Higher Tier - Year 10

Edexcel-Style Year 10 Higher Paper 1

An Edexcel-style Year 10 Higher Paper 1: 80 marks, 90 minutes, non-calculator, matching Edexcel 1MA1's published paper format. Covers the same nine Year 10 topics as the board-agnostic set: number and calculation, fractions, decimals and percentages, ratio and proportion, indices and standard form, linear algebra, graphs and coordinates, sequences, angles and geometrical reasoning, and area, volume and measures.

28 questions - 80 marks

Year 10 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 10, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.

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Questions

Question 1 [1 marks]

Area, Volume and Measures

The diagram shows a circle with centre O. Points A and B lie on the circumference. Radii OA and OB are drawn, and the minor arc AB is marked. The region enclosed by OA, OB and the arc AB is shaded.

Question 2 [1 marks]

Ratio and Proportion

Simplify the ratio 35:14.

Question 3 [1 marks]

Indices and Standard Form

Write the number 68000 in standard form.

Question 4 [1 marks]

Graphs and Coordinates

Find the gradient of the straight line joining the points (2, 3) and (5, 12).

Question 5 [2 marks]

Fractions, Decimals and Percentages

Convert 0.2222... (2 recurring) to a fraction in its simplest form.

Question 6 [1 marks]

Graphs and Coordinates

Plot the point with coordinates (3, -2) on a pair of axes.

Question 7 [2 marks]

Number and Calculation

Find the LCM of 6, 8 and 10.

Question 8 [2 marks]

Linear Algebra

Solve 5x - 7 = 18

Question 9 [2 marks]

Ratio and Proportion

Simplify the ratio 900 m : 2 km, giving your answer in its simplest form.

Question 10 [2 marks]

Angles and Geometrical Reasoning

Work out the value of x.

Question 11 [2 marks]

Linear Algebra

Expand and simplify (x + 6)(x + 4)

Question 12 [2 marks]

Angles and Geometrical Reasoning

A regular polygon has 30 sides. Work out the size of each exterior angle.

Question 13 [3 marks]

Sequences

The sequence 5, 11, 19, 29, 41,... is generated by a quadratic rule. Show that the nth term is n^2 + 3n + 1.

Question 14 [3 marks]

Linear Algebra

Write (x + 1)/3 - (x - 2)/5 as a single fraction in its simplest form.

Question 15 [3 marks]

Angles and Geometrical Reasoning

Describe the locus of points that are 4 cm from point P(0, 0) and also 3 cm from the line y = 5. In your answer, state the geometric place for each locus and then state how many intersection points there can be between them.

Question 16 [4 marks]

Graphs and Coordinates

A line has equation 4x + 2y = 10.

Rearrange the equation into the form y = mx + c.

State the gradient of a line that is parallel to this line.

State the gradient of a line that is perpendicular to this line.

Question 17 [3 marks]

Linear Algebra

Solve 6(2x - 1) - 4(x - 3) = 22 Show your working.

Question 18 [4 marks]

Fractions, Decimals and Percentages

A dripping tap loses water at a constant rate of 5/18 of a litre every minute.

Convert 5/18 to a decimal, stating clearly which digit(s) recur.

Hence work out, in litres correct to 3 decimal places, how much water the tap loses in exactly 12 minutes.

Question 19 [3 marks]

Linear Algebra

P = 2x^2 - 3x + 1. Work out the value of P when x = -4, showing your working.

Question 20 [4 marks]

Graphs and Coordinates

The circle x^2 + y^2 = 100 has a point A with x-coordinate 8 and y positive. (a) Find the equation of the tangent at A. (b) Find the coordinates of the point where this tangent meets the line y = 2x - 5.

Find the coordinates of the intersection point of the tangent and the line y = 2x - 5.

Question 21 [6 marks]

Indices and Standard Form

Simplify each expression fully.

(8x^6)^(1/3)

(9x^4)^(1/2)

(x^3)^-2 * x^4

Question 22 [2 marks]

Linear Algebra

Prove algebraically that the product of any two even numbers is always a multiple of 4.

Question 23 [3 marks]

Angles and Geometrical Reasoning

A student called Aaliyah enlarges a triangle with vertices (2, 0), (4, 0) and (2, 3) by scale factor -2, centre the origin O. Aaliyah's image has vertices (4, 0), (8, 0) and (4, 6).

Explain the error Aaliyah has made.

Write down the correct coordinates of the image vertices.

Question 24 [5 marks]

Ratio and Proportion

A coffee roastery blends Colombian and Kenyan beans. In Batch A the beans are blended in the ratio 7:3 (Colombian : Kenyan). Batch B is made using 630 g of Colombian beans and 260 g of Kenyan beans. Determine, showing your working, whether Batch B has been blended in the same ratio as Batch A.

