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Either paper 2 marks
A stir-fry is made with three different vegetables. The vegetables are chosen from
- carrot (c) - broccoli (b) - pepper (p) - onion (o)
List all the possible options for the three vegetables.
One option has been done for you: \(c b p\)
Show answer
\(b p o\), \(c p o\), \(c b o\), \(c b p\), in any orderEach option has three of the four vegetables, so each option leaves out exactly one of them. Leave out each one in turn.
Leave out carrot (c): \(b p o\)
Leave out broccoli (b): \(c p o\)
Leave out pepper (p): \(c b o\)
Leave out onion (o): \(c b p\)
The option shown leaves out onion (o). Changing the order of the letters in an option does not make a new option, so there are four options in all.
- B1three of the four options, none written twice in a different order (the option shown counts as one of them)
- B1all four options: \(b p o\), \(c p o\), \(c b o\), \(c b p\), with no option written twice in a different order
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Either paper 4 marks
Here are four cards. Each card shows a digit.
\(7,\ 6,\ 9,\ 5\)
Grace chooses two of the cards and puts them side by side to make a two-digit number.
(a)Calculate how many different two-digit numbers she can make.
(b)List all the even two-digit numbers she can make.
(c)State the largest two-digit number she can make.
Show answer
(a) \(12\)
(b) \(56,\ 76,\ 96\)
(c) \(97\)There are \(4\) cards to choose for the tens digit, and then \(4 - 1\) cards left for the units digit.
\(4 \times 3 = 12\)
A number is even when its last digit is even, and only one of the four cards shows an even digit: \(6\).
So every even number she can make ends in \(6\), and any of the other \(3\) cards can go in front of it.
\(56,\ 76,\ 96\)
Put the largest digit in the tens column and the next largest in the units column.
\(9\) then \(7\), giving \(97\).
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Either paper 3 marksHere are the subjects available in Year 12 at a school.
Students choose three subjects. They cannot choose more than one subject from a block. Grace decides to study Maths. Grace decides not to study any of these subjects:Block 1 Block 2 Block 3 Block 4 Maths Geography English Spanish History Drama Physics Biology French Chemistry Computing Art - Drama
- English
- Physics
- Spanish
- Art
Show answer
Subject 1 Subject 2 Subject 3 Maths Geography Computing Maths Chemistry Computing Maths Geography Biology Maths Chemistry Biology Maths Computing Biology Grace has already chosen Maths, which is in block 1, so no other subject from block 1 can be chosen. The subjects ruled out cannot be chosen either. That leaves the allowed subjects in each of the other blocks: Block 2: Geography and Chemistry Block 3: Computing Block 4: Biology The other two subjects must come from two different blocks. Go through the blocks in order: block 2 with block 3, then block 2 with block 4, then block 3 with block 4. Within each pair of blocks, take each allowed subject of the first block with each allowed subject of the second. A group is never made from two subjects in the same block, and no pair of blocks is used twice, so nothing is repeated or missed.
The number of groups from blocks 2 and 3 is \(2\), from blocks 2 and 4 is \(2\) and from blocks 3 and 4 is \(1\). \(2 + 2 + 1 = 5\), so there are five groups.Subject 1 Subject 2 Subject 3 Maths Geography Computing Maths Chemistry Computing Maths Geography Biology Maths Chemistry Biology Maths Computing Biology - B1allowed subjects found in each block: block 2 Geography and Chemistry; block 3 Computing; block 4 Biology
- M1systematic list: Maths with one allowed subject from each of two different blocks, taking the blocks in a fixed order, and no group with two subjects from one block
- A1all 5 groups listed, with no repeats and no extras
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Either paper 3 marks
Put the whole numbers 2 to 13 into the boxes to make the calculations correct.
Only use each number once.
\(\square + \square + \square = 19\)
\(\square + \square + \square = 9\)
\(\square + \square + \square = 36\)
\(\square + \square + \square = 26\)
Show answer
\(5 + 6 + 8 = 19\); \(2 + 3 + 4 = 9\); \(11 + 12 + 13 = 36\); \(7 + 9 + 10 = 26\) (the three numbers in each calculation may be in any order)Start with the totals that leave the fewest choices.
The smallest total is \(9\). The three smallest numbers are \(2\), \(2 + 1\) and \(2 + 2\), which add to \(9\). That is exactly the total, so no other three numbers can make it.
\(2 + 3 + 4 = 9\)
The largest total is \(36\). The three largest numbers are \(13\), \(13 - 1\) and \(13 - 2\), which add to \(36\). That is exactly the total, so no other three numbers can make it.
\(11 + 12 + 13 = 36\)
Cross those six numbers off. The numbers left are \(5,\ 6,\ 7,\ 8,\ 9,\ 10\), and the two totals still to make are \(19\) and \(26\).
