### What are Word Problems?

Word problems present a **calculation** to you **in text**.

These are often the most challenging type of question because you need to really **understand the question** and then you must **create your own calculation** to find the answer.

These questions will test bits of knowledge from lots of maths topics and so more advanced questions are usually of this type. The best way to prepare for them is to know the rest of the lessons really well and then to pay a lot of attention to the question to help you form a sum using your knowledge.

Read through the examples in the next section to see how you can **tackle word problems**.

### Examples

Let’s go through a few examples.

Example 1

**Question**: A spaceship has to be refuelled after each flight. The cost of refuelling is £1018. If it is launched once every month, what will be the total cost of refuelling in one year?

If you read the question carefully you can **prepare your own calculation**. It says the spaceship will be launched once every month, so in one year it will be launched **12** times. Every time it is launched it needs to be refuelled costing **£1018**. So our calculation is:

12 x 1018 = **?**

We need to work this out as a long multiplication:

Our answer is **£12216**

Example 2

**Question**: Terry goes to the shop with a £20 note, he buys 5 packs of crisps at 80p each, 6 pints of milk at 55p per pint, and a chocolate bar costing £1.50. How much change does he get from his £20 note?

We need to make our own calculation based on the information. When money is involved in pence and pounds, as it is here it can help to work with everything in pounds, so 80p becomes £0.80:

He has 5 x 0.80 = £4.00 (crisps)

He has 6 x 0.55 = £3.30 (milk)

And he has a £1.20 chocolate bar. Now we can add these together:

4 + 3.3 + 1.2 = £8.50

The question asks ‘how much change does he get from his £20 note?’ So our final sum is:

20.00 – 8.50 = **£11.50**

Example 3

**Question**: If I take 25% of a number, divide it by 3 and then add 2³ I get 13. What is the number I started with?

When the question asks something like ‘what number did I start with’, we need to work backwards through the sum in steps, reversing each operation.

So the last bit of information we are given is that ‘something’ + 2³ = 13. Let’s set this out as a sum with a ‘?’ for the number we don’t know:

? + 2³ = 13

We can work out that 2³ = 8 (2³ means 2 x 2 x 2). So the sum is:

? + 8 = 13

We can rearrange this to:

? = 13 – 8

So…

? = 5

Now if we look back at the question we need to reverse the next step. The question says that ‘25% of a number **divided by 3**‘ = 5

So lets lay this out as a sum:

? ÷ 3 = 5

Rearrange the sum by moving the division over and changing it to multiplication:

? = 3 x 5

So…

? = 15

Now we know that ‘25% of a number’ = 15:

25% of ? = 15

Well if 15 is 25% of the number, 100% of the number must be 15 x 4 (because there are 4 lots of 25% in 100%):

15 x 4 = **60**

Example 4

**Question**: By how much is 10% of 66 greater than one fifth of 25?

Let’s work out the two values seperately and then compare them.

To find 10% of something you divide it by 10:

66 ÷ 10 = 6.6

To find one fifth of 25 we divide it by 5:

25 ÷ 5 = 5

The question asks us by how much 6.6 is greater than 5. This means we need to subtract 5 from 6.6:

6.6 – 5 = ?

If you can do this in your head then that is great but let’s write it down as a decimal subtraction sum (I am writing 5 as 5.0 because they are the same thing and it helps to make the sum clearer):

Example 5

**Question**: A glass of orange juice weighs 15kg, if the glass weighs 11kg, how much does the orange juice weigh?

If the orange juice and the glass weigh 15kg. And the glass alone weighs 11kg. Then the orange juice weighs the weight of both orange juic and glass minus the weight of the glass:

15 – 11 = **4kg**