 ## Presentation on theme: "Percentages Questions and Answers"— Presentation transcript:

Fractions, Decimals and Percentages Finding Percentages Percentage Increase/Decrease Reverse Percentages You tube playlist LINK

Percentages Find 12% of 500 500 X 0.12 Increase 500 by 12% 500 x 1.12
Decrease (100- 70% 7% 16.5% 23% 5.25% 16% 3% 11% Find 12% of X 0.12 Increase 500 by 12% x 1.12 Decrease 500 by 12% x 0.88

Percentages Find 12% of 500 500 X 0.12 Increase 500 by 12% 500 x 1.12
Decrease (100- 70% 0.7 1.7 0.3 7% 0.07 1.07 0.93 16.5% 0.165 1.165 .835 23% 0.23 1.23 0.77 5.25% 0.0525 1.0525 0.9475 16% 0.16 1.16 .84 3% 0.03 1.03 0.97 11% .11 1.11 0.89 Find 12% of X 0.12 Increase 500 by 12% x 1.12 Decrease 500 by 12% x 0.88

N5.4 Increasing and decreasing by a percentage
Contents N5 Percentages N5.1 Fractions, decimals and percentages N5.2 Percentages of quantities N5.3 Finding a percentage change N5.4 Increasing and decreasing by a percentage N5.5 Reverse percentages N5.6 Compound percentages

Percentage increase The value of Frank’s house has gone up by 20% since last year. If the house was worth £ last year how much is it worth now? There are two methods to increase an amount by a given percentage. Method 1 We can work out 20% of £ and then add this to the original amount. The amount of the increase = 20% of £ Stress that this method requires two calculations. One to find the actual increase and one to add this amount to the original value. This method is most useful when we need to know the actual value of the increase. In most cases, however, we only need to know the end result. Although this method is reliable when done correctly, there is more room for error because it is easy to forget to do the second part of the calculation. Completing the calculation in one step as shown in Method 2 is more efficient. = 0.2 × £ = £30 000 The new value = £ £30 000 = £

Percentage increase Method 2
If we don’t need to know the actual value of the increase we can find the result in a single calculation. We can represent the original amount as 100% like this: 100% 20% Remind pupils that 100% of the original amount is like one times the original amount: it remains unchanged. 100% means ‘all of it’. When we add on 20% we are adding on 20% to the original amount, 100%, to make 120% of the original amount. Ask pupils why might it be better to find 120% than to find 20% and add it on. Establish that it’s quicker because we can do the calculation in one step. Discuss how 120% can be written as a decimal. When we add on 20%, we have 120% of the original amount. Finding 120% of the original amount is equivalent to finding 20% and adding it on.

Percentage increase So, to increase £ by 20% we need to find 120% of £ 120% of £ = 1.2 × £ = £ In general, if you start with a given amount (100%) and you increase it by x%, then you will end up with (100 + x)% of the original amount. To convert (100 + x)% to a decimal multiplier we have to divide (100 + x) by 100. This is usually done mentally.

Percentage increase Here are some more examples using this method:
Increase £50 by 60%. Increase £86 by 17.5%. 160% × £50 = 1.6 × £50 117.5% × £86 = 1.175 × £86 = £80 = £101.05 Increase £24 by 35% Increase £300 by 2.5%. Talk through each example as it appears. Explain the significance of finding a 17.5% increase in relation to VAT. 135% × £24 = 1.35 × £24 102.5% × £300 = 1.025 × £300 = £32.40 = £307.50

Percentage decrease A CD walkman originally costing £75 is reduced by 30% in a sale. What is the sale price? There are two methods to decrease an amount by a given percentage. Method 1 We can work out 30% of £75 and then subtract this from the original amount. 30% of £75 The amount taken off = Stress that, as with the similar method demonstrated for percentage increases, this method requires two calculations. One to find the reduction and one to subtract this amount from the original price. = 0.3 × £75 = £22.50 The sale price = £75 – £22.50 = £52.50

