JAMB 2011 Mathematics · Question 42

T varies inversely as the cube of R. When R = 3, T = 281\frac{2}{81}, find T when R = 2

Options

  1. A.118\frac{1}{18}
  2. B.112\frac{1}{12}
  3. C.124\frac{1}{24}
  4. D.16\frac{1}{6}

Answer: B. 112\frac{1}{12}

Explanation written with AI and checked against the answer key

1. Analyze the question

We are given a relationship of inverse variation: TT varies inversely as the cube of RR. We are provided with an initial set of values (R=3,T=281R=3, T=\frac{2}{81}) and asked to find the value of TT when R=2R=2.

2. Explain the concept

In inverse variation, the relationship is expressed as:

T∝1R3T \propto \frac{1}{R^3}

To turn this into an equation, we introduce a constant of proportionality, kk:

T=kR3T = \frac{k}{R^3}

Our goal is to first find the value of kk using the given values, and then use that kk to solve for the unknown TT.

3. Show the solution

Step 1: Find the constant kk Using the values T=281T = \frac{2}{81} and R=3R = 3:

281=k33\frac{2}{81} = \frac{k}{3^3}
281=k27\frac{2}{81} = \frac{k}{27}

Now, solve for kk by multiplying both sides by 27:

k=2×2781k = \frac{2 \times 27}{81}

Since 81=27×381 = 27 \times 3, we can simplify:

k=23k = \frac{2}{3}

Step 2: Find TT when R=2R = 2 Now that we know k=23k = \frac{2}{3}, our formula is:

T=2/3R3T = \frac{2/3}{R^3}

Substitute R=2R = 2:

T=2/323T = \frac{2/3}{2^3}
T=2/38T = \frac{2/3}{8}
T=23×18T = \frac{2}{3} \times \frac{1}{8}
T=224T = \frac{2}{24}
T=112T = \frac{1}{12}
4. Identify the answer

Comparing our result to the given options: A. 118\frac{1}{18} B. 112\frac{1}{12} C. 124\frac{1}{24} D. 16\frac{1}{6}

The correct option is B.

5. Tip

When dealing with inverse variation, always remember the "Product Rule": if T∝1R3T \propto \frac{1}{R^3}, then T1×R13=T2×R23T_1 \times R_1^3 = T_2 \times R_2^3. This allows you to solve the problem without explicitly finding kk if you prefer!

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