mathematics · Question 17

Make K the subject of the relation T = TK−HK−H\sqrt{\frac{TK - H}{K - H}}

Options

  1. A.K = H(T2−1)T2−T\frac{H(T^2 - 1)}{T^2 - T}
  2. B.K = HT(T−1)2\frac{HT}{(T - 1)^2}
  3. C.K = H(T2+1)T\frac{H(T^2 + 1)}{T}
  4. D.K = H(T−1)T\frac{H(T - 1)}{T}

Answer: A. K = H(T2−1)T2−T\frac{H(T^2 - 1)}{T^2 - T}

Explanation written with AI and checked against the answer key

Note: the relation as written is T = √((TK − H)/(K − H)). Step 1: Remove the square root by squaring both sides T² = (TK − H)/(K − H) Step 2: Cross-multiply T²(K − H) = TK − H Step 3: Expand T²K − T²H = TK − H Step 4: Collect the K terms on one side T²K − TK = T²H − H Step 5: Factorise both sides K(T² − T) = H(T² − 1) Step 6: Divide by (T² − T) K = H(T² − 1)/(T² − T) This matches option A. Check: Take T = 2, H = 1. Then K = 1(4 − 1)/(4 − 2) = 3/2. Substituting back: (TK − H)/(K − H) = (3 − 1)/(1.5 − 1) = 2/0.5 = 4, and √4 = 2 = T ✓. Tip: When the unknown appears on both sides of a relation, square (if needed), cross-multiply, then group all the terms containing the subject on one side and factorise. This is a standard WASSCE "change of subject" pattern.

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