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seek (exhaust) all possibilities look at/for alternative methods

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Presentation on theme: "seek (exhaust) all possibilities look at/for alternative methods"— Presentation transcript:

1 seek (exhaust) all possibilities look at/for alternative methods
some (quick) ways to (probably) make consolidation tasks at least a bit more interesting reverse the question greater generality seek (exhaust) all possibilities look at/for alternative methods

2 solve 3x – 4 = 20 make up some equations with a solution of x = 8: make them as complicated as you can what integer solutions can you find to 3x – 4y = 20? what numbers must ‘n’ be for 3x – n = 20 to have an integer solution? in trying to solve 3x – 4 = 20, what happens if you divide everything by 3 first?

3 what are the coordinates of the mid-point between (2 , 1) and (6, 9)?
which pairs of points have (4, 5) as a mid-point? if (4, 5) is one third of the way along a line what could the two end-points be? what other integer points lie on the line joining (2, 1) to (6, 9)? can you find two (or more) ways to find the mid-point between two points?

4 find several two, 2-digit subtraction sums with an answer of 7
what is 86 – 79? what general properties do the digits in a two, 2-digit subtraction sum, AB – CD have if the result is 7? find all the (positive) options for the result of a subtraction sum using the digits 6, 7, 8 and 9 (without repeats) how can you see that the result of the sum 86 – 79 must be 7 by relating the numbers to multiples of 7?

5 what is the mean average of 0 , 3 and 15?
give sets of three numbers with a mean of 6 can the mean be smaller than the median for a set of three integers? find all integer sets of 4 numbers with a mean of 6 and a range of 6 can you think of (another) way to find the mean of 138, 142, 135, 148 and 137?

6 reflect the rectangle in the line y = x
the rectangle is reflected and ends up at (1, – 2), (5, – 2), (1, – 4) and (5, – 4) what is the mirror line’s equation? what happens to the coordinates of a shape when you reflect it in y = x? reflect the rectangle in the line y = x what happens to the four corner coordinates when the shape is reflected in y = x + 1? can you use a rotation followed by a reflection that has the same effect as reflecting in y = x?

7

8 Simpsons Bart Homer Marge Maggie Lisa Abraham Mr Burns
Gavin 7, Cathy 6, Brenda 4, Errol 5, Dennis 1, Freda 3, Alice 2. Thanks to the Shell Centre

9 a large bag of flour weighs 24kg
it costs £21.50 a sponge cake uses 150g of this flour what are the questions that these calculations find out? 24000 150 24 21.50 2150 24 21.50 24000 (a) (b) (c) (d) × 150

10 the numbers 0 to 9 are arranged randomly probability of
play your cards right the numbers 0 to 9 are arranged randomly probability of 9 ? 7 6 5 H: L: 3 1 9 2 5 L: 3 ? H: ? H: 2 6 L: 4 6 2 5 ? H: 3 5 L: H = higher L = lower

11 can you suggest an experiment where the probability is ¾ ? some subtraction sums with an answer – 5 ? two coordinate pairs with a gradient of ½ between them ? a quadrilateral whose diagonals cut at 90o ? two numbers with a highest common factor of 15 ? a hexagon with rotational but no line symmetry ? numbers that might round to 350 ? numbers with exactly five factors ? the dimensions of a cuboid with a volume of 24 cm3 ? a ratio question with an answer £180 and £240?

12 Egyptian fractions the Ancient Egyptians only used unit numerator fractions they turned other fractions into sums of two or more fractions, all with a numerator of 1 (apart from ⅔) they used fractions with different denominators

13 Egyptian fractions 5 36 1 12 1 18 = + 9 80 1 16 1 20 = +
Ancient Egyptians turned fractions into a sum of fractions with unit numerators can you find other ways to represent these two fractions in this ‘Egyptian’ way? 5 36 1 12 1 18 = + 9 80 1 16 1 20 = +

14 Egyptian fractions 5 36 5 36 4 36 1 36 = + 5 36 1 9 1 36 = +
one way to do this is to split up the numerator into a sum so that the two numbers are both factors of the denominator so that they cancel… 5 36 4 36 1 36 = + 5 36 1 9 1 36 = +

15 Egyptian fractions 2 7 8 28 7 28 1 28 = + 2 7 1 4 1 28 = +
sometimes you cannot do this straight away so, look at an equivalent fraction until you can find two (or more) numbers that are both factors of the denominator which cancel… remember that the denominators must be different 8 28 7 28 1 28 = + 2 7 1 4 1 28 = +

16 Egyptian fractions 5 12 try to make this out of two unit fractions and another way 5 12 7 12 try to make this out of two unit fractions and another way 7 12 11 12 why does this need three unit fractions? try to find two ways to do this

17 try to write these as the sum of two fractions with numerator 1 and different denominators, as the Ancient Egyptians did Egyptian fractions 4 9 7 13 3 8 (1) (8) (15) find two ways for 3 10 2 5 (16) 7 20 find two ways for (2) (9) (17) find three ways to write as the sum of two unitary fractions, with different denominators 10 21 2 9 1 8 (3) (10) 11 18 5 14 (4) (11) (18) find three ways to write as the sum of two unitary fractions, with different denominators 1 10 11 24 4 49 (5) (12) (19) find four ways to write as the sum of two unitary fractions, with different denominators 2 15 5 9 6 23 (6) (13) (20) find three ways to write as the sum of three unitary fractions, with different denominators 2 15 9 50 4 5 (7) (14)

18 2 7 4 19 8 23 2 13 5 21 harder Egyptian fractions
try to write these as the sum of two unit fractions they can all be done 2 7 4 19 8 23 2 13 5 21

19 1 4 1 28 2 7 answers + = 1 5 1 95 4 19 + = 1 3 1 69 8 23 + = 1 7 1 91 2 13 + = 1 6 1 14 5 21 + =

20 Egyptian fractions 2 11 1 6 1 ?? = + ? 11 1 3 1 33 = + 3 11 1 ? 1 ?? =
some of the ‘elevenths’ family why/how do these work in this way? 2 11 1 6 1 ?? = + ? 11 1 3 1 33 = + 3 11 1 ? 1 ?? = + 6 11 1 ? 1 ?? = + 1 11 1 12 1 ??? = +

21 Egyptian fractions 1 ? 1 5 1 20 = + 1 ? 1 8 1 56 = + 1 3 1 ? 1 ?? = +
Egyptian fractions sum to another Egyptian fraction what is a general rule for these? 1 ? 1 5 1 20 = + 1 ? 1 8 1 56 = + 1 3 1 ? 1 ?? = + 1 10 1 ? 1 ?? = + 1 6 1 ? 1 ?? = +

22 1 ? 1 n 1 n(n – 1) = + 2 n 2 n + 1 2 n(n + 1) = + 2 n 2 n + 2 4
substitute some numbers for ‘n’ what happens? try to prove that this will always work 1 ? 1 n 1 n(n – 1) = + substitute some odd numbers for ‘n’ what happens? 2 n 2 n + 1 2 n(n + 1) = + 2 n 2 n + 2 4 n(n + 2) substitute some even numbers for ‘n’ what happens? = + what do these last two statements establish?


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