Daily Maths Revision – Week 14 Walkthrough

If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

If it’s a formula you need, here is the one that is relevant this week.

Area of a Trapezium
Area = \(\frac{1}{2}(a + b)h\)

This week, we’re exploring:

  1. Bound calculations
  2. Equations of parallel lines
  3. Circle theorems
  4. Area of a trapezium
  5. Substituting into functions

Question 1: Bound Calculations

Error intervals are essentially an inequality showing the values that a rounded number could take.

Bound calculations use the maximum and minimum values that rounded numbers could take to find the maximum or minimum value that a calculation could take.

A number, \(x\), is rounded to 1 decimal place. The result is 46.8. Give the error interval for \(x\).

The smallest number that can round up to 46.8 is 46.75 and the smallest number that can round down to 46.8 is just below 46.85 – we use the \(<\) sign to show it can’t equal 46.85. Therefore, the error interval for \(x\) is \( 46.75 ≤ x < 46.85\).

A particle travels 46.8m, to one decimal place, in 34 seconds, to the nearest second. Find the maximum speed of the particle. Give you answer to five decimal places.

To find the maximum speed, we want the highest possible distance travelled in the smallest possible time. In other words, we divide the upper bound of the distance by the lower bound of the time.

Maximum speed = \(\frac{46.85}{33.5}\)
                                = \(1.39851\)m/s (5d.p.)

A particle travels 46.8m, to one decimal place, in 34 seconds, to the nearest second. By considering bounds, find the speed of the particle to a suitable degree of accuracy.

In this case, we need to find the maximum and minimum values for the speed and then find the most accurate number that they both round too.

Minimum speed = \(\frac{46.75}{34.5}\)
                                = \(1.35507\)m/s (5d.p.)

We can see that both the maximum and minimum speed round to 1.4m/s when rounding to one decimal place, so this is the most accurate answer we can be certain of. 

Question 2: Equations of Parallel Lines

Any straight line can be written in the form \(y=mx+c\) where \(m\) is the gradient of the line and \(c\) is the \(y\)-intercept. This week, we aren’t given the gradient of the line. Instead, we are given the equation of a line that is parallel.

If two lines are parallel, then their gradients are the same. So, if we can identify the gradient in one line, we have the gradient of the other line.

A straight line is parallel to the line with equation \(y = 7x – 5\) and passes through the point (2, 3). Work out the equation of the line in the form \(y=mx+c\).

The gradient of the line with equation \(y=7x-5\) is the coefficient of \(x\), 7.  Putting this in the form \(y=mx+c\) gives: \(y=7x+c\)

Since the line passes through the point (2, 3), we can substitute these coordinates in to find the value of \(c\). Remember, coordinates are in the format \((x, y)\) so \(x\) = 2 and \(y\) = 3.

\(\begin{aligned} y &= 7x + c \\ 3 &= 7 \times2+c \\ 3&=14+c\\c&=\text{-}11\end{aligned}\)

Putting this into our equation gives us our final answer: \(y = 7x – 11\).

Question 3: Circle Theorems

These are the two circle theorems that are relevant this week.

The point O is the centre of a circle and the points A, C and D lie on the circumference. The radius OD is a perpendicular to the chord AC and crosses it at the point B.  Given that the length of AC is 12cm, find the length of AB. Give a reason for your answer.

The perpendicular from the centre to a chord bisects the chord. Therefore, AB is simply 12 ÷ 2 = 6cm.

A circle has two tangents at points B and C. The tangents meet at point A. Find the size of angle ABC, giving reasons for your answer.

Two tangents to a circle from a point are of equal length so AB = AC and the triangle ABC is isosceles. Since angles in a triangle add up to 180° and the base angles in an isosceles triangle are equal:

\( \begin{aligned} ABC&=(180-30)÷2\\&= 150÷2\\&=75°\end{aligned}\)

Question 4: Area of a Trapezium

To find the area of a trapezium, we add together the lengths of the parallel sides then multiply the result by the perpendicular height and divide by 2. In other words, if we call the parallel sides \(a\) and \(b\):

Area = \(\frac{1}{2}(a + b)h\)

Find the area of the trapezium below.

\(\begin{aligned} area &= \frac{1}{2}(a + b)h \\ &= \frac{1}{2}(10 + 17) \times 9 \\ &= 121.5cm^2 \end{aligned} \)

Question 5: Substituting into Functions

Substituting into functions is no more complex than substituting in general – it’s just about knowing the notation. \(f(x)\) simply means this is a function with the variable \(x\). So, \(f(3)\) just means substitute \(x\) for 3.

The function \(f\) is defined as \(f(x)=2x^2+4x\). Find the value of f(3).


Don’t forget to read even more of our blogs here and you can find our main Daily Maths Revision Page here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.

Leave a Reply