-
Either paper 4 marks
Amara and Ben share 112 counters between them.
Ben gets 22 more counters than Amara.
Amara gets \(x\) counters.
(a)Write down an equation in terms of \(x\).
(b)Solve your equation to find the value of \(x\).
(c)Write down how many counters Ben gets.
Show answer
(a) \(2x + 22 = 112\)
(b) \(45\)
(c) \(67\)Amara gets \(x\) counters and Ben gets \(x + 22\) counters. Together that is all 112 of them:
\(x + (x + 22) = 112\), which tidies up to \(2x + 22 = 112\)
Subtract 22 from both sides:
\(2x = 90\)
Divide both sides by 2:
\(x = 45\)
Amara gets 45 counters, and Ben gets 22 more than Amara:
\(45 + 22 = 67\)
- (b) M1subtracts 22 to get \(2x = 90\)
- (b) A1\(x = 45\)
-
Either paper 3 marks
Four positive whole numbers add up to 84.
One of the numbers is a multiple of 20.
The other three numbers are equal.
What are the four numbers?
Show answer
\(8, 8, 8\) and \(60\)The three equal numbers add up to a multiple of 3, so take a multiple of 20 away from 84 and test whether what is left divides exactly by 3.
Multiples of 20 that are smaller than 84: 20, 40, 60, 80.
\(84 - 20 = 64\), which leaves remainder \(1\) when divided by 3, so this does not work.
\(84 - 40 = 44\), which leaves remainder \(2\) when divided by 3, so this does not work.
\(84 - 60 = 24\), which leaves remainder \(0\) when divided by 3, so this works.
\(84 - 80 = 4\), which leaves remainder \(1\) when divided by 3, so this does not work.
The multiple is 60, and the other three numbers share \(24\), so each is \(24 \div 3 = 8\).
The four numbers are \(8, 8, 8\) and \(60\).
- M1a multiple of 20 taken from 84, for example \(84 - 20 = 64\)
- M1a remainder tested for division by 3, for example \(64\) leaves remainder \(1\)
- A1three numbers of \(8\) and a multiple of 20 equal to \(60\)
-
Either paper 3 marks
Multiplying \(y\) by 5 gives the same result as adding 20 to \(y\).
(a)Write this as an equation.
(b)Show that the value of \(y\) is not 4.
Show answer
(a) \(5y = y + 20\)
(b) When \(y = 4\), the left side is \(20\) and the right side is \(24\). These are not equal, so \(y\) is not 4.Multiplying \(y\) by 5 gives \(5y\).
Adding 20 to \(y\) gives \(y + 20\).
The two results are the same, so \(5y = y + 20\)
Substitute \(y = 4\) into each side of \(5y = y + 20\).
Left side: \(5 \times 4 = 20\)
Right side: \(4 + 20 = 24\)
\(20\) is not equal to \(24\), so \(y\) is not 4.
(Solving gives \(4y = 20\), so \(y = 5\), which is not 4.)
- (b) M1left side \(5 \times 4 = 20\) (or rearranging to \(4y = 20\))
- (b) A1right side \(4 + 20 = 24\), not equal, so \(y\) is not 4 (or \(y = 5\))
-
Either paper 3 marks
The sum of three consecutive odd numbers is 51.
Work out the three numbers.
Show answer
\(15\), \(17\) and \(19\)Let \(x\) stand for the smallest of the three numbers.
Consecutive odd numbers go up in steps of two, so the other two are \(x + 2\) and \(x + 4\).
\(x + (x + 2) + (x + 4) = 51\)
\(3x + 6 = 51\)
\(3x = 45\)
\(x = 15\)
The three numbers are 15, 17 and 19.
- M1equation for the sum: \(x + (x + 2) + (x + 4) = 51\)
- M1collecting terms: \(3x = 45\)
- A1the three numbers \(15\), \(17\) and \(19\)
-
Either paper 4 marks
The three sides of a triangle have lengths \(2x + 9\) cm, \(3x - 2\) cm and \(x + 3\) cm.
The perimeter of the triangle is 40 cm.
(a)Find the value of \(x\).
(b)Write down the length of the longest side of the triangle.
Show answer
(a) \(5\)
(b) \(19\ \text{cm}\)The perimeter is the total of the three sides, so
\((2x + 9) + (3x - 2) + (x + 3) = 40\)
Collect the \(x\) terms and the numbers separately:
\(6x + 10 = 40\)
\(6x = 30\)
\(x = 5\)
Putting \(x = 5\) into each expression gives sides of 19 cm, 13 cm and 8 cm.
The longest side is 19 cm.
- (a) M1adding the three sides: \((2x + 9) + (3x - 2) + (x + 3) = 40\)
- (a) M1collecting terms: \(6x + 10 = 40\)
- (a) A1\(x = 5\)
-
Either paper 2 marks
\(2a + 2b = 9\) and \(16x + 16y = 136\)
Show that \(8a + 8b\) has a greater value than \(4x + 4y\).
