JAMB 2019 Mathematics · Question 24

If P=(Q(R−T)15)13P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}, make T the subject of the formula.

Options

  1. A.T=15R−QP3T = \frac{15R - Q}{P^3}
  2. B.T=R−15P3QT = R - \frac{15P^3}{Q}
  3. C.T=R−15P3QT = \frac{R - 15P^3}{Q}
  4. D.T=R+P315QT = \frac{R + P^3}{15Q}

Answer: B. T=R−15P3QT = R - \frac{15P^3}{Q}

Explanation written with AI and checked against the answer key

To make TT the subject of the formula, we need to isolate it by performing inverse operations step-by-step.

1. Analyze the equation

We are given:

P=(Q(R−T)15)13P = \left( \frac{Q(R - T)}{15} \right)^{\frac{1}{3}}

Our goal is to move everything away from TT until it stands alone on one side of the equation.

2. Step-by-step solution

Step 1: Remove the cube root (the power of 1/31/3) To eliminate the exponent of 13\frac{1}{3}, we raise both sides of the equation to the power of 3:

P3=[(Q(R−T)15)13]3P^3 = \left[ \left( \frac{Q(R - T)}{15} \right)^{\frac{1}{3}} \right]^3
P3=Q(R−T)15P^3 = \frac{Q(R - T)}{15}

Step 2: Clear the fraction Multiply both sides by 15 to remove the denominator:

15P3=Q(R−T)15P^3 = Q(R - T)

Step 3: Isolate the term containing TT Divide both sides by QQ:

15P3Q=R−T\frac{15P^3}{Q} = R - T

Step 4: Solve for TT To get TT by itself, we can move TT to the left side (making it positive) and move 15P3Q\frac{15P^3}{Q} to the right side (making it negative):

T=R−15P3QT = R - \frac{15P^3}{Q}
3. Conclusion

Comparing our result to the given options, we can see that our derived formula matches option B.

4. Exam Tip

When dealing with "change of subject" questions, always remember the Reverse Order of Operations (SADMEP). Usually, you undo addition/subtraction first, then multiplication/division, then exponents/roots. However, because the term (R−T)(R-T) was trapped inside a bracket with a power, we had to deal with the power first!

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