Ratio and proportion
National Curriculum Objectives, Y6, Ratio and proportion • solve problems involving similar shapes where the scale factor is known or can be found • solve problems involving unequal sharing and grouping using knowledge of fractions and multiples
Fractions, ratio or both? Challenge 1:
Challenge 1 Answer: Maria and Neil share £35 in the ratio 2:5. How much more money does Neil have
Neil has £15 more than Maria.
than Maria?
Challenge 2: Challenge 3:
= £35
Oliver and Petra walk a total of 42 km each week to school. Oliver walks 24 km. What is the ratio of kilometres walked each week by the two pupils in its simplest form?
Assessment:
Stefan decides to make a cake for 8 people. He uses 4 eggs, 200 g flour,
Pupils should be able to argue that the sum of money is divided into 7 equal parts: Maria’s 2 parts and Neil’s 5 parts. They should reason that £35 divided by 7 calculates the value of one part in the ratio as £5. By multiplying the respective parts belonging to Neil and Maria by 5, pupils calculate Neil to have £25 and Maria to have £10; they check that these figures total £35. Finally, they subtract Maria’s total from Neil’s to get the answer to the question. Alternatively, they will see that the ratio shows that Neil has 3 times the value of one part more than Maria.
200 g sugar, 200 g butter and 60 g dark chocolate. The following week he makes enough cake for 18 people. How much of each ingredient does he need?
Challenge 4:
The pie chart shows the results of a football team supporters’ survey in a school. The sum of pupils who support Chelsea and Leicester City is 75%. The ratio of Chelsea to
Pupils will move towards solving this type of problem by adding the numbers in the ratio then dividing the sum (in this case £35) by the result (7). When pupils have identified the value of one part, they will multiply this by the number of parts given in the ratio. They will show that this information needs the final operation of subtraction to find the answer to the question.
Leicester supporters is 7 : 8. The sum of Manchester United and Aston Villa supporters
Challenge 2 Answer:
is 45.
The ratio of kilometres walked is 4 : 3.
Assessment:
Manchester United Aston Villa
Pupils will reason that 42 km in total are covered each week by Oliver and Petra and that they cover a fraction of 42 km each. From a concrete perspective, pupils will reason that the fraction of the total journeys that Petra makes is (42 – Oliver’s journey) out of 42.
Chelsea
24 km
Oliver
= 42 km Leicester City
Petra
?
Pupils should continue to make links to fractions and, in particular, equivalent fractions to reason that if Oliver’s 24 18 this can be simplified to 47 . In the same way, Petra’s journey represents 42 of the proportion of the journey is 42 3 total journey which reduces to 7 . As both journeys are reduced to 7 equal parts, the ratio of distance travelled in its simplest form is Oliver : Petra = 4 : 3.
a)
How many pupils said they preferred Leicester City?
b) What fraction of the survey prefer Chelsea?
Pupils will start to take an algebraic approach to this type of problem by initially seeing the problem as an addition of two fractions with identical denominators, of which one numerator is unknown. 24 42
+
n 42
=
42 42
⇒ 24 + n = 42
From this, they calculate 18 km as Petra’s part of the total journey. They continue this idea by reducing both fractions to their simplest form. Once this is completed they know that the ratio of one to the other is simply the numerator values of the simplified fraction. 24 42
2
+
18 42
=
12 21
+
9 21
=
4 7
+
3 7
⇒4:3
3
Challenge 3 Answer: Stefan uses 9 eggs, 450 g flour, 450 g sugar, 450 g butter and 135 g dark chocolate.