Percentage decrease Method 2
We can use this method to find the result of a percentage decrease in a single calculation. We can represent the original amount as 100% like this: 70% 100% 30% Remind pupils again that 100% of the original amount is like one times the original amount, it remains unchanged. 100% is ‘all of it’. Explain that when we subtract 30% we are subtracting 30% from the original amount, 100%, to make 70% of the original amount. Ask pupils why might it be better to find 70% than to find 30% and take it away. Again, establish that it’s quicker. We can do the calculation in one step. Discuss how 70% can be written as a decimal. Conclude that to decrease an amount by 30% we multiply it by 0.7. Give more verbal examples as necessary. When we subtract 30% we have 70% of the original amount. Finding 70% of the original amount is equivalent to finding 30% and subtracting it.

Percentage decrease So, to decrease £75 by 30% we need to find 70% of £75. 70% of £75 = 0.7 × £75 = £52.50 In general, if you start with a given amount (100%) and you decrease it by x%, then you will end up with (100 – x)% of the original amount. To convert (100 – x)% to a decimal multiplier we have to divide (100 – x) by 100. This is usually done mentally.

Percentage decrease Here are some more examples using this method:
Decrease £65 by 20%. Decrease £320 by 3.5%. 80% × £65 = 0.8 × £65 96.5% × £320 = 0.965 × £320 = £52 = £308.80 Decrease £56 by 34% Decrease £1570 by 95%. Talk through each example as it appears. Pupils should be able to confidently subtract any given amount from 100 in their heads. 66% × £56 = 0.66 × £56 5% × £1570 = 0.05 × £1570 = £36.96 = £78.50

Percentage increase and decrease

Contents N5 Percentages N5.5 Reverse percentages
N5.1 Fractions, decimals and percentages N5.2 Percentages of quantities N5.3 Finding a percentage change N5.4 Increasing and decreasing by a percentage N5.5 Reverse percentages N5.6 Compound percentages

Reverse percentages Sometimes, we are given the result of a given percentage increase or decrease and we have to find the original amount. I bought some jeans in a sale. They had 15% off and I only paid £25.50 for them. What is the original price of the jeans? We can solve this using inverse operations. In this example, establish that we must have multiplied the original amount by 0.85 (100% – 15%) to get the new price. To get back to the original amount we can use inverse operations and divide the new amount by 0.85 to get back to the original amount. Explain this process using equations with p for the unknown original price. Let p be the original price of the jeans. p × 0.85 = £25.50 so p = £25.50 ÷ 0.85 = £30

Reverse percentages Sometimes, we are given the result of a given percentage increase or decrease and we have to find the original amount. I bought some jeans in a sale. They had 15% off and I only paid £25.50 for them. What is the original price of the jeans? We can show this using a diagram: Explain how we can use inverse operations to find the original price before the discount using a diagram. Price before discount. × 0.85% Price after discount. ÷ 0.85%

Reverse percentages Talk through as many examples as necessary.

Reverse percentages We can also use a unitary method to solve these type of percentage problems. For example, Christopher’s monthly salary after a 5% pay rise is £ What was his original salary? The new salary represents 105% of the original salary. 105% of the original salary = £ 1% of the original salary = £ ÷ 105 Some pupils may find it easier to remember a unitary method as shown by this example. 100% of the original salary = £ ÷ 105 × 100 = £1250 This method has more steps involved but may be easier to remember.

N5.6 Compound percentages
Contents N5 Percentages N5.1 Fractions, decimals and percentages N5.2 Percentages of quantities N5.3 Finding a percentage change N5.4 Increasing and decreasing by a percentage N5.5 Reverse percentages N5.6 Compound percentages

Compound percentages A jacket is reduced by 20% in a sale.
Two weeks later the shop reduces the price by a further 10%. What is the total percentage discount? It is not 30%! When a percentage change is followed by another percentage change do not add the percentages together to find the total percentage change. It is very common to assume that when two percentage changes are combined, the total percentage change can be found by adding. Discuss why this is incorrect. The second percentage change is found on a new amount and not on the original amount.