Choose three of the numbers left to make \(19\), and check that the other three make \(26\):
\(5 + 6 + 8 = 19\)
\(7 + 9 + 10 = 26\)
Every number from 2 to 13 has now been used once, so the four calculations are
\(5 + 6 + 8 = 19\); \(2 + 3 + 4 = 9\); \(11 + 12 + 13 = 36\); \(7 + 9 + 10 = 26\)
- B1the smallest total \(9\) made correctly: \(2 + 3 + 4\)
- B1the largest total \(36\) made correctly: \(11 + 12 + 13\)
- B1the other two totals \(19\) and \(26\) made correctly, each number from 2 to 13 used exactly once
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Either paper 3 marksThe numbers 1 to 12 are put in a grid. Six of the numbers are shown. The letters A to F stand for the six numbers that are missing.
Each of the four sides of the grid must add up to 29. Complete the grid using the numbers 1, 2, 3, 4, 6, 9. Use each of them once, and give the number that each letter stands for.11 8 A B C 5 10 D E 7 F 12 Show answer
A = 1, B = 9, C = 2, D = 3, E = 6, F = 4Each side of four numbers must add up to 29. So the blanks on a side must add up to 29 minus the given numbers on that side. A corner belongs to two sides, so a blank corner has to suit both of them.
On the top side, \(A + B = 10\). Of the numbers still unused, only 1 + 9, 4 + 6 make 10. On the right side, \(B + D = 12\). Of the numbers still unused, only 3 + 9 make 12. On the bottom side, \(F + E = 10\). Of the numbers still unused, only 1 + 9, 4 + 6 make 10. On the left side, \(E + C = 8\). Of the numbers still unused, only 2 + 6 make 8.
Putting what the sides allow together, B can only be 9, so \(B = 9\).
The top side has only one blank, A. \(A = 29 - (11 + 8 + 9) = 1\). The right side has only one blank, D. \(D = 29 - (9 + 5 + 12) = 3\).
Putting what the sides allow together, E can only be 6, so \(E = 6\).
The bottom side has only one blank, F. \(F = 29 - (12 + 7 + 6) = 4\). The left side has only one blank, C. \(C = 29 - (6 + 10 + 11) = 2\).
Check each side: top: \(11 + 8 + 1 + 9 = 29\); right: \(9 + 5 + 3 + 12 = 29\); bottom: \(12 + 4 + 7 + 6 = 29\); left: \(6 + 10 + 2 + 11 = 29\).
- M1uses a side total of 29 to find what blanks on a side must add to, or one blank, for example A + B = 10 from the top side
- A1at least four of the six numbers correct
- A1all six numbers correct: A = 1, B = 9, C = 2, D = 3, E = 6, F = 4
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Either paper 6 marks
A shop sells ice creams. Each ice cream has two scoops.
The flavours are peach (p), orange (o), vanilla (v) and coffee (c).
The two scoops can be the same flavour or different flavours.
(a)List all the possible options for the two scoops. The order of the scoops does not matter.
One option, vv, has been written in for you.
(b) In one day the shop sells 144 scoops of ice cream. The table shows the number of scoops of each flavour.
A pie chart is to be drawn to show the data. Work out the angle, in degrees, that stands for one scoop.Flavour Number of scoops Peach 64 Orange 20 Vanilla 24 Coffee 36 (c)The sector for peach has been drawn. The marks round the circle are \(10^\circ\) apart.
Complete the pie chart to show the data in the table in part (b).
You cannot draw on this page. Draw (c) on paper, then show the answer and check your drawing against the model.
Show answer
(a) all 10 options, in any order and either way round: pp, po, pv, pc, oo, ov, oc, vv, vc, cc
(b) \(2.5\)
(c) Sectors of \(50^\circ\) for orange, \(60^\circ\) for vanilla and \(90^\circ\) for coffee, in any order round the chart(c)Pair each flavour with itself and with the flavours listed after it, so that no option is written twice.
p: pp, po, pv, pc
o: oo, ov, oc
v: vv, vc
c: cc
That gives 10 options. The four with the same flavour twice (pp, oo, vv and cc) are part of the list.
The whole circle is \(360^\circ\) and stands for 144 scoops.
\(360 \div 144 = 2.5\)
So one scoop is \(2.5^\circ\).
Each scoop is \(2.5^\circ\), so multiply each number of scoops by \(2.5\).
Orange: \(20 \times 2.5 = 50^\circ\)
Vanilla: \(24 \times 2.5 = 60^\circ\)
Coffee: \(36 \times 2.5 = 90^\circ\)
The sector for peach is \(64 \times 2.5 = 160^\circ\), and the four angles add up to \(160 + 50 + 60 + 90 = 360^\circ\).
Rule each radius by counting the marks, which are \(10^\circ\) apart, and write each name in its sector.
- (a) B1at least six different options, none repeated
- (a) B1all ten options: pp, po, pv, pc, oo, ov, oc, vv, vc, cc
- (c) M1one more sector drawn to the right angle
- (c) A1all three sectors the right size: \(50^\circ\), \(60^\circ\) and \(90^\circ\)
- (c) B1and each sector named correctly
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Either paper 2 marks
Rosa has six banknotes.
The value of each banknote is £10, £20 or £50.
She can make £60 using four of the banknotes.
She can make £70 using three of the banknotes.
She cannot make £50 using any four of the banknotes.