Assessment: Pupils will reason that all of the original amounts and the number of people the cake is made for are divisible by 2 and that this will help them to find the numbers for a cake for 18 people (which is also divisible by 2). Finding the amounts required for a cake for 2 people would be a reasonable starting point. Pupils may see from the original information that it would be possible to make a cake for 2 people by dividing all the ingredient amounts by 4. This can be planned on a chart. If these numbers are then multiplied by 9 there will be enough cake for 18 people. Using this type of labelled table to aid systematic working will be beneficial in working through to the eventual solution. Ingredient
8 people
2 people
18 people (9 × 2 people)
Eggs
4
1
9
Flour
200 g
50 g
450 g
Sugar
200 g
50 g
450 g
Butter
200 g
50 g
450 g
Chocolate
60 g
15 g
135 g
Pupils will continue to take an algebraic approach to this type of problem by scaling up the value of each ingredient by a scale factor 2.25 as 18 is the result of 8 × 2.25. This makes links to the distributive property of multiplication over addition. Ingredient
8 people
18 people (2.25 × 8 people)
Eggs
4
(4 × 2) + (4 × 14 ) = 9
Flour
200 g
(200 × 2) + (200 × 14 ) = 450 g
Sugar
200 g
(200 × 2) + (200 × 14 ) = 450 g
Butter
60 g
(200 × 2) + (200 × 14 ) = 450 g
Chocolate
60 g
(60 × 2) + (60 × 14 ) = 135 g
Challenge 4 Answer: a)
72 pupils prefer Leicester City.
b)
63 180
or
7 20
Assessment: Pupils will reason that the sum of Manchester United and Aston Villa supporters is 14 of those surveyed. They see the ratio as a sum of fractions equal to the whole and the resulting ratio as a sum of the equal parts. 1 4
:
3 4
:: 1 : 3
They see that Manchester United plus Aston Villa supporters is 45 and this figure is equal to 14 of the total pupils questioned. Chelsea plus Leicester City supporters total 135, three times this amount. Pupils then reason that, as the ratio of this secondary sum is 7 : 8, there are 15 equal parts, each part being equal to 9 pupils. Linking this, they calculate that there are 63 Chelsea supporters and 72 Leicester City supporters. In part a) pupils show that adding 135 to 45 calculates the total 63 number of pupils in the survey as 180. This information shows the fraction of Chelsea supporters as 180 or its equivalent. Pupils construct the problem in different but effective ways. Manchester United + Aston Villa : Chelsea + Leceister City : :
1 4
:
3 4
:: 1 : 3
1: 3 ::1( 45) : 3 ( 45) :: 45 :135 45
135
As the total number of pupils in the survey is 180, they convert this information into 180 and 180 . With 75% being Chelsea and Leicester supporters in the ratio 7 : 8, division of 135 by 15 (7 + 8) results in the ratio 7 × 9 : 8 × 9. Pupils will calculate 63 7 63 the fraction for Chelsea supporters to be 180 (reduced this to 20 ). Allow for the misconception of the result of 135 .
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Ratio and Proportion
National Curriculum Objectives, Y6, Ratio and proportion • solve problems involving similar shapes where the scale factor is known or can be found • solve problems involving unequal sharing and grouping using knowledge of fractions and multiples
Ratio in parts Challenge 5:
Challenge 5 Answer: You will need cubes and counters. You can exchange 3 cubes for 12 counters. a)
How many counters can you exchange for 8 cubes?
a)
8 cubes are equivalent to 32 counters.
b)
36 counters are equivalent to 9 cubes.
c)
To convert cubes to counters we multiply the number of cubes by 4. To convert counters to cubes we divide the number of counters by 4.
d)
It is not possible to get an exact number of cubes because 42 is not divisible by 4.
b) How many cubes can you exchange for 36 counters? c)
What is the difference between converting cubes to counters and counters to cubes?
From the concrete model of 3 cubes being equal to 12 counters, pupils should be able to deduce that 1 cube is equal to 4 counters.
d) Is it possible to convert 42 counters into cubes? If not, why not?
Challenge 6:
A horse eats 5 times more hay than a goat each day. a)
What would it cost to feed a horse if it costs £1.80 to feed a goat?
b) If it costs £4 to feed a horse every day how much will it cost to feed a goat? c)
If it costs £3.50 a day to feed a horse, what would be the total cost to feed three horses and five goats?
Challenge 7:
=
=
Two shoppers have a combined total of £50 in their wallets. Shopper A has exactly £8
Importantly, the link to multiplication and division should be made. This will enable a more fluent approach of multiplying the number of cubes by 4 to calculate the equivalence in counters and using the inverse process to calculate the number of cubes from a given number of counters.
Assessment: Pupils can show that 1 cube is equivalent to 4 counters and from this show that 2 cubes are the same in value as 8 counters. They are able to deduce that a multiplication by 4 of the number of cubes calculates the number of counters. Conversely they are able to show that a division by 4 of the number of counters produces the equivalent number of cubes.
less in her wallet than shopper B.