Compound percentages A jacket is reduced by 20% in a sale.
Two weeks later the shop reduces the price by a further 10%. What is the total percentage discount? To find a 20% decrease we multiply by 80% or 0.8. To find a 10% decrease we multiply by 90% or 0.9. A 20% discount followed by a 10% discount is equivalent to multiplying the original price by 0.8 and then by 0.9. original price × 0.8 × 0.9 = original price × 0.72

Compound percentages A jacket is reduced by 20% in a sale.
Two weeks later the shop reduces the price by a further 10%. What is the total percentage discount? The sale price is 72% of the original price. This is equivalent to a 28% discount. A 20% discount followed by a 10% discount A 28% discount

Compound percentages A jacket is reduced by 20% in a sale.
Two weeks later the shop reduces the price by a further 10%. What is the total percentage discount? Suppose the original price of the jacket is £100. After a 20% discount it costs 0.8 × £100 = £80 After an other 10% discount it costs 0.9 × £80 = £72 £72 is 72% of £100. 72% of £100 is equivalent to a 28% discount altogether.

Compound percentages Jenna invests in some shares.
After one week the value goes up by 10%. The following week they go down by 10%. Has Jenna made a loss, a gain or is she back to her original investment? To find a 10% increase we multiply by 110% or 1.1. To find a 10% decrease we multiply by 90% or 0.9. Pupils may wish to discuss this problem by giving the original investment a value, £100 say. If this increases by 10% Jenna will have £110. When £110 is reduced by 10% Jenna will have £99. This is equivalent to a 1% loss overall. original amount × 1.1 × 0.9 = original amount × 0.99 Fiona has 99% of her original investment and has therefore made a 1% loss.

Compound percentages Use this slide to demonstrate the effect of following a percentage change with a second percentage change. Ask pupils to predict whether the final amount is going to be greater than or less than the original amount. Discuss methods to find the result of multiple percentage changes.

Compound interest Jack puts £500 into a savings account with an annual compound interest rate of 6%. How much will he have in the account at the end of 4 years if he doesn’t add or withdraw any money? At the end of each year interest is added to the total amount in the account. This means that each year 5% of an ever larger amount is added to the account. To increase the amount in the account by 5% we need to multiply it by 105% or 1.05. We can do this for each year that the money is in the account.

Compound interest At the end of year 1 Jack has £500 × 1.05 = £525
(These amounts are written to the nearest penny.) We can write this in a single calculation as We can see that the actual increase gets larger each time. In the first year £25 is added on. In the second year it’s £26.25, in the third year it’s £27.56 and in the fourth year it’s £28.94. This is an example of exponential growth. Ask pupils to verify that £500 × = £ using the xy key on their calculators. £500 × 1.05 × 1.05 × 1.05 × 1.05 = £607.75 Or using index notation as £500 × = £607.75

Compound interest How much would Jack have after 10 years?
After 10 years the investment would be worth £500 × = £ (to the nearest 1p) How long would it take for the money to double? Using trial and improvement, Again ensure that pupils can use the xy key on their calculators to complete these calculations. £500 × = £ (to the nearest 1p) £500 × = £ (to the nearest 1p) It would take 15 years for the money to double.

Compound interest Use this activity to demonstrate the effect of compound interest.

Repeated percentage change
We can use powers to help solve many problems involving repeated percentage increase and decrease. For example, The population of a village increases by 2% each year. If the current population is 2345, what will it be in 5 years? To increase the population by 2% we multiply it by 1.02. After 5 years the population will be 2345 × = 2589 (to the nearest whole) Ask pupils to calculate each answer using the xy key on their calculators. What will the population be after 10 years? After 5 years the population will be 2345 × = 2859 (to the nearest whole)

Repeated percentage change
The value of a new car depreciates at a rate of 15% a year. The car costs £ in 2005. How much will it be worth in 2013? To decrease the value by 15% we multiply it by 0.85. There are 8 years between 2005 and 2013. Explain that a percentage depreciation is equivalent to a percentage decrease. After 8 years the value of the car will be £ × = £ (to the nearest pound)

