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Theresa bought a house on 2 January 1970 for £8000 - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 2

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Theresa bought a house on 2 January 1970 for £8000. The house was valued by a local estate agent on the same date every 10 years up to 2010. The valuations are sho... show full transcript

Worked Solution & Example Answer:Theresa bought a house on 2 January 1970 for £8000 - AQA - A-Level Maths Pure - Question 8 - 2019 - Paper 2

Step 1

Show that $V = pq$ can be written as $\log_{10} V = \log_{10} p + \log_{10} q$

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Answer

To show that V=pqV = pq can be written as log⁡10V=log⁡10p+log⁡10q\log_{10} V = \log_{10} p + \log_{10} q, we start by taking the logarithm of both sides:

  1. Taking logarithm on both sides:
    log⁡10V=log⁡10(pq)\log_{10} V = \log_{10} (pq)
  2. Using the property of logarithms that states log⁡10(ab)=log⁡10a+log⁡10b\log_{10} (ab) = \log_{10} a + \log_{10} b, we can rewrite the right-hand side:
    log⁡10V=log⁡10p+log⁡10q\log_{10} V = \log_{10} p + \log_{10} q

Thus, it is demonstrated that V=pqV = pq can be expressed as log⁡10V=log⁡10p+log⁡10q\log_{10} V = \log_{10} p + \log_{10} q.

Step 2

The values in the table of $\log_{10} V$ against $t$ have been plotted

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Answer

From the graph plotted with values of log⁡10V\log_{10} V against tt, we observe a linear relationship between the two variables. To determine the line of best fit:

  1. We identify two points from the given data that correspond to their respective tt values and log⁡10V\log_{10} V values.
  2. Calculate the gradient of the line using the formula: slope=(y2−y1)(x2−x1)\text{slope} = \frac{(y_2 - y_1)}{(x_2 - x_1)}, which represents the change in log⁡10V\log_{10} V per change in tt.
  3. Substitute the points (0, 3.90) and (30, 5.31) into this formula to find the slope: slope=5.31−3.9030−0=1.4130≈0.047\text{slope} = \frac{5.31 - 3.90}{30 - 0} = \frac{1.41}{30} \approx 0.047.
  4. The y-intercept can be calculated by substituting one of the points into the equation of a line in the form: y=mx+cy = mx + c to determine cc.

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