Daily Maths Revision – Week 6 Walkthrough

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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 are the two that are relevant this week. You aren’t given these so you will need to remember them.

volume = \pi r^2h       
surface area = 2\pi r^2 + \pi dh       

This week, we’re exploring:

  1. Probability Tree Diagrams
  2. Cylinders
  3. Lowest Common Multiple with Prime Factors
  4. Exchange Rates
  5. Expanding 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 two 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: Cylinders

The volume of a cylinder can be found in the same way as the volume of a prism. You find the cross-sectional area (the area of the circle) and multiply it by the height. If you prefer a formula, volume = \pi r^2h where r is the radius of the cylinder and h is the height of the cylinder.

Let’s find the volume of the cylinder below. Give your answer in terms of \boldsymbol{\pi} .

radius = 12 ÷ 2
            = 6cm

height = 18cm

volume = \pi × 62 × 18
               = 648\boldsymbol \pi cm3

Now, let’s find the surface area of the cylinder, giving the answer to 2 decimal places.

diameter = 12cm

radius = 12 ÷ 2
            = 6cm

height = 18cm

surface area = 2 × \pi × 62 + \pi × 12 × 18
                        = 288\pi
                        = 904.78cm2

Question 3: Lowest Common Multiple with Prime Factors

For large numbers, finding the prime factors is the easiest way to find the lowest common multiple. We use a Venn diagram to organise the prime factors and see which are common. 

Find the prime factors of 180 and 460 and, hence, find the lowest common multiple of 180 and 460.

Let’s start by finding the prime factors. This was covered in week 1, so you should be relatively confident with it. Otherwise, go back and have a look at that week’s blog.

Now, we place each of the factors in a Venn diagram. If a factor is in both numbers, then we place it in the middle (the intersection). Otherwise, we just put it on the relevant side:

The lowest common multiple is the product of these numbers.

lowest common multiple = 2 × 2 × 3 × 3 × 5 × 23
                                               = 22 × 32 × 5 × 23
                                               = 4140

Any of these three answers would get full marks on a question like this. The first two give the answer as the product of prime factors.

Question 4: Exchange Rates

Exchange rates are a way to convert between two different currencies or two different units of measurement (such as inches and centimetres).

Luke is travelling to New York. He wants to take £450 for spending money. The current exchange rate is £1 = $1.22. How much money, in dollars, will Luke get?

As with proportion, the best way to tackle exchange rates is by setting it up in two columns and making sure you keep the values balanced.

× 450× 450

So, Luke gets $549.

At the end of his holiday, Luke has $79 left. He gets the same exchange rate to change the money back to pounds. How much, in pounds, is Luke left with?

When working in the other direction, start by calculating how much 1 dollar is in pounds.

÷ 1.22÷ 1.22
× 79× 79

So, Luke gets £64.75. Remember, money is always given to 2 decimal places.  

Question 5: Expanding Double Brackets

There are a lot of different methods to expand double brackets. We will look at the grid method as this is the simplest method.

Expand and simplify:

\huge{ \boldsymbol{(2x+1)(3x-7)}}

We start by drawing a grid. Write one of the brackets across the top and the other down the side – it doesn’t matter which way round.

Now, we simply multiply the term in each row by each column. Make sure you are careful with the negatives.

\begin{aligned} (2x+1)(3x-7)&=6x^2+3x-14x-7\\&=\boldsymbol{6x^2-11x-7} \end{aligned}

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