Reverse Bought a car 1 year ago and it has lost 45% of its value and is worth £ 3000 now, what did it cost me? ? X .55 = £ so ? = 3000/0.55 = £

Compound Invest £ 5000 for 5 years earns 3% compound interest
5000 x 1.03^5

Percentages Find 12% of 500 500 X 0.12 Increase 500 by 12% 500 x 1.12
Decrease (100- 70% 0.7 1.7 0.3 7% 0.07 1.07 0.93 16.5% 0.165 1.165 .835 23% 0.23 1.23 0.77 5.25% 0.0525 1.0525 0.9475 16% 0.16 1.16 .84 3% 0.03 1.03 0.97 11% .11 1.11 0.89 Find 12% of X 0.12 Increase 500 by 12% x 1.12 Decrease 500 by 12% x 0.88

Fractions, Decimals and Percentages
75% 10% 20% 35% 42% 0.7 0.25 0.3 0.15 0.05 60% 70% 8% 27% 80% 33/100 51/100 4/5 1/5 0.4 0.9 0.74 0.03 7/10 3/5 11/50 7/20 21/50 Home

divide by 10 divide by 2 divide by 4 half the answer double multiply by 9 or subtract 10% from original quantity half 50% 75 200 33 63 154 292.5 35.2 78.72 601.6 41.6 59.4 50.88 Some percentages I can find easily by doing a single sum, what single sums can I do to find: 10% b. 50% c.25% If I know 10% how can I find: 5% b. 1% c. 20 % d. 90% If I know 50% how can I find: 5% b. 25% Find: 30% of 250 b. 40% of c. 15% of d. 75% of 84 35% of 440 b. 65% of c. 16% of d. 82% of 96 94% of 640 b. 8% of c. 27% of d. 53% of 96 Compare you methods for the questions above with a partner, where they the same ? Home

Percentage Increase/Decrease
ANSWERS 319.5 434.52 177.5 636.16 727.04 134.40 157.20 186 195 239.88 73.50 77.91 91.49 162 220.32 90 322.56 368.64 97.20 81.60 25.20 31.80 69.60 333.91 Explain how you would use a calculator to increase an amount by a given percent. Increase the following amounts by 42% £225 £306 £125 £448 £512 A TV costs £120, how much will it cost if its price is increased by: 12% 31% 55% 62.5% 99.9% Simon puts £70 in a bank, each year the money in his bank increase by 5.5%, how much does he have in: 1 year 2 years 5 years? 5. Explain how you would use a calculator to decrease an amount by a given percent. 6. Decrease the following amounts by 28% £225 £306 £125 £448 £512 7. A TV costs £120, how much will it cost if its price is decreased by: 19% 32% 79% 73.5% 42% 8. A car bought for £6, 500 depreciates in value by 12.5% each year, how much will it be worth after: 1 year 2 years 5 years? Home

Reverse Percentages Home
Answers 1.15 1.25 1.04 1.005 1.135 45 65 50 80 240 0.85 0.75 0.96 0.995 0.865 110 60 15 150 3450 What would you multiply an amount by to increase it by: 15% 25% 4% 0.5% 13.5% Find the original prices of these prices that have been increased by the given percentage: Cost= £49.5 after 10% increase Cost= £74.75 after 15% increase Cost= £61 after 22% increase Cost= £104 after 30% increase Cost= £120 after 50% increase I have £252 in my bank account; this is due to me earning 5% interest on what I originally had put in. How much money did I have originally in my bank account? 4. What would you multiply an amount by to decrease it by: 15% 25% 4% 0.5% 13.5% 5. Find the original prices of these items that have been decreased by the given percentage: Cost= £72 after 10% decrease Cost= £93.5 after 15% decrease Cost= £39 after 35% decrease Cost= £4 9fter 40% decrease Cost= £67.50 after 55% decrease 6. A Cars value has dropped by 11.5% it is now worth £ , what was it worth when it was new? Home