Pupils reason that there cannot be a whole number of equivalent cubes, unless the number of counters is a multiple of 4.
a)
They solve problems by multiplying or dividing by 4 depending on whether cubes or counters are being calculated. They are able to apply the reasoning that underpins this to solve similar problems involving numbers other than 4.
Can you work out how much is in each wallet?
b) What percentage of the total money does each shopper have? c)
If the total in both wallets was different and shopper A had 75% of it, leaving shopper B with £32, how much would shopper A have?
Challenge 6 Answer: a)
£9
b)
£0.80
c)
£14
The bar method would be useful to help solve the particular type of ratio problem in part a).
Challenge 8:
Anna has 12 more cards than Barry. Clare has twice as many cards as Anna. Altogether
goat
= £1.80
they have 92 cards. How many cards do they each have? horse
= £9.00
Assessment: Pupils can show that the operations needed to convert numerical values of two given linked variables are multiplication and division. They are able to identify when multiplication or division should be used for a given variable. In part b) pupils know that the cost of a horse is five times that of a goat and that division by 5 will give the correct result. Pupils check their answers by looking at the inverse process and deciding what presents the most reasonable answer. For example, in part a) if they divided £1.80 by 5 giving a result of £0.36 as the cost of feeding a horse, they would see this was not a reasonable answer as horses cost more than goats to feed.
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In part c) pupils should be able to reason that although there is a process of addition in the question, the multiplicative element within ratio is necessary and ultimately the most important to solve the problem. Pupils solve problems using multiplication and division appropriately depending upon the context of the question. In part d) they use mental mathematics to change the problem by converting the cost of 5 goats to that of 1 horse then multiplying this by 4 to get the required result of £14.
Challenge 7 Answer: a)
Shopper A has £21 and shopper B has £29
b) Shopper A has 42% and shopper B has 58%
c)
£96
Assessment: The bar method would be useful to understand how the problem looks from a concrete perspective. In part a), pupils reason that subtracting the extra £8 that shopper B has will result in both shoppers having an equal amount, but that the total would be £8 less than £50. Dividing by 2 at this stage will provide the amount of money that shopper A starts with and adding £8 results in the amount of money shopper B has. Pupils check their answers by adding both amounts to see if the total is £50.
Shopper A = £50 £8
Shopper B
Subtract £8 making both shoppers equal and total £42
Shopper A = £42 Shopper B = £21
21 29 : 50 . By using their In part b) pupils can reason that the ratio of shopper A : B is 21 : 29 which is the same as 50 42 58 : 100 which can be written as 42% : 58%. knowledge of equivalent fractions, the ratio can also be expressed as 100
In part c) they should be able to reason that, if shopper A has 75% of the total, then 25% represents the £32 of shopper B. A simple multiplication by 3 therefore calculates the amount of money in the wallet of shopper A. Pupils solve these types of ratio problems efficiently by making links to algebra. In part a) they will see the problem as 2x (an amount both shoppers have) + 8 (extra that shopper B has) = 50 in order to solve the value of x (which is interpreted in this question as shopper A). They will reason that the outcome of part c) could also be interpreted as 25% = £32, therefore 4 (25%) = £128 and from this 3 (25%) = £96.
Challenge 8 Answer: Anna has 26 cards, Barry has 14 and Clare has 52 cards.
Assessment: Pupils should use the bar method and start with Barry. From this they add 12 to represent Anna and then double this to represent Clare. They add these results and get a total of 92.
Barry
By subtracting 36 from both sides, pupils calculate that 4 shaded bars are equal to 56 and a further division by 4 results in 1 bar being equal to 14. This is the number of cards Barry has. Pupils can then calculate how many the other two have from the information given.
A+B+C
Anna Clare
12 12 12 12 12 12
= 92 = 56 = 14
Barry
= 14
They should reason that Barry is the important variable as the results of the other two depend upon the number of cards he has. They see the problem in an algebraic format and reason that, if the unknown quantity of cards belonging to Barry is n, then Anna’s quantity is n + 12 and Clare’s is 2n + 24. They solve the problem by gathering like terms and choosing mathematical operations on both sides of the equation to calculate the value of n (Barry’s total).
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