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Oh! Hi everyone! In this video we're going to learn how to increase and decrease percentage amounts.
Hi everyone! I'm going to use the bar model to help visualise a percentage decrease.
What is the missing percentage from this bar model? Well, if a bar model equals 100%, decreasing by 30%, finds 70% of my original amount.
It's like saying, 100% subtract 30% equals 70%.
Let's have a look at a percentage increase using a bar model to help us visualise what's happening.
If I increase something that means I'm going to add onto it.
So if I add on 43%, what percentage is the entire bar model? 100% plus 43% finds me 143%.
The bar model on the right represents an increase of 43%.
In this lesson we're going to increase and decrease amounts by decimal multipliers.
If I wanted to decrease a number by 30% using a decimal multiplier, I would say I'm going to multiply it by 0.
7.
Remember, 100% subtract 30% finds you 70% and 70% as a decimal is 0.
7.
If you look at the example on the right, if I want to increase a number by 43%, that means I would say one add 0.
43 my decimal multiplier would be 1.
43.
Can you help me to match up these percentage increases and decreases with their corresponding multipliers? Have a quick look at those questions.
So let's look at question one.
If I want to increase by 30%, then my decimal multiplier would be 1.
3 if I wanted to decrease by 3%, now the decimal equivalent of 3% is 0.
03, so a decrease of 3% would be 0.
97.
If I wanted to increase by 3%, that's what I would find.
And if I wanted to decrease by 30% We've done that already, you would find 0.
7 as your multiplier.
And then finally to increase by 300%, the decimal equivalent 300% is three.
So I just multiply it three times.
Make it three times bigger.
Okay! If you can use decimal multipliers with confidence, we can decrease 1,870 by 30% quite easily.
Here's a bar model.
100% of this bar model is worth 1,870.
Imagine that.
Now we want to decrease the whole amount, 100% by 30% that means we're going to look for 70% of the total amount.
The total amount is 1,870 and I'm going to use my multiplier, I'm going to multiply it by 0.
7, that gives me the answer 1,309.
70% of 1,870 is 1,309.
So let's increase 37 pounds by 43%.
100% of this bar model is worth 37 pounds.
We want to find how much it's worth with the extra 43%.
So if I said 143% of 37 is the same as saying 37 times by the multiplier of 1.
43 and that finds me an answer of 52 pounds and 91 pence.
Here's an example for you to try, pause the video, and return to check your answer.
Here's the solutions to question number one.
You don't have to use the multiplier method you might have been taught other methods, and that's okay if they work and find the right answers, use them.
The multiplier method is nice and quick.
Let's look at an exam-style question.
Teddy earns 25,000 pounds per year.
He receives a 12% pay rise.
Work out his new salary.
If he is getting 12% pay rise, that means we are going to increase 25,000 pounds by 12%.
Using a decimal multiplier, 100% plus 12% gives you 112%, its decimal equivalent is 1.
12, multiply the original amount by that multiplier, will find you his new salary of 28,000 pounds.
Teddy's off on a holiday.
Let's look at a second example.
A phone costs 870 pounds, its price is reduced by 23% in a sale.
Work out the sale price.
So it's reduced by 23%, that's like saying 100% subtract 23%, so I want to find 77% of the original price.
Let's decrease 870 by 23%, we're going to use the decimal multiplier 0.
77 and that finds me the sale price for the phone of 669 pounds and 90 pence.
Here's an example for you to try, pause the video and return to check your answer.
Here are the solutions to question number two, if you add, uh, calculated 110 times by the multiplier of 1.
15, you would have found the right solution there, okay? Top marks for getting this far, uh, here is your last question.
Pause the video and return to check your answers.
Here's the solutions to question number three.
Uh, in your exams, show your working out.
Make annotations to help your examiner find the correct solutions easily.