She cannot make £130 using any four of the banknotes.
List the values of the six banknotes.
Show answer
£10, £10, £20, £20, £20, £50 (in any order)Start with the "can" clues, because each one can be made in only one way.
£70 with three banknotes: £10 + £10 + £50 = £70
£60 with four banknotes: £10 + £10 + £20 + £20 = £60
All of these banknotes are among the six. The two groups use three and four banknotes, which is more than six, so they share some. Take each value as many times as it is needed in either group, and count a shared banknote once:
£10, £10, £20, £20, £50
That is five banknotes, so there is one more banknote to find. It is £10, £20 or £50. Test each against the "cannot" clues.
A sixth banknote of £10 would mean Rosa could make £10 + £10 + £10 + £20 = £50 using four banknotes, but that is a total Rosa cannot make. So the sixth banknote is not £10.
A sixth banknote of £50 would mean Rosa could make £10 + £20 + £50 + £50 = £130 using four banknotes, but that is a total Rosa cannot make. So the sixth banknote is not £50.
The sixth banknote is £20, so the six banknotes are
£10, £10, £20, £20, £20, £50
- B1six banknotes that include £10, £10, £20, £20, £50, the five that the two "can" clues need
- B1all six correct: £10, £10, £20, £20, £20, £50
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Either paper 2 marksIn a game, three stars are hidden on this board. Each star is behind a different square.
The stars are behind three consecutive squares. The squares are in one row or one column. One of the squares is b2. It is shaded on the board. Write down all the possible pairs for the other two squares.a b c d e 1 2 3 4 Show answer
(a2, c2), (c2, d2), (b1, b3), (b3, b4)The three stars make a line of three squares that includes b2. Check the row and the column, one at a time. A line of three that would need a square beyond the edge of the board is not allowed.
Along row 2, the lines of three that stay on the board are: [a2, b2, c2], [b2, c2, d2].
Leaving out b2 gives the pairs (a2, c2), (c2, d2).
Down column b, the lines of three that stay on the board are: [b1, b2, b3], [b2, b3, b4].
Leaving out b2 gives the pairs (b1, b3), (b3, b4).
That is 2 pairs along the row and 2 pairs down the column, so 4 possible pairs in all: (a2, c2), (c2, d2), (b1, b3), (b3, b4).
- B1every pair along row 2: (a2, c2), (c2, d2)
- B1every pair along column b: (b1, b3), (b3, b4)
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Either paper 3 marks
A code has five different digits.
Each digit is a whole number that is at least \(1\) and less than \(8\).
The digits are written in order, starting with the smallest.
The fourth digit is the only multiple of 3.
The first digit is the only square number.
Work out the code.
Show answer
\(12567\)"The only" means that exactly one digit has that property, so the other four digits do not.
The fourth digit is the only multiple of 3, so the other four digits are not multiples of 3.
The first digit is the only square number, so the other four digits are not square numbers.
Three of the digits in the code are neither multiples of 3 nor square numbers. The digits from \(1\) to \(7\) that are neither are
\(2, 5, 7\)
The fourth digit has to be a multiple of 3 that is not also a square number: \(3, 6\).
The first digit has to be a square number that is not also a multiple of 3: \(1, 4\).
The digits go in order from the smallest. The fourth digit has 3 digits smaller than it and the first digit has no digits smaller than it. The only choice that does this is the fourth digit \(6\) and the first digit \(1\), with the rest from \(2, 5, 7\).
The code is \(12567\).
Check: the only multiple of 3 in \(1, 2, 5, 6, 7\) is \(6\), in the fourth place, and the only square number is \(1\), in the first place.
- M1uses "the only" to rule digits out: the digits that are neither multiples of 3 nor square numbers are \(2, 5, 7\)
- M1the fourth digit comes from \(3, 6\) (multiples of 3 that are not square numbers) and the first digit from \(1, 4\) (square numbers that are not multiples of 3)
- A1the code \(12567\)
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Either paper 3 marks
Erin has a 6-digit code.
Erin remembers it as three 2-digit numbers written side by side.
The first number is between 11 and 21.
The second number is 6 times the first number.
The third number is 5 times the first number.
All six digits are different.
Work out the code.
Show answer
\(137865\)Try each possible first number from 11 to 21. Work out both multiples and look for a repeated digit.
Only \(13\) works: \(13\), \(78\) and \(65\) are three 2-digit numbers and the six digits are all different. The code is \(137865\).First number \(\times 6\) \(\times 5\) Check 11 66 55 a digit is repeated 12 72 60 a digit is repeated 13 78 65 all six digits different 14 84 70 a digit is repeated 15 90 75 a digit is repeated 16 96 80 a digit is repeated 17 102 85 a multiple has three digits 18 108 90 a multiple has three digits 19 114 95 a multiple has three digits 20 120 100 a multiple has three digits 21 126 105 a multiple has three digits - M1both multiples worked out for first numbers in the range, for example \(13 \times 6 = 78\) and \(13 \times 5 = 65\)
- M1first numbers rejected because a multiple has three digits or a digit is repeated, leaving only \(13\)
- A1code \(137865\)