Question 25 [4 marks]

Linear Algebra

Show that 3/(x - 1) + 4/(x + 2) = (7x + 2)/[(x - 1)(x + 2)], stating any value(s) of x for which the expression is not defined.

Question 26 [5 marks]

Graphs and Coordinates

Line L1 has equation kx + 3y = 12. Line L2 has equation 2x - y = 7. Given that L1 is perpendicular to L2, find the value of k.

Question 27 [4 marks]

Linear Algebra

Simplify fully [1/(x + 2) - 1/(x - 2)] divided by [1/(x + 2) + 1/(x - 2)]

Question 28 [5 marks]

Graphs and Coordinates

A circle has centre O(0, 0) and passes through the point P(6, 8). Find the equation of the tangent to the circle at the point P. Give your answer in the form y = mx + c.

Model solutions

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or working	Marks
B (Sector)	B1
Final answer: B

Mark scheme for Question 2 [1 mark]
Question 2[1 mark]
Answer or working	Marks
5:2	B1cao

Mark scheme for Question 3 [1 mark]
Question 3[1 mark]
Answer or working	Marks
6.8 x 10^4	B1oe

Mark scheme for Question 4 [1 mark]
Question 4[1 mark]
Answer or working	Marks
3	B1cao

Mark scheme for Question 5 [2 marks]
Question 5[2 marks]
Answer or working	Marks
x = 0.2222... and 10x = 2.2222... written (or equivalent valid method)	M1
2/9	A1cao

Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or working	Marks
point plotted at (3, -2) within tolerance (correct quadrant and position)	B1oe
Final answer: (3, -2)

Mark scheme for Question 7 [2 marks]
Question 7[2 marks]
Answer or working	Marks
6 = 2 x 3, 8 = 2^3 and 10 = 2 x 5 all shown	M1
120	A1cao

Mark scheme for Question 8 [2 marks]
Question 8[2 marks]
Answer or working	Marks
5x = 25	M1oe
x = 5	A1cao

Mark scheme for Question 9 [2 marks]
Question 9[2 marks]
Answer or working	Marks
convert 2 km to 2000 m	M1oe
9:20	A1cao

Mark scheme for Question 10 [2 marks]
Question 10[2 marks]
Answer or working	Marks
3x + 5 = 65 oe (corresponding angles are equal)	M1
x = 20	A1cao

Mark scheme for Question 11 [2 marks]
Question 11[2 marks]
Answer or working	Marks
x^2 + 4x + 6x + 24, or at least 3 of the 4 terms correct	M1oe
x^2 + 10x + 24	A1cao

Mark scheme for Question 12 [2 marks]
Question 12[2 marks]
Answer or working	Marks
360 / 30 soi	M1
12 (degrees)	A1cao
Final answer: 12 degrees

Mark scheme for Question 13 [3 marks]
Question 13[3 marks]
Answer or working	Marks
second difference = 2 so a = 1 (2a = 2)	M1oe
use n = 1 and n = 2 to find b and c (1 + b + c = 5 and 4 + 2b + c = 11)	M1oe
nth term = n^2 + 3n + 1 cao	A1cso
Final answer: n^2 + 3n + 1

Mark scheme for Question 14 [3 marks]
Question 14[3 marks]
Answer or working	Marks
common denominator 15 used	M1
numerator 5(x + 1) - 3(x - 2) formed and expanded	M1
(2x + 11)/15 oe	A1cao
Final answer: (2x + 11)/15

Mark scheme for Question 15 [3 marks]
Question 15[3 marks]
Answer or working	Marks
identify locus 1 as a circle centre (0,0) radius 4 and locus 2 as the two lines parallel to y = 5 at distance 3 (y = 8 and y = 2)	M1
state that a circle and a horizontal line can intersect in 0, 1 or 2 points	M1
conclusion: there can be up to 2 intersection points	A1cao
Final answer: Locus 1: a circle centre (0,0) radius 4 cm. Locus 2: the set of points 3 cm from the line y = 5 are the two lines y = 8 and y = 2. A horizontal line and a circle can meet in 0, 1 or 2 points, so there can be up to 2 intersection points in total.

Mark scheme for Question 16 [4 marks]
Question 16[4 marks]
Answer or working	Marks
2y = -4x + 10	M1oe
y = -2x + 5	A1
-2 ft from part (a)	B1
1/2 ft from part (a)	B1
Final answer: y = -2x + 5 \| -2 \| 1/2

Mark scheme for Question 17 [3 marks]
Question 17[3 marks]
Answer or working	Marks
12x - 6 - 4x + 12 = 22 (expand both brackets correctly, with correct signs)	M1
8x + 6 = 22 (dep, simplify the left-hand side)	dM1
x = 2	A1cao

Mark scheme for Question 18 [4 marks]
Question 18[4 marks]
Answer or working	Marks
0.27777... seen (or awrt 0.2778)	B1
correctly states that only the digit 7 recurs (not the 2)	B1oe
12 x 5/18 oe, or 10/3 seen	M1
awrt 3.333 litres	A1cao
Final answer: See individual parts above.