Show answer
\(8a + 8b = 36\) and \(4x + 4y = 34\), so \(36 > 34\)There is one equation for each pair of letters, so \(a\) and \(b\) (or \(x\) and \(y\)) cannot be found separately. Instead, scale each given equation to match the expression asked about.
\(2a + 2b = 9\). Multiply both sides by 4:
\(8a + 8b = 36\)
\(16x + 16y = 136\). Divide both sides by 4:
\(4x + 4y = 34\)
So \(8a + 8b = 36\) and \(4x + 4y = 34\).
\(36 > 34\), so \(8a + 8b\) has the greater value.
- M1value of the greater expression, \(36\), found by scaling its given relation
- A1value of the other expression, \(34\), found by scaling its given relation, and the comparison \(36 > 34\)
-
Either paper 6 marks
Marcus buys 4 metres of linen at £13.10 per metre.
He also buys 6 metres of cotton.
The total cost is £112.40.
(a)Work out the cost of one metre of cotton.
(b)Buttons cost 53p each.
The greatest number of buttons Marcus can buy with £9 is 16.
Marcus says, "The greatest number of buttons I can buy with £18 is 32 because £18 is double £9."
Is Marcus correct?
You must show your working.
Show answer
(a) \(£10.00\)
(b) No: with £18 Marcus can buy 33 buttons, and \(2 \times 16 = 32\), which is not equal to 33The linen costs \(4 \times 13.10 = £52.40\).
Take the linen from the total:
\(112.40 - 52.40 = 60.00\), so the cotton costs £60.00.
Divide by the 6 metres of cotton:
\(60.00 \div 6 = 10.00\)
One metre of cotton costs £10.00.
Double the money is £18, which is 1800p.
Divide by the price of a button:
\(1800 \div 53 \approx 33.96\)
Only whole buttons can be bought, so the greatest number is 33.
Check: \(33 \times 53 = 1749\), which is no more than 1800.
Twice the first number of buttons is \(2 \times 16 = 32\), which is not equal to 33, so the answer is No.
- (a) M1cost of the linen \(4 \times 13.10 = £52.40\)
- (a) M1cost of the cotton \(112.40 - 52.40 = £60.00\)
- (a) A1dividing by 6 to get £10.00 per metre
- (b) M1£18 is 1800p, and \(1800 \div 53 \approx 33.96\)
- (b) B1the greatest number of whole buttons is 33
- (b) A1No, comparing 33 with \(2 \times 16 = 32\)
-
Either paper 4 marks
Kofi is twice as old as Ella.
In 8 years' time the sum of their two ages will be 61.
Work out how old Ella is now.
Show answer
\(15\ \text{years}\)Let \(x\) stand for Ella's age now. Kofi is twice as old, so Kofi is \(2x\).
In 8 years' time Ella will be \(x + 8\) and Kofi will be \(2x + 8\).
\((x + 8) + (2x + 8) = 61\)
\(3x + 16 = 61\)
\(3x = 45\)
\(x = 15\)
Ella is 15 years old now. (Check: Kofi is 30, and \(15 + 8 + 30 + 8 = 61\).)
- P1older age as \(2x\) and younger age as \(x\)
- M1equation \((x + 8) + (2x + 8) = 61\)
- M1\(3x = 45\)
- A1\(x = 15\)
-
Either paper 4 marks
£195 is shared between Maya, Nico and Orla.
Nico gets three times as much as Maya.
Orla gets £10 more than Maya.
Work out how much Nico gets.
Show answer
\(\text{£}111\)Let \(x\) stand for the number of pounds Maya gets.
Nico gets \(3x\) and Orla gets \(x + 10\), and the three shares use up the whole £195:
\(x + 3x + (x + 10) = 195\)
\(5x + 10 = 195\)
\(5x = 185\)
\(x = 37\)
Maya gets £37, so Nico gets \(3 \times £37 = £111\).
- P1shares \(x\), \(3x\) and \(x + 10\)
- M1equation \(x + 3x + (x + 10) = 195\)
- M1\(5x = 185\) leading to \(x = 37\)
- A1\(111\)
-
Either paper 5 marks
Beacon and Kingsway are two swimming clubs.
Beacon charges a joining fee of £60 and then £21 for each month of membership.
Kingsway charges a joining fee of £108 and then £17 for each month of membership.
(a)Calculate the number of months for which the total cost at Beacon is the same as the total cost at Kingsway.
(b)Work out that total cost.
Show answer
(a) \(12\ \text{months}\)
(b) \(\text{£}312\)Let \(x\) stand for the number of months of membership.
Total cost at Beacon: \(21x + 60\) pounds.
Total cost at Kingsway: \(17x + 108\) pounds.
The two totals are equal, so
\(21x + 60 = 17x + 108\)
Take \(17x\) from both sides:
\(4x + 60 = 108\)
Subtract 60 from both sides:
\(4x = 48\)
\(x = 12\)
At Beacon: \(21 \times 12 + 60 = 252 + 60 = 312\)
So the cost is £312 at either of them.
- (a) M1costs \(21x + 60\) and \(17x + 108\)
- (a) M1equates the two costs
- (a) M1\(4x = 48\)
- (a) A1\(x = 12\)