Daily Maths Revision – Week 5 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, this is the relevant one this week.

Arc Length

\text{arc length} = \frac{\theta}{360} \times \pi d

This week, we’re exploring:

  1. Probability Tree Diagrams
  2. Interquartile Range
  3. Arcs
  4. Similar Shapes
  5. Factorising into Double Brackets

Question 1: Probability Tree Diagrams

Probability trees are a brilliant way to organise different outcomes and their probabilities in a clear and easy-to-use way.

Let’s say the probability of Hugo winning a game is 0.7. He is going to play the game twice. What is the probability that he wins at least one game?

We should always use a probability tree to answer questions like this. As your teacher probably tells you, use a ruler or similar to make nice straight lines because neat diagrams are easier to follow. Even using your hand as a ruler is better than nothing!

Draw your tree similar to the one below. For each outcome in the first game, we draw a branch. Then we do the same for the second game, branching out from the outcomes of the first game.

We write the probabilities along the relevant branches. So, for the branches that lead to a win, we write 0.7. To find the probability of a loss, we subtract 0.7 from 1. The probability of Hugo losing is:
1 – 0.7 = 0.3.  

Now, we can use the probability tree to find the chance of each outcome. To do this, we multiply each of the probabilities on the branches to a particular outcome.

So, the probability that Hugo wins both games is the probability he wins the first one multiplied by the probability he wins the second one. We do the same for each of the others.

Now we have these values, there are two methods to find the probability that Hugo wins at least one game. Remember, “at least one” means one or more games.

First, we could add together all the probabilities in which Hugo wins one or two games.

P(wins at least once) = P(WW) + P(WL) + P(LW)
                                        = 0.49 + 0.21 + 0.21
                                        = 0.91

Alternatively, we could subtract the probability that Hugo doesn’t win at all from 1.

P(wins at least once) = 1 – 0.09
                                        = 0.91

Question 2: Interquartile Range

When we have the upper and lower quartile, we can use them to find the interquartile range. The interquartile range is the range of the middle 50% of the data; it will give us an idea as to how spread out the data it. Unlike the range, this won’t be affected by any extreme values.

Find the interquartile range of the data below.

4.7              3.2             4.5             5.1             2.7             5.6

We covered how to find the upper and lower quartile in week 2. We start by ordering the numbers, starting with the smallest:

2.7              3.2              4.5              4.7              5.1              5.6

Lower quartile = 3.2
Upper quartile = 5.1

The interquartile range is simply the upper quartile subtract the lower quartile.

IQR = 5.1 – 3.2
       = 1.9

Question 3: Arcs

A sector is simply a fraction of a circle – you can think of it as a pizza slice. The arc length is the curved section of this sector – you can think about it like the crust of the pizza slice! To find the arc length, we need to find what fraction of the whole circle it is.  

Let’s find the perimeter of this sector. We will give our answer to 1 decimal place.

To find the arc length, we need the diameter. Remember, this is double the radius.

Diameter = 2 × 9.2
                = 18.4cm

Here are two methods – the first follows a simple procedure and the second jumps straight in with a formula. In an exam, either of the methods will get you the marks.

For the first method, we will find the diameter of the whole circle, then find the arc length of a 1° slice and finally, find the arc length of the sector itself.

Sector AngleArc Length
360°π × 18.4 = 57.8…
÷ 360÷ 360
57.8… ÷ 360 = 0.160…
× 117× 117
117°0.160… × 117 = 18.7…

So, the arc length is 18.8cm to 1 decimal place. However, the question asks for the perimeter, not the arc length. So, we need to add the two straight edges to this.

Perimeter = 18.7… + 9.2 + 9.2
                   = 37.2cm (1d.p.)

If you prefer to use the formula, you need to start by identifying the different variables. Remember, \theta represents the angle of the sector.

\theta = 117 ^\text{o}
\theta = 18.4\text{cm}

\begin{aligned} \text{Area} &= \frac{\theta}{360}\times\pi d \\ &= \frac{117}{360} \times \pi \times 18.4 \\ &= 18.7...\text{cm} \end{aligned}

Again, we need to add this to the straight edges to find the perimeter.

\begin{aligned} \text{Perimeter} &= 18.7... + 9.2 + 9.2 \\ &= 37.2\text{cm   (1d.p.)} \end{aligned}

Question 4: Similar Shapes

Two shapes are similar if one is an enlargement of the other. The key to finding missing sides is finding the scale factor.

The two shapes below are similar. The height of the smaller shape is 10cm. Find the height of the larger shape.

First, we compare the two lengths we’re given. By dividing the larger length by the smaller length, we can find the scale factor.

Scale factor for length = 6 ÷ 4
                                           = 1.5

So, if we multiply any of the lengths on the smaller shape by 1.5, we will get the corresponding length on the larger shape.

Height of the larger shape = 10 × 1.5
                                                   = 15cm

Question 5: Factorising into Double Brackets

We first looked at factorising into double brackets in week 2. In that case, all the expressions were of the form x^2+bx+c.

In this case however, the expressions are of the form ax^2+bx+c. These expressions will still factorise into two brackets but the process of factorising them is a little different. We will use the grid method to factorise.

Fully factorise \boldsymbol{6x^2-19x+10}.

First, we need to identify the values of a , b and c . Every quadratic expression can be written in the form ax^2+bx+c . In this case, a=6 , b = \text{-}19 and c=10 .

When we have this, we find the value of ac and the value of b:

\begin{aligned} ac&= 6 \times 10 \\&=60\\b&=\text{-}19 \end{aligned}

Now, we are looking for a pair of numbers that multiply together to give 60 and add together to give -19.

For two numbers to multiply to give a positive but add to give a negative, both numbers must be negative. So, we write out the negative factors of 60:

-1         -60
-2         -30
-3         -20
-4         -15
-5         -12
-6         -10

The only pair of factors that adds to give -19 is -4 and -15. We use these values to split the middle term, \text{-}19x , into two terms, \text{-}4x and \text{-}15x . We put these terms in the top right and bottom left of our grid.

In the top left, we put 6x^2 and in the bottom right we put +10 .

Now we find the largest common factor of 6x^2 and \text{-}4x , which is 2x . We write this on the left-hand side of the grid.

Then, we consider what we need to multiply 2x by to get each of 6x^2 and +4x . In this case, its 3x and \text{-}2 respectively. These terms go across the top of the grid.

Finally, we consider what we multiply 3x by to get \text{-}15x , the result (\text{-}5) is written on the left-hand side.

We can check our answer by multiplying \text{-}5 and\text{-}2, which is +10 . This is the value in the bottom right, so our answer is probably correct.

The two terms across the top make up one of the brackets and the two terms down the side make up the other bracket:

6x^2 -19x+10 = \boldsymbol{(3x-2)(2x-5)}

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