Mark scheme for Question 19 [3 marks]
Question 19[3 marks]
Answer or working	Marks
2 x (-4)^2 (= 32) seen	M1oe
-3 x (-4) (= 12) seen, correctly signed	dM1
45	A1cao

Mark scheme for Question 20 [4 marks]
Question 20[4 marks]
Answer or working	Marks
find y: y^2 = 100 - 64 so y = 6 and slope -x/y = -8/6 = -4/3	M1oe
y = -4/3 x + 50/3	A1cao
solve -4/3 x + 50/3 = 2x - 5	M1oe
intersection (13/2, 8)	A1cao
Final answer: y = -4/3 x + 50/3 \| (13/2, 8)

Mark scheme for Question 21 [6 marks]
Question 21[6 marks]
Answer or working	Marks
8^(1/3) = 2 and x^6 raised to power 1/3 gives x^2, both found	M1
2x^2	A1cao
9^(1/2) = 3 and x^4 raised to power 1/2 gives x^2, both found	M1
3x^2	A1cao
(x^3)^-2 = x^-6 found	M1
1/x^2 oe (x^-2)	A1
Final answer: 2x^2 \| 3x^2 \| x^-2 (= 1/x^2)

Mark scheme for Question 22 [2 marks]
Question 22[2 marks]
Answer or working	Marks
Even numbers written as 2n and 2m, product formed as 2n x 2m = 4nm	M1oe
Conclusion that 4nm is a multiple of 4 for integers n and m	A1cso
Final answer: 2n x 2m = 4nm, which is a multiple of 4 for all integers n and m.

Mark scheme for Question 23 [3 marks]
Question 23[3 marks]
Answer or working	Marks
states Aaliyah has used scale factor +2 (ignored the negative sign), placing the image on the same side as the object instead of the opposite side	B1oe
multiplies each vertex by -2	M1
(-4,0), (-8,0) and (-4,-6) all correct	A1cao
Final answer: Aaliyah used scale factor +2 instead of -2, so her image is on the same side of the centre as the object instead of the opposite side. \| (-4, 0), (-8, 0) and (-4, -6)

Mark scheme for Question 24 [5 marks]
Question 24[5 marks]
Answer or working	Marks
find the scale factor for Batch A's ratio using the Colombian mass, 630/7=90	M1oe
use the scale factor to find the Kenyan mass Batch A's ratio requires, 3x90	M1oe
270 g	A1cao
compare 270 g with Batch B's actual 260 g of Kenyan beans	M1
correct conclusion with valid reason, e.g. 'No - Batch B is not blended in the same ratio; it has less Kenyan beans (260 g) than the 270 g Batch A's ratio requires, so Batch B is relatively more Colombian' oe	A1cao
Final answer: No - Batch B is not in the same ratio as Batch A (Batch B has relatively more Colombian beans).

Mark scheme for Question 25 [4 marks]
Question 25[4 marks]
Answer or working	Marks
common denominator (x - 1)(x + 2) used	M1
numerator 3(x + 2) + 4(x - 1) expanded correctly	M1
(7x + 2)/[(x - 1)(x + 2)] cso, answer printed	A1
states x not equal to 1 and x not equal to -2	B1
Final answer: (7x + 2)/[(x - 1)(x + 2)] (shown); x not equal to 1 or -2

Mark scheme for Question 26 [5 marks]
Question 26[5 marks]
Answer or working	Marks
rearrange L2 to y = 2x - 7 oe, gradient of L2 = 2 identified	M1
rearrange L1 to y = -k/3 x + 4	M1oe
perpendicular condition set up: (-k/3) x 2 = -1	M1oe
-2k/3 = -1 solved to 2k = 3	M1oe
k = 3/2 oe (1.5)	A1cao
Final answer: k = 3/2

Mark scheme for Question 27 [4 marks]
Question 27[4 marks]
Answer or working	Marks
combines the numerator over a common denominator (x + 2)(x - 2) to get -4/(x^2 - 4)	M1
combines the denominator over a common denominator (x + 2)(x - 2) to get 2x/(x^2 - 4)	M1
divides the two fractions (multiplies by the reciprocal), cancelling (x^2 - 4)	M1
-2/x oe	A1cao
Final answer: -2/x

Mark scheme for Question 28 [5 marks]
Question 28[5 marks]
Answer or working	Marks
gradient of OP = (8 - 0) / (6 - 0)	M1oe
gradient of OP = 4/3	A1
tangent gradient = -3/4 identified (radius is perpendicular to tangent)	M1
8 = -3/4 (6) + c oe substituted	M1
y = -3/4 x + 25/2 oe cao (accept y = -0.75x + 12.5)	A1
Final answer: y = -3/4 x + 25/2 (y = -0.75x + 12.5)