Unit 1 Review Exercises

Numbers, Approximation and the GDC

These questions mix the ideas from the whole unit. Paper 1 marks a short question and Paper 2 marks a longer, structured question. In Applications and Interpretation, a graphic display calculator (GDC) is allowed on both papers.

1 Paper 1 The mass of one grain of sand is \(6.7 \times 10^{-4}\) g.

(a) Find the number of grains of sand in \(1\) kg of sand. Give your answer in the form \(a \times 10^k\), where \(1 \le a < 10\) and \(k \in \mathbb{Z}\).

(b) A student writes the answer to (a) as 1.49E6. Explain why this would not get full marks.

(a) \(1 \text{ kg} = 1000\) g. \(\;\dfrac{1000}{6.7 \times 10^{-4}} = 1\,492\,537\ldots \approx 1.49 \times 10^{6}\) grains.

(b) 1.49E6 is calculator notation. It must be written as \(1.49 \times 10^{6}\). See 1.1.

2 Paper 1 Write \(14.0739\) correct to

(a) 1 decimal place

(b) 2 decimal places

(c) 3 significant figures

(d) 1 significant figure

(a) \(14.1\)

(b) \(14.07\)

(c) \(14.1\)

(d) \(10\) \(\quad\) See 1.2.

3 Paper 1 A photo is \(9.5\) cm long and \(6.3\) cm wide. Both measurements are correct to 1 decimal place.

(a) Write down the lower and upper bounds of the length.

(b) Find the lower and upper bounds of the area of the photo.

(a) \(9.45 \le l < 9.55\).

(b) LB \(= 9.45 \times 6.25 = 59.0625 \text{ cm}^2\). \(\quad\) UB \(= 9.55 \times 6.35 = 60.6425 \text{ cm}^2\). See 1.2.

4 Paper 1 A recipe needs \(375\) g of flour. A cook does not weigh the flour and uses \(400\) g. Find the percentage error.

\[ \begin{aligned} \varepsilon &= \left| \dfrac{400 - 375}{375} \right| \times 100\% \\ &= 6.666\ldots\% \\ &\approx 6.67\% \end{aligned} \] See 1.3.

5 Paper 1 A square floor tile has sides of \(30\) cm, correct to the nearest centimetre. Find the maximum percentage error in the area of the tile.

Measured area: \(900 \text{ cm}^2\). Side: \(29.5 \le s < 30.5\), so the area is between \(870.25\) and \(930.25 \text{ cm}^2\).

\(\left| \dfrac{900 - 870.25}{870.25} \right| \times 100\% = 3.418\ldots\%\) \(\quad\) \(\left| \dfrac{900 - 930.25}{930.25} \right| \times 100\% = 3.251\ldots\%\)

Maximum percentage error: \(3.42\%\). See 1.3.

6 Paper 1 (a) By rounding each number to 1 significant figure, estimate the value of \(\dfrac{61.8 \times 0.49}{2.07}\).

(b) Use your GDC to find the value correct to 3 significant figures, and use (a) to check that it is reasonable.

(a) \(\dfrac{60 \times 0.5}{2} = 15\).

(b) \(14.628\ldots \approx 14.6\). This is close to the estimate \(15\), so it is reasonable. See 1.3.

7 Paper 2 A market stall sells apples, bananas and cherries by the kilogram.

  • Customer 1 buys \(2\) kg of apples, \(1\) kg of bananas and \(3\) kg of cherries for \(\$48.50\).
  • Customer 2 buys \(1\) kg of apples, \(3\) kg of bananas and \(1\) kg of cherries for \(\$24.50\).
  • Customer 3 buys \(3\) kg of apples, \(2\) kg of bananas and \(0.5\) kg of cherries for \(\$24.05\).

(a) Write down a system of three equations for the prices per kilogram.

(b) Find the price per kilogram of each fruit.

(c) Find the cost of \(1\) kg of each fruit together.

(a) Let \(a\), \(b\) and \(c\) be the prices per kg, in dollars.

\[ \begin{aligned} 2a + b + 3c &= 48.50 \\ a + 3b + c &= 24.50 \\ 3a + 2b + 0.5c &= 24.05 \end{aligned} \]

(b) GDC: \(a = 4.2\), \(b = 2.6\), \(c = 12.5\). Apples cost \(\$4.20\), bananas \(\$2.60\) and cherries \(\$12.50\) per kg.

(c) \(4.20 + 2.60 + 12.50 = \$19.30\). See 1.4.

8 Paper 2 The path of a jet of water from a fountain is modelled by \(y = ax^2 + bx + c\), where \(x\) m is the horizontal distance from the fountain and \(y\) m is the height above the ground. The jet passes through \((0, 0.8)\), \((1, 2.3)\) and \((2, 2.8)\).

(a) Find \(a\), \(b\) and \(c\).

(b) Find the horizontal distance from the fountain at which the water hits the ground.

(a) \(c = 0.8\), \(\;a + b + c = 2.3\), \(\;4a + 2b + c = 2.8\). GDC: \(a = -0.5\), \(b = 2\), \(c = 0.8\).

(b) Solve \(-0.5x^2 + 2x + 0.8 = 0\). GDC: \(x = -0.366\) or \(x = 4.37\). A distance cannot be negative, so the water hits the ground \(4.37\) m from the fountain. See 1.4 and 1.5.

9 Paper 2 A water tank is a cylinder with radius \(r\) m and height \((r + 2)\) m. Its volume is \(50 \text{ m}^3\).

(a) Show that \(\pi r^3 + 2\pi r^2 - 50 = 0\).

(b) Find the radius of the tank.

(a) \(V = \pi r^2 h = \pi r^2 (r + 2) = \pi r^3 + 2\pi r^2\). So \(\pi r^3 + 2\pi r^2 = 50\), which gives the result.

(b) Find Roots of Polynomial with coefficients \(\pi\), \(2\pi\), \(0\), \(-50\) (or the intersection of \(y = \pi x^2(x + 2)\) and \(y = 50\)). The only real root is \(r = 1.995\,76\ldots\) The radius is \(2.00\) m (3 s.f.). See 1.5.

10 Paper 2 A car travels \(150\) km, correct to the nearest \(10\) km, in \(1.8\) hours, correct to the nearest \(0.1\) hour.

(a) Find the upper and lower bounds of the average speed of the car.

(b) Calculate the average speed from the rounded values.

(c) Find the maximum percentage error in your answer to (b).

Distance: \(145 \le d < 155\). \(\quad\) Time: \(1.75 \le t < 1.85\).

(a) UB \(= \dfrac{155}{1.75} = 88.57\ldots \approx 88.6 \text{ km h}^{-1}\). \(\quad\) LB \(= \dfrac{145}{1.85} = 78.37\ldots \approx 78.4 \text{ km h}^{-1}\).

(b) \(\dfrac{150}{1.8} = 83.33\ldots \approx 83.3 \text{ km h}^{-1}\).

(c) The exact speed could be at either bound:

\(\left| \dfrac{83.33\ldots - 78.37\ldots}{78.37\ldots} \right| \times 100\% = 6.32\%\) \(\quad\) \(\left| \dfrac{83.33\ldots - 88.57\ldots}{88.57\ldots} \right| \times 100\% = 5.91\%\)

The maximum percentage error is \(6.32\%\). See 1.2 and 1.3.